Gardner Differential geometry and relativity (lecture notes, web draft, 2004) (198s) PGr

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Differential Geometry (and Relativity) - Summer

2000

Classnotes

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Chapter 1: Introduction.

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PS.

DVI.

Section 1-1: Curves.

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PS.

DVI.

Section 1-2: Gauss Curvature.

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PS.

DVI.

Section 1-3: Surfaces in E

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DVI.

Section 1-4: First Fundamental Form.

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DVI.

Section 1-5: Second Fundamental Form.

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DVI.

Section 1-6: The Gauss Curvature in Detail.

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DVI.

Section 1-7: Geodesics.

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PS.

DVI.

Section 1-8: The Curvature Tensor and the Theorema Egregium.

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DVI.

Section 1-9: Manifolds.

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DVI.

Chapter 2: Special Relativity: The Geometry of Flat Spacetime.

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DVI.

Section 2-1: Inertial Frames of Reference.

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DVI.

Section 2-2: The Michelson Morley Experiment.

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DVI.

Section 2-3: The Postulates of Relativity.

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DVI.

Section 2-4: Relativity of Simltaneity.

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DVI.

Section 2-5: Coordinates.

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DVI.

Section 2-6: Invariance of the Interval.

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DVI.

Section 2-7: The Lorentz Transformation.

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DVI.

Section 2-8: Spacetime Diagrams.

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DVI.

Section 2-9: Lorentz Geometry.

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DVI.

Section 2-10: The Twin Paradox.

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DVI.

Section 2-11: Temporal order and Causality.

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DVI.

Chapter 3: General Relativity: The Geometry of Curved Spacetime.

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DVI.

Section 3-1: The Principle of Equivalence.

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DVI.

Section 3-2: Gravity as Spacetime Curvature.

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DVI.

Section 3-3: The Consequences of Einstein's Theory.

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DVI.

Section 3-6: Geodesics.

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DVI.

Section 3-7: The Field Equations.

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DVI.

Section 3-8: The Schwarzschild Solution.

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DVI.

Section 3-9: Orbits in General Relativity.

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DVI.

Section 3-10: The Bending of Light.

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DVI.

Black Holes.

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DVI.

Return to

Bob Gardner's home page

Chapter 1. Surfaces and the

Concept of Curvature

Notation. We shall denote the familiar three dimensional Euclidean

space (tradiationally denoted R

) as E

Recall. The Euclidean metric on E

x = (x, y, z) =

+ y

+ z

1.1 Curves

Definition. A curve in E

is a vector valued function of the parameter

α(t) = (x(t), y(t), z(t)).

Note. We assume the functions x(t), y(t), and z(t) have continuous

second derivatives.

Definition. The derivative vector of curve α is

(t) = (x

(t), y

(t), z

(t)).

If α(t) is the position of a particle at time t, then α

(t) is the velocity

vector of the particle and α

(t) is the acceleration vector of the particle.

The speed of the particle is the scalar function α

(t).

Note. According to Newton’s Second Law of motion, the force acting

on a particle of mass m and position α(t) is

F (t) = mα

(t).

Definition. The length (or arclength) of the curve α(t) for t ∈ [a, b] is

S =

(t)dt.

Note. If

β(t) is a curve for t ∈ [a, b], then β can be written as a

function of arclength (which we will denote α(s)) as follows. First,

S(t) =

(t)dt

(that is, S(t) is an antiderivative of speed which satisﬁes S(a) = 0).

Therefore S is a one to one function and S

−1

exists. S

−1

gives the time

at which the particle has travelled along

β(t) a (gross) distance s. S o

we denote this as t = S

−1

(s). Second, we make the substitution for t:

β(t) = β(S

−1

(s)) ≡ α(s).

However, it may be algebraically impossible to calculate t = S

−1

(s)

(see page 11, number 5).

Recall. If f is diﬀerentiable on an interval I and f

is nonzero on I,

then f

−1

exists (i.e. f is one-to-one on I) on f (I) and f

−1

is diﬀeren-

tiable of I. In addition,




−1




x=f (a)

x=a

or f

−1

(f (a)) =

(a)

Note. If

β(t) is parameterized as α(s) as above, then

β(t) = β(S

−1

(s)) = α(s)

and

dα

dβ

−1

(s))

−1

(s))

(t)

Notice

dα

= α

(s) is a unit vector in the direction of the velocity vector

β(t).

Definition. If α(s) is a curve parameterized in terms of arclength s,

then the unit tangent vector of α(s) is α

(s) =

T (s). (α(s) is called a

unit speed curve since α

(s) = 1.)

Example 3 (page 6). Consider the circular helix

β(t) = (a cos t, a sin t, bt)

(see Figure I-3, page 6). Parameterize

β(t) in terms of arclength α(s)

and calculate

T (s).

Solution. We have

(t) = (−a sin t, a cos t, b). With S(t) the total

arclength travelled by a particle along the helix at time t, we have

(t) =

√

+ b

Therefore, S(t) = t

√

+ b

(taking S(0) = 0). Hence

t = S

−1

(s) =

√

+ b

and

α(s) = β(t) = β(S

−1

(s)) =

√

+ b

a cos

√

+ b

, a sin

√

+ b

√

+ b

Also,

T(s) = α

(s) =

√

+ b

−a sin

√

+ b

, a cos

√

+ b

, b

Notice that

T =

(t)

−1

(s))

−1

(s))

Note.

T (s) always has unit length. The only way T(s) can change is

in direction. Notice that this corresponds to a change in the direction

of travel of a particle along the path α(s). Since

T (s) · T(s) = 1, we

have

(s) ·

T (s) = 0 (by the product rule) and so T

(s) = α

(s) is

orthogonal to

T (s) = α

(s).

Definition. The curvature of α(s) (denoted k(s)) is

k(s) = T

(s) = α

(s).

(s) = 0 (and therefore curvature is nonzero) then the unit vector

in the direction of

(s) is the principal normal vector, denoted

N(s).

Notice.

N(s) =

(s)

k(s)

Example 3, page 6 (cont.). Calculate the curvature k(s) and prin-

cipal normal vector

N(s) for the helix

β(t) = (a cos t, a sin t, bt).

Solution. From above, we have

(s) = α

(s) =

−1

+ b

a cos

√

+ b

, a sin

√

+ b

, 0

and so k(s) =

(s) =

|a|

+ b

(a constant). Now

N(s) =

(s)

k(s)

= −

cos

√

+ b

, sin

√

+ b

, 0

if a > 0.

Notice that in terms of t,

N(t) = −(cos t, sin t, 0).

That is,

N(t) is a vector that points from the particle at β(t) =

(a cos t, a sin t, bt) back to the z−axis (that is,

N(t) is a unit vector

from

β(t) to (0, 0, bt)).

Note. If we take b = 0 in Example 3, we just get

β(t) to trace out

a circle of radius a in the xy−plane. The curvature of this circle is
k(s) =

+ b

1
a

. Therefore, circles of “small” radius have “large”

curvature and circles of “large” radius have “small” curvature (and the

curvature of a straight line is 0). See Figure I-4.

Definition. For a given value of s, the circle of radius

k(s)

which

is tangent to α and which lies in the plane of

T (s) and

N(s) is the

osculating circle of α at point α(s). The center of the osculating circle

is the center of curvature of α at point α(s), denoted c(s). The plane

containing the osculating circle is the osculating plane. See Figure I-6.

Note. c(s) is calculated by going from point α(s) a distance

k(s)

the direction

N(s). That is,

c(s) = α(s) +

k(s)

N(s).

Example (p. 13, # 12). Consider the helix above parameterized in

terms of s:

α(s) =

a cos

√

+ b

, a sin

√

+ b

√

+ b

Find c(s) and show that it is also a helix.

Solution. The center of curvature is

c(s) = α(s) +

k(s)

N(s),

where, from Example 3, k(s) =

+ b

and

N(s) =

− cos

√

+ b

, − sin

√

+ b

, 0

c(s) =




a cos

√

+ b

−

+ b

cos

√

+ b

a sin

√

+ b

−

+ b

sin

√

+ b

√

+ b







−

cos

√

+ b

, −

sin

√

+ b

√

+ b




If we let A =

−b

and t =

√

+ b

, then

c(t) = (A cos t, A sin t, bt) ,

which is a circular helix.

Note. The curvature k(s) of a curve α(s) gives an idea of how a

curve “twists” but does not provide a complete description of the curves

“gyrations” (as the text puts it - see page 8). There is information in

how the osculating plane tilts as s varies.

Recall. A plane in E

is determined by a point (x

, y

, z

) and a nor-

mal vector n = (A, B, C) (we will not notationally distinguish between

points and vectors). If (x, y, z) is a point in the plane, then a vector

from (x

, y

, z

) to (x, y, z) is perpendicular to n and so

n · (x − x

, y − y

, z − z

) = (A, B, C) · (x − x

, y − y

, z − z

)

= A(x − x

) + B(y − y

) + C(z − z

) = 0.

This can be rearranged as Ax + By + Cz = D for some constant D.

Notice that “twistings” of the plane would be reﬂected in changes in

the direction of the normal vector.

Example. Find the equation of the plane through the points (1, 2, 3),

(−2, 3, 3) and (1, 2, 4).

Solution. The vectors a = (1 − (−2), 2 − 3, 3 − 3) = (3, 1, 0) and

b = (1 − 1, 2 − 2, 4 − 3) = (0, 0, 1) both lie in the desired plane. Recall

that in E

, a and b are both orthogonal to a × b (provided a is not a

scalar multiple of b - See Appendix A for more details). So we can take

n = a ×b as a normal vector for the desired plane.

n = a ×b =

i j k

3 1 0

0 0 1

= i − 3j + 0k = (1, −3, 0).

So the desired plane satisﬁes 1(x − 1) − 3(y − 2) + 0(z − 3) = 0 or
x − 3y = −5. Expressed parametrically,

x = −5 + 3t

y = t
z = z (a “free variable”).

(Again, we make no notational distinction between a vector and a point.

There is a diﬀerence, though: points have locations, but vectors don’t

[nonzero vectors have a length and a direction, but no position].)

Definition. The binormal vector is

B = T ×

Note.

B is orthogonal to both T and

N (and therefore to the osculating

plane). The derivative of

B is

= (

T ×

N + T ×

(where represents derivative with respect to whatever the variable of

parameterization is). Since

= k

N, T

N = 0 and so B

T ×

Since

N is a unit vector,

is perpendicular to

N (as argued above for

T). Also, T is perpendicular to N since N = (1/k)T

. S o both T and

are perpendicular to

N and so B

T ×

is a multiple of

N (since

we are in 3-dimensions), say

= −τ

N. (Notice that if T and

are

multiples of each other, then τ = 0.)

Definition. The torsion of α at α(s) is the function τ (s) where

(s) =

−τ(s)

Note. The torsion measures the twisting (or turning) of the osculating

plane and therefore describes how much α “departs from being a plane

curve” (as the text says - see page 9).

Example. Calculate

B, B

and τ (s) for the helix above.

Solution. We have

T(s) =

√

+ b

−a sin

√

+ b

, a cos

√

+ b

, b

and

N(s) = −

cos

√

+ b

, sin

√

+ b

, 0

B = T × N =

√

+ b

−a sin

√

a cos

√

− cos

√

− sin

√

+ b

b sin

√

+ b

, −b cos

√

+ b

, a

and

+ b

cos

√

+ b

, sin

√

+ b

, 0

Therefore, since

= −τ

N, we have τ(s) =

+ b

Note. Notice that the torsion is a constant in the previous example.

This makes sense since the osculating plane tilts at a constant rate as a

particle travels (uniformly) up the helix (or “spring”). With b = 0, the

helix is, in fact, a circle in the xy−plane and so the osculating plane

does not change and τ = 0. (If τ (s) = 0 for all s, then α(s) is planar -

see page 13 #13.)

Note. We will see that the shape of a curve is completely determined

by the curvature k(s) and torsion τ (s).

Example (Excercise 14, page 14). Prove

= −k

T + τ B.

Proof. Since

N, T, and B are mutually orthogonal, and each is a unit

vector, we can write

α = (α ·

N + (α · T)T + (α · B) B.

Diﬀerentiating with respect to s:

= (α

N + α ·

)

N + (α ·

+(α

· T + α · T

)

T + (α · T)T

+ (α

· B + α · B

)

B + (α · B) B

= α

+ (α ·

)

N + (α · k

N)T + (α · (−τ

N)) B

+(α ·

+ (α ·

T )k

N + (α · B)(−τ

using the ﬁrst and third Serret-Frenet formulas. Now

[1] =

N ·

= 2

N ·

= 0.

N and

are orthogonal. Equating multiples of

N in the above

equation (since

T and B are also orthogonal to

N):

α · (

+ k

T − τ B) = 0.

So either

= −k

T + τ B, or

+ k

T − τ B is orthogonal to α. In the

second case, it must be that α = a(s)

N = a

N (since

T and B are

all orthogonal to

N). Then α

= a

N + a

T implies that T = a

and a

N = 0. So a

= 0 and a(a) = a is constant. Therefore α(s) lies

on a sphere of radius |a|. Now

B = T ×

N so

N + T ×

= (k

N) ×

N + (a

) ×

= 0.

But

= −τ

N so τ = 0. Therefore, as commented in Exercise 1.1.13,

α(s) is planar. So α(s) is a circle of radius |a|. In this case, α = a

and k = 1/a. S ince τ = 0,

k T − τ B = k T =

1
a

T = 1

) =

In either case, the result holds. (Note: The second case occurs in the

case of circular motion: consider α(t) + (a cos t, a sin t, 0).)

1.2 Gauss Curvature

(Informal Treatment)

Recall. If f (x, y, z) is a (scalar valued) function, then for c a constant,
f(x, y, z) = c determines a surface (we assume all second partials of f
are continuous and so the surface is smooth). The gradient of f is

∇f =

∂f
∂x

∂f

∂y

∂f

∂z

If v

is a vector tangent to the surface f (x, y, z) = c at point

, y

, z

), then ∇f (x

, y

, z

) is orthogonal to v

(and so ∇f is orthog-

onal to the surface). The equation of a plane tangent to the surface

can be calculated using ∇f as the normal vector for the plane.

Definition. Let v be a unit vector tangent to a smooth surface M ⊂ E

at a point

P (again making no distinction between a vector and a point).

Let

U be a unit vector normal (perpendicular) to M at point P . The

plane through point

P which contains vectors v and U intersects the

surface in a curve α

called the normal section of M at

P in the direction

v. See Figure I-10.

Example. Find the normal section of M : x

+ y

= 1 (an inﬁnitely

tall right circular cylinder of radius 1) at the point

P = (1, 0, 0) in the

direction v = (0, 1, 0).

Solution. A normal vector to M at

P is

∇(x

+ y

) = (2x, 2y, 0)|

(1,0,0)

= (2, 0, 0).

Therefore, we take

U = (1, 0, 0). The plane containing U and v has as

a normal vector

U × v =

i j k

1 0 0

0 1 0

= (0, 0, 1).

Therefore the equation of this plane is

0(x − 1) + 0(y − 0) + 1(z − 0) = 0

or z = 0. The intersection of this plane and the surface is

= {(x, y, z) | x

+ y

= 1, z = 0}.

Note. Each normal vector α

to a surface can be approximated by

a circle (as in the previous section). Recall that if a plane curve has

a curvature k at some point

P , then this osculating circle has radius

1/k and its center is located 1/k units from

P in the direction of the

principal normal vector

Definition. Let α

be a normal section to a smooth surface M at point

P in the direction v. Let U be a unit normal to M at P (−U is also

a unit normal to M at

P ). The normal curvature of M at P in the v

direction with respect to

U, denoted k

(v), is

(v) =

U · N

R(v)

where

N is the principal normal vector of α

P and R(v) is the

radius of the osculating circle to α

P . If α

has zero curvature at

P, we take k

(v) = 0.

Note. If

U and

N are parallel, then

(v) =

R(v)

and if

U and

N are antiparallel (i.e. point in opposite directions) then

(v) =

−1

R(v)

So,

(v)

is just the curvature of α

P . The text does not include

the vector

U in its notation, but our approach is equivalent to its.

Example. What is k

(v) for the cylinder x

+ y

= 1 at

P = (1, 0, 0)

in the direction v = (0, 1, 0) with respect to

U = (1, 0, 0)?

Solution. As we saw in the previous example,

= {(x, y, z) | x

+ y

= 1, z = 0}.

We can parameterize α

α(s) = (cos s, sin s, 0)

where s ∈ [0, 2π]. Then

T(s) = α

(s) = (− sin s, cos s, 0)

and

(s) = α

(s) = (− cos s, − sin s, 0).

At point

P , s = 0, so the principal normal vector at P is

N(0) = T

(0)/

(0) = (−1, 0, 0).

Therefore

(v) =

U · N

R(v)

−1

= −1.

Note. We will see in Section 6 that the normal curvature of M at

P in

the v direction with respect to

U assumes a maximum and a minimum

in directions v

and v

(respectively) which are orthogonal.

Definition. The directions v

and v

described above are the principal

directions of M at

P . Let k

and k

be the maximum and minimum

values (respectively) of k

(v) at

P (we can take v = v

and v = v

respectively). Then k

and k

are the principal curvatures of M at

P .

The product k

is the Gauss curvature of M at

P , denoted K( P):

K( P) = k

Note. Even though k

(v) depends on the choice of

U, K( P) is inde-

pendent of the choice of

U (if we use −U for the normal to the surface

instead of

U, we change the sign of k

(v) and so the product k

remains the same).

Example. Evaluate K(

P) for the right circular cylinder x

+ y

= 1

P = (1, 0, 0).

Solution. At

P , α

is an ellipse with semi-minor axis 1, unless v =

(0, 0, ±1):

With

U = (1, 0, 0) (and

N = (−1, 0, 0) which implies nonpositive

(v)) we have a minimum value of normal curvature of k

= −1

(as given in the previous example - this value is attained when α

is a

circle). Now with v = (0, 0, ±1) we get that α

is a pair of parallel lines

and then k

(v) = 0 (recall the curvature of a line is 0). So

K( P) = (−1)(0) = 0.

Note. The Gauss curvature of a cylinder is 0 at every point. This

is also the case for a plane. An INFORMAL reason for this is that a

cylinder can be cut and peeled open to produce a plane (and conversley)

without stretching or tearing (other than the initial cut) and without

aﬀecting lengths (such an operation is called an isometry).

Example. A sphere of radius r has normal curvature at every point

of k

(v) = ±1/r (depending on the choice of

U) and so the Gauss

curvature is K = 1/r

Note. A surface M has positive curvature at point

P if, in a deleted

neighborhood of

P on M, all points lie on the same side of the plane

tangent to M at

P . If for all neighborhoods of P on M, some points

are on one side of the tangent plane and some points are on the other

side, then the surface has negative curvature (this will be made more

rigorous latter).

Example 5, p. 18. The hyperbolic paraboloid

z =

1
2

− x

)

has negative curvature at each point.

Example 6, p. 19. A torus has some points with positive curvature,

some with negative curvature and some with 0 curvature.

Note. We have deﬁned curvature as an extrinsic property of a surface

(using things external to the surface such as normal vectors). We will

see in Gauss’ Theorema Egregium that we can redeﬁne curvature as

an intrinsic property which can be measured only using properties of

the surface itself and not using any properties of the space in which

the surface is embedded. This will be important when we address the

questions as to whether the universe is open or closed (and whether it

has positive, zero, or negative curvature).

1.3 Surfaces in E

Note. A surface M may be described as the image of a subset D of

under a vector valued function of two variables

X(u, v) = (x(u, v), y(u, v), z(u, v)).

When using this notation, we assume x, y, z have continuous partial

derivatives up to the third order.

Definition. A surface given as above is regular if the vectors

(u, v) =

∂

∂u

∂x
∂u

∂y

∂u

∂z

∂u

(u, v) =

∂

∂v

∂x

∂v

∂y
∂v

∂z
∂v

are linearly independent for each (u, v) ∈ D.

Note.

and

are linearly independent on D is equivalent to the

property:

= 0 for all (u, v) ∈ D.

Note. The condition of regularity insures that

X is one-to-one and has

a continuous inverse.

Example (Exercise 1.3.1(a)). If a smooth curve of the form α(u) =

(f (u), 0, g(u)) in the xz−plane is revolved about the z−axis, the re-

sulting surface of revolution is given by

X(u, v) = (f(u) cos v, f(u) sin v, g(u)).

Show that

X is regular provided f(u) = 0 and α

(u) = 0 for all u.

Solution. Well,

(u, v) =

∂

∂u

= (f

(u) cos v, f

(u) sin v, g

(u))

(u, v) =

∂

∂v

= (−f (u) sin v, f (u) cos v, 0).

If α

(u) = 0 for all u, then for a given u, either g

(u) = 0 or f

(u) = 0.

If g

(u) = 0 then

and

are linearly independent (in the third

component). If f

(u) = 0 and g

(u) = 0, consider:

= f

(u)f (u)(cos

v + sin

v)k = f

(u)f (u)k.

Since f (u) = 0 for all u, f

(u)f (u) = 0 and so

= 0 and

and

are linearly independent.

Definition. A vector v is a tangent vector to surface M at point

P if

there is a curve on M which passes through

P and has velocity vector

v at P. The set of all tangent vectors to M at P is the tangent plane

of M at

P , denoted T

Note. T

M is a 2-dimensional vector space with {

, v

)}

as a basis, where

X(u

, v

) =

P .

Definition. The curve

X(u, v

) is a u−parameter curve and

X(u

, v)

is a v−parameter curve of surface M (u

and v

are constants).

Note.

, v

) is a velocity vector of

X(u, v

) and

, v

) is a

velocity vector of

X(u

, v).

Example (Exercise 1.3.1(b)). Consider the surface of revolution

of Exercise 1.3.1(a). Describe the u− and v−parameter curves and

show they intersect orthogonally (the u−parameter curves are called

meridians and the v−parameter curves are called parallels).

