
#1
Dec712, 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 



#2
Dec712, 11:02 AM

Sci Advisor
Thanks
P: 3,846

My reference (DeWitt's lectures) does have a factor of 4. Plus there are several terms involving the Ricci tensor.




#3
Dec712, 11:05 AM

P: 3





#4
Dec712, 11:38 AM

Sci Advisor
Thanks
P: 3,846

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):
L_{3} ≡ √g R_{μνστ}R^{μνστ} δL_{3}/δg_{μν} = √g (4R^{μν}_{;σ}^{σ} 2R_{;}^{μν} 2R^{μ}_{στρ}R^{νστρ} +½g^{μν}R_{στρλ}R^{στρλ} 4R^{μσντ}R_{στ} +4R^{μ}_{σ}R^{νσ}) 



#5
Dec712, 11:43 AM

P: 3





#6
Dec712, 12:44 PM

Sci Advisor
Thanks
P: 3,846

Without plowing through the algebra, only a vague comment. When you write F^{μν}F_{μν} = g^{μα}g^{νβ}F_{μν}F_{αβ} you know you've factored out all the gdependence, and it's easy to see there are two g's. But when you write R^{μνστ}R_{μνστ} = g^{μα}g^{νβ}g^{σγ}g^{τδ}R_{μνστ}R_{αβγδ}, the R's still contain gdependence, so it's not going to be just a factor of four.



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