Proving Timelike Killing Vectors for g_{ab},0 = 0

[befj0001]

If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.

How can one prove that?

 

[WannabeNewton]

Consider the vector field $\xi^{\mu}$ which in the coordinates for which $\partial_0 g_{\mu\nu} = 0$ takes the form $\xi^{\mu} = \delta^{\mu}_0$.

Then $\mathcal{L}_{\xi}g_{\mu\nu} = \partial_0 g_{\mu\nu} = 0$ so $\xi^{\mu}$ is a time-like Killing field for this metric.

But this isn't the correct way to think about the time-like Killing field. Rather one first looks for the existence of a time-like vector field $\xi^{\mu}$ satisfying $\nabla_{(\mu}\xi_{\nu)} = 0$. If such a vector field exists then there necessarily exists a coordinate system $\{x^{\mu}\}$ in which $\partial_0 g_{\mu\nu} = 0$ and $\xi^{\mu} = \delta^{\mu}_0$.

 

[Bill_K]

But this isn't the correct way to think about the time-like Killing field. Rather one first looks for the existence of a time-like vector field $\xi^{\mu}$ satisfying $\nabla_{(\mu}\xi_{\nu)} = 0$. If such a vector field exists then there necessarily exists a coordinate system $\{x^{\mu}\}$ in which $\partial_0 g_{\mu\nu} = 0$ and $\xi^{\mu} = \delta^{\mu}_0$.

Why is this the wrong way to think of it?? It seems to me a perfectly reasonable way to think of it.

Someone hands you an unfamiliar metric. What's the first thing you do -- write out Killing's Equation and try to solve it?

Or do you say, aha, I see it doesn't depend on one of the coordinates ζ, so there's an obvious Killing vector.

 

[PeterDonis]

If g_{ab},0 = 0 (g does not depend on time), then the manifold must have a timelike killing vector.

You're assuming that the "0" coordinate is timelike. That's not always the case. Consider, for example, Schwarzschild coordinates on Schwarzschild spacetime inside the horizon; there the "0" (i.e., $t$) coordinate is spacelike, not timelike, but the metric is still independent of $t$, so $\partial / \partial t$ is still a KVF--just not a timelike one.

 

[sardar]

To prove that a manifold has a timelike killing vector, we must first understand the properties of a timelike killing vector. A timelike killing vector is a vector field that preserves the metric tensor of a manifold, meaning that it does not change the distances or angles between points on the manifold. Additionally, it must also be tangent to the worldlines of observers moving through the manifold at a constant rate.

Now, if we are given that g_{ab},0 = 0, we can use this information to show that the manifold must have a timelike killing vector. This is because if g_{ab},0 = 0, it means that the metric tensor is independent of time. In other words, the metric tensor does not change as time passes. This implies that the manifold is static, meaning it does not undergo any changes over time.

In a static manifold, the worldlines of observers do not change, as there is no time component to alter their trajectories. This aligns with the properties of a timelike killing vector, as it must be tangent to the worldlines of observers moving at a constant rate. Therefore, we can conclude that a timelike killing vector exists in a manifold where g_{ab},0 = 0.

To further support this, we can also use the Killing equation, which states that for a vector field V, if the metric tensor g_{ab} satisfies the equation g_{ab;c} + g_{ac;b} = 0, then V is a killing vector. In our case, since g_{ab},0 = 0, the only non-zero term in the Killing equation would be g_{00;c} + g_{0c;0} = 0. This implies that g_{00;c} = 0, which means that the vector field V is a timelike killing vector.

In conclusion, by understanding the properties of a timelike killing vector and using the information that g_{ab},0 = 0, we can prove that a manifold must have a timelike killing vector. This further supports our understanding of the relationship between the metric tensor and the existence of a timelike killing vector in a manifold.