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If we take a variation of a covariant derivative, we must take into account the connections, so we get:

$$ \delta (\nabla_\beta T_{\mu \nu}) = \nabla_\beta \delta(T_{\mu \nu}) -\delta (\Gamma_{\beta \mu}^i)T_{i \nu} - \delta(\Gamma^i_{\beta \nu}) T_{\mu i}$$

$$ = \nabla_\beta \delta(T_{\mu \nu}) - \frac{g^{i \lambda}}{2} \left[ (\nabla_\beta h_{\mu \lambda} + \nabla_\mu h_{\beta \lambda} - \nabla_\lambda h_{\beta \mu}) T_{i \nu} + (\nabla_\beta h_{\nu \lambda} + \nabla_\nu h_{\beta \lambda} - \nabla_\lambda h_{\beta \nu} ) T_{\mu i} \right]$$

The question I have is whether the covariant derivatives of the Christoffel symbols affect to what I have on its right side. That is if I have to consider this: $$ \frac{-1}{2} g^{i \lambda} \nabla_\beta (h_{\mu \lambda}) T_{i \nu}$$ or this: $$ \frac{-1}{2} g^{i \lambda} \nabla_\beta (h_{\mu \lambda} T_{i _\nu})$$

I guess it should be the former but I want to be sure about it.

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The answer to your specific question is no. Covariant derivatives there act only on $h$, not $T$.

Also, your formulas are not correct. You want the first order variation of $\nabla_\beta T_{\mu \nu}$, which you explicitly write as

$$ \nabla_\beta T_{\mu \nu} = \partial_\beta T_{\mu \nu} -\Gamma_{\beta \mu}^\alpha T_{\alpha \nu} - \Gamma^\alpha_{\beta \nu} T_{\mu \alpha}.$$

Then varying this (wrt metric $g_{\mu \nu}$) to first order means that you collect all terms that are of $\mathcal{O}(h)$. Here, the full metric is some background plus perturbation: $g_{\mu \nu} = \bar{g}_{\mu \nu} + h_{\mu \nu}$. Essentially, you have the following contributions:

$$ \delta(\nabla_\beta T_{\mu \nu}) = \partial_\beta \delta(T_{\mu \nu}) -\delta(\Gamma_{\beta \mu}^\alpha) T_{\alpha \nu} -\Gamma_{\beta \mu}^\alpha \delta(T_{\alpha \nu}) - \delta(\Gamma^\alpha_{\beta \nu}) T_{\mu \alpha} - \Gamma^\alpha_{\beta \nu} \delta(T_{\mu \alpha}),$$

where the expression for $\delta \Gamma$ is what you wrote. There, $\nabla$ acts only on $h$, as you can see from above.

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  • $\begingroup$ Thanks for the comment. What I did is to rearrange $\partial_\beta \delta(T_{\mu \nu}) - \Gamma^{\alpha}_{\beta \mu} \delta(T_{\alpha \nu}) - \Gamma^{\alpha}_{\beta \nu} \delta (T_{\mu \alpha}) = \nabla_\beta \delta(T_{\mu \nu})$. Is not possible to do that? In case not, why? $\endgroup$ – cramirpe Aug 7 at 10:07
  • $\begingroup$ @cramirpe Ahh yeah okay, I missed that. I prefer expanding everything to not miss any terms. Yeah, what you wrote is ok. You can just define a new tensor $\delta(T_{\mu \nu}) \equiv S_{\mu \nu}$ and then you write $\nabla_\beta S_{\mu \nu}$. But you have to be careful when you do this. For example the trace of $S_{\mu \nu} = g^{\mu \nu} S_{\mu \nu}$ is $g^{\mu \nu} \delta(T_{\mu \nu})$ and not $\delta(g^{\mu \nu} T_{\mu \nu})$. Raising and lowering the indices of $T$ happens inside the variation $\delta ( \ )$. $\endgroup$ – Avantgarde Aug 7 at 10:21

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