Understanding killing vectors and transformations of metric

[dwd40physics]

so I get $\tilde{g}_{\mu\nu} (x + \xi) = (\delta^{\rho}{ }_{\mu} - \partial{_{\mu}}\xi^{\rho})(\delta^{\sigma}{ }_{\nu} - \partial{_{\nu}}\xi^{\sigma})g_{\rho\sigma}(x)$?

 

[PeterDonis]

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]

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

 

[PeterDonis]

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.

 

[dwd40physics]

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.

 

[PeterDonis]

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]

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.

 

[ergospherical]

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 $D_b \xi_a + D_a \xi_b = 0$.

 

[haushofer]

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?

 

[samalkhaiat]

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 $$g_{ad}\partial_{b}\chi^{d} = \partial_{b}\chi_{a} - \chi^{d}\partial_{b}g_{ad}.$$

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 $\bar{g}_{ab}(x) = g_{ab}(x) - \epsilon \ \delta g_{ab}$.

 

[brotherbobby]

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?