Solution. A u−parameter curve is of the form (f (u) cos v

, f(u) sin v

g(u)) and has direction

(u, v

) = (f

(u) cos v

, f

(u) sin v

, g

(u)). A

v−parameter curve is of the form (f(u

) cos v, f (u

) sin v, g(u

)) and has

direction

, v) = (−f(u

) sin v, f (u

) cos v, 0). If a u−parameter

curve and a v−paramter curve intersect at (u

, v

) then at this point

of intersection

, v

) ·

, v

) = −f (u

) cos v

sin v

+f (u

) cos v

sin v

+ g

) × 0 = 0.

Therefore the u−parameter and v−parameter curves are orthogonal

when they intersect.

Example (Exercise 1.3.2(e)). For the surface of revolution

X(u, v) =

(a sinh u cos v, a sinh u sin v, b cosh u), u = 0, sketch the profile curve

(v = 0) in the xz−plane, and then sketch the surface. Prove that

X is

regular and give an equation for the surface of the form g(x, y, z) = 0.

Solution. For the profile, with v = 0 we have x = a sinh u and z =
b cosh u. Since cosh

u − sinh

u = 1, we have

−

= 1,

x = 0.

So the profile and surface are:

Next,

∂

∂u

= (a cosh u cos v, a cosh u sin v, b sinh u)

∂

∂v

= (−a sinh u sin v, a sinh u cos v, 0).

and

are linearly independent since b sinh u = 0 for u = 0.

Therefore

X is regular. Since

x = a sinh u cos v

y = a sinh u sin v
z = b cosh u

then g(x, y, z) = a

− (b

+ b

) − a

= 0 is the equation of the

surface.

1.4 The First Fundamental Form

Note. Suppose M is a surface determined by

X(u, v) ⊂ E

and

suppose α(t) is a curve on M, t ∈ [a, b]. Then we can write α(t) =

X(u(t), v(t)) (then (u(t), v(t)) is a curve in R

whose image under

is α). Then

(t) =

∂

∂u

∂

∂v

= u

+ v

If s(t) represents the arc length along α (with s(a) = 0) then

s(t) =

(r)dr

and

= α

(t)

= α

(t)

= α

· α

= (u

+ v

) · (u

+ v

)

= u

(

) + 2u

(

) + v

(

Following Gauss’ notation (brieﬂy) we denote

E =

F =

G =

and have

= E

+ 2F

+ G

or in diﬀerentialnotation

= E(du)

+ 2F (du dv) + G(dv)

Definition. Let M be a surface determined by

X(u, v). The first

fundamental form (or more commonly metric form) of M is

(ds)

as deﬁned above.

Definition. A property of a surface which depends only on the metric

form of the surface is an intrinsic property.

Note. The idea of an intrinsic property is that a “resident” of the sur-

face can detect such a property without appealling to a “larger space”

in which the surface is imbedded. Certainly an inhabitant of a surface

can measure distance within the surface.

Example 10, page 32. Consider the xy−plane described as

X(u, v) =

(u, v, 0) where u ∈ R and v ∈ R. Then

= (1, 0, 0) and

= (0, 1, 0).

E =

= 1,

F =

= 0,

G =

= 1.

Then the ﬁrst fundamentalform is

or, in terms of x and y:

Of course, this is the “usual” expression for the diﬀerential of arclength

from Calculus 2.

Definition. The matrix of the first fundamental form of a surface M

determined by

X(u, v) is








E F
F G








≡















where E, F , G are as deﬁned as above.

Note. This matrix determines dot products of tangent vectors. If

v = a

+ b

and

w = c

+ d

are vectors tangent to a surface M

at a given point, then

v · w = (a

+ b

) · (c

+ d

) = Eac + F (ad + bc) + Gbd

= (a, b)








E F
F G






















Notation. We now replace the parameters u and v with u

and u

We then have

= g

(du

)

+ 2g

+ g

(du

)

i,j

where the summation is taken (throughout this chapter) over the set
{1, 2}. In Chapter 3, we will sum over {1, 2, 3, 4}. If v is a vector
tangent to M at a point

P and v = (v

, v

) in the basis {

} for

the tangent plane at

P , then we have

v =

If α(t) is a curve on M where α is represented by

X(u

(t), u

(t)) then

(t) = u

(t)

+ u

(t)

Notation. We denote the ij entry of (g

)

−1

as g

. Therefore (g

)(g

) =

I and

= δ

(the ik entry of I) where












1 if i = k

0 if i = k

Example (Exercise 1.4.3(c)). For the surface

X(u, v) = (u cos v, u sin v, bv)

(the helicoid of Example 9), compute the matrix (g

), its determinate

g, the inverse matrix (g

) and the unit normalvector

Solution. Well

∂

∂u

= (cos v, sin v, 0)

∂

∂v

= (−u sin v, u cos v, b)

and so

= cos

v + sin

v + 0 = 1

= u

sin

v + u

cos

v + b

= u

+ b

= −u cos v sin v + u cos v sin v + 0 = 0 = g

Therefore

G =






















0 u

+ b








and g = det(g

) = u

+ b

. Then

−1















1
g








−g








+ b








+ b






















Now the unit normalvector is

U =

and

cos v

sin v

−u sin v u cos v b

= (b sin v, −b cos v, u cos

v + u sin

v) = (b sin v, −b cos v, u).

Now

sin

v + b

cos

v + u

√

+ u

Therefore

U =

b sin v

√

+ u

−b cos v

√

+ u

√

+ u

Definition. Suppose Ω is a closed subset of the u

−plane and that

X : Ω → E

is smooth (i.e. has continuous ﬁrst partials), is one-to-one

and regular (i.e.

and

are linearly independent) on the interior

of Ω. Then the area of the surface

X(Ω) is

A =

Ω

√

gdu

(See page 37 of the text for motivation of this deﬁnition.)

Example (Exercise 1.4.6). (a) Show that the area A of the surface

of revolution

X(u, v) = (f(u) cos v, f(u) sin v, g(u)) where u ∈ [a, b] and

v ∈ [0, 2π] is given by

A = 2π

|f(u)|

(u)

+ g

(u)

du.

(b) Show that the area of the surface obtained by revolving the graph
y = f(x) for x ∈ [a, b] about the x−axis is given by

A = 2π

|f(x)|

1 + f

(x)

dx.

Solution. (a) Consider the surface area of the surface of revolution

X(u, v) = (f(u) cos v, f(u) sin v, g(u)). We have (from Exercise 1.4.5)

= |f(u)|

(u)

+ g

(u)

and so

A =

Ω

du dv

(see page 37)

2π

|f(u)|

(u)

+ g

(u)

dv du

= 2π

|f(u)|

(u)

+ g

(u)

du.

(b) If y = f (x), x ∈ [a, b] where a ≥ 0 is revol ved about the x−axis,

then we have:

This is equivalent to taking

X(u, v) = (f(u), 0, u) (that is, the curve

x = f(z) in the xz−plane) and revolving it about the z−axis):

Then by Exercise 1.3.1, the surface is

X(u, v) = (f(u) cos v, f(u) sin v, u).

So by part (a), the surface area is

A = 2π

|f(u)|

(u) + 1 du = 2π

|f(x)|

(x) + 1 dx.

1.5 The Second Fundamental Form

Notation. We adopt the Einstein summation convention in which any

expression that has a single index appearing both as a subscript and a

superscript is assumed to be summed over that index.

Example. We denote

as v

Example. We denote

i,j

as g

Note. We have treated a path α(t) along a surface M as if it were the

trajectory of a particle in E

. We then interprete α

(t) as the accelera-

tion of the particle. Well, a particle can accelerate in two diﬀerent ways:

(1) it can accelerate in the direction of travel, and (2) it can accelerate

by changing its direction of travel. We can therefore decompose α

into two components, α

(representing acceleration in the direction of

travel) and α

(representing acceleration that changes the direction of

travel). You may have dealt with this in Calculus 3 by taking α

the component of α

in the direction of α

(computed as







and α

as the “remaining component” of α (that is, α

= α

− α

This is reminiscent of the Frenet formulas or the Frenet frame (

N, B)

of Exercise 1.1.14).

Note. With α parameterized in terms of arc length s, α = α(s) =

X(u

(s), u

(s)) we have the unit tangent vector

T (s) = α

(s) = u

We saw in Section 1.1 that α

(s) =

(s) is a vector normal to α

(

= k

N - see Exercise 1.1.14). In this section, we again decompose

into two orthogonal components, but this time we make explicit use

of the surface M. We wish to write

= α

tan + α

nor

where α

tan is the component of α

tangent to M and α

nor is the com-

ponent of α

normal to M. Notice that α

tan will be a linear combina-

tion of

and

(they are a basis for the tangent plane, recall) and

nor will be a multiple of the unit normal vector to M,

U (calculated

U =

Note. Since α(s) =

X(u

(s), u

(s)) and α

= u

(here, means

d/ds), then

= u

+ u

= u

+ u

Now u

is part of α

tan, but u

may also have a component in the

tangent plane. Well,

(s), u

(s))

∂

∂u

∂

∂u

∂

∂u

∂

∂u

∂

∂u

If we denote

∂

∂u

(we have assumed continuous second par-

tials, so the order of diﬀerentiation doesn’t matter) then we have

. So acceleration becomes

= u

+ u

We now need only to write

in terms of a component in the tangent

plane (and so in terms of

and

) and a component normal to the

tangent plane (which will be a multiple of

U).

Definition. With the notation above, we deﬁne the formulae of Gauss

= Γ

+ L

That is we deﬁne L

as the projection of

in the direction

U. Notice,

however, that Γ

may not be the projection of

onto

since the

’s are not orthonormal.

Note. Since projections are computed from dot products, we immedi-

ately have that

· U =

Note. We therefore have

= α

tan + α

nor =

+ Γ

Definition. The second fundamental form of surface M is the matrix















(Notice this diﬀers from the text’s deﬁnition on page 44.) We denote

the determinate of this matrix as L.

Note. The second fundamental form is a function of u and v. Also,

since we have

, it follows that L

= L

and so (L

) is a

symmetric matrix.

Note. We will see that the second fundamental form reﬂects the ex-

trinsic geometry of surface M (that is, the way M is imbedded in E

“how it curves relative to that space” as the text says).

Example (Exercise 1.5.2). Compute the second fundamental form

of the surface of revolution

X(u, v) = (f(u) cos v, f(u) sin v, g(u)).

Solution. Well

∂

∂u

= (f

(u) cos v, f

(u) sin v, g

(u))

∂

∂v

= (−f (u) sin v, f (u) cos v, 0)

and so (from Exercise 1.4.5)

U =

f(u)

|f(u)|

(u)

+ g

(u)

(−g

(u) cos v, −g

(u) sin v, f

(u)).

Next,

∂

= (f

(u) cos v, f

(u) sin v, g

(u))

∂

= (−f (u) cos v, −f (u) sin v, 0)

∂

∂u∂v

= (−f

(u) sin v, f

(u) cos v, 0) =

· U =

f(u)

|f(u)|

(u)

+ g

(u)

(−f

(u)g

(u) cos

−f

(u)g

(u) sin

v + f

(u)g

(u))

f(u)(f

(u)g

(u) − f

(u)g

(u))

|f(u)|

(u)

+ g

(u)

· U =

(f (u)g

(u) cos v sin v − f (u)g

(u) cos v sin v + 0)f (u)

|f(u)|

(u)

+ g

(u)

= 0 = L

· U =

f(u)

|f(u)|

(u)

+ g

(u)

(f (u)g

(u) cos

+f (u)g

(u) sin

v + 0)

f(u)

(u)

|f(u)|

(u) + g

(u)

|f(u)|g

(u)

+ g

(u)

Therefore the Second Fundamental Form is

L = det L

= L

− L

f(u)(f

(u)g

(u) − f

(u)g

(u))

|f(u)|

(u)

+ g

(u)

|f(u)|g

(u)

+ g

(u)

f(u)g

(u)(f

(u)g

(u) − f

(u)g

(u))

(u)

+ g

(u)

Definition. Let v = v

be a unit vector tangent to M at

P . The

normal curvature of

M at P in the direction v, denoted k

(v) is

(v) = L

where v = (v

, v

Example (Exercise 1.5.5). Find the normal curvature of the surface
z = f(x, y) at an arbitrary point, in the direction of a unit tangent
vector (a, b, c) at that point.

Solution. We have

∂

∂u

= (1, 0,

∂f
∂u

(u, v)) = (1, 0, f

)

∂

∂v

= (0, 1,

∂f

∂v

(u, v)) = (0, 1, f

i j k

1 0 f

0 1 f

= (−f

, −f

, 1)

and

)

+ (f

)

+ 1. Therefore

U =

)

+ (f

)

+ 1

(−f

, −f

, 1).

Now

∂

∂u

= (0, 0, f

)

∂

∂u ∂v

= (0, 0, f

) =

∂

∂v

= (0, 0, f

)

and so

· U =

)

+ (f

)

+ 1

· U =

)

+ (f

)

+ 1

· U =

)

+ (f

)

+ 1

= L

Now v = v

= v

(1, 0, f

) + v

(0, 1, f

) = (a, b, c), implying that

= a and v

= b. Hence

(v) = L

= L

+ 2L

+ L

)

+ (f

)

+ 1

+ 2abf

+ b

Note. If α =

X(u

(s), u

(s)) is a curve on M,

P is a point on M

with α(s

) =

P and v = α

) then α

) = u

)

), u

))

and so v

= u

) (see page 35 for representation of a tangent vector:

v = v

). Therefore

(v) = L

= L

Now α

= u

(equation (16), page 43), and

U =

· U =

+ u

= 0 + u











= u

Hence k

(v) = α

· U. This equation is used in Exercise 1.5.6.

1.6 The Gauss Curvature in Detail

Note. We have deﬁned the normal curvature of a surface at a point

in the direction v: k

(v). Therefore, for a given point on a surface, there

are an inﬁnite number of (not necessarily distinct) curvatures (one for

each “direction”). We can think of k

(v) as a function mapping the

vector space T

(M) (the plane tangent to surface M at point

P ) into

R. That is k

: T

(M) → R. We need v to be a unit vector, so the

domain of k

is {v ∈ T

(M) | v = 1}. Therefore, k

is a continuous

functions on a compact set and by the Extreme Value Theorem (for

metric spaces), k

assumes a maximum and a minimum value.

Definition. Let M be a surface and

P a point on the surface. Deﬁne

=max k

(v) and k

=min k

(v) where the maximum and minimum

are taken over the domain of k

. k

and k

are called the principal cur-

vatures of M at

P , and the corresponding directions are called principal

directions. The product K = K(P ) = k

is the Gauss curvature of

M at P .

Theorem I-5. The Gauss curvature at any point

P of a surface M is

K( P) = L/g where L =det(L

) and g =det(g

Proof. First, if v = v

then

+ v

= (v

)

+ 2(v

)(v

)

+ (v

)

= g

(recall g

, see page 35).

Therefore ﬁnding extrema of k

(v) for v =1 is equivalent to ﬁnding

extrema of

k = k

(v) =

for v ∈ T

(M) and v = 0. If k

(v) is an extreme value of k, where

v = v

, then

∂k

∂v

∂k

∂v

= 0 at v (that is, the gradient of k is 0 -

however, this gradient is computed in a (v

, v

) coordinate system, not

(x, y)). Now

∂k

∂v

[2L

](g

) − (L

)[2g

]

)

for r = 1, 2 (the derivatives in the numerator follow from Exercise

1.5.1). Now k =

, so replacing L

with kg

gives

∂k

∂v

) − (kg

)2g

)

− 2kg

2(L

− kg

for r = 1, 2. So at an extreme value, (L

−kg

= 0 for i = 1, 2. This

is two linear equations in two unknowns (v

and v

). Since v is nonzero,

the only way this system can have a solution is for det(L

− kg

) = 0.

That is

det








− kg








= 0

or (L

− kg

)(L

− kg

) − (L

− kg

)(L

− kg

) = 0

or L

− kL

+ k

−L

+ kL

− k

= 0

or k

− g

) − k(g

+ g

−g

− g

) + (L

− L

) = 0

or k

g − k(g

+ g

− 2g

) + L = 0

since L

= L

, L =det(L

), and g =det(g

). So for extrema of k we

need

− k

+ g

− 2g

= 0.

Since k

and k

are known to be roots of this equation, this equation

factors as (k − k

)(k − k

) = k

− (k

+ k

)k + k

= 0. Therefore, the

Gauss curvature is k

= L/g.

Note. L is the Second Fundamental form and g is the determinate

of the First Fundamental Form. We now see good evidence for these

being called “Fundamental” forms.

Example (Example 14, page 45 and Example 16, page 51).

Consider the surface

X(u, v) = (u, v, f(u, v)). Then

=(1, 0, f

=(0, 1, f

=(0, 0, f

), and

(0, 0, f

). With g

we have

) =








1 + f








and so g =det(g

) = 1 + f

+ f

. Now

i j k

1 0 f

0 1 f

= (−f

, −f

, 1)

and

U =

√g =

√g(−f

, −f

, 1).

Next, L

· U, so

√gf

Therefore L =det(L

) =

1
g

− (f

)

). So the Gauss Curvature

− (f

)

− (f

)

(1 + f

+ f

)

Note. You may recall from Calculus 3 that a critical point of z =
f(x, y) was tested to see if it was a local maximum or minimum by
considering D = f

− (f

)

at the critical point. If D < 0, the

surface has a saddle point. If D > 0 and f

> 0, it has a local

minimum. If D > 0 and f

< 0, it has a local maximum. This all

makes sense now in the light of curvature!

Theorem. If v and

w are principal directions for surface M at point

P corresponding to k

(maximum normal curvature at

P ) and k

(min-

imum normal curvature at

P ) respectively, then if k

= k

we have v

and

w orthogonal.

Proof. Let v = v

and

w = w

. As in Theorem I-5 (equation (24),

page 50)

− k

= 0 for i = 1, 2, and

− k

= 0 for i = 1, 2.

The ﬁrst of these equations is equivalent to

= k

for j = 1, 2

and since L

= L

and g

= g

= k

for j = 1, 2.

(25)

The second of these equations implies

− k

= 0

(we now sum over i = 1, 2). So

− k

= 0

and from (25) we have

− k

= 0

− k

= 0.

Now v ·

w = g

(see page 35). Since k

− k

=0, it must be that

v · w =0.

Note. We are now justiﬁed in refering to “two” principal directions.

When we consider the Gauss curvature at a point, we deal with the

normal curvature k

(v) at this point, where v = v

(i takes on the

values 1 and 2). So our collection of directions is a two dimensional

space. Since we have shown (for k

= k

) that the direction in which

(v) equals k

and the direction in which k

(v) equals k

are orthogo-

nal, there can be ONLY ONE direction in which k

(v) equals k

(well,

. . . plus or minus) and similarly for k

. In the event that k

= k

, we

choose two directions v and

w as principal directions where v · w = 0.

Definition. Suppose

P =

X(u

, u

) and let Ω be a neighborhood

of (u

, u

) on which

X is one-to-one with a continuous inverse

−1

X(Ω) → Ω. Deﬁne U(u

, u

) to be a unit normal vector to the surface

M determined by

X at point

X(u

, u

) (recall that

U =

). Therefore U :

X(Ω) → S

U is called the sphere mapping or

Gauss mapping of

X(Ω). The image of

X(Ω) under U (a subset of S

)

is the spherical normal image of

X(Ω).

Example (Exercise 9 (d), page 57). The spherical normal image

of a torus (see Example 12, page 34) is the whole sphere S

(there is a

normal vector pointing in any direction - in fact, the sphere mapping

is two-to-one).

Lemma I-6.

× U

= K(

Proof. Deﬁne L

= L

, u

) = L

for i, j = 1, 2. Notice

= (L

= L

(recall (g

) is the inverse of (g

)). Since

U · U =1, U · U

=0 (product

rule) and so

is tangent to M. Therefore

is a linear combination

and

= a

for j = 1, 2

for some coeﬃcients a

. Since

U is normal to M and

is tangent to

M (at a given point) then U ·

=0. Diﬀerentiating this equation with

respect to u

gives

U ·

= 0 and so

= −

U ·

= −L

(this last equality follows from equation (2), page 44). So

−L

= a

for j, k = 1, 2 (recall the deﬁnition of g

). We now solve these four

equations (j, k = 1, 2) in the four unknowns a

−L

= a

(j, k = 1, 2)

−g

= a

(i, j = 1, 2).

Therefore (by the deﬁnition of L

) a

= −L

. We now see how U

and

relate:

= −L

for j = 1, 2.

From these relationships:

× U

= (−L

) × (−L

)

= (−L

− L

) × (−L

− L

)

= (L

− L

)

(recall v × v =0)

=det(L

)

Since L

= L

, then det(L

) = det(L

) det(g

) and since (g

) is

the inverse of (g

det(g

) =

det(g

)

1
g

and so

det(L

) =

det(L

)

det(g

)

= K.

Therefore,

× U

= K(

Definition. For a surface determined by

X(u

, u

), with

and

deﬁned as above, the equations

= −L

for j = 1, 2 are the

equations of Weingarten.

Note. For Ω a neigborhood of (u

, u

) on which

X is one-to-one with

a continuous inverse, the set

X(Ω) is a connected region on M. The

spherical normal image of

X(Ω), U(Ω) is a region on S

(see Figure

I-26, page 52). If the curvature of

X(Ω) varies little then the area of

U(Ω) will be small. In fact, if X(Ω) is part of a plane, then the area of
U(Ω) is zero. In fact, for Ω small, the ratio of the area of U(Ω) to the

area of

X(Ω) approximates the curvature of M on Ω.

Note. The tangent plane to S

U(u

, u

), T

, is parallel (that is,

has the same normal vector) to the tangent plane to M at

X(u

, u

M. If U

× U

= 0 (i.e. if U

and

are linearly independent)

then

× U

and

U are both unit normal vectors to S

at the point

U and do can diﬀer at most in sign. That is, U = ±

× U

= ±

× U

or U · U

× U

= ±

× U

(recall U · U =1).

