- #36
dwd40physics
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so I get ##\tilde{g}_{\mu\nu} (x + \xi) = (\delta^{\rho}{ }_{\mu} - \partial{_{\mu}}\xi^{\rho})(\delta^{\sigma}{ }_{\nu} - \partial{_{\nu}}\xi^{\sigma})g_{\rho\sigma}(x)##?
That's equation 2.25 for the forward transformation. Is that what you intended?dwd40physics said:so I get ##\tilde{g}_{\mu\nu} (x + \xi) = (\delta^{\rho}{ }_{\mu} - \partial{_{\mu}}\xi^{\rho})(\delta^{\sigma}{ }_{\nu} - \partial{_{\nu}}\xi^{\sigma})g_{\rho\sigma}(x)##?
I have explained how in a couple of posts now. That should be enough for you to give it a try.dwd40physics said:I don't know how to do the backward transform.
I'm being honest in that your explanation does not actually explain how to do an inverse transformation. It just says do it. I've fudged something to get the answer I need now but not sure I understand how I got it to to be honest.PeterDonis said:I have explained how in a couple of posts now. That should be enough for you to give it a try.
Have you tried to follow the steps I described? I did describe steps. I'll describe them again:dwd40physics said:your explanation does not actually explain how to do an inverse transformation
Actually, this by itself might be sufficient to see how the substitution of ##- \xi^\lambda \partial_\lambda g_{\mu \nu}## for ##\xi^\lambda \partial_\lambda \tilde{g}_{\mu \nu}## is justified, at least in these notes. The inverse transformation might not be necessary.PeterDonis said:do you see how to get from equation 2.25 to equation 2.26 for the forward transformation?
Why do you put that prime on x on the left hand side?ergospherical said:Another viewpoint - a Killing transformation is a coordinate transformation that preserves the functional form of the metric, i.e. ##g'(x') = g(x)##.
This is not correct. Use [tex]g_{ad}\partial_{b}\chi^{d} = \partial_{b}\chi_{a} - \chi^{d}\partial_{b}g_{ad}.[/tex]ergospherical said:Let's do some work on the RHS, [tex]
- g_{ad} \partial_b \xi^d - g_{cb} \partial_a \xi^c = -\partial_b \xi_a - \partial_a \xi_b [/tex]
You made the same mistake in here.ergospherical said:Remember that\begin{align*}
\Gamma^c_{ab} \xi_c &= \frac{1}{2} \xi_c g^{cd}(\partial_a g_{db} + \partial_b g_{ad} - \partial_d g_{ab}) \\
&= \frac{1}{2} \xi_c(\partial_a \delta^c_b + \partial_b \delta^c_a - \partial^c g_{ab}) \\
&= \frac{1}{2} \xi_c (0 + 0 - \partial^c g_{ab}) \\
&= -\frac{1}{2} \xi^c \partial_c g_{ab}
\end{align*}