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

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TL;DR Summary
What are the effects on a stationary observer at a specific distance from a Kerr Black Hole?
A Kerr Black Hole (BH) is a spinning BH. There is an Event Horizon (EH) which is $$r_H^\pm = \frac{r_S \pm \sqrt{r_S^2 -4a^2}}{2}$$ where ##a=\frac{J}{Mc}## and ##r_S## is the Schwarzschild radius. My question is, suppose I'm in a spacecraft, not in orbit, but stationary at a distance ##r##. I want to have a comprehensive understanding of the effects of that BH on the spacecraft, and by extension on me. I would like to know, if I'm subjected to the effects of:

1. Time Dilation: If this is present, the formula should be derived from the ##g_{tt}## component of the Kerr metric, which is $$\gamma = \frac{1}{\sqrt{1-\frac{r_S r}{r^2 + a^2 cos^2 \theta}}}$$
2. Tidal Force: What is the formula for this in a Kerr metric? Normally the equation is $$\frac{2GMd}{r^3}$$ with ##d## being in the instance of a human being, about 2 meters.
3. Acceleration due to the gravitational attraction of the BH at that distance (surface gravity from an arbitrary distance). Normally the formula is $$\frac{GM}{r^2}$$ But I've read here that this should be multiplied by ##\gamma##
4. Escape velocity I need for my spacecraft from this distance ##r##; normally the formula is $$\sqrt{\frac{2GM}{r}}$$
Can the above four quantities be derived from the metric?
To prevent ambiguity, let me edit the question with the following (please note that I'm currently dealing in SI units):
1. Kerr BH mass is: ##10^8## times that of Solar mass (Solar Mass ##=1.98847e+30 kg##)
2. Spin Parameter: ##0.999999999999986673399999999999954##
3. Distance ##r##: ##252083284394.398975225334674476303## m
4. ##\theta##: ##45^0##
G = ##6.6743e-11## ## m^3.kg^{-1}.s^{-2}##

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Dale
Vick said:
Can the above four quantities be derived from the metric?
Yes, but it's somewhat more complicated than in Schwarzschild spacetime.

The time dilation factor is derived just as you suggest, it's the reciprocal of the square root of ##g_{tt}##.

The tidal force (or more precisely the factor that multiplies ##d##, the size of the object) has to be computed from the relevant components of the Riemann tensor; for an object with constant ##r##, ##\theta##, ##\phi## this is ##R^r_{t r t}##. That is not a simple computation, but it can be done.

The gravitational acceleration (or more precisely the proper acceleration for the static observer) is ##(1/2) g^{rr} \partial_r g_{tt}##. For Schwarzschild spacetime that gives ##\gamma GM / r^2##, as you say. It's a more complicated formula for Kerr spacetime, but it's straightforward to compute.

The escape velocity is the hardest one in Kerr spacetime because it is direction dependent. Off the top of my head I'm not sure what the simplest way to compute it is, but it will not be a single formula that depends only on your position. It will also depend on the direction in which you try to escape.

Last edited:
Dale and vanhees71
PeterDonis said:
has to be computed from the relevant components of the Riemann tensor; for an object with constant r, θ, ϕ this is Rtrtr.
I can compute the Riemann tensor! But are you suggesting that the Tidal force is derived only from the Riemann Tensor component ##R^r_{trt}## ? That component is shown below:

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Dale
Vick said:
are you suggesting that the Tidal force is derived only from the Riemann Tensor component ?
The tidal force on a stationary observer. That's what you said you were interested in.

Vick said:
That component is shown below
You can simplify it further by assuming that the stationary observer is in the equatorial plane, i.e., ##\theta = \pi / 2##. That eliminates all the ##\cos \theta## terms and leaves us with

$$\frac{2m}{r^3} \left( 1 - \frac{2m}{r} + \frac{3}{2} \frac{a^2}{r^2} \right)$$

which makes clearer how it compares with the Schwarzschild value. Note also that the Schwarzschild value for this Riemann tensor component, in Schwarzschild coordinates (which are what the Boyer-Lindquist coordinates you appear to be using for Kerr spacetime reduce to for ##a = 0##), is

$$\frac{2m}{r^3} \left( 1 - \frac{2m}{r} \right)$$

Computing the actual tidal force felt requires contracting twice with the observer's 4-velocity, which gets rid of the extra factor of ##1 - 2m / r## in Schwarzschild spacetime. In Kerr spacetime that exact cancellation does not occur and the formula for the tidal force felt is more complicated.

