Classical field theory  exercise no. 1
Pawel Laskoś-Grabowski
October 30, 2007
1 Content
Prove that for the action of the form S = d4x L("�ŚI, ŚI), the equations of motion are
&!
"L "L
- "� = 0. (1)
"ŚI "("�ŚI)
2 Solution
Let s examine the variation of a bivariate function expressed as a Taylor series (where we
omit terms of non-linear order in �x, �y by assumption that variations of the variables are
small):
"f "f
�f(x, y) = f(x + �x, y + �y) - f(x, y) = �x + �y. (2)
"x "y
Now let s substitute f = L, x = "�Ś, y = Ś (we omit the index I for clarity, as it is
insignificant for the calculations) and use the last result to express �S:
"L "L
�S = d4x �L = d4x �("�Ś) + �Ś . (3)
"("�Ś) "Ś
&! &!
We ll prove that �("�Ś) = "�(�Ś)  in the following, let f be function mapping x to
x + �x (it is clear that "�f = 1 as �x is arbitrary and thus does not depend on x):
"�(�Ś) = "�(Ś(x + �x) - Ś(x)) = "�((Ś ć% f)(x) - Ś(x))
= ("�Ś)(f(x)) � ("�f)(x) - ("�Ś)(x) = ("�Ś)(x + �x) - ("�Ś)(x)
= �("�Ś). (4)
Let s express an instance of Leibniz law:
"L "L "L
"� �Ś = "�(�Ś) + "� �Ś. (5)
"("�Ś) "("�Ś) "("�Ś)
Under the integral in �S, we add lhs and subtract rhs of the last result, and after one
obvious (from (4)) term cancellation we obtain:
"L "L "L
�S = d4x - �Ś + �Ś + d4x "� �Ś
"("�Ś) "Ś "("�Ś)
&! &!
"L "L "L
= d4x - + �Ś + d3x �Ś. (6)
"("�Ś) "Ś "("�Ś)
&! "&!
1
The last term vanishes, as we assume �Ś = 0 on the edge ("&!) of the space-time volume of
integration. To obtain equations of motion we assume �S = 0, and if �Ś is arbitrary, then
by the fundamental lemma of calculus of variations we get that the term in parentheses
is 0, which is the desired result.
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