Understanding killing vectors and transformations of metric

• A
• dwd40physics
In summary: Jacobian matrix, but more general...the "tensor" of the transformation), which, if you assume ##\tilde{g}_{\mu \nu} = g_{\mu \nu}##, gives you the Killing equation (albeit with a ##+## sign on the right-hand side, which is just a matter of convention). The reason for the minus sign is kind of...I don't know why they have it that way, but somehow it's natural to have it that way, so that it's "the same" as equation 2.4. That's my take on it, anyway.In summary, the
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)##?

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)##?
That's equation 2.25 for the forward transformation. Is that what you intended?

you asked if I could do the forward transformation so I answered that part, I don't know how to do the backward transform.

dwd40physics said:
I don't know how to do the backward transform.
I have explained how in a couple of posts now. That should be enough for you to give it a try.

PeterDonis said:
I have explained how in a couple of posts now. That should be enough for you to give it a try.
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.

dwd40physics said:
your explanation does not actually explain how to do an inverse transformation
Have you tried to follow the steps I described? I did describe steps. I'll describe them again:

(1) Invert the transformation equation. The forward transformation is ##y^\mu = x^\mu + \xi^\mu##. Inverting this just means rearranging it so it's an equation for ##x^\mu## in terms of ##y^\mu## instead of ##y^\mu## instead of ##x^\mu##. (The answer is already in post #14.)

(2) Invert equation 2.4; that just means switching ##x## and ##y## in the equation. This gives an equation for ##g(x)## in terms of partial derivatives of ##y## with respect to ##x## and ##\tilde{g}(y)##.

(3) Follow the same procedure that is used for the forward transformation to go from equation 2.4 to equation 2.25, to obtain an inverted version of equation 2.25 from the inverted version of equation 2.24.

(4) Follow the same procedure that is used for the forward transformation to go from equation 2.25 to equation 2.26, to obtain an inverted version of equation 2.26.

Then you should be able to just look at the inverted version of equation 2.26 and compare it with the forward version.

PeterDonis said:
do you see how to get from equation 2.25 to equation 2.26 for the forward 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.

Another viewpoint - a Killing transformation is a coordinate transformation that preserves the functional form of the metric, i.e. ##g'(x') = g(x)##. Under the coordinate change ##x' = x + \xi## (or ##x = x' - \xi##),\begin{align*}
g'_{ab}(x') &= \frac{\partial x^c}{\partial x'^{a}} \frac{\partial x^d}{\partial x'^{b}} g_{cd}(x) \\
&= (\delta^c_a - \partial_a \xi^c)(\delta^d_b - \partial_b \xi^d) g_{cd}(x) \\
&= g_{ab}(x) - g_{ad}(x) \partial_b \xi^d - g_{cb}(x) \partial_a \xi^c + O(\xi^2)
\end{align*}To first order in ##\xi## the Taylor expansion of the LHS is \begin{align*}
g'_{ab}(x') = g'_{ab}(x+\xi) = g'_{ab}(x) + \xi^e \partial_e g'_{ab}(x)
\end{align*}so to first order in ##\xi##,\begin{align*}
g'_{ab}(x) + \xi^e \partial_e g'_{ab}(x) &= g_{ab}(x) - g_{ad}(x) \partial_b \xi^d - g_{cb}(x) \partial_a \xi^c \\
\implies \xi^e \partial_e g_{ab}(x) &= - g_{ad}(x) \partial_b \xi^d - g_{cb}(x) \partial_a \xi^c
\end{align*}since ##g'_{ab}(x) = g_{ab}(x)## from the Killing condition. Let's do some work on the RHS,\begin{align*}
- g_{ad} \partial_b \xi^d - g_{cb} \partial_a \xi^c &= -\partial_b \xi_a - \partial_a \xi_b \\
&= -D_b \xi_a - \Gamma^c_{ab} \xi_c - D_a \xi_b - \Gamma^c_{ba} \xi_c \\
&= -(D_b \xi_a + D_a \xi_b) - 2\Gamma^c_{ab} \xi_c
\end{align*}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*}so that ##-2\Gamma^c_{ab} \xi_c = \xi^c \partial_c g_{ab}##, and overall\begin{align*}
\xi^e \partial_e g_{ab} &= -(D_b \xi_a + D_a \xi_b) + \xi^c \partial_c g_{ab}
\end{align*}which gives you ##D_b \xi_a + D_a \xi_b = 0##.

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)##.
Why do you put that prime on x on the left hand side?

ergospherical said:
Let's do some work on the RHS, $$- g_{ad} \partial_b \xi^d - g_{cb} \partial_a \xi^c = -\partial_b \xi_a - \partial_a \xi_b$$
This is not correct. Use $$g_{ad}\partial_{b}\chi^{d} = \partial_{b}\chi_{a} - \chi^{d}\partial_{b}g_{ad}.$$
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*}
You made the same mistake in here.
Using the following relations in the calculus of infinitesimals:
$$\epsilon \ \bar{g}_{ab}(x) \approx \epsilon \ g_{ab}(x), \ \ \ \epsilon \ \partial_{c}\bar{g}_{ab}(x) \approx \epsilon \ \partial_{c}g_{ab}(x) ,$$$$\epsilon \ \chi (\bar{x}) \approx \epsilon \ \chi (x), \ \ \ \epsilon \ \frac{\partial}{\partial \bar{x}} \chi (\bar{x}) \approx \epsilon \ \frac{\partial}{\partial x} \chi (x) ,$$ you can show, for an arbitrary infinitesimal transformation $\bar{x}^{a} = x^{a} + \epsilon \chi^{a}(x)$, the following general expression holds: $$\delta g_{ab} = \chi^{c}D_{c}g_{ab} + D_{a}\chi_{b} + D_{b}\chi_{a} ,$$ where $\bar{g}_{ab}(x) = g_{ab}(x) - \epsilon \ \delta g_{ab}$.

Last edited:
vanhees71 and dextercioby
I read through the two pages that the OP @dwd40physics provided and they were very good. May I ask the OP if it would be possible for him/her to send me privately the entire notes?

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