Note. If (

× U

)(u

, u

) = 0 then

U is regular at (u

, u

) (by deﬁni-

tion) and therefore (by the comment on page 24)

U is one-to-one with

a continuous inverse on suﬃciently small Ω, a neighborhood of (u

, u

Also, with Ω suﬃciently small,

U · U

× U

will be the same multiple of

× U

(namely +1 or −1). By equation (13), page 37,

Area U(Ω) =

Ω

× U

Area

X(Ω) =

Ω

Now

U · X

· (

) =

Also, we refer to

Ω

U · U

× U

as the signed area of

U(Ω)

(recall it is ±area of

U(Ω)). Therefore

signed area

U(Ω) =

Ω

U · U

× U

area

X(Ω) =

Ω

U · X

Note. If (

× U

)(u

, u

) = 0, then notice that

U · U

× U

may change

sign and

U may not be one-to-one over Ω and

Ω

U · U

× U

then represents a “net area” of

U(Ω). In all these cases, we denote

Ω

U · U

× U

as “Area

U(Ω)” even though this is a bit of a misnomer.

Theorem. Suppose M is a surface determined by

X(u

, u

) and

P =

X(u

, u

) is a point on M. Let Ω be a neighborhood of (u

, u

) on which

X is one-to-one with continuous inverse. Let U(Ω) be the spherical
normal image of

X(Ω). Then

K(P ) =lim

Ω→(u

)

Area

U(Ω)

Area

X(Ω)

Here “Area

U(Ω)” is as discussed above. The limit is taken in the sense

that

sup{ dist (ω, (u

, u

)) | ω ∈ Ω}

approaches zero.

Proof. Let ! > 0. Then there exists δ

> 0 such that for Ω a ball with

center (u

, u

) and radius δ

we have

Area

U(Ω) − (U · U

× U

)(

P) Area(Ω)

Ω

U · U

× U

− (U · U

× U

)(

P ) Area(Ω)

< !

(since

U · U

× U

is continuous and Ω is connected). A similar result

holds for Area

X(Ω). Therefore, for Ω suﬃciently small,

Area

U(Ω)

Area

X(Ω)

−

U · U

× U

U · X

(

P )

< !.

That is,

lim

Ω→(u

)

Area

U(Ω)

Area

X(Ω)

U · U

× U

U · X

By Lemma I-6,

U · U

× U

U · X

U · K( X

)

U · X

= K

and the result follows.

Example (Exercise 8 (a), page 56). Let

X =

X(u, v) where (u, v) ∈

D be a parameterization of a surface M. The (signed) area of the
spherical normal image of M,

U · U

× U

du dv, is called the total

curvature of M (assuming the integral, which may be improper, exists).

Show that the total curvature of M is

√

g du dv (remember, K

and g are functions of u and v).

Solution. By Lemma I-6,

× U

= K(

). Therefore

U · U

× U

U · K(

Now the unit normal vector is

U =

, so

U · U

× U

) · (

)

= K

By equation (10), page 35,

√g = X

. Therefore

U · U

× U

= K

√

and the total curvature of M over D is

U · U

× U

du dv =

√

g du dv.

Example (Exercise 9 (d), page 57). Compute the total curvature

of the torus

X(u, v) = ((R + r cos u) cos v, (R + r cos u) sin v, r sin u).

Solution. From Example 12, page 34, and Exercise 1.4.3 (d), page 38,

U = (− cos u cos v, − cos u sin v, − sin u).

∂ U

∂u

=(sin u cos v, sin u sin v, − cos u)

∂ U

∂v

=(cos u sin v, − cos u cos v, 0).

Therefore

× U

= (− cos

u cos v, − cos

u sin v, − sin u cos u cos

− sin u cos u sin

= (− cos

u cos v, − cos

u sin v, − sin u cos u)

and

U · U

× U

=cos

u cos

v + cos

u sin

v + sin

u cos u

=cos

u + sin

u cos u.

So the total curvature is

−π

(cos

u + sin

u cos u) du dv.

Now cos

u+sin

u cos u is an even function, so the integral is 0 and the

total curvature is 0.

1.7 Geodesics

Note. A curve α(s) on a surface M can curve in two diﬀerent ways.

First, α can bend along with surface M (the “normal curvature” dis-

cussed above). Second, α can bend within the surface M (the “geodesic

curvature” to be deﬁned).

Recall. For curve α on surface M, α

can be written as components

tangent and normal to M as α

= α

tan + α

nor where

tan = (u

+ Γ

)

nor = (L

)

and the parameters on the right hand side are deﬁned in Section 5.
α

nor reﬂects the curvature of α due to the bending of M and α

tan

reﬂects the curvature of α within M. Now

tan ·

U =

+ Γ

)

· U = 0

(recall

U =

X ×

) and

tan · α

= α

tan · α

+ 0 = α

tan · α

+ α

nor · α

(recall α

= u

and

· U = 0)

= (α

tan + α

nor) · α

= α

· α

= 0

(recall α

= α

(s) = 1 and = d/ds).

Therefore α

tan is orthogonal to both

U and α

. If we deﬁ ne

w as the

unit vector

w = U × α

, then α

tan is a multiple of

w (and w is a vector

tangent to M).

Definition I-7. Let α(s) be a curve on M where s is arc length. The

geodesic curvature of α at α(s) is the function k

= k

(s) deﬁ ned by

tan = k

w = k

(

U × α

Recall. The scalar triple product of three vectors (in R

) satisﬁes:

(

A × B) · C = ( B × C) ·

A = ( C ×

A) · B.

Theorem. The geodesic curvature k

of curve α in surface M can be

calculated as

U · α

× α

Proof. Since k

w = α

tan we have

w · w = α

tan ·

w = α

tan · (

U × α

)

= (α

tan + α

nor) ·

U × α

(since α

nor is parallel to

= α

· U × α

U · α

× α

Definition I-8. Let α = α(s) be a curve on a surface M. Then α is a

geodesic if α

tan =

0 ( or equivalently, if α

= α

nor) at every point of

α.

Note. A geodesic on a surface is, in a sense, as “straight” as a curve

can be on the surface. That is, α has no curvature within the surface.

For example, on a sphere the geodesics are great circles.

Note. If α is a geodesic on M then

+ Γ

= 0

for r = 1, 2 and

U · α

× α

= 0.

(We’ll use these LOTS!)

Example (Exercise 1.7.4(a)). Prove that on a surface of revolution,

every meridian is a geodesic.

Proof. Suppose

X(u, v) = (f(u) cos v, f(u) sin v, g(u)).

Let

m(s) = (f(s) cos v, f(s) sin v, g(s)) be a meridian of the surface

(where we assume the curve has been parameterized in terms of ar-

clength s). Then

(s) = (f

(s) cos v, f

(s) sin v, g

(s))

= (f

(s) cos v, f

(s) sin v, g

(s))

× m

= ((f

(s)g

(s) − f

(s)g

(s)) sin v, (−f

(s)g

(s) + f

(s)g

(s)) cos v, 0).

Now

= (−f (s)g

(s) cos v, −f (s)g

(s) sin v, f

(s)f (s))

and so

U =

(−f (s)g

(s) cos v, −f (s)g

(s) sin v, f

(s)f (s))

(f (s)g

(s))

+ (f

(s)f (s))

Therefore

U · m

× m

(f (s))

{(g

(s))

+ (f

(s))

}

×{(f

(s)g

(s) − f

(s)g

(s))(−f (s)g

(s)) cos v sin v

+(−f

(s)g

(s) + f

(s)g

(s))(−f (s)g

(s)) cos v sin v, 0)}

(f (s))

{(g

(s))

+ (f

(s))

}

(0) = 0.

Therefore

m(s) is a geodesic.

Definition. Let

X(u

, u

) be a surface and let g

(see page 34) and

(see page 43) be as deﬁned in Sections 4 and 5. The Christoﬀel

symbols of the ﬁrst kind are

ijk

= Γ

for i, j, k = 1, 2.

Definition. The Γ

deﬁned in Section 1.5 are the Cristoﬀel symbols

of the second kind.

Note. Since Γ

= Γ

(see (17), page 43) then Γ

ijk

= Γ

jik

. Also, since

)

−1

= (g

), we have Γ

= Γ

ijk

Theorem. Let

X(u

, u

) be a surface and let g

and Γ

be as deﬁned

in Sections 4 and 5. Then

ijk

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

and

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

Proof. Since

= Γ

+ L

U (by deﬁnition, see page 43) then

= Γ

+ (L

U) · X

= Γ

+ 0 = Γ

ijk

establishing the ﬁrst identity (recall g

). Next,

∂g

∂u

∂

∂u

[

] =

= Γ

ijk

+ Γ

kji

Permuting the indices:

∂g

∂u

= Γ

jki

+ Γ

ikj

and

∂g

∂u

= Γ

kij

+ Γ

jik

Now

ijk

1
2

(2Γ

ijk

) =

1
2

(Γ

ijk

+ Γ

jik

)

1
2

(Γ

ijk

+ Γ

kji

− Γ

kji

+ Γ

kij

− Γ

kij

+ Γ

jik

)

1
2

{(Γ

ijk

+ Γ

kji

) + (Γ

kij

+ Γ

jik

) − (Γ

jki

+ Γ

ikj

)}

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

and the second identity is established. Finally, multiplying this identity

on both sides by g

, summing over k and using the deﬁnition of Γ

have

= Γ

ijk

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

(recall (g

= (g

)

−1

), and the third identity is established.

Note. Since the Christoﬀel symbols depend only on the metric form

(or First Fundamental Form), they are part of the intrinsic geometry

of the surface M.

Definition. Let

X(u

, u

) be a surface. Then the coordinates

and

are orthogonal if g

= g

= 0. (This makes sense since g

Corollary. Let

X(u

, u

) be a surface and let g

and Γ

be as deﬁned

in Sections 4 and 5. If

and

are orthogonal coordinates, then

∂g

∂u

∂g

∂u

−

∂g

∂u

(no sums over any of i, j, r).

Proof. Since g

= g

= 0, then g

= g

= 0 and g

= 1/g

. The result follows from the above theorem.

Corollary. With the hypotheses of the previous corollary (with i, j, r =

1, 2), when j = r

∂g

∂u

1
2

∂

∂u

[ln g

]

and when i = j = r

−

∂g

∂u

Proof. Follows from g

= g

= 0.

Note. By symmetry, Γ

= Γ

, and so the previous two corollaries

cover all possible cases when i, j, r ∈ {1, 2} (i.e. when we deal with

two dimensions). In dimensions 3 and greater (in particular, in the 4

dimensional spacetime of Chapter III) we have a third case which we

state now, and address in detail later:

Theorem. In dimensions 3 and greater, if coordinates are mutually

orthogonal, then for i, j, r all distinct, Γ

= 0. (In the event that one

or more of i, j, r are equal, the above corollaries apply.)

Note. In the case of orthogonal coordinates, if we return to Gauss’

notation:

= E,

= g

= F = 0,

= G

we have the First Fundamental Form (or metric form) ds

= Edu

Gdv

on surface

X(u, v). In this notation, the Christoﬀel symbols are

then

= Γ

= −

Example 17, page 62. In the Euclidean plane, ds

= du

+ dv

Therefore E = G = 1 and all the Christoﬀel symbols are 0. Therefore

a geodesic α satisﬁes

+ Γ

= 0

for r = 1, 2, or u

= 0 for r = 1, 2. That is, u

= u

= 0 and

= v

= 0. Therefore u(s) = as + b and v(s) = cs + d for some

a, b, c, d. Therefore, geodesics in the Euclidean plane are straight lines.

Note. We will show in Theorem I-9 that the shortest path on a surface

joining two points is a geodesic. This theorem, combined with the pre-

vious example PROVES that the shortest distance between two points

in a plane is a straight line. Oddly enough, you’ve probably never seen

this PROVEN before!

Example 18, page 62. Consider a sphere of radius r with “geographic

coordinates” (like latitude and longitude) u and v. Then the sphere is

given by

X(u, v) = (r cos u cos v, r sin u cos v, r sin v)

(see Example 7, page 23). The metric form is (see page 33) ds

cos

vdu

+ r

(since there is no du dv term, F = g

= g

= 0

and these coordinates are orthogonal). Therefore E = r

cos

v and

G = r

(a constant). Then E

= G

= 0 and the nonzero

Christoﬀel symbols are

= Γ

−2r

cos v sin v

cos

= − tan v

−E

cos v sin v

= cos v sin v.

It is shown (and not trivially) in Exercise 14 that this implies geodesics

are great circles.

Note. In Example 19 page 62, it is shown that the Euclidean plane

when equipped with polar coordinates (which are orthogonal coordi-

nates) yields geodesics which are lines (as expected).

Note. In general, to determine the geodesics for a surface, requires

that one solve diﬀerential equations. This can be diﬃcult (sometimes

impossible to do in terms of elementary functions). In Chapter III

we will compute some geodesics in 4-dimensional spacetime (in fact,

planets and light follow geodesics if 4-D spacetime).

Theorem I-9. Let α(s), s ∈ [a, b] be a curve on the surface M :

X(u

, u

), where s is arclength. If α is the shortest possible curve on

M connecting its two end points, then α is a geodesic.

Idea of Proof. We will vary α(s) by a slight amount . Then com-

paring the arclength of α from α(a) to α(b) with the arclength of the

slightly varied curve from α(a) to α(b) and assuming α to yield the

minimal arclength, we will show that α satisﬁes equation (32a) (page

59) and is therefore a geodesic.

Proof. Let α(s) =

X(u

(s), u

(s)). Consider the family of curves α

(s)

of the form

(s, ) = u

(s) + v

(s)

for i = 1, 2, s ∈ [a, b] where v

are smooth functions with v

(a) = v

(b) =

0 for i = 1, 2 (so (U

, U

) still joins α(a) and α(b)), (U

, U

) ⊂ M, but

otherwise v

are arbitrary.

Let L( ) denote the length of α

L( ) =

λ(s, ) ds

where

λ(s, ) =






, U

)

∂U

∂s

∂U

∂s






1/2

(the square root of the metric form of M along α

). Now L has a

minimum at = 0 so

[L( )] =

λ(s, ) ds

∂

[λ(s, )] ds

(since λ and ∂λ/∂ are continuous) satisﬁes

(0) =

∂

[λ(s, 0)] ds = 0.

Now

∂λ

∂









, U

)

∂U

∂s

∂U

∂s




1/2





1
2

(λ(s, ))

−1






∂

, U

)]

∂U

∂s

∂U

∂s

, U

)

∂




∂U

∂s




∂U

∂s

+ g

, U

)

∂U

∂s

∂




∂U

∂s









2λ(s, )









∂

∂U

, U

)]

∂U

∂

∂U

, U

)]

∂U

∂




∂U

∂s

∂U

∂s

+ 2g

, U

)

∂U

∂s

∂




∂U

∂s









2λ(s, )









∂

∂U

, U

)]

∂U

∂




∂U

∂s

∂U

∂s

+2g

, U

)

∂U

∂s

∂

∂ ∂s






2λ(s, )






∂g

∂U

∂s

∂U

∂s

+ 2g

∂U

∂s

∂

∂ ∂s






since

∂U

∂

= v

. With = 0,

∂U

∂

= v

and λ(s, 0) = 1 (s is arclength

on α = α

) we have

∂λ

∂

(s, 0) =

1
2

∂g

∂U

+ 2g

and since = 0 implies U

= u

, then

∂λ

∂

(s, 0) =

1
2

∂g

∂u

+ 2g

and so

(0) =

1
2

∂g

∂u

+ 2g

ds = 0.

Now by Integration by Parts

Let u = 2g

and dv = v

ds.

Then du =

∂

∂s

[2g

]ds and v =

(s)ds = v

−

∂

∂s

[2g

= 0 −

∂

∂s

[2g

ds since v

(a) = v

(b) = 0.

Therefore

(0) =

1
2

∂g

∂u

−

∂

∂s

[2g

1
2

∂g

∂u

−

∂

∂s

[2g

]

= 0.

Since the integral must be zero for all arbitrary v

, then the remaining

part of the integrand must be zero:

1
2

∂g

∂u

−

∂

∂s

] = 0

for k = 1, 2. Now when = 0, U

= u

and

∂

∂s

] =

∂

∂s

, u

]




∂g

∂u

∂g

∂u




+ g

, u

)

∂g

∂u

∂g

∂u

+ g

, u

∂g

∂u

+ g

∂g

∂u

+ g

Therefore

1
2

∂g

∂u

−

∂

∂s

] = 0

for k = 1, 2 implies

1
2

∂g

∂u

−

∂g

∂u

− g

= 0

for k = 1, 2, or using the notation of equation (35) (page 60)

1
2

(Γ

ikj

+ Γ

jki

) − (Γ

kji

+ Γ

ijk

)

− g

= 0

(∗)

for k = 1, 2. Since Γ

ikj

= Γ

jki

(interchanging dummy variables

i and j) and Γ

kji

= Γ

jki

(symmetry in the ﬁrst and second coordinates)

then

1
2

(Γ

ikj

+ Γ

jki

) − Γ

kji

1
2

(Γ

jki

+ Γ

jki

) − Γ

jki

= 0

and the above equation (∗) becomes

ijk

+ g

= 0

for k = 1, 2. Multiplying by g

and summing over k:

ijk

+ g

= 0

for r = 1, 2 or

+ u

= 0

for r = 1, 2. This is equation (32a) and therefore α is a geodesic of M.

Note. Again, Theorem I-9 along with Example 17 shows that the

shortest distance between two points in the Euclidean plane is a “straight

line.” Theorem I-9 along with Example 18 show that the shortest dis-

tance between two points on a sphere is part of a great circle (explaining

apparently unusual routes on international airline ﬂights).

Note. The converse of Theorem I-9 is not true. That is, there may be a

geodesic joining points which does not minimize distance. (Recall that

we set L

(0) ≡ 0, but did not check L

(0) - we may have a maximum

of L!) For example, we can travel the six miles from Johnson City to

Jonesboro (along a very small piece of a geodesic), or we can travel

in the opposite direction along a very large piece of a geodesic (∼

24, 000 miles) and travel around the world to get to Jonesboro (NOT

a minimum distance).

Note. Not all surfaces may allow one to create a geodesic joining

arbitrary points. For example, the Euclidean plane minus the origin

does not admit a geodesic from (1, 1) to (−1, −1).

Note. In the next theorem, we prove that for any point on a surface,

there is a unique (directed) geodesic through that point in any direction.

Theorem I-10. Given a point

P on a surface M and a unit tangent

vector v at

P , there exists a unique geodesic α such that α(0) = P and

(0) = v.

Proof. Let

P =

X(u

, u

) and v = v

, u

). We need two functions

(t), r = 1, 2 where












+ Γ

= 0

for r = 1, 2

(0) = u

, u

(0) = v

for r = 1, 2.

This is a system of two ordinary diﬀerential equations in two unknown

functions, each with two initial conditions. Such a system of IVPs has

a unique solution (check out the chapter of an ODEs book entitled

“Existence and Uniqueness Theorems”) u

(t) for r = 1, 2. We now

only need to establish that t represents arclength. With s equal to

arclength,

= E

+ 2F

+ G

= g

≡ f(t)

is the metric form and if we show this quantity is 1, then |t| = s and
t equals arclength (we need α(0) = v to eliminate the negative sign -
this is insured by the initial conditions). Well,

f(0) = g

, u

(0)u

(0) =

, u

) ·

, u

(0)u

(0)

, u

) ·

, u

)

, u

)

= v · v = v = 1.

Next,

(t) =

∂g

∂u

+ g

Since

∂g

∂u

= Γ

ikj

+ Γ

jki

(equation (35b), page 60)

= Γ

+ Γ

(equation (33), page 59)

= g

+ g

(symmetry of g

)

then

(t) = (g

+ g

= g

+ Γ

) + g

+ Γ

)

= 0 (from the conditions of the ODE).

Therefore f (t) is a constant and f (t) = 1. Hence

= f (t) = 1

and t = s (that is, t is arclength). Therefore α(s) =

X(u

(s), u

(s)) is

the desired geodesic.

Example (Exercise 1.7.14(a)). If M has metric form ds

= Edu

Gdv

with E

= G

= 0, then a geodesic on M satisﬁes

√

E − h

for some constant h. Use this above equation to show that a geodesic

on the geographic sphere

X(u, v) = (R cos u cos v, R sin u cos v, R sin v)

satisﬁes

h sec

√

− h

sec

h sec

√

− h

tan

where h is a constant.

Solution. First,

= (−R sin u cos v, R cos u cos v, 0)

= (−R cos u sin v, −R sin u sin v, R cos v)

E = g

= R

cos

G = g

= R

sin

v + R

cos

v = R

Then

√

cos

√

cos

v − h

R cos v

√

cos

v − h

since v ∈ (−π/2, π/2)

h sec v

cos v

√

− h

sec

h sec

− h

(1 + tan

h sec

√

− h

tan

Example (Exercise 1.7.14(b)). Substitute w = h tan v and integrate

the above equation to obtain cos(u − u

) + γ tan v = 0 where u

and γ

are constants.

Solution. With w = h tan v, dw = h sec

v dv and so

u =

sec

√

− h

tan

= −

−1

√

− h

− w

dw = − cos

−1

√

− h

+ u

Therefore

cos(u − u

) =

√

− h

h tan v

√

− h

With γ = −h/

√

− h

we have

cos(u − u

) + γ tan v = 0.

1.8 The Curvature Tensor and the

Theorema Egregium

Recall. A property of a surface which depends only on the metric

form is an intrinsic property. We have shown (Theorem I-5) that the

Gauss curvature at a point

P is K( P) = L/g where L is the Second

Fundamental Form and g is the determinate of the matrix of the First

Fundamental Form (or metric form). Therefore, to show that curvature

is an intrinsic property of a surface, we need to show that L is a function

of the g

(and their derivatives) which make up the metric form.

