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Easy Variation Question?

by sergenyalcin
Tags: variation
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sergenyalcin
#1
Dec7-12, 10:30 AM
P: 3
I am considering the variation of

[itex] \delta ( \sqrt{g} R_{abcd} R^{abcd} ) [/itex]

and I know the answer is

[itex] - \frac12 \sqrt{g} g_{\mu\nu}R_{abcd} R^{abcd} +\sqrt{g} R_{( \mu}{}^{bcd} R_{\nu ) bcd} + \ldots [/itex]

what i do not understand is the coefficient of the last term. For example, when we evaluate the Maxwell Action

[itex] \sqrt{g} F_{ab} F^{ab} [/itex]

what we do is to write down as

[itex] \sqrt{g} g^{\mu\nu} g^{\alpha\beta} F_{\mu\alpha} F_{\nu\beta}[/itex]

so when we vary the action, we get

[itex] -\frac12 \sqrt{g} g_{\mu\nu} F^2 + 2 \sqrt{g} F_{(\mu}{}^{\sigma} F_{\nu ) \sigma}[/itex]

why is it not working with Riemann Tensor? How come there is no factor of 4 on the front of the last term in the variation of Riemann squared action?

Thanks in advance
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Bill_K
#2
Dec7-12, 11:02 AM
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My reference (DeWitt's lectures) does have a factor of 4. Plus there are several terms involving the Ricci tensor.
sergenyalcin
#3
Dec7-12, 11:05 AM
P: 3
Quote Quote by Bill_K View Post
My reference (DeWitt's lectures) does have a factor of 4. Plus there are several terms involving the Ricci tensor.
do you have a link for these lecture notes?

Bill_K
#4
Dec7-12, 11:38 AM
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Easy Variation Question?

Sorry, my mistake, the additional factor is 2 not 4. The reference is "Dynamical Theory of Groups and Fields", which is apparently not available online, although many libraries have it. Although DeWitt claims the calculation is easy, it is not! So let me quote his result in full (He must be using the opposite sign convention):

L3 ≡ √g RμνστRμνστ
δL3/δgμν = √g (4Rμνσ -2R;μν -2RμστρRνστρ +gμνRστρλRστρλ -4RμσντRστ +4RμσRνσ)

sergenyalcin
#5
Dec7-12, 11:43 AM
P: 3
Quote Quote by Bill_K View Post
Sorry, my mistake, the additional factor is 2 not 4. The reference is "Dynamical Theory of Groups and Fields", which is apparently not available online, although many libraries have it. Although DeWitt claims the calculation is easy, it is not! So let me quote his result in full (He must be using the opposite sign convention):

L3 ≡ √g RμνστRμνστ
δL3/δgμν = √g (4Rμνσ -2R;μν -2RμστρRνστρ +gμνRστρλRστρλ -4RμσντRστ +4RμσRνσ)

You are absolutely right about the factor of 2! i am sorry for the typo. but i still do not understand why the factor is not 4, but 2?
Bill_K
#6
Dec7-12, 12:44 PM
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Without plowing through the algebra, only a vague comment. When you write FμνFμν = gμαgνβFμνFαβ you know you've factored out all the g-dependence, and it's easy to see there are two g's. But when you write RμνστRμνστ = gμαgνβgσγgτδRμνστRαβγδ, the R's still contain g-dependence, so it's not going to be just a factor of four.


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