Gravity Theory  Homework #1
Paweł Laskoś-Grabowski
10 czerwca 2008
Á
Verify that "X("Y Zµ) - "Y ("XZµ) - "[X,Y ]Zµ = Rµ ZÄ…X½Y .
Ä…½Á
We ll use the following identities:
"µX½ = "µX½ + “½ XÁ
µÁ
µ µ
"XY = X½"½Y
½ µ
[X, Y ]½ = Xµ"µY - Y "µX½
Plugging all these into the lhs of the identity in question and expanding we obtain the
following:
Á ½
LHS = X½"½(Y "ÁZµ) - Y "½(XÁ"ÁZµ) - [X, Y ]½"½Zµ =
Á Á ½
= X½("½(Y "ÁZµ) + “µ (Y "ÁZÃ)) - Y ("½(XÁ"ÁZµ) + “µ (XÁ"ÁZÃ)) - [X, Y ]½"½Zµ
ý ý
Á Á Á
= X½("½Y )"ÁZµ + X½Y "½"ÁZµ + X½“µ Y "ÁZÃ
ý
½ ½ ½
- Y ("½XÁ)"ÁZµ - Y XÁ"½"ÁZµ - Y “µ XÁ"ÁZÃ
ý
½ Á
- (XÁ"ÁY - Y "ÁX½)"½Zµ
Now we see that the last term cancels the first and fourth. We need to expand " s in the
remaining terms.
Á Á Á
LHS = X½Y "½"ÁZµ + X½Y "½(“µ ZÃ) + X½“µ Y ("ÁZà + “à ZÄ…)
Áà ý ÁÄ…
½ ½ ½
- Y XÁ"½"ÁZµ - Y XÁ"½(“µ ZÃ) - Y “µ XÁ("ÁZÃ + “Ã ZÄ…)
Áà ý ÁÄ…
First terms in each line cancel out, for "½, "Á commute.
Á Á Á Á
LHS = X½Y ("½“µ )Zà + X½Y “µ "½Zà + X½“µ Y "ÁZà + X½“µ Y “à ZÄ…
Áà Áà ý ý ÁÄ…
½ ½ ½ ½
- Y XÁ("½“µ )Zà - Y XÁ“µ "½Zà - Y “µ XÁ"ÁZà - Y “µ XÁ“à ZÄ…
Áà Áà ý ý ÁÄ…
Second term in the first line cancels the third in the second line and vice versa. After
renaming indices in the remaining terms, we obtain the desired result.
Á Á Á Á
LHS = ("½“µ )ZÄ…X½Y - ("Á“µ )ZÄ…X½Y + “µ “à ZÄ…X½Y - “µ “à ZÄ…X½Y
ÁÄ… ½Ä… ý ÁÄ… ÃÁ ½Ä…
Á Á
= ("½“µ - "Á“µ + “µ “à - “µ “à )ZÄ…X½Y = Rµ ZÄ…X½Y .
ÁÄ… ½Ä… ½Ã ÁÄ… Áà ½Ä… Ä…½Á
Verify that for geodesic x(Ä), "vv a" "‹‹ = 0.
By simply rewriting the expression we recover the geodesic equation.
(chain rule)
"‹‹ = ‹µ("µ‹½ + “½ ‹Á) = ć½ + “½ ‹µ‹Á = 0.
µÁ µÁ