Recall. For a surface M determined by

X(u

, u

) the coeﬃcients of

the Second Fundamental Form are

· U =

(equation (20), page 44)

and

= L

(equation (27), page 54)

and the Christoﬀel symbols are

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

(equation (37), page 60). Also recall the formulas of Gauss

= Γ

+ L

U (equation (17), page 43)

and the formulas of Weingarten

= −L

(equation (28), page 55).

Lemma. The coeﬃcients of the Second Fundamental Form and the

Christoﬀel symbols are related as follows (for h = 1, 2):

∂Γ

∂u

−

∂Γ

∂u

+ Γ

− Γ

= L

− L

Proof. Diﬀerentiating the formulas of Gauss:

∂

∂u

∂Γ

∂u

+ Γ

∂

∂u

∂L

∂u

U + L

∂ U

∂u

or by deﬁning ∂/∂u

with a subscript of j

ikj

∂Γ

∂u

+ Γ

∂L

∂u

U + L

Using the formulas of Gauss and Weingarten to rewrite

and

get

ikj

∂Γ

∂u

+ Γ

(Γ

+ L

U) + ∂L

∂u

U + L

(−L

)

or (by interchanging h and r in the second term [since we are summing

over both])

ikj

∂Γ

∂u

+ Γ

(Γ

+ L

U) + ∂L

∂u

U + L

(−L

)




∂Γ

∂u

+ Γ

− L




∂L

∂u

U. (49)

Interchanging j and k gives

ijk




∂Γ

∂u

+ Γ

− L




∂L

∂u

U (50)

(so we have

ijk

broken into a component normal to surface M and

components which lie in the tangent plane to M at a given point -

namely the components in directions

and

). We have assumed

that

X is suﬃciently continuous that

ikj

ijk

and so

ikj

−

ijk

= 0.

Subtracting (50) from (49) and using the fact that the coeﬃcients of

and

in the resultant are 0 we have

∂Γ

∂u

−

∂Γ

∂u

+ Γ

− Γ

− L

+ L

= 0

for h = 1, 2 and the result follows.

Definition. For a surface M with Christoﬀel symbols as above, deﬁne

ijk

∂Γ

∂u

−

∂Γ

∂u

+ Γ

− Γ

These make up the Riemann-Christoﬀel curvature tensor (with h =

1, 2).

Note. Since the Christoﬀel symbols (Γ

’s) are intrinsic properties of

surface M, the Riemann-Christoﬀel curvature tensor is also an intrinsic

property of M.

Note. Interchanging j and k we trivially have R

ijk

= −R

ikj

Theorem I-11. Gauss’ Theorema Egregium.

The Gauss curvature of a surface is an intrinsic property. That is,

the Gauss curvature of a surface is a function of the coeﬃcients of the

metric form and their derivatives.

Proof. From the lemma and deﬁnition of R

ijk

we have

ijk

= L

− L

(54)

Now deﬁne R

mijk

= g

ijk

= g

ijk

. Then R

ijk

= g

mijk

. Now the

Riemann-Christoﬀel curvature symbols R

ijk

are intrinsic and therefore

mijk

are also intrinsic. Multiplying (54) by g

gives (summing over

h = 1, 2)

ijk

= g

− g

= g

− g

or R

mijk

= L

− L

since g

= L

(page 54, line after

equation (27)). In particular, with m, j = 1 and i, k = 2

1212

= L

− L

= L

− L

(since L

= L

- see equation (20), page 44)

= det(L

) = L.

Therefore, since R

mijk

are intrinsic, then L is intrinsic and K = L/g =

1212

/g is intrinsic!

Note. We now give an explicit equation for K in terms of the metric

form.

Corollary. For a surface M determined by

X(u, v) =

X(u

, u

) the

curvature is given by

K =

1
g

−

1
2

−

1
2

+ (Γ

− Γ

where

= E

= F = g

= G

g = det(g

)

and

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

where (g

)

−1

= (g

Proof. Since R

mijk

= g

ijk

(equation (55), page 76) and

ijk

∂Γ

∂u

−

∂Γ

∂u

+ Γ

− Γ

(equation (52), page 75) then

ijk

= g

∂Γ

∂u

+ g

− g

∂Γ

∂u

− g

mijk

= g

∂Γ

∂u

+ g

− g

∂Γ

∂u

− g

(since g

= g

)

= g

∂Γ

∂u

+ Γ

rjm

− g

∂Γ

∂u

− Γ

rkm

(∗)

since Γ

ijk

= Γ

(equation (33), page 59). Now, interchanging the

indices in equation (33) we have g

= Γ

ikm

or diﬀerentiating with

respect to u

∂g

∂u

+ g

∂Γ

∂u

∂Γ

ikm

∂u

∂Γ

∂u

∂Γ

ikm

∂u

− Γ

∂g

∂u

(∗∗)

Now using (∗∗) in (∗) we have

mijk

∂Γ

ikm

∂u

− Γ

∂g

∂u

+ Γ

rjm

−

∂Γ

ijm

∂u

− Γ

∂g

∂u

−Γ

rkm

Replacing r by h in the products of Γ’s gives

mijk

∂Γ

ikm

∂u

− Γ

∂g

∂u

+Γ

hjm

−

∂Γ

ijm

∂u

− Γ

∂g

∂u

−Γ

hkm

Now

ikm

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

(equation (36), page 60)

and

∂g

∂u

= Γ

hjm

+ Γ

mjh

(equation (35), page 60)

= Γ

hjm

+ Γ

(equation (33), page 59)

mijk

∂

∂u

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

+Γ

hjm

− Γ

(Γ

hjm

+ Γ

)

−

∂

∂u

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

−Γ

hkm

+ Γ

(Γ

hkm

+ Γ

)

1
2




∂

∂u

∂

∂u

−

∂

∂u




+Γ

hjm

− Γ

(Γ

hjm

+ Γ

)

−

1
2




∂

∂u

∂

∂u

−

∂

∂u




−Γ

hkm

+ Γ

(Γ

hkm

+ Γ

)

1
2




∂

∂u

−

∂

∂u

∂

∂u

−

∂

∂u




+(Γ

− Γ

So with m = j = 1 and i = k = 2

1212

1
2




∂

∂u

−

∂

∂u

∂

∂u

−

∂

∂u




+(Γ

− Γ

1
2

− E

+ F

− G

) + (Γ

− Γ

Since K = R

1212

/g (equation (57), page 76), and the result follows.

Corollary. For a surface M determined by

X(u, v) with orthogonal

coordinates (

= F = 0) the curvature is

K =

√

∂

∂u

√

∂

∂v

√

Proof. With F = 0 and equation (40) of page 62 (which gives the

Christoﬀel symbols in an orthogonal coordinate system in terms of E

and G) we have

K =

−

1
2

−

1
2

+ ((Γ

)

− Γ

+ ((Γ

)

− Γ

(since g

= g

= 0 and det(g

) = g

= EG)






−

1
2

−

1
2




−








−















−

1
2

−

1
2







E +












= −






1
2

−

+ EE

−

+ E






−1

2EG

√






√

EGE

√

EGG

−

√




+ E




−

√




+ E









−1

2EG

√






√

EGE

√

EGG

−

+ E

√

−

+ EE

√






−1

2EG

√






√

EGE

√

EGG

−

(EG

+ E

√

−

(EG

+ E

√






−1

√










√

EGG

−

(EG

√

EGE

−

(EG

√










−1

√

∂

∂u

√

∂

∂v

√

Note. The equation given in the previous corollary will be useful in

the exercises in this section.

Note. Some symmetry relations in R

mijk

are given at the end of the

section.

Example (Exercise 2, page 80). Let

X(u, v) = (f(u) cos v, f(u) sin v, g(u))

be a surface of revolution whose proﬁle curve α(u) = (f (u), 0, g(u)) has

unit speed. Show that K = −f

/f.

Solution. By Exercise 1.4.5, page 39, E = g

= (f

(u))

+ (g

(u))

F = g

= g

= 0 (coordinates are orthogonal), G = g

= (f (u))

. So

by equation (59), page 78,

K =

−1

(f (u))

{(f

(u))

+ (g

(u))

}






∂

∂u




2f (u)f

(u)

(f (u))

{(f

(u))

+ (g

(u))

}




∂

∂v

[0]






Now assuming α

(u))

+ (g

(u))

= 1 and f (u) ≥ 0:

K =

−1

2f (u)

∂

∂u

[2f

(u)]

−f

(u)

f(u)

Example (Exercise 5 (b), page 81). The pseudosphere may be

represented as the surface of revolution

X(u, v) =

a sin u cos v, a sin u sin v, a

cos u + ln

tan

for u ∈ (0, π/2). Show that K = −1/a

(and so the pseudosphere has

constant negative curvature).

Solution. In Exercise 5 (a), you will show that E = a

cot

u and

G = a

sin

u. Therefore by equation (59), page 78:

K =

−1

√

cot

u sin




∂

∂u




sin u cos u

√

cot

u sin




∂

∂v

[0]




−1

cot

u sin

∂

∂u




sin u cos u

cot

u sin




since u ∈ (0, π/2)

−1

cot

u sin

∂

∂u

[2 sin u]

−1

cot

u sin

(2 cos u) =

− cot u

cot u

−1

Note. In the zx−plane, the proﬁle curve of the pseudosphere is

We get the point (z, x) = (0, 1) for u = π/2. Let’s calculate the ar-

clength s for u ranging from π/2 to u

∗

s = −

∗

π/2

(u))

+ (z

(u))

du (since u

∗

< π/2)

π/2

∗

cos

(u) +




− sin u +

sec

(u/2)

2 tan(u/2)




π/2

∗

cos

u +




− sin u +

2 sin(u/2) cos(u/2)




π/2

∗

cos

u +

− sin u +

sin u

π/2

∗

cos

u + sin

u − 2 + csc

u d u

π/2

∗

√

csc

u − 1 du =

π/2

∗

| cot u| du

π/2

∗

cot u d u = ln(sin u)

π/2

∗

= − ln(sin u

∗

Therefore

exp(−arclength) = e

−s

= e

−(− ln(sin u

∗

))

= sin u

∗

= x

∗

So we have x = e

−s

where s is arclength. This curve is called a tractrix.

It can be generated by placing a box at point (0, 1) and dragging it

by attaching a 1 unit rope and pulling along the z−axis (therefore the

tangent line at any point meets the z−axis 1 unit from the point of

tangency):

1.9 Manifolds

Note. In this section, we extend the ideas of tangents, metrics, geo-

desics, and curvature to “manifolds” (in a sense, “n−dimensional sur-

faces”) without appealling to how they are imbedded in a higher di-

mensional space.

Definition. Let M be a non-empty set whose elements we call points.

A coordinate patch on M is a one-to-one function

X : D → M (contin-

uous and regular) from an open subset D of E

(or more generally E

)

into M.

Note. In the following deﬁnition, by “domain” of a function we mean

the largest set on which the function is deﬁned. By “smooth” we mean

suﬃciently diﬀerentiable for our purposes. Afunction whose domain is

empty is considered smooth.

Definition I-12. An abstract surface or 2−manifold (more generally,
n−manifold) is a set M with a collection C of coordinate patches on M
satisfying:

(a) M is the union of images of the patches in C (that is, if C = {

}

and

is deﬁned on set D

, then M =

)).

(b) The patches of C overlap smoothly, that is if

: D

→ M and

→ M are two patches in C, then (

)

−1

◦

and (

)

−1

◦

have open domains and are smooth.

(c) Given two points

and

of M, there exist coordinate patches

: D

→ M and

: D

→ M in C such that P

∈

) and

) ∩

) = ∅ (this is the Hausdorﬀ

property).

(d) The collection C is maximal. That is, any coordinate patch on

M which overlaps smoothly with every patch of C is itself in C.
(Notice that two disjoint coordinate patches “overlap smoothly”

by convention).

Definition. The collection C is called a diﬀerentiable structure on M

and patches in C are called admissible patches.

Note. If properties (a), (b), and (c) of Deﬁnition I-12 are satisﬁed by a

collection C

then we can adjoin to C

all patches that overlap smoothly

with the patches of C

to create a collection C which satisﬁes (a), (b),

(c), (d). In this case, C

is said to generate C.

Example (Exercise 1.9.1). Let M be the plane with Cartesian coor-

dinates. The identity mapping of M onto itself is a coordinate patch. A

diﬀerentiable structure on M is obtained by adjoining to this mapping

all patches in M which overlap smoothly with this mapping. The polar

coordinate patch

u = r cos θ

v = r sin θ

(r, θ) ∈ D

overlaps smoothly with the identity patch IF D is of the form

D = {(r, θ) | r > 0, θ ∈ (a, b), b − a ≤ 2π}

(an open sector).

Solution. Let

X : D → M and X : D → M. By deﬁnition,

X and

X overlap smoothly if ( X)

−1

◦ X (which maps D → M → D) and

(

−1

◦

X (which maps D → M → D) have open domains and are

smooth.

First,

X and X are one-to-one and so are invertible. Explicitly,

(

−1

◦ X(r, θ) = (x, y) = (r cos θ, r sin θ).

∂

∂r

[(

−1

◦ X] = (cos θ, sin θ)

and

∂

∂θ

[(

−1

◦ X] = (−r sin θ, r cos θ).

Therefore, (

−1

◦ X is smooth (the ﬁrst partials are continuous... in

fact, it is inﬁnitely diﬀerentiable). Similarly, (

−1

◦

X(x, y) = (r, θ)

where r =

√

+ y

and tan θ = y/x where we choose θ such that

θ ∈ (a, b),

θ is in Quadrant I if x > 0, y > 0,
θ is in Quadrant II if x < 0, y > 0,
θ is in Quadrant III if x < 0, y < 0,
θ is in Quadrant IV if x > 0, y < 0,

(and similar choices are made if x = 0 or y = 0). So θ = tan

−1

(y/x) +

constant

(so θ is a continuous function of (x, y), even though tan

−1

(y/x)

is not continuous — this is how we choose the θ to associate with (x, y)).

We then have

∂

∂x

[(

−1

◦

X)] =




√

+ y

−y/x

1 + (y/x)




and

∂

∂y

[(

−1

◦

X)] =




√

+ y

1/x

1 + (y/x)




therefore (since (x, y) = (0, 0)) (

−1

◦

X is smooth (in fact, inﬁnitely

diﬀerentiable). Next, the domain of (

−1

◦

X : D → D is D itself and

D is open (by deﬁnition). The domain of (

−1

◦ X : D → D is the set

of all (x, y) ∈ M such that

√

+ y

= r > 0 and tan

−1

(y/x) ∈ (a, b)

(where tan

−1

(y/x) is calculated as described above). Therefore the

domain of (

−1

◦

X is open. Hence,

X and X overlap smoothly.

Definition. An admissible patch

X : D → M associates with each

point

P of

X(D) a unique ordered pair (or in general, ordered n−tuple)

, u

) =

−1

(

P ) called a local coordinate of P with respect to

Note. Apoint

P can have diﬀerent local coordinates with respect to

diﬀerent admissible patches. Suppose, for example,

P =

X(u

, u

) =

X(u

, u

). Then (

−1

◦ X(u

, u

) = (u

, u

) and (

−1

◦

, u

) =

, u

Definition. In the above setting, the equations (

−1

◦ X(u

, u

) =

, u

) and (

−1

◦

X(u

, u

) = (u

, u

) are changes of coordinates. See

Figure I-29, page 83. In terms of local coordinates:

−1

◦

X is given by u

= u

, u

), i = 1, 2

(61a)

−1

◦ X is given by u

= u

, u

), i = 1, 2.

(61b)

Definition I-13. Aset Ω ⊂ M is a neighborhood of a point

P ∈ M if

there exists an admissible patch

X : D → M such that P ∈

X(D) and

X(D) ⊂ Ω. Asubset of M is open if it is a neighborhood of each of its
points.

Definition. Let Ω be an open subset of the 2-manifold (or generally
n−manifold) M. Afunction f : Ω → R is smooth if f ◦

X is smooth

for every admissible patch

X in M (notice f ◦

X maps E

[or more

generally E

] to Ω and then to R - so the idea of diﬀerentiability is

clearly deﬁned). For f : Ω → R smooth and

X : D → M an admissible

patch whose image intersects Ω, deﬁne

∂f

∂u

X(D) ∩ Ω → R for i = 1, 2 (or generally i = 1, 2, . . . , n)

∂f

∂u

∂(f ◦

∂u

◦

−1

This is called the partial derivative of f with respect to u

Note. For

P ∈

X(D) ∩ Ω:

∂f

∂u

(

P ) =

∂(f ◦

∂u

(

−1

(

P )).

The mappings are:

−1

∂(f ◦

∂u

P ∈ M −→ (u

, u

)

−→ R.

The usual product rules hold:

∂

∂u

(fg) =

∂f

∂u

g + f

∂g

∂u

where f and g have common domain.

Definition. For

P ∈

X(D), deﬁne an operator on the collection of

functions smooth in a neighborhood of

P as

∂

∂u

(

P )[f] =

∂f

∂u

(

P ).

Notation. Asuperscript which appears in the denominator, such as
∂/∂u

, counts as a subscript and therefore will impact the Einstein

summation notation. (The motivation is that partial diﬀerentiation is

usually denoted with subscripts.)

Note. If

X : D → M and X : D → M are admissible patches, then

on the overlap

X(D) ∩ X(D) we have from equation (61), page 83, the

operator identities

∂

∂u

∂

∂u

for i = 1, 2

(63a)

∂

∂u

∂

∂u

for k = 1, 2

(63b)

Definition I-14. Let m ∈ Z

and suppose O is an open subset of E

Afunction f : O → M is smooth if

−1

◦ f (which maps E

to E

) is

smooth for every admissible patch

X on M. If O is not open, we say

f : O → M is smooth if f is smooth on an open set containing O. A
curve in M is a smooth function from an interval (a connected subset

of R) into M.

Note. Now for tangent vectors and planes. We replace the idea of

vectors as arrows, with the idea of vectors as operators. Remember

that a vector is something which satisﬁes the properties given in the

deﬁnition of a vector space! The “arrows” idea is just (technically) an

aid in visualization!

Definition I-15. Let α : I → M be a curve on a 2-manifold (or

generally, n−manifold) M. For t ∈ I, deﬁne the velocity vector of α at
α(t) as the operator

(t)[f ] = (f ◦ α)

(t) =

[f (α(t))]

for each smooth f which maps an open neighborhood of α(t) into R.

Definition I-16. Let

P be a point of the 2−manifold M. An operator

v which assigns a real number v[f] to each smooth real-valued function

f on M is called a tangent vector to M at P if there exists a curve in M
which passes through

P and has velocity v at P. The set of all tangent

vectors to M at

P is called the tangent plane of M at P, denoted T

Note. The previous two deﬁnitions are independent of the choice of

coordinate patch (although we may do computations in some coordinate

patch).

Theorem. Let

P be a point on manifold M and let

X be an admissible

coordinate patch such that

P =

X(u

), u

)). If v is a tangent

vector to M at

P then v is a linear combination of

∂

∂u

(

P ) and

∂

∂u

(

P ).

Proof. With v a tangent vector, there is a curve α(t) in M such that
α(t

) =

P and α

) = v. Let f be a smooth real-valued function.

Then with α(t) =

X(u

(t), u

(t)),

v(t)[f] = α

(t)[f ] =

[(f ◦ α)(t)]

[f ◦

X(u

(t), u

(t))] =

∂(f ◦

∂u

(t), u

(t))

∂(f ◦

∂u

(

−1

◦ α(t))

∂f

∂u

(α(t))u

(t) (by deﬁnition)

= u

(t)

∂f

∂u

(α(t)).

So as an operator, α

(t) = u

(t)

∂

∂u

(α(t)), or simply

= u

∂

∂u

(64)

A t point

P ,

v = α

) = u

)

∂

∂u

(

P ) = v

∂

∂u

(

P )

where v

= u

Note. The vector

∂

∂u

(

P ) and

∂

∂u

(

P ) are linearly independent (con-

sider their behavior on functions of the form f (u

, u

) = u

and g(u

, u

) =

... although this argument is weak!). So the vectors form a basis for

a 2-dimensional vector space, the tangent plane to M at

P , T

M. In

general, a tangent plane to an n−manifold is an n−dimensional vector

space (a “hyperplane”).

Note. The converse of the Theorem also holds: If v is a linear com-

bination of

∂

∂u

(

P ) and

∂

∂u

(

P ), then v is a tangent vector to M at

Note. Suppose

X : D → M and X : D → M are overlapping admissi-

ble patches at

P . Then tangent vector v has two coordinate represen-

tations:

v = v

∂

∂u

(

P ) = v

∂

∂u

(

P ).

From equation (63a), page 85, we have

∂

∂u

∂

∂u

for i = 1, 2

and so

v = v

∂

∂u

(

P ) = v




∂u

∂

∂u




(

P ) =




∂u




∂

∂u

(

P )

and so

= v

∂u

for j = 1, 2

(67a)

(remember the linear independence of the ∂/∂

’s). Similarly

= v

∂

∂u

(

P ) for i = 1, 2.

This gives us a relationship between the coordinates of tangent vectors.

Notice that all these ideas extend to higher dimensions.

Note. We now introduce an inner product which generalizes the idea of

a dot product and use this to carry over several of the ideas developed

earlier for surfaces to manifolds.

Definition I-17. Let V be a vector space with scalar ﬁeld R. A n

inner product on V is a mapping ·, · : V × V → R such that for all

v,v

, w, w

∈ V and for all a, a

∈ R:

(a) v,

w = w,v (symmetry).

(b) av + a

, w = av, w + a

, w and v, aw + a

= av, w +

v, w

(bilinear).

(c) v, v ≥ 0 for all v ∈ V and v, v = 0 if and only if v = 0 (positive

deﬁnite).

