How Do You Compute Killing Vectors for a Given Metric?

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SUMMARY

This discussion focuses on computing Killing vectors for the metric defined by the equation \(g_{ab} = \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\). The Killing equation \(L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} = 0\) is utilized to derive the Killing vectors \(\vec{X} = (X^0, X^1)\). Participants emphasize the importance of calculating the partial derivatives of the metric components, specifically \(\partial_0 g_{00} = 2x\) and \(\partial_1 g_{11} = 2xy\), to substitute into the Killing equation and solve for the Killing vectors.

PREREQUISITES
  • Understanding of Killing vectors in differential geometry
  • Familiarity with metric tensors and their components
  • Knowledge of partial derivatives and their application in tensor calculus
  • Experience with the Killing equation and its significance in general relativity
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  • Study the derivation and applications of the Killing equation in various metrics
  • Learn about the significance of Killing vectors in the context of symmetries in general relativity
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This discussion is beneficial for mathematicians, physicists, and students specializing in general relativity, differential geometry, or theoretical physics, particularly those interested in the properties of Killing vectors and their applications in understanding symmetries of spacetime metrics.

LCSphysicist
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I want to find all the killing vectors of the metric ##x²dx² + xdy²##. We could guess somethings by intuition and check it, but i decided to use the equation itself. Unfortunatelly, i realized that i am not sure how to manipulate the equation

$$L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} =^{?} 0$$

When i wrote this equation, i was not sure how the index should work here, so i guessed that we need to check it for all possible combinations of ab. ##(a,b) = [(0,0),(0,1),(1,0),(1,1)]##$. So, seeing this way, we need to find ##\vec X = (X^{0},X^{1})##? THe answer is something in partial derivatives. How would i got that just with the equation? I would appreciate if you give a tip of how to really start the computations, because as you can see, i am stuck right at the start.
 
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The Killing equation for a metric $g_{ab}$ is given by$$L_{\chi}g_{ab} = g_{ad}\partial_{b} X^{d}+g_{bd}\partial_{a} X^{d}+ X^{e}\partial_{e} g_{ab} = 0.$$In this case, you are trying to find the Killing vectors $\vec{X}=(X^0,X^1)$ for the metric $$g_{ab}=\begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}.$$ To solve for $\vec{X}$, begin by computing the partial derivatives of the metric components with respect to each coordinate:$$\partial_0 g_{00} = \partial_0 (x^2) = 2x,$$$$\partial_1 g_{00} = \partial_1 (x^2) = 0,$$$$\partial_0 g_{01} = \partial_0 (0)=0,$$$$\partial_1 g_{01} = \partial_1 (0) = 0,$$$$\partial_0 g_{11} = \partial_0 (xy^2) = y^2,$$$$\partial_1 g_{11} = \partial_1 (xy^2) = 2xy.$$Now, substitute these partial derivatives into the Killing equation and solve for $\vec{X}$:\begin{align*}L_\chi g_{ab} &= g_{ad}\partial_b X^d + g_{bd}\partial_a X^d + X^e \partial_e g_{ab} \\&= \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\begin{bmatrix}\partial_0 X^0 & \partial_1 X^0 \\ \partial_0 X^1 & \partial_1 X^1\end{bmatrix} + \begin{bmatrix}x^2 & 0 \\ 0 & xy^2\end{bmatrix}\begin{bmatrix}\partial_0 X^0 & \partial_1 X^0 \\ \partial
 

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