What are the effects on a stationary observer for Kerr metric?

[PeterDonis]

For the Tidal force, I think that the Weyl Tensor is a better component than the Riemann one.

For the case you're computing, they're the same. The Kerr metric is a vacuum solution, so its Ricci tensor is zero. So you can call it the Riemann tensor or the Weyl tensor, it doesn't matter.

For the escape velocity, I do think that this particular formula works for all

If you mean the Schwarzschild formula also works for the Kerr metric, this is known to be false.

as it was correct for the Kerr-Newman metric

What are you basing this on?

The time dilation factor gamma is also correct after taking reciprocal of the sqrt of the $g_{00}$ metric component.

For a stationary observer, yes, this works for both the Schwarzschild and the Kerr metrics. The specific formulas are different but the method is the same.

The only trouble I have is Proper acceleration from the Christoffel component. I don't know how to make the vectors for 4-velocity!

The 4-velocity for a stationary observer has only one nonzero component, $u^t$. That, and the fact that the vector must have unit norm, should be enough for you to compute the formula for $u^t$.

 

[member 657093]

If you mean the Schwarzschild formula also works for the Kerr metric, this is known to be false.

I am not talking about the Schwarzschild formula! I talked about an Escape velocity formula that involves the $\gamma$ factor that works for all the BH metrics because it is based on their respective $\gamma$.

For the Weyl Tensor, they give different values than the Riemann ones because I took the valence (0,4) instead.

And I finally succeeded in using vector field in the BL coordinates system and finding its covariant derivative to find the nonzero component of the acceleration. From there I just used the Christoffel components with it.

But thanks for the clues, even though they come with a lot of attitudes!!!

 

[PeterDonis]

I talked about an Escape velocity formula that involves the $\gamma$ factor that works for all the BH metrics because it is based on their respective $\gamma$.

What formula is that? The only escape velocity formula I see posted anywhere in this thread is the Schwarzschild formula you posted in the OP.

For the Weyl Tensor, they give different values than the Riemann ones because I took the valence (0,4) instead.

Do you mean you lowered an index? That doesn't make the Weyl tensor different from the Riemann tensor; for any vacuum spacetime they are automatically the same. It just means you lowered an index.

And I finally succeeded in using vector field in the BL coordinates system and finding its covariant derivative to find the nonzero component of the acceleration. From there I just used the Christoffel components with it.

What did you come up with?

even though they come with a lot of attitudes!!!

Statements like this are off topic. Please refrain from them.

 

[member 657093]

The only escape velocity formula I see posted anywhere in this thread is the Schwarzschild formula you posted in the OP.

In post number 25.

Yes I lowered the first index of the Weyl tensor!

$((a^2*m*cos(\theta)^2-m*r^2)/(a^4*cos(\theta)^4+2*a^2*r^2*cos(\theta)^2+r^4))$ is the nonzero component of the acceleration for the Kerr metric.

 

[PeterDonis]

In post number 25.

Ah, sorry, I missed it. So you mean this:

By the way, for escape velocity, I've read on wikipedia for the Kerr-Newman metric, that they are using $c * \frac {\sqrt{\gamma^2 - 1}}{\gamma}$ Would this work for the Kerr metric?

The Wikipedia page you refer to explicitly says that the formula is for radial escape velocity. We're not just talking about the radial case in this thread.

 

[PeterDonis]

Yes I lowered the first index of the Weyl tensor!

Which, given that this is a vacuum spacetime, is equivalent to lowering the first index of the Riemann tensor, since they are identical.

 

[PeterDonis]

I'm talking about this one

Ok. Yes, your point is valid regarding that post. @Vick, @Yukterez is correct that the 4-velocity of a stationary observer is not $(1, 0, 0, 0)$, it is $(1 / \sqrt{g_{tt}}, 0, 0, 0)$.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After a thread ban and some cleanup, the thread is reopened.