Definition I-18. A Riemannian metric (or simply metric) on an

2−manifold M is an assignment of an inner product to each tangent

plane of M. For each coordinate patch

X : D → M, we require the

functions g

X(D) → R deﬁned as

(

P ) =

∂

∂u

(

P ),

∂

∂u

(

P )

for i, j = 1, 2, . . . , n to be smooth. An n−manifold with such a Rie-

mannian metric is called a Riemannian n−manifold.

Example. R

is a Riemannian manifold where the tangent planes are

themselves R

(since R

is “ﬂat”) and the inner product is the usual

dot product in R

Example. All the surfaces we dealt with earlier are examples of Rie-

mannian 2-manifolds (well... technically, a manifold does not have a

boundary, so we might have to throw out some of the examples [such as

the pseudosphere], although we could include in a study the so called

“manifolds with a boundary”).

Definition. Avector space V with a mapping ·, · : V × V → R

satisfying (a) and (b) given above along with

) If v,

w = 0 for all w ∈ V, then v = 0 (nonsingular).

is a semi-Riemannian n−manifold (again, we require g

(

P ) to be smooth).

Note. Condition (c

) is weaker than condition (c) (and so every Rie-

mannian n−manifold is also a semi-Riemannian n−manifold). Con-

dition (c

) allows lengths of vectors to be negative. We will see that

spacetime is a semi-Riemannian 4-manifold.

Note. If

X : D → M and X : D → M are overlapping admissible

patches then

= g

∂u

for m, n = 1, 2,

= g

∂u

for i, j = 1, 2.

(You will verify these as homework.)

Theorem. If v and

w are tangent vectors at P to a semi-Riemannian

n−manifold M, and if

X : D → M, X : D → M are admissible patches

with

P ∈

X(D) ∩ X(D) then

= g

Therefore g

is called an invariant.

Proof. We have v = v

∂

∂u

and

w = w

∂

∂u

, so

v, w =

∂

∂u

, w

∂

∂u

= v

∂

∂u

, w

∂

∂u

= v

∂

∂u

∂

∂u

= g

Similarly, with v = v

∂

∂u

and

w = w

∂

∂u

we have v,

w = g

Therefore g

= g

. (This is consistent with the fact that inner

products are independent of the choice of coordinates).

Note. We see from the above theorem, that the g

’s determine inner

products of tangent vectors to a manifold just as the g

’s of Section

1.4 determined dot products of tangent vectors to a surface.

Definition. Let v be a tangent vector to a semi-Riemannian n−manifold.

Then deﬁne v = v, v

1/2

. For α(t), a ≤ t ≤ b a curve in M, deﬁne

the arclength of α as

L =

(t) dt.

Note. Let s(t) = s denote the arc length along the curve from α(a) to
α(t). Then

s(t) =

∗

) dt

∗

and so s

(t) = α

(t) and

(t))

= α

(t)

= α

(t), α

(t).

Let

X : D → M be an admissible coordinate patch deﬁned in a neigh-

borhood of α(t). Then α

= α

∂

∂u

= u

∂

∂u

(by equation (64), page

86) and as in the above Theorem

(t), α

(t) =

∂

∂u

, u

∂

∂u

= u

∂

∂u

∂

∂u

= g

(71)

Since expressions of the form g

are invariant from one coordinate

system to another, arclength and expression (71) are invariant.

Definition. Let M be a semi-Riemannian manifold. The expression

= g

(which is invariant from one “coordinate patch” to another) is the met-

ric form or the fundamental form of the manifold.

Note. We now mimic earlier sections and give a number of deﬁnitions.

Definition. Create the matrix (g

) and deﬁne (g

)

−1

= (g

). For

each coordinate system,

X(u

, u

, . . . , u

) deﬁne the Christoﬀel symbols

of the ﬁrst kind as

ijk

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

and the Christoﬀel symbols of the second kind as

1
2

∂g

∂u

∂g

∂u

−

∂g

∂u

Definition I-19. If α = α(s) is a curve in a semi-Riemannian n−manifold
M, where s is arclength, then α is a geodesic if in each local coordinate
system deﬁned on part of α

+ Γ

= 0

for r = 1, 2, . . . , n. (compare this to equation (29), page 58.)

Note. Theorems I-9 and I-10 carry over to semi-Riemannian n−manifolds.

In particular, the shortest distance between two points is along a geo-

desic.

Definition. For a semi-Riemannian n−manifold, deﬁne the Riemann-

Christoﬀel curvature tensor as

ijk

∂Γ

∂u

−

∂Γ

∂u

+ Γ

− Γ

for h, i, j, k = 1, 2, . . . , n. Deﬁne

mijk

= g

ijk

Note. The curvature tensor has n

entries (although there is some

symmetry). When n = 2 the only nonzero entries are

1212

= R

2121

= −R

2112

= −R

1221

and for 2-manifolds (as in Section 1.8), curvature is K = R

1212

/g.

However, things are much more complicated in higher dimensions!

Note. The curvature tensor R

ijk

for an n−manifold has n

− 1)/12

independent components (so sayeth the text, page 90). Therefore

curvature for an n−manifold is NOT determined by a single number

when n > 2!

Example (Exercise 1.9.4). Suppose a Riemannian metric on M (an

open subset of R

) is given by

(du

+ dv

)

where γ = γ(u, v) is a smooth positive-valued function. Then M has

Gauss curvature

K = γ(γ

+ γ

) − (γ

+ γ

Proof. First, we have E = 1/γ

= G and F = 0. So we have from

Exercise 1.8.3

K =

−1

√






∂

∂u




√

∂

√

∂u




∂

∂v




√

∂

√

∂v









Now

√

E =

√

G = 1/γ and so

K = −γγ






∂

∂u




∂[1/γ]

∂u




∂

∂v




∂[1/γ]

∂v









= −γ

∂

∂u

−1

]

∂

∂v

−1

= −γ

∂

∂u

−γ

∂

∂v

−γ

= −γ






(−γ

)γ − (−γ

)(γ

)

(−γ

)γ − (−γ

)(γ

)






= −γ






−γγ

+ (γ

)

− γγ

+ (γ

)






= γ(γ

+ γ

) − ((γ

)

+ (γ

)

Example (Exercise 1.9.7). Let M be the subset of R

: M = {(u, v) |

+ v

< 4k

} (where k > 0). Introduce the metric

(du

+ dv

)

where γ(u, v) = 1 −

+ v

. This is called the Poincare Disk. Then

K = −1/k

Proof. From Exercise 1.9.4,

K = γ(γ

+ γ

) − (γ

+ γ

Well,

−u

, γ

−v

, γ

−1

, γ

−1

Therefore,

K =




1 −

+ v




−1

−




−u

−v

2






1 −

+ v




−1

−


















2





−(4k

− u

− v

)

−

−1

Chapter 2. Special Relativity: The

Geometry of Flat Spacetime

Note. Classically (i.e in Newtonian mechanics), space is thought of as

1. unbounded and infinite,

2. 3-dimensional and explained by Euclidean geometry, and

3. “always similar and immovable” (Newton, Principia Mathematica,

1687).

This would imply that one could set up a system of spatial coordinates

(x, y, z) and describe any dynamical event in terms of these spatial

coordinates and time t.

Note. Newton’s Three Laws of Motion:

1. (The Law of Inertia) A body at rest remains at rest and a body in

motion remains in motion with a constant speed and in a straight

line, unless acted upon by an outside force.

2. The acceleration of an object is proportional to the force acting upon

it and is directed in the direction of the force. That is,

F = ma.

3. To every action there is an equal and opposite reaction.

Note. Newton also stated his Law of Universal Gravitation in Prin-

cipia:

“Every particle in the universe attracts every other particle

in such a way that the force between the two is directed

along the line between them and has a magnitude propor-

tional to the product of their masses and inversely propor-

tional to the square of the distance between them.” (See

page 186.)

Symbolically, F =

GMm

where F is the magnitude of the force, r the

distance between the two bodies, M and m are the masses of the bodies

involved and G is the gravitational constant (6.67 × 10

−8

cm./(g sec

)).

Assuming only Newton’s Law of Universal Gravitation and Newton’s

Second Law of Motion, one can derive Kepler’s Laws of Planetary Mo-

tion.

2.1 Inertial Frames of Reference

Definition. A frame of reference is a system of spatial coordinates and

possibly a temporal coordinate. A frame of reference in which the Law

of Inertia holds is an inertial frame or inertial system. An observer at

rest (i.e. with zero velocity) in such a system is an inertial observer.

Note. The main idea of an inertial observer in an inertial frame is that

the observer experiences no acceleration (and therefore no net force).

If S is an inertial frame and S

is a frame (i.e. coordinate system)

moving uniformly relative to S, then S

is itself an inertial frame (see

Exercise II-1). Frames S and S

are equivalent in the sense that there is

no mechanical experiment that can be conducted to determine whether

either frame is at rest or in uniform motion (that is, there is no pre-

ferred frame). This is called the Galilean (or classical) Principle of

Relativity.

Note. Special relativity deals with the observations of phenomena by

inertial observers and with the comparison of observations of inertial

observers in equivalent frames (i.e. NO ACCELERATION!). General

relativity takes into consideration the eﬀects of acceleration (and there-

fore gravitation) on observations.

2.2 The Michelson-Morley Experiment

Note. Sound waves need a medium though which to travel. In 1864

James Clerk Maxwell showed that light is an electromagnetic wave.

Therefore it was assumed that there is an ether which propagates light

waves. This ether was assumed to be everywhere and unaﬀected by

matter. This ether could be used to determine an absolute reference

frame (with the help of observing how light propagates through the

ether).

Note. The Michelson-Morley experiment (circa 1885) was performed

to detect the Earth’s motion through the ether as follows:

The viewer will see the two beams of light which have traveled along

diﬀerent arms display some interference pattern. If the system is ro-

tated, then the inﬂuence of the “ether wind” should change the time

the beams of light take to travel along the arms and therefore should

change the interference pattern. The experiment was performed at

diﬀerent times of the day and of the year. NO CHANGE IN THE

INTERFERENCE PATTERN WAS OBSERVED!

Example (Exercise 2.2.2). Suppose L

is the length of arm #1 and

is the length of arm#2. The speed of a photon (relative to the

source) on the trip“over” to the mirror is c − v and so takes a time of
L

/(c − v). On the return trip, the photon has speed of c + v and so

takes a time of L

/(c + v). Therefore the round triptime is

c − v

c + v

(c + v) + L

(c − v)

− v

2cL

− v

1 − v




1 −




−1

The photon traveling along arm #2 must follow a path (relative to a

“stationary” observer) of

We need the time the photon travels (t

) and the angle at which the

photon leaves the mirror to be such that

(the photon must travel with a component of velocity “upstream” to

compensate for the wind). Then

−

= t




−




√

− v

1 − v

Now

1 − x

∞

n=0

, so

≈




1 +




Also,

(1 + x)

= 1 + mx +

m(m − 1)

m(m − 1)(m − 2)

+ · · ·

and with m = −1/2 and x = −v

≈




1 +

−1





−v







1 +




For the Earth’s orbit around the sun, v/c ≈ 10

−4

so the approximation

is appropriate. Now the rays recombine at the viewer separated by

∆t = t

− t

≈




− L

−




Now suppose the apparatus is rotated 90

◦

so that arm #1 is now trans-

verse to the ether wind. Let t

and t

denote the new round triplight

travel times. Then (as above, replacing L

with L

in t

to determine

and replacing L

with L

in t

to determine t




1 +







1 +




Then

∆t

= t

− t

− L

) +

− 2L

)

and

∆t−∆t

−L

−






− L

) +

− 2L

)






+ L

This is the time change produced by rotating the apparatus.

Note. In 1892, Fitzgerald proposed that an object moving through the

ether wind with velocity v experiences a contraction in the direction of

the ether wind of

1 − v

. That is, in the diagram above, L

is con-

tracted to L

1 − v

and then we get t

= t

when L

= L

, p oten-

tially explaining the results of the Michelson-Morley experiment. This

is called the Lorentz-Fitzgerald contraction. Even under this assump-

tion, “it turns out” (see the following example) that the Michelson-

Morley apparatus with unequal arms will exhibit a pattern shift over a

6 month period as the Earth changes direction in its orbit around the

Sun. In 1932, Kennedy and Thorndike performed such an experiment

and detected no such shift.

Example (Exercise 2.2.4). Suppose in the Michelson-Morley appa-

ratus that ∆L = L

−L

= 0 and that there is a contraction by a factor

1 − v

in the direction of the ether wind. Then show

∆t =

∆L




1 +




Solution. As in Exercise 2.2.1, we have

1 − v

and so

∆t = t

− t

1 − v

−

1 − v

− L

)

1 − v

≈

− L

)




1 +

−1





−v




− L

)




1 +




Note. Since the equation in the above exercise expresses ∆t as a

function of v only (c, L

, and L

being constant - although they may

be contracted, but this is taken care of in the computations), we would

see ∆t vary with a period of 6 months (as mentioned above).

Note. Another suggestion to explain the negative result of the Michelson-

Morley experiment was the idea that the Earth “drags the ether along

with it” as it orbits the sun, galactic center, etc. This idea is rejected

because of stellar aberration, discovered by James Bradley in 1725.

Example (Exercise 2.2.9). Light rays from a star directly overhead

enter a telescope. Suppose the Earth, in its orbit around the Sun, is

moving at a right angle to the incoming rays, in the direction indicated

in the ﬁgure on page 108. In the time it takes a ray to travel down

the barrel to the eyepiece, the telescope will have moved slightly to the

right. Therefore, in order to prevent the light rays from falling on the

side of the barrel rather than on the eyepiece lens, we must tilt the

telescope slightly from the vertical, if we are to see the star. Conse-

quently, the apparent position of the star is displaced forward somewhat

from the actual position. Show that the angle of displacement, θ, is (in

radians)

θ = tan

−1

(v/c) ≈ v/c

where v is the Earth’s orbital velocity. Verify that θ ≈ 20.6

(roughly

the angle subtended by an object 0.1 mm in diameter held at arm’s

length).

Solution. Clearly

θ = tan

−1

v∆t

c∆t

= tan

−1

Now

tan

−1

x =

∞

n=0

(−1)

2n+1

2n + 1

for |x| < 1

and so tan

−1

≈

. We have the Earth’s orbital velocity implying

v/c = 10

−4

(see Exercise 2.2.2). So

θ = 10

−4

= 10

−4

360

◦

2π





◦










≈ 20.6

Note. As the Earth revolves around the Sun in its nearly circular

annual orbit, the apparent position of the star will trace a circle with

angular radius 20.6

. This is indeed observed. If the Earth dragged

a layer of ether along with it, the light rays, upon entering this layer,

would aquire a horizontal velocity component matching the forward

velocity v of the telescope. There would then be no aberration eﬀect.

Conclusion. The speed of light is constant and the same in all direc-

tions and in all inertial frames.

2.3 The Postulates of Relativity

Note. Albert Einstein published “Zur Elektrodynamik bewegter K¨orper”

(On the Electrodynamics of Moving Bodies) in Annalen der Physik (An-

nals of Physics) 17 (1905). In this paper, he established the SPECIAL

THEORY OF RELATIVITY! I quote (from “The Principles of

Relativity” by H. A. Lorenz, A. Einstein, H. Minkowski, and H. Weyl,

published by Dover Publications):

“...the same laws of electrodynamics and optics will be valid

for all frames of reference for which the equations of me-

chanics hold good. We raise this conjecture (the purport of

which will hereafter be called the “Principle of Relativity”)

to the status of a postulate, and also introduce another

postulate, which is only apparently irreconcilable with the

former, namely, that light is always propagated in empty

space with a definite velocity c which is independent of the

state of motion of the emitting body.”

In short:

P1. All physical laws valid in one frame of reference are equally valid

in any other frame moving uniformly relative to the first.

P2. The speed of light (in a vacuum) is the same in all inertial frames

of reference, regardless of the motion of the light source.

From these two simple (and empirically varified) assumptions arises the

beginning of the revolution that marks our transition from classical to

modern physics!

2.4 Relativity of Simultaneity

Note. Suppose two trains T and T

pass each other traveling in op-

posite directions (this is equivalent to two inertial frames moving uni-

formly relative to one another). Also suppose there is a flash of lighten-

ing (an emission of light) at a certain point (see Figure II-3). Mark the

points on trains T and T

where this flash occurs at A and A

respec-

tively. “Next,” suppose there is another flash of lightning and mark

the points B and B

. Suppose point O on train T is midway between

points A and B, AND that point O

on train T

is midway between

points A

and B

. An outsider might see:

Suppose an observer at point O sees the flashes at points A and B occur

at the same time. From the point of view of O the sequence of events

is:

(1) Both flashes occur,

A, O, B opposite
A

, O

, B

, resp.

(2) Wavefront from

meets O

(3) Both wavefronts

meet O

(4) Wavefront from

meets O

From the point of view of an observer at O

, the following sequence of

events are observed:

(1) Flash occurs

at BB

(2) Flash occurs

at AA

(3) Wavefront from

meets O

(4) Wavefronts from

and BB

meet O

(5) Wavefront from

meets O

Notice that the speed of light is the same in both frames of reference.

However, the observer on train T sees the flashes occur simultaneously,

whereas the observer on train T

sees the flash at BB

occur before the

flash at AA

. Therefore, events that appear to be simultaneous in one

frame of reference, may not appear to be simultaneous in another. This

is the relativity of simultaneity.

Note. The relativity of simultaneity has implications for the mea-

surements of lengths. In order to measure the length of an object, we

must measure the position of both ends of the object simultaneously.

Therefore, if the object is moving relative to us, there is a problem.

In the above example, observer O sees distances AB and A

equal,

but observer O

sees AB shorter than A

. Therefore, we see that

measurements of lengths are relative!

2.5 Coordinates

Definition. In 3-dimensional geometry, positions are represented by

points (x, y, z). In physics, we are interested in events which have both

time and position (t, x, y, z). The collection of all possible events is

spacetime.

Definition. With an event (t, x, y, z) in spacetime we associate the

units of cm with coordinates x, y, z. In addition, we express t (time) in

terms of cm by multiplying it by c. (In fact, many texts use coordinates

(ct, x, y, z) for events.) These common units (cm for us) are called

geometric units.

Note. We express velocities in dimensionless units by dividing them

by c. S o for velocity v (in cm/sec, say) we associate the dimensionless

velocity β = v/c. Notice that under this convention, the speed of light

is 1.

Note. In an inertial frame S, we can imagine a grid laid out with a

clock at each point of the grid. The clocks can by synchronized (see

page 118 for details). When we mention that an object is observed in

frame S, we mean that all of its parts are measured simultaneously

(using the synchronized clocks). This can be quite diﬀerent from what

an observer at a point actually sees.

Note. From now on, when we consider two inertial frames S and S

moving uniformly relative to each other, we adopt the conventions:

1. The x− and x

−axes (and their positive directions) coincide.

2. Relative to S, S

is moving in the positive x direction with velocity

β.

3. The y− and y

−axes are always parallel.

4. The z− and z

−axes are always parallel.

We call S the laboratory frame and S

the rocket frame:

Assumptions. We assume space is homogeneous and isotropic, that

is, space appears the same at all points (on a suﬃciently large scale)

and appears the same in all directions.

Lemma. Suppose two inertial frames S and S

move uniformly relative

to each other. Then lengths perpendicular to the direction of motion

are the same for observers in both frames (that is, under our convention,

there is no length contraction or expansion in the y or z directions).

Proof. Suppose there is a right circular cylinder C of radius R (as

measured in S) with its axis along the x−axis. Similarly, suppose there

is a right circular cylinder C

of radius R

(as measured in S

) with its

axis lying along the x

−axis. Suppose the cylinders are the same radius

when “at rest.” Since space is assumed to be isotropic, each observer

will see a circular cylinder in the other frame (or else, there would be

directional asymmetry to space). Now suppose the lab observer (S)

measures a smaller radius r < R for cylinder C

. Then he will see

cylinder C

pass through the interior of his cylinder C (see Figure II-7,

page 120). Now if two points are coincident (at the same place) in one

inertial frame, then they must be coincident in another inertial frame

(they are, after all, at the same place). So if the lab observer sees C

inside C, then the rocket observer must see this as well. However, by

the Principle of Relativity (P

), the rocket observer must see C inside

. This contradiction yields r ≥ R. Similarly, there is a contradiction

if we assume r > R. Therefore, r = R and both observers see C and
C

as cylinders with radius R. Therefore, there is no length change in

the y or z directions.

Note. In the next section, we’ll see that things are much diﬀerent in

the direction of motion.

2.6 Invariance of the Interval

Note. In this section, we deﬁne a quantity called the “interval” be-

tween two events which is invariant under a change of spacetime coor-

dinates from one inertial frame to another (analogous to “distance” in

geometry). We will also derive equations for time and length dilation.

Note. Consider the experiment described in Figure II-8. In inertial

frame S

a beam of light is emitted from the origin, travels a distance

L, hits a mirror and returns to the origin. If ∆t

is the amount of time

it takes the light to return to the origin, then L = ∆t

/2 (recall that t

is multiplied by c in order to put it in geometric units). An observer in

frame S sees the light follow the path of Figure II-8b in time ∆t. Notice

that the situation here is not symmetric since the laboratory observer

requires two clocks (at two positions) to determine ∆t, whereas the

rocket observer only needs one clock (so the Principle of Relativity does

not apply). In geometric units, we have: (∆t/2)

= (∆t

/2)

+ (∆x/2)

or (∆t

)

= (∆t)

− (∆x)

with β the velocity of S

relative to S, we

have β = ∆x/∆t and so ∆x = β∆t and (∆t

)

= (∆t)

− (β∆t)

∆t

1 − β

∆t.

(78)

Therefore we see that under the hypotheses of relativity, time is not

absolute and the time between events depends on an observer’s motion

relative to the events.

Note. You might be more familiar with equation (78) in the form:

∆t =

√

1 − β

∆t

where ∆t

is an interval of time in the rocket frame and ∆t is how the

laboratory frame measures this time interval. Notice ∆t ≥ ∆t

so that

time is dilated (lengthened).

