A Help with 2nd Order Ricci: Indexes in GR

[GrimGuy]

In my studies trying to get the Ricci tensor of 2° order i stuck in this expression:

​

$h_{\mu}^{\ \sigma},_{\lambda}h_{\sigma}^{\ \lambda},_{\nu}=h_{\mu \lambda},^{\sigma}h_{\sigma}^{\ \lambda},_{\nu}$

So, to complete my calculations those quantities should be the same, but i don't understand what happens that make them equal. I don't know how to work with lowering an upering index to modify it.

 

[Ibix]

Welcome to PF. I've modified your LaTeX to fix the index placement - is this correct?$$h_{\mu}{}^{\sigma}{}_{,\lambda}
h_{\sigma}{}^{\lambda}{}_{,\nu}
=h_{\mu \lambda,}{}^{\sigma}
h_{\sigma}{}^{\lambda}{}_{,\nu}$$
(Edit: corrected above following correction of OP - see below). What's $h$ here? Is it a small deviation of the metric from flat? I.e. $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$?

 

[Ibix]

On a second look, your free indices don't match. The left hand side has $\mu$ and $\sigma$ free and the right hand side has $\mu$ and $\nu$. Have you made a transcription error? If not, your problems run deep. (Edit: now corrected.)

 

[GrimGuy]

Welcome to PF. I've modified your LaTeX to fix the index placement - is this correct?$$h_{\mu}{}^{\nu}{}_{,\lambda}
h_{\sigma}{}^{\lambda}{}_{,\nu}
=h_{\mu \lambda,}{}^{\sigma}
h_{\sigma}{}^{\lambda}{}_{,\nu}$$
What's $h$ here? Is it a small deviation of the metric from flat? I.e. $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$?

Ohhh, i realize i made a mistake (the first up $\mu$ is actually $\sigma$), I edit myself to correct it.
Yes the $h$ is the small deviation of the flat space.

 

[GrimGuy]

On a second look, your free indices don't match. The left hand side has $\mu$ and $\sigma$ free and the right hand side has $\mu$ and $\nu$. Have you made a transcription error? If not, your problems run deep.

A little transcription mistake, now it's in the correct form.

 

[Ibix]

Right. Do you not just regard anything proportional to $h$ as small and anything proportional to $(h)^2$ as negligible? Or does the "second order" in your title mean you are keeping $(h)^2$ and neglecting $(h)^3$?

 

[GrimGuy]

Right. Do you not just regard anything proportional to $h$ as small and anything proportional to $(h)^2$ as negligible? Or does the "second order" in your title mean you are keeping $(h)^2$ and neglecting $(h)^3$?

Yeah, I'm keeping the $(h)^2$ and neglecting $(h)^3$.

 

[Ibix]

I'd suggest swapping the up/down $\sigma$s so that your first $h$ is all lower on both sides, note that the second $h$ is the same on both sides, then think about what you sum over and see if any of the terms cancel. Write the summation out explicitly if you don't see it.

 

[etotheipi]

@GrimGuy did you just want to show they're equal? If so, using $\eta_{ab}$ and $\eta^{ab}$ to raise and lower indices, you have$$
\begin{align*}

(\partial_{\lambda} {h_{\mu}}^{\sigma}) (\partial_{\nu} {h_{\sigma}}^{\lambda}) & = \eta^{\sigma \alpha} \eta_{\lambda \beta} (\partial^{\beta} h_{\mu \alpha})(\partial_{\nu} {h_{\sigma}}^{\lambda}) \\

&= (\partial^{\beta} h_{\mu \alpha}) (\partial_{\nu} {h^{\alpha}}_{\beta}) \\

&= (\partial^{\beta} h_{\mu \alpha}) (\partial_{\nu} {h_{\beta}}^{\alpha})\end{align*}
$$where we used the symmetry in $h_{ab} = g_{ab} - \eta_{ab}$

 

[GrimGuy]

@GrimGuy did u just want to show they're equal? if so, using $\eta_{ab}$ and $\eta^{ab}$ to raise and lower indices, u have$$
\begin{align*}

(\partial_{\lambda} {h_{\mu}}^{\sigma}) (\partial_{\nu} {h_{\sigma}}^{\lambda}) & = \eta^{\sigma \alpha} \eta_{\lambda \beta} (\partial^{\beta} h_{\mu \alpha})(\partial_{\nu} {h_{\sigma}}^{\lambda}) \\

&= (\partial^{\beta} h_{\mu \alpha}) (\partial_{\nu} {h^{\alpha}}_{\beta}) \\

&= (\partial^{\beta} h_{\mu \alpha}) (\partial_{\nu} {h_{\beta}}^{\alpha})\end{align*}
$$where we used the symmetry in $h_{ab} = g_{ab} - \eta_{ab}$

Nice approach, understood perfectly, except for the symmetry identity. I know it is symmetrical $h$, but i can't put it in the calculus.
How can i explicitly the symmetry property in this solution.

 

[GrimGuy]

I'd suggest swapping the up/down $\sigma$s so that your first $h$ is all lower on both sides, note that the second $h$ is the same on both sides, then think about what you sum over and see if any of the terms cancel. Write the summation out explicitly if you don't see it.

it worked, but it was hard work

 

[etotheipi]

Nice approach, understood perfectly, except for the symmetry identity. I know it is symmetrical $h$, but i can't put it in the calculus.
How can i explicitly the symmetry property in this solution.

Come again; are you asking why $h_{ab}$ is symmetric? the difference between two symmetric tensors is symmetric, i.e. $C_{ab} = A_{ab} - B_{ab} = A_{ba} - B_{ba} = C_{ba}$.

 

[GrimGuy]

come again - are you asking why $h_{ab}$ is symmetric? the difference between two symmetric tensors is symmetric, i.e. $C_{ab} = A_{ab} - B_{ab} = A_{ba} - B_{ba} = C_{ba}$

Yes, i 'm not sure how to explicit the symmetric property in that solution. Maybe i lack of information.

 

[etotheipi]

$g$ and $\eta$ are both symmetric, by the definition of a metric.