I Schwarzschild metric not dependent on time

[PeterDonis]

You are arriving at this conclusion by assuming KrK^r vanishes, which is not a priori thing.

Yes, here I am only considering the case where $K^t$ is the only nonzero component. As I said at the end of that post, I'm going to refrain for now from explicitly considering the other cases and showing how they are eliminated.

By checking the non trivial field of my post #42 you will see that it satisfies the Killing equations.

Ok, let's check. Since you wrote it with lower indexes, we can leave out the index lowering. So we have:

$$
K_t = \sqrt{1- \frac{2m}{r}} \frac{mt}{r^2}
$$

$$
K_r = \sqrt{ \frac{1}{1- \frac{2m}{r}}}
$$

The third Killing equation becomes

$$
\partial_t K_r + \partial_r K_t - 2 \Gamma^t{}_{tr} K_t = 0 + \sqrt{ \frac{1}{1- \frac{2m}{r}}} \left( \frac{m}{r^2} \right) \frac{mt}{r^2} - \sqrt{1- \frac{2m}{r}} \frac{mt}{r^3} - 2 \frac{m}{r^2 \left( 1 - \frac{2m}{r} \right)} \sqrt{1- \frac{2m}{r}} \frac{mt}{r^2}
$$

This does not all cancel out to zero, so the Killing equation is not satisfied.

I strongly suggest that you very, very carefully check your work. The fact that $(1, 0)$ is the only Killing vector field in the $t-r$ plane of Schwarzschild spacetime has been well known for decades and proving it is often assigned as a textbook problem. You should not be persisting in claiming that there are others; you should be looking for where you made a mistake.

 

[stevendaryl]

I'm going to check the last one I posted.
The non vanishing connection coefficients are $$\Gamma^t{}_{rt} = \Gamma^t{}_{tr} = \frac{-m}{2mr - r^2} \\ \Gamma^r{}_{tt} = \frac{m(r - 2m)}{r^3} \\ \Gamma^r{}_{rr} = \frac{m}{2mr - r^2}$$ The Killing equations are $$0 = \nabla_r K_r = \partial_r K_r - \Gamma^t{}_{rr} K_t - \Gamma^r{}_{rr} K_r = 0 \\ 0 = \nabla_t K_t = \partial_t K_t - \Gamma^t{}_{tt} K_t - \Gamma^r{}_{tt} K_r = 0 \\ 0 = \nabla_t K_r + \nabla_r K_t = \partial_t K_r + \partial_r K_t - 2 \Gamma^r{}_{tr} K_r - 2 \Gamma^t{}_{tr} K_t = \frac{-2m}{r^2} - \frac{2m}{2mr -r^2} \bigg(1 - \frac{2m}{r} \bigg) = 0$$

You need three equations:

$\nabla_r K_r = 0$
$\nabla_r K_t + \nabla_t K_r = 0$
$\nabla_t K_t = 0$

 

[kent davidge]

You need three equations:

$\nabla_r K_r = 0$
$\nabla_r K_t + \nabla_t K_r = 0$
$\nabla_t K_t = 0$

There are indeed three equations on my post

The fact that $(1, 0)$ is the only Killing vector field in the $t-r$ plane of Schwarzschild spacetime has been well known for decades and proving it is often assigned as a textbook problem

I've actually seen this through explicit calculation. Is there an other way of knowing it?

 

[PeterDonis]

Is there an other way of knowing it?

Yes, there certainly is. And it is a good idea to consider such a question before even trying an explicit calculation. I think it was John Wheeler who said that you should never try to calculate something unless you already know the answer.

The key is to consider the geometric meaning of a Killing vector field: it is a vector field that generates an isometry. That means that along any integral curve of the KVF, the metric does not change.

This has the following implications if we consider the components of the KVF in a coordinate chart:

(1) The KVF cannot have a component in the direction of any coordinate that the metric depends on (because the metric would have to change along any such component). In the case under discussion, this means a KVF cannot have an $r$ component, since the metric depends on $r$.

(2) The KVF cannot depend on any coordinate that changes along its integral curves (because if it did, you would be able to distinguish points along an integral curve by a geometric invariant--the norm of the KVF--and you can't distinguish points along the integral curves of an isometry by any geometric invariant). In the case under discussion, since we know from #1 just above that the KVF cannot have an $r$ component, it can only have a $t$ component; and therefore it cannot depend on $t$, since $t$ must change along any integral curve.

(3) There cannot be two distinct KVFs that have the same integral curves (because there can only be one isometry that has a given set of integral curves). This means that, once we have found one KVF of the form $(K^t, 0)$, that must be the only one, because any KVF with only a $t$ component will have the same integral curves.

Doing an explicit calculation will show how the above considerations work out in the details of the math.

 

[PeterDonis]

I've actually seen this through explicit calculation.

Do you mean you have just seen this now, or that you had seen it at some time previously before this thread was started?

 

[kent davidge]

Do you mean you have just seen this now, or that you had seen it at some time previously before this thread was started?

that I went to check it after your post #51

 

[PeterDonis]

that I went to check it after your post #51

Ok, good.

 

[kent davidge]

$(1, 0)$ is the only Killing vector field in the $t-r$ plane of Schwarzschild spacetime has been well known for decades and proving it is often assigned as a textbook problem

Is it proven the way you did in post #54?

 

[PeterDonis]

Is it proven the way you did in post #54?

The textbook problems I remember asked you to crank through the detailed math using the Killing equations in component form in Schwarzschild coordinates. My reasoning in post #54 could probably be worked into a proof using coordinate-free math, but I don't remember being asked to do it that way as an actual proof.