Note. Since β = v/c, for v c, β ≈ 0 and ∆t

≈ ∆t.

Definition. Suppose events A and B occur in inertial frame S at

, x

, y

, z

) and (t

, x

, y

, z

), respectively, where y

= y

and z

. Then deﬁne the interval (or proper time) between A and B as

∆τ =

(∆t)

− (∆x)

where ∆t = t

− t

and ∆x = x

− x

Note. As shown above, in the S

frame

(∆t

)

− (∆x

)

= (∆t)

− (∆x)

(recall ∆x

= 0). So ∆τ is the same in S

. That is, the interval is

invariant from S to S

. As the text says “The interval is to spacetime

geometry what the distance is to Euclidean geometry.”

Note. We could extend the deﬁnition of interval to motion more com-

plicated than motion along the x−axis as follows:

∆τ = {(∆t)

− (∆x)

− (∆y)

− (∆z)

}

1/2

(interval)

= (time separation)

− (space separation)

Note. Let’s explore this “time dilation” in more detail. In our example,

we have events A and B occuring in the S

frame at the same position

(∆x

= 0), but at diﬀerent times. Suppose for example that events A

and B are separated by one time unit in the S

frame (∆t

= 1). W e

could then represent the ticking of a second hand on a watch which is

stationary in the S

frame by these two events. An observer in the S

frame then measures this ∆t

= 1 as

∆t =

√

1 − β

∆t

That is, an observer in the S frame sees the one time unit stretched

(dilated) to a length of

√

1 − β

≥ 1 time unit. So the factor

√

1 − β

shows how much slower a moving clock ticks in comparison to a sta-

tionary clock. The Principle of Relativity implies that on observer in

frame S

will see a clock stationary in the S frame tick slowly as well.

However, the Principle of Relativity does not apply in our example

above (see p. 123) and both an observer in S and an observer in S

agree that ∆t and ∆t

are related by

∆t =

√

1 − β

∆t

So both agree that ∆t ≥ ∆t

in this case. This seems strange ini-

tially, but will make more sense when we explore the interval below.

(Remember, ∆x = 0.)

Definition. An interval in which time separation dominates and (∆τ )

0 is timelike. An interval in which space separation dominates and

(∆τ )

< 0 is spacelike. An interval for which ∆τ = 0 is lightlike.

Note. If it is possible for a material particle to be present at two events,

then the events are separated by a timelike interval. No material object

can be present at two events which are separated by a spacelike interval

(the particle would have to go faster than light). If a ray of light can

travel between two events then the events are separated by an interval

which is lightlike. We see this in more detail when we look at spacetime

diagrams (Section 2.8).

Note. If an observer in frame S

passes a “platform” (all the train talk

is due to Einstein’s original work) of length L in frame S at a speed of
β, then a laboratory observer on the platform sees the rocket observer
pass the platform in a time ∆t = L/β. As argued above, the rocket

observer measures this time period as ∆t

= ∆t

√

1 − β

. Therefore, the

rocket observer sees the platform go by in time ∆t

and so measures

the length of the platform as

= β∆t

1 − β

= L

1 − β

Therefore we see that the time dilation also implies a length contraction:

= L

1 − β

(83)

Note. Equation (83) implies that lengths are contracted when an ob-

ject is moving fast relative to the observer. Notice that with β ≈ 0,
L

≈ L.

Example (Exercise 2.6.2). Pions are subatomic particles which de-

cay radioactively. At rest, they have a half-life of 1.8 × 10

−8

sec. A pion

beam is accelerated to β = 0.99. According to classical physics, this

beam should drop to one-half its original intensity after traveling for

(0.99)(3 × 10

)(1.8 × 10

−8

) ≈ 5.3m. However, it is found that it drops

to about one-half intensity after traveling 38m. Explain, using either

time dilation or length contraction.

Solution. Time is not absolute and a given amount of time ∆t

one inertial frame (the pion’s frame, say) is observed to be dilated in

another inertial frame (the particle accelerator’s) to ∆t = ∆t

√

1 − β

So with ∆t

= 1.8 × 10

−8

sec and β = 0.99,

∆t =

√

1 − .99

(1.8 × 10

−8

sec) = 1.28 × 10

−7

sec.

Now with β = .99, the speed of the pion is (.99)(3 × 10

m/sec) =

2.97 × 10

m/sec and in the inertial frame of the accelerator the pion

travels

(2.97 × 10

m/sec)(1.28 × 10

−7

sec) = 38m.

In terms of length contraction, the accelerator’s length of 38m is con-

tracted to a length of

= L

1 − β

= (38m)

√

1 − .99

= 5.3m

in the pion’s frame. With v = .99c, the pion travels this distance in

5.3m

(.99)(3 × 10

m/sec)

= 1.8 × 10

−8

sec.

This is the half-life and therefore the pion drops to 1/2 its intensity

after traveling 38m in the accelerator’s frame.

2.7 The Lorentz Transformation

Note. We seek to ﬁnd the transformation of the coordinates (x, y, z, t)

in an inertial frame S to the coordinates (x

, y

, z

, t

) in inertial frame

. Throughout this section, we assume the x and x

axes coincide, S

moves with velocity β in the direction of the positive x axis, and the

origins of the systems coincide at t = t

= 0. See Figure II-9, page 128.

Note. Classically, we have the relations

x = x

+ βt

y = y

z = z

t = t

Definition. The assumption of homogeneity says that there is no pre-

ferred location in space (that is, space looks the same at all points [on

a suﬃciently large scale]). The assumption of isotropy says that there

is no preferred direction in space (that is, space looks the same in every

direction).

Note. Under the assumptions of homogeneity and isotropy, the rela-

tions between (x, y, z, t) and (x

, z

, t

) must be linear (throughout,

everything is done in geometric units!):

x = a

+ a

y = a

+ a

z = a

+ a

t = a

+ a

If not, say y = ax

, then a rod lying along the x−axis of length x

− x

would get longer as we moved it out the x−axis, contradicting homo-

geneity. Similarly, relationships involving time must be linear (since the

length of a time interval should not depend on time itself, nor should

the length of a spatial interval).

Note. We saw in Section 2.5 that lengths perpendicular to the direction

of motion are invariant. Therefore

y = y

z = z

Note. x does not depend on y

and z

. Suppose not. Suppose there is a

ﬂat plate at rest in S

and perpendicular to the x

−axis. Since the above

equations are linear, an observer in S would see the plate tilted (but still

ﬂat) if there is a dependence on y

or z

. However, this implies a “special

direction” in space violating the assumption of isotropy. Therefore, the

coeﬃcients a

and a

are 0. Similarly, isotropy implies a

= a

= 0.

We have reduced the system of equations to

x = a

+ a

(85)

t = a

+ a

(86)

Note. Recall Figure II-8, page 122. A beam of light is emitted from

the origin of S

at time t = t

= 0 (when x = x

= 0), bounced oﬀ a

mirror and reﬂected back to the S

origin (x

= 0) at time t

= ∆t

. In

S, the light returns to the origin (x

= 0) at time t = ∆t = 1/

√

1 − β

(equation (78), page 123). Also, in S with x

= 0 we have (from

equation (86)) that t = a

. Therefore a

= 1/

√

1 − β

. Next, with

= 0 and t = t

√

1 − β

, since the point x

= 0 occurs in the S frame

at x = βt (due to the relative motion), we have from equation (85):

x = βt =

√

1 − β

= a

and so a

√

1 − β

. So equations (85) and (86) give

x = a

√

1 − β

(87)

t = a

√

1 − β

(88)

Note. Now consider a ﬂash of light emitted at the origins of S and S

at t = t

= 0. This produces a sphere of light in each frame (according

to the constancy of the speed of light). And so

= x

+ y

+ z

= x

+ y

+ z

Since y = y

and z = z

, we have t

− x

= t

− x

. From equations

(87) and (88) we get

− x

√

1 − β

−

√

1 − β

= t

− x

Expanding

)

√

1 − β

− (a

)

−2

√

1 − β

−

1 − β

= t

− x

− a

) + x

√

1 − β

− 2

√

1 − β




1 − β

−

1 − β




= t

− x

Comparing coeﬃcients, we need

− a

= −1

√

1 − β

− 2

√

1 − β

= 0

− a

= −1 and a

− βa

= 0.

Solving this system: a

= βa

and so

− a

= (βa

)

− a

= −1

1 − β

and a

= ±

√

1 − β

From equation (87) with β = 0 we see that x =

β=0

and we

want x = x

in the event that β = 0. Therefore, we have

√

1 − β

and a

√

1 − β

We now have the desired relations between (x, y, z, t) and (x

, y

, z

, t

Definition. The transformation relating coordinates (x, y, z, t) in S to

coordinates (x

, y

, z

, t

) in S

given by

x =

+ βt

√

1 − β

y = y

z = z

t =

βx

+ t

√

1 − β

is called the Lorentz Transformation.

Note. With β 1 and β

≈ 0 we have

x = x

+ βt

t = t

(in geometric units, x and x

are small compared to t and t

[see page

117 - remember time gets multiplied by c to express it in units of length]

and βx

is negligible compared to t

, but βt

is NOT negligible compared

to x

Note. By the Principle of Relativity, we can invert the Lorentz Trans-

formation simply by interchanging x and t with x

and t

, respectively,

and replacing β with −β!

Note. If we deal with pairs of events separated in space and time, we

denote the diﬀerences in coordinates with ∆’s to get

∆x =

∆x

+ β∆t

√

1 − β

(91a)

∆t =

β∆x

+ ∆t

√

1 − β

(91b)

With ∆x

= 0 in (91b) we get the equation for time dilation. With a

rod of length L = ∆x in frame S, the length measured in S

requires a

simultaneous measurement of the endpoints (∆t

= 0) and so from (91a)

L = L

√

1 − β

or L

= L

√

1 − β

, the equation for length contraction.

Example (Exercise 2.7.2). Observer S

seated at the center of a

railroad car observes two men, seated at opposite ends of the car, light

cigarettes simultaneously (∆t

= 0). However for S, an observer on the

station platform, these events are not simultaneous (∆t = 0). If the

length of the railroad car is ∆x

= 25m and the speed of the car relative

to the platform is 20m/sec (β = 20/3 × 10

), ﬁnd ∆t and convert your

answer to seconds.

Solution. We have ∆t

= 0, ∆x

= 25m, and β = 20/3 × 10

≈

6.67 × 10

−8

. So by equation (91b)

∆t =

β∆x

+ ∆t

√

1 − β

(6.67 × 10

−8

)(25m)

1 − (6.67 × 10

−8

)

≈ 1.67 × 10

−6

or in seconds

∆t =

1.67 × 10

−6

3 × 10

m/sec

= 5.56 × 10

−15

sec.

Example (Exercise 2.7.14). Substitute the transformation Equation

(91) into the formula for the interval and verify that

(∆t)

− (∆x)

− (∆y)

− (∆z)

= (∆t

)

− (∆x

)

− (∆y

)

− (∆z

)

Solution. With ∆y = ∆z = 0 we have

(∆t)

−(∆x)

− (∆y)

− (∆z)

(∆t)

− (∆x)




β∆x

+ ∆t

√

1 − β




−




∆x

+ β∆t

√

1 − β




(∆x

)

+ 2β∆x

∆t

+ (∆t

)

− (∆x

)

− 2β∆x

∆t

− β

(∆t

)

1 − β

(∆x

)

(β

− 1) + (∆t

)

(1 − β

)

1 − β

(∆t

)

− (∆x

)

= (∆t

)

− (∆x

)

− (∆y

)

− (∆z

)

since ∆y

= ∆z

= 0.

2.8 Spacetime Diagrams

Note. We cannot (as creatures stuck in 3 physical dimensions) draw

the full 4 dimensions of spacetime. However, for rectilinear or planar

motion, we can depict a particle’s movement. We do so with a spacetime

diagram in which spatial axes (one or two) are drawn as horizontal axes

and time is represented by a vertical axis. In the xt−plane, a particle

with velocity β is a line of the form x = βt (a line of slope 1/β):

Two particles with the same spacetime coordinates must be in collision:

Note. The picture on the cover of the text is the graph of the orbit of

the Earth as it goes around the Sun as plotted in a 3-D spacetime.

Definition. The curve in 4-dimensional spacetime which represents the

relationships between the spatial and temporal locations of a particle

is the particle’s world-line.

Note. Now let’s represent two inertial frames of reference S and S

(considering only the xt−plane and the x

−plane). Draw the x and t

axes as perpendicular (as above). If the systems are such that x = 0

and x

= 0 coincide at t = t

= 0, then the point x

= 0 traces out the

path x = βt in S. We define this as the t

axis:

The hyperbola t

−x

= 1 in S is the same as the “hyperbola” t

−x

1 in S

(invariance of the interval). So the intersection of this hyperbola

and the t

axis marks one time unit on t

. Now from equation (90b)

(with t

= 0) we get t = βx and define this as the x

axis. Again we

calibrate this axis with a hyperbola (x

− t

= 1):

We therefore have:

and so the S

coordinate system is oblique in the S spacetime diagram.

Note. In the above representation, notice that the larger β is, the more

narrow the “first quadrant” of the S

system is and the longer the x

and t

units are (as viewed from S).

Note. Suppose events A and B are simultaneous in S

They need not

be simultaneous in S. Events C and D simultaneous in S need not be

simultaneous in S

Note. A unit of time in S is dilated in S

and a unit of time in S

dilated in S.

Note. Suppose a unit length rod lies along the x axis. If its length

is measured in S

(the ends have to be measured simultaneously in S

)

then the rod is shorter. Conversely for rods lying along the x

axis.

Example (Exercise 2.8.4). An athlete carrying a pole 16m long runs

toward the front door of a barn so rapidly that an observer in the barn

measures the pole’s length as only 8m, which is exactly the length of

the barn. Therefore at some instant the pole will be observed entirely

contained within the barn. Suppose that the barn observer closes the

front and back doors of the barn at the instant he observes the pole

entirely contained by the barn. What will the athlete observe?

Solution. We have two events of interest:

A = The front of the pole is at the back of the barn.

B = The back of the pole is at the front of the barn.

From Exercise 2.8.3, the observer in the barn (frame S) observes these

events as simultaneous (each occuring at t

= 3.08 × 10

−8

sec after the

front of the pole was at the front of the barn). However, the athlete

observes event A after he has moved the pole only 4m into the barn.

So for him, event A occurs when t

= 1.54 × 10

−8

sec. Event B does not

occur until the pole has moved 16m (from t

= 0) and so event B occurs

for the athlete when t

= 6.16 × 10

−8

sec. Therefore, the barn observer

observes the pole totally within the barn (events A and B), slams the

barn doors, and observes the pole start to break through the back of

the barn all simultaneously. The athlete first observes event A along

with the slamming of the back barn door and the pole starting to break

through this door (when t

= 1.54 × 10

−8

sec) and THEN observes the

front barn door slam at t

= 6.16 × 10

−8

sec. The spacetime diagram

is:

A occurs at
t

= 1.54 × 10

−8

sec

B occurs at
t = 3.08 × 10

−8

sec

Since the order of events depends on the frame of reference, the appar-

ent paradox is explained.

Example (Exercise 2.8.5). Using a diagram similar to Figure II-15,

show that (a) at time t = 0 in the laboratory frame, the rocket clocks

that lie along the positive x−axis are observed by S to be set behind

the laboratory clocks, with the clocks further from the origin set further

behind, and that (b) at time t

= 0 in the rocket frame, the laboratory

clocks that lie along the positive x

−axis are observed by S

to be set

further ahead.

Solution. (a) At t = 0, clocks in S

that lie along the positive x−axis

are observed by S to be behind the S clocks, with the clocks further

from the origin set further behind:

So clocks c

arranged as in the figure read times 0 > t

> t

> · · ·.

(b) Conversely, in the S

frame:

So clocks c

arranged as in the figure read times 0 < t

< t

< · · ·.

2.9 Lorentz Geometry

Note. We wish to extend the idea of arclength to 4-dimensional space-

time. We do so by replacing the idea of “distance” (

(∆x

)

) by the

interval. The resulting geometry is called Lorentz geometry.

Definition. Let α be a curve in spacetime. The spacetime length (or

proper time) of α is

L(α) −

dτ =

(dt)

− (dx)

− (dy)

− (dz)

Note. Since ∆τ (and so dτ ) is an invariant from one inertial frame to

another, then so is L(α). L(α) may be viewed as the actual passage

of time that would be recorded for a clock with world-line α (this is

certainly clear when dx = dy = dz = 0).

Definition. R

with the semi-Riemannian metric

dτ

= (dt)

− (dx)

− (dy)

− (dz)

(93)

is called Minkowski space. (This may seem a bit unusual to see (dτ )

referred to as the “metric,” but of course it does determine a way

to measure the distance between points - although the square of this

“distance” may be negative).

Note. We parameterized curves with respect to arclength in Chapter

1. It is convenient to parameterize timelike curves (those curves for

which (dτ /dt)

> 0) in terms of proper time.

Example. Suppose a free particle travels with constant speed and

direction, so that

= a,

= b,

= c

for constants a, b, c. Deﬁne β =

√

+ b

+ c

(the particle’s speed).

From equation (93),

dτ

= 1 −

−

= 1 − β

Since dτ /dt is constant, τ is a monotone function of t and so dt/dτ =

√

1 − β

(well ±)

dx
dτ

dτ

√

1 − β

dy
dτ

dτ

√

1 − β

dz
dτ

dτ

√

1 − β

Notice that each of these derivatives is constant and so the particle

follows a straight line in spacetime. If we calculate second derivatives,

we see that a free particle satisﬁes:

dτ

= 0.

In fact, free particles follow geodesics in the spacetime of special rela-

tivity (in which geodesics are straight lines).

2.10 The Twin Paradox

Note. Suppose A and B are two events in spacetime separated by a

timelike interval (whose y and z coordinates are the same). Joining

these events with a straight line produces the world-line of an inertial

observer present at both events. Such an observer could view both

events as occuring at the same place (say at x = 0) and could put these

two events along his t−axis.

Note. Oddly enough, in a spacetime diagram under Lorentz geom-

etry, a straight line gives the longest distance (temporally) between

two points. This can be seen by considering the fact that the interval

(∆τ )

= (∆t)

− (∆x)

is invariant. Therefore, if we follow a trajectory

in spacetime that increases ∆x, it MUST increase ∆t. Figure II-19b

illustrates this fact:

That is, the non-inertial traveler (the one undergoing accelerations and

therefore the one not covered by special relativity) from A to B ages

less than the inertial traveler between these two events.

Example (Jack and Jill). We quote from page 152 of the text: “Let

us imagine that Jack is the occupant of a laboratory ﬂoating freely in

intergalactic space. He can be considered at the origin of an inertial

frame of reference. His twin sister, Jill, ﬁres the engines in her rocket,

initially alongside Jack’s space laboratory. Jill’s rocket is accelerated

to a speed of 0.8 relative to Jack and then travels at that speed for

three years of Jill’s time. At the end of that time, Jill ﬁres powerful

reversing engines that turn her rocket around and head it back toward

Jack’s laboratory at the same speed, 0.8. After another three-year

period, Jill returns to Jack and slows to a halt beside her brother.

Jill is then six years older. We can simplify the analysis by assuming

that the three periods of acceleration are so brief as to be negligible.

The error introduced is not important, since by making Jill’s journey

suﬃciently long and far, without changing the acceleration intervals,

we could make the fraction of time spent in acceleration as small as

we wish. Assume Jill travels along Jack’s x−axis. In Figure II-20(see

below), Jill’s world-line is represented on Jack’s spacetime diagram. It

consists of two straight line segments inclined to the t−axis with slopes

+0.8 and −0.8, respectively. For convenience, we are using units of

years for time and light-years for distance.”

Note. Because of the change in direction (necessary to bring Jack and

Jill back together), no single inertial frame exists in which Jill is at rest.

But her trip can be described in two diﬀerent inertial frames. Take the

ﬁrst to have t

axis x = βt = 0.8t (in Jack’s frame). Then at t = 5 and

= 3, Jill turns and travels along a new t

axis of x = −0.8t + 10(in

Jack’s frame). We see that upon the return, Jack has aged 10years,

but Jill has only aged 6 years. This is an example of the twin paradox.

Note. One might expect that the Principle of Relativity would imply

that Jack should also have aged less than Jill (an obvious contradiction).

However, due to the asymmetry of the situation (the fact that Jack is

inertial and Jill is not) the Principle of Relativity does not apply.

Note. Consider the lines of simultaneity for Jill at the “turning point”:

So our assumption that the eﬀect of Jill’s acceleration is inconsequential

is suspect! Jill’s “turning” masks a long period of time in Jacks’s frame

(t = 1.8 to t = 8.2).

Note. Now suppose that Jill emits a ﬂash of light at the end of each

(in her frame) year. Consider the spacetime diagram:

The ﬂash of light emitted at event B travels along a 45

◦

line (recall the

units) until it intersects the t−axis at point C and at time t

. Now

distance AC equals distance AB (since CBA is a 45-45-90). So

= t

+ AC = t

+ AB = t

+ x

With t

= 1 and x

= 0, equation (89), page 131, we have

βx

+ t

√

1 − β

√

1 − β

and

+ βt

√

1 − β

√

1 − β

We therefore have

= t

+ x

√

1 − β

√

1 − β

1 + β

√

1 − β

√

1 + β

√

1 + β

√

1 − β

√

1 + β

1 + β
1 − β

With β = 0.8 we have t

= 3. Similarly, the second ﬂash is observed

by Jack at t = 6, and the third ﬂash is observed at t = 9. Now the

above argument is general and we can show that if a light signal is

emitted by Jill every T units of time (in her frame), then Jack receives

the signals every

1 + β
1 − β

T units of time (in his frame). This change

in frequency is called the Doppler eﬀect and results in a redshift (that

is, lengthening of wavelength) for β > 0and a blueshift (that is, a

shortening of wavelength) for β < 0. Notice that the ﬂashes Jill emits

at t

= 3, t

= 4 and t

= 5 are observed by Jack at t = 9, t = 9

, and

t = 9

, respectively.