PeterDonis said:
That's what you said you were interested in.
I only meant that I didn't realized it was only derived from one single component of the Riemann Tensor. I'll have to check this with numerical calculations as my question has ##\theta = 45^0##.

For the gravitational acceleration, I didn't quite understand your reply: is this 0.5 * ##g_{rr}## * partial derivative of ##g_{tt}## with respect to ##r## ?

Vick said:
For the gravitational acceleration, I didn't quite understand your reply: is this 0.5 * ##g_{rr}## * partial derivative of ##g_{tt}## with respect to ##r## ?
Not ##g_{rr}##, ##g^{rr}##. The inverse metric component.

Vick said:
my question
Is this a homework question?

PeterDonis said:
Is this a homework question?
No. It's personal stuff!

Vick said:
partial derivative of ##g_{tt}## with respect to ##r## ?
Yes. Though now that I think about it, there might be more terms in there in Kerr spacetime if you're not in the equatorial plane. You would need to check the connection coefficients to see if there are any others of the form ##\Gamma^\mu_{tt}## (the ##\mu = r## component is the only nonzero one in Schwarzschild spacetime, but it might not be in Kerr spacetime) that are nonzero (possibly only for values of ##\theta## other than ##\pi / 2##).

Vick said:
For example for the Time dilation, we took the ##g_{tt}## component which is already an inversed g!!
No, ##g_{tt}## is not an inverse metric component. It has lower indexes. It is a component of the metric, not its inverse. The inverse metric has upper indexes.

Vick said:
I did no such thing. See above.

Vick said:
So that means we are taking the g after inverse and then multiplying with the partial derivative of the g before inverse??
The formula I wrote was for the ##r## component of the proper acceleration. In that formula, ##g_{tt}## is a component of the metric (the t-t component) and ##g^{rr}## is a component of the inverse metric (the r-r component).

If you are not familiar with these notations, you really need to take the time to get familiar with them before trying to solve GR problems.

Dale
PeterDonis said:
The formula I wrote was for the ##r## component of the proper acceleration. In that formula, ##g_{tt}## is a component of the metric (the t-t component) and ##g^{rr}## is a component of the inverse metric (the r-r component).

If you are not familiar with these notations, you really need to take the time to get familiar with them before trying to solve GR problems.
I know about indexing for the other tensors. Yeah I made a mistake with the metric. The inverse metric is used to compute the Christoffel symbols, Ricci Scalar and Weyl and Kretschmann Tensors (Anyway those that I used)

But I'm not getting the correct answer for even the Schwarzschild metric! I did ## 0.5 . g^{rr} . d_r g_{tt} ##

So that's ## 0.5 * (- \frac{2m-r}{r}) * (-\frac{2m}{r^2})## Is that correct?

Vick said:
The inverse metric is used to compute the Christoffel symbols, Ricci Scalar and Weyl and Kretschmann Tensors
Yes, but the ordinary metric itself is also used. You need both to do the computation.

Vick said:
I'm not getting the correct answer for even the Schwarzschild metric!
You need to be careful to distinguish between the Christoffel symbol ##\Gamma^r_{tt}## itself, the radial component of the proper acceleration (which is the only nonzero one in Schwarzschild spacetime), and the actual measured value of the proper acceleration.

The Christoffel symbol is what I wrote down the formula for; it is

$$\Gamma^r_{tt} = \frac{1}{2} g^{rr} \partial_r g_{tt} = \frac{1}{2} \left( 1 - \frac{2m}{r} \right) \partial_r \left( 1 - \frac{2m}{r} \right) = \frac{m}{r^2} \left( 1 - \frac{2m}{r} \right)$$

Then the radial component of the proper acceleration is ##a^r = \Gamma^r_{tt} u^t u^t##, where ##u^t## is the ##t## component of the observer's 4-velocity, and the actual measured proper acceleration is the norm of the acceleration 4-vector, which in this case is ##a^r \sqrt{g_{rr}}##.

Dale
Moderator's note: Thread level changed to "I".

