Physics Asked on May 5, 2021
Question 10(a) on pages 469-470 in the book "Spacetime and Geometry" by Sean Carroll asks:
Suppose that two metrics are related by an overall conformal transformation of the form:
$$ tilde{g}_{mu nu} = exp(alpha(x))g_{mu nu} $$
Show that if $xi^{mu}$ is a Killing vector for the metric ${g}_{mu nu}$, then it is a conformal Killing vector for the metric $tilde{g}_{mu nu}$· A conformal Killing vector obeys the equation
$$ nabla_{mu} xi_{nu} + nabla_{nu} xi_{mu} = g_{mu nu} xi^{lambda} nabla_{lambda} alpha $$
My solution starts with the statement
$$ mathcal{L}_{xi} g_{mu nu} = 0 implies g_{mu lambda}nabla_{nu} xi^{lambda} + g_{nu lambda}nabla_{mu} xi^{lambda} + xi^{lambda}nabla_{lambda}g_{mu nu} = 0 $$
We can subsitiute $g_{mu nu} = exp(-alpha(x))tilde g_{mu nu}$ on the third term and lower the indices of $xi$ on the first and second terms to get the desired result. But isn’t the third term $0$ because of the Levi Civita connection? And I’ve googled and found the equation of the conformal Killing vector to have a factor of $frac{2}{n}$ on the right-hand side, where $n$ is the number of dimensions on the manifold. Where am I going wrong?
EDIT:
I started with the transformation of $tilde g’_{mu nu}(x)= tilde g_{mu nu}(x) – epsilon(tilde g_{mu sigma}(x) partial_{nu} xi^{sigma} + tilde g_{rho nu}(x) partial_{mu} xi^{rho} + xi^{lambda} partial_{lambda} tilde g_{mu nu}(x)) + mathcal{O}(epsilon^{2})$
Now we set $tilde g_{mu nu}(x) = exp(alpha(x))g_{mu nu}(x)$ to get:
$$ g’_{mu nu}(x)= g_{mu nu}(x) – epsilon( g_{mu sigma}(x) partial_{nu} xi^{sigma} + g_{rho nu}(x) partial_{mu} xi^{rho} + xi^{lambda} partial_{lambda} g_{mu nu}(x) – xi^{lambda} (partial_{lambda} alpha)g_{mu nu}(x)) + mathcal{O}(epsilon^{2}) $$
Since $xi^{mu}$ is a Killing vector for $g_{mu nu}$, we must have
$$ g_{mu sigma}(x) partial_{nu} xi^{sigma} + g_{rho nu}(x) partial_{mu} xi^{rho} + xi^{lambda} partial_{lambda} g_{mu nu}(x) – xi^{lambda} (partial_{lambda} alpha)g_{mu nu}(x) = 0$$
This leads us to the desired equation.
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