2.11 Temporal Order and Causality

Note. Suppose a flash of light is emitted at the origin of a spacetime

diagram. The wavefront is determined by the lines x = t and x = −t

where t > 0 (we use geometric units). We label the region in the upper

half plane that is between these two lines as region F . Extending

the lines into the lower half plane we similarly define region P . The

remaining two regions we label E.

Note. Events in F are separated from O by a timelike interval. So
O could influence events in F and we say O is causally connected to
the events in F . In fact, if A is an event in the interior of F , then

there is an inertial frame S

in which O and A occur at the same place.

The separation between O and A is then only one of time (and as we

claimed, O and A are separated by a timelike interval). The point A

will lie in the “future” relative to O, regardless of the inertial frame.

Therefore, region F is the absolute future relative to O.

Note. Similarly, events in P can physically influence O and events

in P are causally connected to O. The region P is the absolute past

relative to O.

Note. Events in region E are separated from O by a spacelike interval.

For each event C in region E, there is an inertial frame S

in which C

and O are separated only in space (and are simultaneous in time). This

means that the terms “before” and “after” have no set meaning between
O and an event in E. The region E is called elsewhere.

Note. We can extend these ideas and represent two physical dimen-

sions and one time dimension. We then find the absolute future relative

to an event to be a cone (called the future light cone). The past light

cone is similarly defined. We can imagine a 4-dimensional version where

the absolute future relative to an event is a sphere expanding in time.

Chapter 3. General Relativity:

The Geometry of Curved

Spacetime

Note. As we have seen, the Special Theory of Relativity deals only

with inertial (unaccelerated) observers. Such observers cannot be under

the influence of a gravitational field, and one might say that special

relativity describes the mechanics in a massless universe!

Note. In order to deal with accelerating frames of reference or frames of

reference under the influence of gravity, the Special Theory of relativity

has to be extended. This was accomplished by Einstein in his “Die

Grundlage der allgemeinen Relativit¨atstheorie” (“The Foundation of

the General Theory of Relativity”) in Annelan der Physik (Annals of

Physics) 49, 1916. As we will see, this was accomplished by considering

gravity not as a force, but as a curvature of spacetime. Falling objects,

planets in orbits, and rays of light then are observed to follow geodesics

in curved spacetime. (Surprisingly, the picture on the cover of the text

is a “straight” geodesic in a curved spacetime!)

Note. After some introductory material, we will discuss geodesics in

the semi-Riemannian 4-manifold of spacetime (Section 6) and “outline”

the reasoning which lead Einstein to his field equations (Section 7). We

will then solve the field outside an isolated sphere of mass M. Finally

(as time permits), we’ll explore orbits and the “bending of light” under

the General Theory of Relativity.

3.1 The Principle of Equivalence

Note. Newton’s Second Law of Motion (

F = ma) treats “mass” as

an object’s resistance to changes in movement (or acceleration). This

is an object’s inertial mass. In Newton’s Law of Universal Gravitation

(F = GMm/r

), an object’s mass measures its response to gravitational

attraction (called its gravitational mass). Einstein was bothered by the

dichotomy in the idea of mass:

inertial mass

acceleration

gravitational mass

gravitational acceleration

As we’ll see, he resolved this by putting gravity and acceleration on an

equivalent footing.

Note. Consider an observer in a sealed box. First, if this box is in

free fall in a gravitational ﬁeld, then the observer in the box will think

that he is weightless in an inertial (unaccelerating) frame. There is no

experiment he can perform (entirely within the box) to detect the accel-

eration or the presence of the gravitational ﬁeld. Second, if the observer

is out in deep space and under no gravitational inﬂuence BUT is accel-

erating rapidly, then he will interpret the acceleration as the presence

of a gravitational ﬁeld. Again, there is no experiment he can perform

(within the box) which will reveal that he is accelerating rapidly, versus

being stationary with respect to a gravitational ﬁeld. Einstein resolved

these observations (and the inertial versus gravitational mass question)

in the Principle of Equivalence.

Definition. Principle of Equivalence.

There is no way to distinguish between the eﬀects of acceleration and

the eﬀects of gravity - they are equivalent.

Note. A consequence of the Principle of Equivalence is that light

is “bent” when in a gravitational ﬁeld. Consider the observer in an

accelerating box. If a ray of light enters a hole in one side of the box,

it hits the other side of the box at a point slightly lower:

Now consider an observer in a box under the inﬂuence of gravity. By

the Principle of Equivalence, he must observe the same thing:

Example. Consider a collection of n particles of masses m

, m

, . . . , m

which interact (gravitationally, say) with force

between particle i

and j (and so

= −

) for i = j. Suppose observer #1 uses co-

ordinates x, y, z, t and observer #2 uses coordinates x

, y

, z

, t

. We

assume low relative velocities and so t = t

. Let

be the location of

particle i in observer #1’s coordinates and let

be the location of

particle i in observer #2’s coordinates. We assume low relative veloci-

ties and therefore ignore relativistic eﬀects and have t = t

. Therefore

we use

X = (x, y, z) and

= (x

, y

, z

) for the spatial coordinate vec-

tors. Suppose observer #1 believes that he and the particles are in the

presence of a gravitational ﬁeld g (and interpretes that he is stationary

experiencing the gravity and the particles are in free fall) and so he sees

all particles moving away from him with acceleration g. The equations

of motion for the particles for observer #1 are

= m

g +

j=i

for i = 1, 2, . . . , n.

Next, Suppose observer #2 is moving relative to observer #1 in such

a way that

X −

(Frame #2 is accelerating relative to Frame

#1 and conversely). By diﬀerentiating twice with respect to t:

− g

+ g.

Therefore, for the ith particle

+ g.

Substituting into the above equation gives

j=i

Therefore, the gravitational ﬁeld has been “transformed” away!

• Observer #1 thinks that observer #2 (along with the particles)

is in free fall and that is why he does not “see” the gravitational

ﬁeld.

• Observer #2 thinks there is no gravitational ﬁeld. He thinks that

he is an inertial observer and that observer #1 is accelerating away

from observer #2 and the particles.

The Principle of Equivalence states that both observers are “right.”

Therefore the Principle of Equivalence puts all frames (inertial or not)

on the “same footing.” In particular, gravitational force is equivalent

to a force created by acceleration.

3.2 Gravity as Spacetime Curvature

Note. Thus far we have only considered uniform gravitational ﬁelds.

In nature, gravitational “force” varies from point to point and so the

acceleration due to this force is not uniform. First consider two parti-

cles “side by side” in a gravitational ﬁeld (see Figure III-1, page 172)

directed towards a point (or the center of a sphere). As the particles

fall, they will be drawn closer together. Second, consider two particles

in a gravitational ﬁeld which are separated vertically. This time, the

diﬀerence in the forces produces a growing separation between the par-

ticles as they fall. In both cases, the behavior is an example of tidal

eﬀects. Therefore, a freely falling “space capsule” does not behave ex-

actly like an inertial frame. However, over short spans of time it is a

good approximation of an inertial frame.

Note. We rephrase the Principle of Equivalence as:

For each spacetime point (i.e. event), and for a given degree

of accuracy, there exists a frame of reference in which in a

certain region of space and for a certain interval of time,

the eﬀects of gravity are negligible and the frame is inertial

to the degree of accuracy speciﬁed.

Such a frame is called a locally inertial frame (at the given event) and

an observer in such a frame is a locally inertial observer.

Note. The convergence and divergence of particles as described above

has a geometric analog. On a positively curved surface, “parallel” geo-

desics converge, and on a negatively curved surface, “parallel” geodesics

diverge (see page 175).

Note. Einstein proposed that gravity is not a force, but a curvature

of spacetime! He hypothesized that free particles (and photons) follow

geodesics in a curved spacetime.

Definition. Deﬁne the matrix of the Lorentz metric as

(η

) =








0 -1








Note. A locally inertial observer in a locally inertial frame can set up

a system of coordinates where the interval satisﬁes

dτ

≈ η

= (du

)

− (du

)

− (du

)

− (du

)

More precisely, a coordinate system (u

) can be set up at a point

P in

a locally inertial frame such that

dτ

= g

and the functions g

satisfy

(

P ) = η

∂g

∂u

= 0.

These two conditions imply that the metric is the Lorentz metric at

P ,

and that the metric diﬀers little from the Lorentz metric near

P (i.e.

the rate of change of the g

’s is small near

P ).

Definition. A coordinate system such that dτ

= g

where g

are functions of u

(i = 0, 1, 2, 3) satisfying

(

P ) = η

∂g

∂u

= 0

for i, j, k = 0, 1, 2, 3, at some point

P is a locally Lorentz coordinate

system at

P .

Note. We are again back to the idea of a manifold. The Principle of

Equivalence tells us that in the 4-manifold of general relativity, small

local neighborhoods look like the “ﬂat” 4-manifold of special relativity.

The departure (on a larger scale) of dτ

from the Lorentz metric is due

to the nonuniformity of gravity and is the result (as we will see) of the

curvature of spacetime!

3.3 The Consequences of Einstein’s Theory

Note. Since we postulate that gravity is a curvature of spacetime

and that photons follow geodesics in spacetime, we ﬁnd that “gravity

bends light” (precisely, its the spacetime which is bent). The eﬀect

was experimentally veriﬁed in the famous 1919 eclipse expedition of

Arthur Eddington. During a total eclipse of the Sun, the position of a

star very near the Sun’s limb was measured. The star’s position was

found to be shifted by an amount predicted by the general theory. See

Figure III-4 on page 183. This experiment played a big role in making

Einstein the “science genius” and public ﬁgure that he was to become

in the 20’s, 30’s and 40’s. This experiment has been reproduced a

number of times using radio sources. A more contemporary example

which is also a consequence of this “bending of light” is gravitational

lensing. If a very distant galaxy is precisely along our line of site with

a massive foreground object, then we will see multiple images of the

background galaxy as the foreground object “focuses” the light rays.

In some situations, the image appears curved and is a segment of the

so called Einstein ring.

Note. A second example of experimental evidence for the general the-

ory is the precession of the orbit of Mercury. Mercury orbits the Sun

in an elliptical orbit (e ≈ .2) and therefore experiences diﬀerent accel-

erations due to the Sun. This results in a precession (or shifting) of

the perihelion (point of the orbit furthest from the Sun) over consec-

utive orbits (see Figure III-5, page 185). The observed precession is

43.11 ± 0.45

per century and general relativity predicts a precession of

43.03

per century (see Section 3.9 and Table III-2, page 230).

Note. Another prediction is the gravitational redshift of a photon in a

strong gravitational ﬁeld. We’ll explore this in Sections 3.7 and 3.8.

3.6 Geodesics

Note. We now view spacetime as a semi-Riemannian 4-manifold such

that for each coordinate system (x

, x

)

(dτ )

= g

µν

where the g

µν

are functions of the coordinates.

Definition. A vector v = v

∂

∂x

is timelike, lightlike, or spacelike if

v,v = g

µν

is positive, zero, or negative, respectively.

Definition. A spacetime curve α is a geodesic if it has a parameteri-

zation x

(ρ) satisfying

dρ

+ Γ

µν

dρ

= 0

(120)

for λ = 0, 1, 2, 3.

Note. “It can be shown” that the deﬁnition of geodesic is independent

of a choice of coordinate system (says the text, page 198).

Note. If α is a geodesic, then

, α

dτ
dρ

= g

µν

dρ

is constant (with the same proof as given at the bottom of page 68).

Definition. A geodesic α is timelike, lightlike, or spacelike according

to whether α

, α

is positive, zero, or negative.

Note. If a geodesic α is timelike, then dτ /dρ = constant, and we have
ρ = aτ +b for some a and b. We then see that equation (120) still holds
when we replace ρ with τ .

Note. If a geodesic α is lightlike then

, α

dτ
dρ

= 0

and τ is constant along α. Therefore proper time τ cannot be used to

reparameterize α.

Note. If a geodesic α is spacelike, then dτ /dρ is imaginary. The proper

distance

dσ =

(dx)

+ (dy)

+ (dz)

− (dt)

= idτ

can be used to parameterize α. We have dσ/dρ a real constant and so
ρ = aσ + b for some a and b. We see that equation (120) still holds
when we replace ρ with σ.

Definition. A curve α is timelike if α

, α

> 0 at each of its points.

Theorem III-2. Let α be a timelike curve which extremizes spacetime

distance (i.e. the quantity ∆τ ) between its two end points. Then α is

a geodesic.

Idea of the proof. The curve can be parameterized in terms of τ (as

remarked above). The proof then follows as did the proof of Theorem

I-9.

Theorem III-3. Given an event

P and a nonzero vector v at P , then

there exists a unique geodesic α such that α(0) =

P and α

(0) = v.

Note. Theorem III-3 implies that all particles in a gravitational ﬁeld

will fall with the same acceleration dependent only on initial position

and velocity. That is, we don’t see heavier objects fall faster!

Note. In the absence of gravity, a particle follows a path

dρ

= 0 (that

is, the particle follows a straight line!). We can therefore interprete the

Christoﬀel symbols as the components of the gravitational ﬁeld.

3.7 The Field Equations

Note. We now want a set of equations relating the metric coeﬃcients
g

µν

which determine the curvature of spacetime due to the distribution

of matter in spacetime. Einstein accomplished this in his “Die Grund-

lage der allgemeinen Relativit¨atstheorie”(The Foundation of the Gen-

eral Theory of Relativity) in Annalen der Physik (Annals of Physics)

in 1916.

Note. Consider a mass M at the origin of a 3-dimensional system. Let

X = (x, y, z) = (x(t), y(t), z(t)), and

X =

√

+ y

+ z

= r. Let u

be the unit radial vector

X/r. Under Newton’s laws, the force F on a

particle of mass m located at

X is

F = −Mm

= m

Therefore

= −

Definition. For a particle at point (x, y, z) in a coordinate system

with mass M at the origin, deﬁne the potential function Φ = Φ(r) as

Φ(r) = −

where r =

√

+ y

+ z

Theorem. The potential function satisﬁes Laplace’s equation

∇

Φ =

∂

∂x

∂

∂y

∂

∂z

= 0

at all points except the origin.

Proof. First

∂r

∂x

∂

∂x

[(

X ·

1/2

] =

X ·

1/2

and

∂Φ

∂x

∂Φ

∂r

∂x

Therefore

−∇Φ = −

∂Φ

∂x

∂Φ

∂y

∂Φ

∂z

= −

z
r

= −

Comparing components,

= −

∂Φ

∂x

(122)

Diﬀerentiating the relationship

∂Φ

∂x

∂

∂x

−

∂

∂x




−M

((x

)

+ (x

)

+ (x

)

1/2




−1(−1/2)M(2x

)

((x

)

+ (x

)

+ (x

)

3/2

gives

∂

(∂x

)

= M




− x

[(3/2)r(2x

)]




= M

− 3r(x

)

− 3(x

)

Summing over i = 1, 2, 3 gives

∇

Φ =

{(r

− 3(x

)

) + (r

− 3(x

)

) + (r

− 3(x

)

)} = 0.

Note. In the case of a ﬁnite number of point masses, the Laplace’s

equation still holds, only Φ is now a sum of terms (one for each particle).

Note. In general relativity, we replace equation (122) with

dτ

+ Γ

µν

dτ

= 0

(125)

where the Christoﬀel symbols are

µν

1
2

λβ

∂g

µβ

∂x

∂g

νβ

∂x

−

∂g

µν

∂x

Note. Comparing equations (122) and (125), we see that

∂Φ

∂x

and Γ

µν

dτ

play similar roles. As the text says, “in a sense then, the metric coef-

ﬁcients play the role of gravitational potential functions in Einstein’s

theory.”

Note. Trying to come up with a result analogous to Laplace’s equation

and treating the g

µν

’s as a potential function, we might desire a ﬁeld

equation of the form G = 0 where G involves the second partials of the
g

µν

’s.

Note. “It turns out”that the only tensors that are constructible from

the metric coeﬃcients g

µν

and their ﬁrst and second derivatives are

those that are functions of g

µν

and the components of R

µνσ

of the

curvature tensor.

Note. We want the ﬁeld equations to have the ﬂat spacetime of special

relativity as a special case. In this special case, the g

µν

are constants

and so we desire R

µνσ

= 0 for each index ranging from 0 to 3 (since the

partial derivatives of the g

µν

are involved). However, “it can be shown”

that this system of PDEs (in the unknown g

µν

’s) implies that the g

µν

’s

are constant (and therefore that we are under the ﬂat spacetime of

special relativity... we could use some details to verify this!).

Definition. The Ricci tensor is obtained from the curvature tensor by

summing over one index:

µν

= R

µνλ

∂Γ

µλ

∂x

−

∂Γ

µν

∂x

+ Γ

β
µλ

νβ

− Γ

µν

βλ

Note. Einstein chose as his ﬁeld equations the system of second order

PDEs R

µν

= 0 for µ, ν = 0, 1, 2, 3. More explicitly:

Definition. Einstein’s field equations for general relativity are the

system of second order PDEs

µν

∂Γ

µλ

∂x

−

∂Γ

µν

∂x

+ Γ

µλ

νβ

− Γ

µν

βλ

= 0

where

µν

1
2

λβ

∂g

µβ

∂x

∂g

νβ

∂x

−

∂g

µν

∂x

Therefore, the ﬁeld equations are a system of second order PDEs in the

unknown function g

µν

(16 equations in 16 unknown functions). The g

µν

determine the metric form of spacetime and therefore all intrinsic prop-

erties of the 4-dimensional semi-Riemannian manifold that is spacetime

(such as curvature)!

Note. The text argues that in a weak static gravitational ﬁeld, we

need

= 1 + 2Φ.

(135)

See pages 204-206 for the argument. We will need this result in the

Schwarzschild solution of the next section.

Lemma III-4. For each µ,

λβ

∂g

λβ

∂x

1
g

∂g

∂x

∂

∂x

[ln |g|].

Proof. See pages 207-208. We will use this result in the derivation of

the Schwarzschild solution.

3.8 The Schwarzschild Solution

Note. In this section, we solve Einstein’s ﬁeld equations for the grav-

itational ﬁeld outside an isolated sphere of mass M assumed to be at

rest at the (spatial) origin of our coordinate system.

Note. We convert to spherical coordinates ρ, φ, θ:

x = ρ sin φ cos θ

y = ρ sin φ sin θ
z = ρ cos φ.

In the event of ﬂat spacetime, we have the Lorentz metric

dτ

= dt

− dx

− dy

− dz

= dt

− dρ

− ρ

dφ

− ρ

sin

φ dθ

(144)

Note. As the book says, “the derivation that follows is not entirely

rigorous, but it does not have to be - as long as the resulting metric

form is a solution to the ﬁeld equations.”

Note. We have a static gravitational ﬁeld (i.e. independent of time)

and it is spherically symmetric (i.e. independent of φ and θ) so we look

for a metric form satisfying

dτ

= U(ρ)dt

− V (ρ)dρ

− W (ρ)(ρ

dφ

+ ρ

sin

φ dθ

)

(145)

where U, V, W are functions of ρ only. Let r = ρ

W (ρ) then (145)

becomes

dτ

= A(r)dt

− B(r)dr

− r

dφ

− r

sin

φ dθ

(146)

for some A(r) and B(r). Next deﬁne functions m = m(r) and n = n(r)

where

A(r) = e

2m(r)

= e

and B(r) = e

2n(r)

= e

Then (146) becomes

dτ

= e

− e

− r

dφ

− r

sin

φ dθ

(147)

Since dτ

= g

µν

in general, if we label x

= t, x

= r, x

= φ,

= θ we have

µν

) =








−e

−r

sin








and g = det(g

) = −e

2m+2n

sin

φ. If we ﬁnd m(r) and n(r), we will

have a solution!

Note. We need the Christoﬀel symbols

µν

1
2

λβ

∂g

µβ

∂x

∂g

νβ

∂x

−

∂g

µν

∂x

(126)

Since g

µν

= 0 for µ = ν, we have g

µµ

= 1/g

µµ

and g

µν

= 0 if µ = ν. So

the coeﬃcient g

λβ

is 0 unless β = λ and we have

µν

λλ

∂g

µλ

∂x

∂g

νλ

∂x

−

∂g

µν

∂x

We need to consider three cases:

Case 1. For λ = ν:

µν

νν

∂g

µν

∂x

∂g

νν

∂x

−

∂g

µν

∂x

νν

∂g

νν

∂x

1
2

∂

∂x

[ln(g

νν

)]

Case 2. For µ = ν = λ:

µµ

λλ

∂g

µλ

∂x

∂g

µλ

∂x

−

∂g

µµ

∂x

−1

λλ

∂g

µµ

∂x

since g

µλ

= 0 in this case.

Case 3. For µ, ν, λ distinct: Γ

µν

= 0 since g

µλ

= g

νλ

= g

µν

= 0 in this

case.

Note. With the g

µν

’s given above (in terms of m, n, r and φ) we can

calculate the nonzero Christoﬀel symbols to be:

= Γ

= m

2m−2n

= n

= −re

−2n

= Γ

1
r

= −re

−2n

sin

= Γ

1
r

= Γ

= cot φ

= − sin φ cos φ

where = d/dr.

Note. W e have

ln |g|

1/2

1
2

ln(e

2m+2n

sin

φ) = m + n + 2 ln r + ln(sin φ).

We saw in Lemma III-4 that

λβ

∂g

λβ

∂x

1
g

∂g

∂x

∂

∂x

[ln |g|].

