Easy Variation Question?

sergenyalcin

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

Bill_K

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

sergenyalcin

do you have a link for these lecture notes?

Bill_K

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 R_μνστR^μνστ
δL₃/δg_μν = √g (4R^μν_;σ^σ -2R_;^μν -2R^μ_στρR^νστρ +½g^μνR_στρλR^στρλ -4R^μσντR_στ +4R^μ_σR^νσ)

sergenyalcin

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

Without plowing through the algebra, only a vague comment. When you write F^μνF_μν = g^μαg^νβF_μνF_αβ 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 R^μνστR_μνστ = g^μαg^νβg^σγg^τδR_μνστR_αβγδ, the R's still contain g-dependence, so it's not going to be just a factor of four.