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

• I
• member 657093
Vick said:
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.

Vick said:
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.

Vick said:
as it was correct for the Kerr-Newman metric
What are you basing this on?

Vick said:
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.

Vick said:
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##.

PeterDonis said:
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!!!

Vick said:
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.

Vick said:
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.

Vick said:
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?

Vick said:
even though they come with a lot of attitudes!!!
Statements like this are off topic. Please refrain from them.

PeterDonis said:
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.

Vick said:
In post number 25.
Ah, sorry, I missed it. So you mean this:

Vick said:
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?

Vick said:
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.

Yukterez said:
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)##.

• Special and General Relativity
Replies
19
Views
2K
• Special and General Relativity
Replies
11
Views
579
• Special and General Relativity
Replies
8
Views
2K
• Special and General Relativity
Replies
16
Views
1K
• Special and General Relativity
Replies
18
Views
2K
• Special and General Relativity
Replies
1
Views
1K
• Special and General Relativity
Replies
10
Views
2K
• Special and General Relativity
Replies
18
Views
1K
• Special and General Relativity
Replies
2
Views
1K