1
$\begingroup$

I am learning about gauge transformations in GR. As I understand, we write the metric (to first order) as $g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$ where $\eta_{\mu\nu}$ is Minkowski and $h_{\mu\nu}$ is a small perturbation. Upon making a change of coordinates \begin{equation} x'^{\mu}\rightarrow x^{\mu} + \xi^\mu \end{equation} the metric in the primed coordinates becomes \begin{equation} g'_{\mu\nu}=\eta_{\mu\nu}' + h'_{\mu\nu}=\eta_{\mu\nu} +\delta_{\mu\nu} + h_{\mu\nu}. \end{equation} Due to the new (primed) coordinates, $\eta_{\mu\nu}$ will not necessarily be Minkowski anymore ($\eta_{\mu\nu}\rightarrow\eta'_{\mu\nu}$), but because the perturbation is small, I can absorb the perturbation of $\eta_{\mu\nu}$ away from Minkowski (which I called $\delta_{\mu\nu}$) into $h'_{\mu\nu}$. Also, because perturbations away from $h_{\mu\nu}$ will be second order in small quantities, we can leave leave h_{\mu\nu} as it is.

  1. Is this reasoning correct? Is there anything that I am missing or a better way to think about this?

  2. I am trying to work through demonstrating that $h'_{\mu\nu}=h_{\mu\nu}-\partial_\mu\xi_\nu-\partial_\nu\xi_\mu$. When I write $g'_{\mu\nu} = g_{\lambda\sigma}\frac{\partial x^\lambda}{\partial x'^\mu}\frac{\partial x^\sigma}{\partial x'^\nu}$, and plug in $x^\lambda = x'^\lambda - \xi^\lambda(x')$, I get a lot of terms contain derivatives that look like $\frac{\partial x^\lambda}{\partial x'^\mu}=\frac{\partial x'^\lambda}{\partial x'^\mu}-\frac{\partial \xi^\lambda}{\partial x'^\mu}$. How should I treat/think about the $\frac{\partial x'^\lambda}{\partial x'^\mu}$ terms?

$\endgroup$
2
  • $\begingroup$ I suspect that $\frac{\partial x'^\lambda}{\partial x'^\mu}$ might have something to do with $\delta^\lambda\,_\mu$, but am unsure of how to demonstrate this (if it is even correct). $\endgroup$ – Bob Aug 25 '17 at 18:00
  • $\begingroup$ After writing this out, can I say that $\frac{\partial x'^\lambda}{\partial x'^\mu}=\delta^\lambda_\mu$ because for $\lambda=\mu$, the partial derivative is trivial, and if the coordinates are orthogonal (since we are considering only up to first order in small terms, and the partial derivative is next to another small term ($\frac{\partial \xi^\lambda}{\partial x'^\mu}$)), then $\frac{\partial x'^\lambda}{\partial x'^\mu}$ the partial derivative will be zero for $\lambda\neq\mu$? $\endgroup$ – Bob Aug 25 '17 at 18:02
2
$\begingroup$

If you have $x'^{\mu}=x^\mu+\epsilon\xi^\mu$ (where only $\epsilon$ needs to be small - it is better to separate 'smallness" this way imo), then $$ \partial x'^{\mu'}/\partial x^\mu=\delta^{\mu'}_{\mu}+\epsilon\partial_\mu\xi^{\mu'}. $$

You should learn Lie derivatives however, because then you'd reaize that the transformation $x'^\mu=x^\mu+\epsilon\xi^\mu$ is an example of a smooth 1-parameter family of point tranformations and the behaviour of tensor fields under this transformation is given by the Lie derivative.

For your first question, the reasoning is essentially correct, however I have found that (in my opinion) the best way to look at perturbations in GR/differential geometry is by how it is detailed in Stewart's Advanced General Relativity.

Basically, imagine you have a background spacetime $(M^0,g^0)$, and you have a one-parameter family of spacetimes that "deviate only slightly" from this background spacetime, let's denote them by $(M^\epsilon,g^\epsilon)$. Because this is a small deviation, we assume the two manifolds are diffeomorphic and one such diffeo is given by $\phi^\epsilon:M^0\rightarrow M^\epsilon$. It should be understood that $\epsilon$ is variable, and "smallness of the deviation" is measured by the fact that the pullback $(\phi^\epsilon)^*g^\epsilon$ is only different to first order from $g^0$, eg. you can approximate well as $$ g=(\phi^\epsilon)^*g^\epsilon=g^0+\epsilon g^{(1)}. $$

I have called this quantity $g$ because in practice, when dealing with perturbations, you want to work on the unperturbed spacetime $M^0$ via pullbacks, so $g$ will be the working metric.

However, the diffeomorphism $\phi^\epsilon$ is not unique. Let $\psi^\epsilon:M^0\rightarrow M^\epsilon$ also be a diffeo, and let $$ g'=(\psi^\epsilon)^*g^\epsilon=g^0+\epsilon g'^{(1)} $$ be the perturbed metric given by this map.

The difference between $\phi$ and $\psi$ can be quantified on $M^0$ by letting $\chi^\epsilon:M^0\rightarrow M^0$ be a 1-parameter family of diffeos on $M^0$ given by $$ \psi^\epsilon=\phi^\epsilon\circ\chi^\epsilon $$ or explicitly, as $$ \chi^\epsilon=(\phi^\epsilon)^{-1}\circ\psi^\epsilon. $$

Then we can express the perturbed metric $g'$ with the other perturbed metric $g$ by noting that $$ g'=(\psi^\epsilon)^*g^\epsilon=(\phi^\epsilon\circ\chi^\epsilon)^*g^\epsilon=(\chi^\epsilon)^*(\phi^\epsilon)^*g^\epsilon=(\chi^\epsilon)^*g. $$

But 1) we are expanding to first order in $\epsilon$, 2) $\chi^0=\text{Id}$ by continuity, so we have $$ g'=g+\epsilon\frac{d}{d\epsilon}(\chi^\epsilon)^*g|_{\epsilon=0}=g+\epsilon\frac{d}{d\epsilon}(\chi^\epsilon)^*(g^0+\epsilon g^{1})|_{\epsilon=0}=g+\epsilon\mathcal{L}_X(g^0+\epsilon g^{(1)}), $$ where $X$ is the vector field which generates the 1-parameter map $\chi^\epsilon$, so we have $$ g'=g+\epsilon\mathcal{L}_Xg^{0}+O(\epsilon^2)=g^0+\epsilon(g^{(1)}+\mathcal{L}_X g^{0}). $$

So different perturbations differ by Lie derivatives of the unperturbed metric with respect to an arbitrary vector field, and $X$ is indeed arbitrary, since any smooth $\chi^\epsilon$ will describe the difference between two diffeomorphisms of $M^0$ and $M^\epsilon$.

In your case, $g^0=\eta$, $g^{(1)}=h$ and $X=\xi$, so we have $$ h_{\mu\nu}'=h_{\mu\nu}+\mathcal{L}_\xi\eta_{\mu\nu}, $$ where $$ \mathcal{L}_\xi\eta_{\mu\nu}=\xi^\sigma\partial_\sigma\eta_{\mu\nu}+\eta_{\mu\sigma}\partial_\nu\xi^\sigma+\eta_{\sigma\nu}\partial_\mu\xi^\sigma=\partial_\mu\xi_\nu+\partial_\nu\xi_\mu. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.