Now

∂

∂x

[ln |g|

1/2

] =

1
2

∂

∂x

[ln |g|] =

1
2

λµ

∂g

λµ

∂x

1
2

λλ

∂g

λλ

∂x

Also, from (126)

µν

1
2

λβ

∂g

µβ

∂x

∂g

νβ

∂x

−

∂g

µν

∂x

we have with µ = β, ν = λ and δ the dummy variable:

βλ

1
2

λδ

∂g

βδ

∂x

∂g

λδ

∂x

−

∂g

βλ

∂x

1
2

λλ

∂g

βλ

∂x

∂g

λλ

∂x

−

∂g

βλ

∂x

1
2

λλ

∂g

λλ

∂x

Therefore we have

∂

∂x

[ln |g|

1/2

] = Γ

βλ

Similarly

∂

∂x

[ln |g|

1/2

] = Γ

µλ

. Therefore the ﬁeld equations imply

µν

∂

∂x

[ln |g|

1/2

] −

∂Γ

µν

∂x

+ Γ

µλ

νβ

− Γ

µν

∂

∂x

[ln |g|

1/2

] = 0.

Note. We ﬁnd that




−m

+ m

− m

−




2m−2n

= m

− m

+ m

−

= e

−2n

(1 + rm

− rn

) − 1

= R

sin

All other R

µν

are identically zero. Next, the ﬁeld equations say that

we need each of these to be zero. Therefore we need:




−m

+ m

− m

−




= 0

− m

+ m

−

= 0

−2n

(1 + rm

− rn

) − 1 = 0

sin

φ = 0

Adding the ﬁrst two of these equations, we ﬁnd that m

= 0, and so

m + n = b, a constant. However, by the boundary conditions both
m and n must vanish as r → ∞, since the metric (147) must approach
the Lorentz metric at great distances from the mass M (compare (147)

and (144)). Therefore, b = 0 and n = −m. The third equation implies:

1 = (1 + 2rm

= (re

)

Hence we have re

= r + C for some constant C, or g

= e

= 1 +

C/r. But as commented in the previous section, we need g

= 1−2M/r

where the field is weak. We therefore have C = −2M. Therefore we

have the solution:

dτ

1 −

−

1 −

−1

− r

dφ

− r

sin

φ dθ

Note. This solution was derived a few months after Einstein published

his paper in 1916. Notice that we have r = ρ and W (ρ) = 1.

3.9 Orbits in General Relativity

Note. We now calculate the geodesic which describes the orbit of

an object about another object of mass M located at the (spatial)

origin. This means that we desire to describe a geodesic under the

“Schwarzschild metric” of the previous section.

Note. We start with the Schwarzschild metric

dτ

1 −

−

1 −

−1

− r

dφ

− r

sin

φ dθ

and describe the path of a planet by a timelike geodesic

(τ ), x

(τ ))

where

dτ

+ Γ

µν

dτ

= 0

for λ = 0, 1, 2, 3. As in the previous section, we take x

= t, x

= r,

= φ, and x

= θ. The resulting Christoﬀel symbols are given in

equation (153), page 214.

Note. With λ = 2 we have from the geodesic condition

dτ

+ Γ

µν

dτ

= 0

and since the only nonzero Γ’s with a superscript of 2 are Γ

, Γ

, and

we have

dτ

+ Γ

dτ

+ Γ

dτ

+ Γ

dτ

= 0

dτ

+ 2

1
r

dτ

dφ

dτ

+ (− sin φ cos φ)

dθ

dτ

= 0.

We orient our axes such that when τ = 0, we have φ = π/2 and
dφ/dτ = 0. So the planet starts in the plane φ = π/2 and due to
symmetry remains in this plane. So we henceforth take φ = π/2. Now

with λ = 0:

dτ

+ Γ

µν

dτ

= 0

and since the only nonzero Γ’s with a superscript of 0 are Γ

and Γ

we have

dτ

+ Γ

dτ

+ Γ

dτ

= 0

dτ

+ 2

dτ

= 0.

(159a)

With λ = 1 :

dτ

+ Γ

µν

dτ

= 0

and since the only nonzero Γ’s with a superscript of 1are Γ

, Γ

and Γ

, we have

dτ

+ Γ

dτ

+ Γ

dτ

+ Γ

dτ

+ Γ

dτ

= 0

dτ

+ m

2m−2n

dτ

+ n

dτ

+(−re

−2n

)

dφ

dτ

+ (−re

−2n

sin

φ)

dθ

dτ

= 0.

(159b)

(since we have φ = π/2, sin

φ ≡ 1). With λ = 3

dτ

+ Γ

µν

dτ

= 0

and since the only nonzero Γ’s with a superscript of 3 are Γ

, Γ

and Γ

, we have

dτ

+ Γ

dτ

+ Γ

dτ

+ Γ

dτ

+ Γ

dτ

= 0

dτ

+ 2

1
r

dτ

dθ

dτ

+ 2

cot φ

dφ

dτ

dθ

dτ

= 0

or since φ = π/2

dτ

+ 2

1
r

dτ

dθ

dτ

= 0.

(159c)

Note. Now by the Chain Rule m

dτ

, and dividing

(159a) by dt/dτ gives

t/dτ

dt/dτ

+ 2

dτ

= 0

dτ

= −2

dτ

Integration yields

dτ

= −2m + constant

dτ

= be

−2m

(160)

where b is some positive constant and we deﬁne γ = e

Note. Equation (159c) can be integrated (see page 63 for the process)

to yield

dθ

dτ

= h

(161)

where h is a positive constant.

Note. From the Schwarzschild metric with φ = π/2, dφ = 0, and
γ = e

= 1 − 2M/r we have

dτ

= γdt

− γ

−1

− r

dθ

1 = γ

dτ

− γ

−1

dτ

− r

dθ

dτ

= γ

− γ

−1

dr
dθ

− r

(162)

by equation (160), the fact that

dτ

dr
dθ

dθ

dτ

dr
dθ

(by the Chain Rule and equation (161)) and by equation (161). Multi-

plying (162) by γ yields

γ = b

−

dr
dθ

− γ

or since γ = 1 − 2M/r:

1 −

= b

−

dr
dθ

−

1 −

dr
dθ

= b

− 1 +

Let u = 1/r so that

dθ

= −

dr
dθ

and the previous equation yields

dr
dθ

− 1

dr
dθ

+ u

− 1

2Mu

+ 2Mu

dθ

+ u

− 1

2Mu

+ 2Mu

Diﬀerentiation with respect to θ yields

dθ

+ 2u

dθ

+ 6Mu

dθ

+ u =

+ 3Mu

(163)

where u = 1/r and h = r

dθ/dτ (constant).

Note. A similar analysis in the Newtonian setting yields

dθ

+ u =

(114)

(see page 193). So the only diﬀerence is the 3Mu

term in (163) and we

can think of this as the “relativistic term.” Equation (114) has solution

u =

(1+ e cos θ).

Note. We can view (163) as a linear ODE (considering the left hand

side) set equal to a nonhomogeneous term

+ 3Mu

. Now the term

3Mu

is “small” as compared to M/h

(see page 226). We perturb this

equation by replacing u with the approximate solution M/h

(1+e cos θ)

on the right hand side of (163) and consider

dθ

+ u =

+ 3M

(1+ e cos θ)

(1+ 2e cos θ + e

cos

θ).

A solution to this (linear) ODE will then be an approximate solution

to (163). The ODE is

dθ

+ u =

cos θ +

cos(2θ). (165)

(The last two terms follow from the fact that cos

θ = (1 + cos(2θ))/2.)

Lemma III-5. Let A ∈ R. Then

1. u = A is a solution of d

u/dθ

+ u = A.

2. u = (A/2)θ sin θ is a solution of d

u/dθ

+ u = A cos θ.

3. u = (−A/3) cos(2θ) is a solution of d

u/dθ

+ u = A cos(2θ).

(The proof follows by simply diﬀerentiating.)

Note. Equation (165) is

dθ

+ u =










cos θ







cos(2θ)




and by Lemma III-5

u =










θ sin θ







−M

cos(2θ)







1 +




1 +




θ sin θ −

cos(2θ)




is a (particular) solution to (165). Now the general solution to the

homogeneous ODE

dθ

+ u = 0

is a linear combination of u = sin θ and u = cos θ. Therefore, we can

add any linear combination of these functions to the above particular

solution to get another solution. We choose to add

e cos θ (for rea-

sons to be discussed shortly - we want to compare this solution to the

Newtonian solution). We have

u =




1 +




1 +




+ e cos θ +

θ sin θ −

cos(2θ)




Note. The term




1 +




is small compared to 1(8 × 10

−8

for

Mercury, see page 227). If we let

α = 1 +




1 +




and deﬁne e

= e/α then

Mα

(1+e

cos θ) =

(α+e cos θ) =




1 +




1 +




+ e cos θ




Since α ≈ 1, e

≈ e and we see that this part of the u function is

approximately the same as the Newtonian solution. A similar argu-

ment shows that the “cos(2θ)” term causes little deviation from the

Newtonian solution. Therefore we have that

u ≈




1 + e cos θ +

θ sin θ




Although the “θ sin θ” may be very small initially, as θ increases “the

term will have a cumulative eﬀect over many revolutions” (see page

228). This eﬀect is the observed perihelial advance.

Note. Since M

is small (≈ 10

−8

for Mercury, see page 228) we

approximate

cos







≈ 1,

sin







≈

and we get






1 + e cos




θ −














1 + e




cos θ cos







+ sin θ sin












≈






1 + e cos θ + e

θ sin θ






= u.

So u has a maximum (and r = 1/u has a minimum - i.e. we are at

perihelion) when cos




θ −




is at a maximum. This occurs for

θ = 0 and

θ =

2π

1 − 3M

≈ 2π




1 +




(since (1 − x)

−1

≈ 1 + x for x ≈ 0). So the perihelion advances (in the

direction of the orbital motion) by an amount 6πM

per revolution.

Note. If we want the orbital precession per century, we have

∆θcent = n∆θ =

6πM

6πMn

a(1 − e

)

(since h

= a(1 − e

) by equation (119)) where n is the number of

orbits of the Sun that a planet makes per century.

Note. Table III-2 gives the calculated precessions as observed and as

predicted for 4 solar system objects. The last two columns represent

precessions measured in seconds per century:

Planet

a(÷10

cm)

n General Relativity

Observed

Mercury

57.91

0.2056 415

43.03

43.11 ± 0.45

Venus

108.21

0.0068 149

8.6

8.4 ± 4.8

Earth

149.60

0.0167 100

3.8

5.0 ± 1.2

Icarus

161.0

0.827

10.3

9.8 ± 0.8

3.10 The Bending of Light

Note. A photon of light follows a lightlike geodesic for which dτ = 0.

Such a geodesic, therefore, cannot be parameterized in terms of τ . So

we parameterize in terms of some ρ where dτ /dρ = 0. As in the previous

section, let the geodesic be (x

(ρ), x

(ρ)) and we have:

dρ

+ Γ

µν

dρ

= 0

for λ = 0, 1, 2, 3.

Note. As in the previous section, we desire equations (159a-c) with τ

replaced with ρ (again, we assume the photon is restricted to the plane
φ = π/2).

Note. As in the previous section

dτ

1 −

−

1 −

−1

− r

dφ

− r

sin

φ dθ

(this is equation (157)) and so

dτ

= γdt

− γ

−1

− r

dθ

(since φ = π/2 and dφ = 0) and




dτ

dρ




= γ

dρ

− γ

−1

dr
dρ

− r

dθ
dρ

where γ = 1 − 2M/r. Now dτ /dρ = 0 so with the notation of the

previous section (see equation (162)) we have

0 = γ

− γ

−1

dr
dθ

− r

0 = b

−

dr
dθ

− γr

dr
dθ

+ γ

= b

Now (as on page 226) with u = 1/r and

dθ

−1

dr
dθ

we get

dθ

1 −

= b

dθ

+ u

2Mu

+ 2Mu

Diﬀerentiating with respect to θ implies

dθ

+ 2u

dθ

= 0 + 6Mu

dθ

or (dividing by 2du/dθ):

dθ

+ u = 3Mu

(168)

(compare to (163)).

Note. We orient our coordinate system such that the closest approach

of the geodesic occurs at θ = 0. If M = 0, the general solution to (168)

(a homogeneous equation under this condition) is u = α cos θ + β sin θ.

If we let R be the minimum distance of the geodesic from M (assumed

to occur at θ = 0) we have α = 1/R. With M = 0, geodesics are

“straight lines” and so β = 0 and u = (1/R) cos θ.

Note. We modify (168) by substituting u = (1/R) cos θ in the right

hand side to produce (as in the previous section)

dθ

+ u ≈ 3M

cos θ

cos

(1 +cos(2θ)) .

By Lemma III-5, a particular solution is

u =

1 −

1
3

cos(2θ)

or (since cos(2θ) = 2 cos

θ − 1)):

u =

(2 − cos

θ).

Now adding the homogeneous solution u = (1/R) cos θ to this particular

solution we get

u =

1
r

(2 − cos

θ) +

cos θ. (170)

Note. From Figure III-10, we have for r → ∞, θ approaches ±(π/2 +

∆θ/2). Since ∆θ ≈ 0, cos

θ ≈ 0 and so for r → ∞ we have from (170)

that

u =

1
r

≈ 0 ≈

cos θ +

(2 − cos

θ)

→

cos

∆θ

− 0.

−1

cos

∆θ

sin

∆θ

≈

∆θ

Therefore

∆θ

≈

and ∆θ ≈

Note. If we assume a photon undergoes a Newtonian acceleration, we

ﬁnd that Newtonian mechanics implies that a photon follows a hyper-

bolic trajectory with ∆θ =

when grazing the Sun. This is half

the displacement predicted by general relativity.

Note. As indicated previously, experiment agrees strongly with Ein-

stein’s theory! It’s relative!

Special Topic: Black Holes

Primary Source: A Short Course in General Relativity, 2nd Ed., J.

Foster and J.D. Nightingale, Springer-Verlag, N.Y., 1995.

Note. Suppose we have an isolated spherically symmetric mass M with

radius r

which is at rest at the origin of our coordinate system. Then

we have seen that the solution to the ﬁeld equations in this situation is

the Schwarzschild solution:

dτ

1 −

−

1 −

−1

− r

dφ

− r

sin

φ dθ.

Notice that at r = 2M, the metric coeﬃcient g

is undeﬁned. Therefore

this solution is only valid for r > 2M. Also, the solution was derived

for points outside of the mass, and so r must be greater than the radius

of the mass r

. Therefore the Schwarzschild solution is only valid for

r > max{2M, r

Definition. For a spherically symmetric mass M as above, the value
r

= 2M is the Schwarzschild radius of the mass. If the radius of the

mass is less than the Schwarzschild radius (i.e. r

< r

) then the object

is called a black hole.

Note. In terms of “traditional” units, r

= 2GM/c

. For the Sun,

= 2.95 km and for the Earth, r

= 8.86 mm.

Note. Since the coordinates (t, r, φ, θ) are inadequate for r ≤ r

, we

introduce a new coordinate which will give metric coeﬃcients which are

valid for all r. In this way, we can explore what happens inside of a

black hole!

Note. We keep r, φ, and θ but replace t with

v = t + r + 2M ln

− 1

(∗)

Theorem. In terms of (v, r, φ, θ), the Schwarzschild solution is

dτ

= (1 − 2M/r)dv

− 2dv dr − r

dφ

− r

sin

φ dθ.

(∗∗)

These new coordinates are the Eddington-Finkelstein coordinates.

Proof. Homework! (Calculate dt

in terms of dv and dr, then substi-

tute into the Schwarzschild solution.)

Note. Each of the coeﬃcients of dτ

in Eddington-Finkelstein coordi-

nates is deﬁned for all nonzero r > r

. Therefore we can explore what

happens for r < r

in a black hole. We are particularly interested in

light cones.

Note. Let’s consider what happens to photons emitted at a given

distance from the center of a black hole. We will ignore φ and θ and

take dφ = dθ = 0. We want to study the radial path that photons

follow (i.e. radial lightlike geodesics). Therefore we consider dτ = 0.

Then (∗∗) implies

(1 − 2M/r)dv

− 2dv dr = 0,

(1 − 2M/r)

− 2

dv
dr

= 0,

dv
dr

(1 − 2M/r)

dv
dr

− 2

= 0.

Therefore we have a lightlike geodesic if

dv
dr

= 0 or if

dv
dr

1 − 2M/r

Note. First, let’s consider radial lightlike geodesics for r > r

. Diﬀer-

entiating (∗) gives

dv
dr

+ 1 +

r/2M − 1

r/2M

r/2M − 1

1 − 2M/r

With the solution dv/dr = 0, we ﬁnd

−1

1 − 2M/r

Notice that this implies that dt/dr < 0 for r > 2M. Therefore for
dv/dr = 0 we see that as time (t) increases, distance from the origin
(r) decreases. Therefore dv/dr = 0 gives the ingoing lightlike geodesics.

With the solution dv/dr = 2/(1 − 2M/r), we ﬁnd

1 − 2M/r

−

1 − 2M/r

Notice that this implies that dt/dr > 0 for r > 2M. Therefore for
dv/dr = 2/(1 − 2M/r), we see that as time (t) increases, distance from
the origin (r) increases. Therefore dv/dr = 2/(1 − 2M/r) gives the

outgoing lightlike geodesics. Therefore for r > 2M, a ﬂash of light at

position r will result in photons that go towards the black hole and

photons that go away from the black hole (remember, we are only

considering radial motion).

Note. Second, let’s consider radial lightlike geodesics for r < 2M. In-

tegrating the solution dv/dr = 0 gives v = A (A constant). Integrating

the solution dv/dr = 2/(1 − 2M/r) gives

v =

dv
dr

dr =

2 dr

1 − 2M/r

2r dr

r − 2M

= 2

1 +

r − 2M

dr = 2r + 4M ln |r − 2M| + B,

B constant. In the following ﬁgure, we use oblique axes and choose A
such that v = A gives a line 45

◦

to the horizontal (as in ﬂat spacetime).

The choice of B just corresponds to a vertical shift in the graph of
v = 2r + 4M ln |r − 2M| and does not change the shape of the graph
(so we can take B = 0).

The little circles represent small local lightcones. Notice that a photon

emitted towards the center of the black hole will travel to the center

of the black hole (or at least to r

). A photon emitted away from the

center of the black hole will escape the black hole if it is emitted at
r > r

= 2M. However, such photons are “pulled” towards r = 0 if

they are emitted at r < r

. Therefore, any light emitted at r < r

will

not escape the black hole and therefore cannot be seen by an observer

located at r > r

. Thus the name black hole. Similarly, an observer

outside of the black hole cannot see any events that occur in r ≤ r

and the sphere r = r

is called the event horizon of the black hole.

Note. Notice the worldline of a particle which falls into the black hole.

If it periodically releases a ﬂash of light, then the outside observer will

see the time between the ﬂashes taking a longer and longer amount of

time. There will therefore be a gravitational redshift of photons emitted

near r = r

(r > r

). Also notice that the outside observer will see the

falling particle take longer and longer to reach r = r

. Therefore the

outside observer sees this particle fall towards r = r

, but the particle

appears to move slower and slower. On the other hand, if the falling

particle looks out at the outside observer, it sees things happen very

quickly for the outside observer. All this action, though, is compressed

into the brief amount of time (in the particles frame) that it takes to

fall to r = r

Note. Notice that the radial lightlike geodesics determined by dv/dr =

2(1−2M/r) have an asymptote at r = r

. This will result in lightcones

tilting over towards the black hole as we approach r = r

(Figure 46, page 93 of Principles of Cosmology and Gravitation, M.

Berry, Cambridge University Press, 1976.) Again, far from the black

hole, light cones are as they appear in ﬂat spacetime. For r ≈ r

and

r > r

, light cones tilt over towards the black hole, but photons can

still escape the black hole. At r = r

, photons are either trapped at

r = r

(those emitted radially to the black hole) or are drawn into

the black hole. For r < r

, all worldlines are directed towards r = 0.

Therefore anything inside r

will be drawn to r = 0. All matter in a

black hole is therefore concentrated at r = 0 in a singularity of inﬁnite

density.

Note. The Schwarzschild solution is an exact solution to the ﬁeld

equations. Such solutions are rare, and sometimes are not appreciated

in their fullness when introduced. Here is a brief history from Black

Holes and Time Warps: Einstein’s Outrageous Legacy by Kip Thorne

(W.W. Norton and Company, 1994):

1915 Einstein (and David Hilbert) formulate the ﬁeld equations (which

Einstein published in 1916).

1916 Karl Schwarzschild presents his solution which later will describe

nonspinning, uncharged black holes.

1916 & 1918 Hans Reissner and Gunnar Nordstr¨om give their solu-

tions, which later will describe nonspinning, charged black holes.

(The ideas of black holes, white dwarfs, and neutron stars did

not become part of astrophysics until the 1930’s, so these early

solutions to the ﬁeld equations were not intended to address any

questions involving black holes.)

1958 David Finkelstein introduces a new reference frame for the Schwarz-

schild solution, resolving the 1939 Oppenheimer-Snyder paradox

in which an imploding star freezes at the critical (Schwarzschild)

radius as seen from outside, but implodes through the critical ra-

dius as seen from outside.

1963 Roy Kerr gives his solution to the ﬁeld equations.

1965 Boyer and Lindquist, Carter, and Penrose discover that Kerr’s

solution describes a spinning black hole.

Some other highlights include:

1967 Werner Israel proves rigorously the ﬁrst piece of the black hole

“no hair” conjecture: a nonspinning black hole must be precisely

spherical.

1968 Brandon Carter uses the Kerr solution to show frame dragging

around a spinning black hole.

1969 Roger Penrose describes how the rotational energy of a black hole

can be extracted.

1972 Carter, Hawking, and Israel prove the “no hair” conjecture for

spinning black holes. The implication of the no hair theorem is

that a black hole is described by three parameters: mass, rotational

rate, and charge.

1974 Stephan Hawking shows that it is possible to associate a tem-

perature and entropy with a black hole. He uses quantum theory

to show that black holes can radiate (the so-called Hawking radi-

ation).

1993 Hulse and Taylor are awarded the Nobel Prize for an indirect

detection of gravitational waves from a binary pulsar.

2002-2005 Gravity waves are directly detected with the Laser Inter-

ferometer Gravity Wave Observatory (LIGO).

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