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

[PeterDonis]

Yes, which I then contrasted to Kerr.

Ok. I'll see if I can find a more detailed analysis for Schwarzschild spacetime (I'm pretty sure I've seen one in the past that my remarks in this thread are based on), or cobble together one myself if I can't find one.

 

[PeterDonis]

So what exactly is the math operation to be done here: Do I just multiply the relevant Connection component with what?

I specifically used the word "contraction". Do you understand what that is? It's a basic operation in tensor calculus, and a basic familiarity with tensor calculus is essential to do calculations in GR.

 

[PAllen]

Ok. I'll see if I can find a more detailed analysis for Schwarzschild spacetime (I'm pretty sure I've seen one in the past that my remarks in this thread are based on), or cobble together one myself if I can't find one.

Don't bother unless you want to. I convinced myself from analyzing the Newtonian case that my 'limiting intuition' was wrong, and that if a velocity is directed in a possible escape direction, then the escape speed is not dependent on direction. The photon sphere is then analogous to a Newtonian direction choice that fails because the trajectory would hit the planet (after possibly a partial orbit). But as long as the direction is a possible escape direction, the initial escape speed required is independent of direction. I am convinced this would carry over to the Shwarzschild case.

 

[PeterDonis]

I am convinced this would carry over to the Shwarzschild case.

This was what I remembered reading somewhere, but unfortunately I haven't been able to dig up the reference--that the Newtonian result carries over to Schwarzschild, with the proviso that in Schwarzschild the directions are more limited (in the Newtonian case any free-fall trajectory in any direction at escape velocity that doesn't actually hit the central mass will escape, but in Schwarzschild the possible escape directions form a narrower and narrower angle around radially outward as $r = 2M$ is approached).

For the Schwarzschild case there is a fairly simple intuitive argument for the result: that an escape trajectory corresponds to one with energy at infinity per unit rest mass equal to one (i.e., energy at infinity exactly equal to rest mass), and energy at infinity per unit mass is given by $E = \left( 1 - 2M / r \right) \left( dt / d\tau \right)$. That means that $dt / d\tau$ for an escape orbit depends only on $r$, and since $dt / d\tau$ is the same as the gamma factor of the escape orbit relative to a static observer, that gamma factor (and therefore the escape velocity relative to a static observer) can only depend on $r$.

An analogous argument for the Kerr case shows that the escape velocity, depends on both $\theta$ and the direction of the orbit, as well as $r$.

 

[member 657093]

For the Tidal force, I think that the Weyl Tensor is a better component than the Riemann one. The latter is computing the change on the volume of object, whereas the former is computing on the distortion of the shape of object, which is a rather closer definition to my 2nd question. (the component is the Weyl Tensor with valence (0,4) $C_{0,1,1,0}$ )

For the escape velocity, I do think that this particular formula works for all; I tested it on the Schwarzschild metric, and as it was correct for the Kerr-Newman metric, it is also correct for the Kerr metric.

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

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

 

[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.