PeterDonis said:
The Christoffel symbol is what I wrote down the formula for
In your first answer you wrote only the ## \Gamma^r_{tt}## as answer which is really not enough!!

Vick said:
In your first answer you wrote only the ## \Gamma^r_{tt}## as answer which is really not enough!!
That's because at the time I made that post, I expected you to have enough background knowledge to already be aware of how to get from the Christoffel symbol to the actual measured acceleration. You originally marked this thread as "A" level, which indicates graduate level background knowledge of the subject. For anyone with that background knowledge, knowing the Christoffel symbol would have been enough.

I have now significantly downgraded my estimate of your background knowledge (and I have now changed the thread level to "I" for this reason), so I am giving more information.

PeterDonis said:
knowing the Christoffel symbol would have been enough.
So does this apply to the Riemann component for Tidal force as well?

Vick said:
So does this apply to the Riemann component for Tidal force as well?
Yes. I have already described how you need to contract that with the observer's 4-velocity (twice).

PeterDonis said:
Yes. I have already described how you need to contract that with the observer's 4-velocity (twice).
Can you please show me the resultant formulae for tidal force and proper acceleration for a static observer for the Kerr Metric?

Vick said:
Can you please show me the resultant formulae for tidal force and proper acceleration for a static observer for the Kerr Metric?
I've already given you enough information to deduce them.

PeterDonis said:
I've already given you enough information to deduce them.
But I thought you said you reduced the level of the thread because you are providing more information!! And normally the whole purpose to ask a question on thread is for other more knowledgeable people to help figure out an answer. The fact that I'm asking means I don't know otherwise I wouldn't ask. Now since you have answered very lightly at the first and then proceeded to downgrade and tell me that I don't know which is true, you might as well go the full way and spoon-feed me!!

Vick said:
I thought you said you reduced the level of the thread because you are providing more information!!
That doesn't mean I'm going to spoon feed you the entire answer. You still need to do some of the work yourself.

Vick said:
Now since you have answered very lightly at the first and then proceeded to downgrade and tell me that I don't know which is true, you might as well go the full way and spoon-feed me!!
Not at all. See above.

Vick said:
normally the whole purpose to ask a question on thread is for other more knowledgeable people to help figure out an answer
Which is exactly what I've done. Helping you figure out an answer is not the same as spoon feeding you.

PeterDonis said:
Which is exactly what I've done. Helping you figure out an answer is not the same as spoon feeding you.
Well I've computed the ##\Gamma^{r}_{tt}## and the ##R^{r}_{trt}## for the Kerr metric already. So what is the next calculations to do to get Proper acceleration and tidal force for d respectively?

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?

On another note, why do you have to reply in 2 or 3 posts for something I wrote in 1 post (posts 10,11,12 and recently posts 23 & 24)?

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Vick said:
Well I've computed the ##\Gamma^{r}_{tt}## and the ##R^{r}_{trt}## for the Kerr metric already. So what is the next calculations to do to get Proper acceleration and tidal force for d respectively?

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?
As I said, in the Kerr metric escape velocity is direction dependent. So that formula can't possibly be correct.

Vick said:
On another note, why do you have to reply in 2 or 3 posts for something I wrote in 1 post (posts 10,11,12 and recently posts 23 & 24)?
Because that was the easiest way for me to do it.

It is pointless to ask about irrelevancies like this; it just clutters the discussion to no purpose. Please stay focused on the actual technical points under discussion.

PeterDonis said:
Only in the sense that escape is impossible in some directions. But at a given point in Schwarzschild spacetime, for all the directions in which escape is possible, the escape velocity is the same. I don't think that's the case for Kerr spacetime.
I haven't done the math for this, but it seems doubtful to me that at epsilon above the photon sphere, tangential escape velocity is same as radial. What I think changes for Kerr is that angle relative to spin makes a difference, as well as angle relative to radial direction.

PAllen said:
it seems doubtful to me that at epsilon above the photon sphere, tangential escape velocity is same as radial
Do you mean in Schwarzschild spacetime?

PeterDonis said:
ask specific questions about specific steps I described if you don't understand them.
Ok! 4 velocity with the Christoffel component: I have only calculated tensors based on metric so far, but now little by little I'm doing things with components to find other stuffs as I see a need for it! So what exactly is the math operation to be done here: Do I just multiply the relevant Connection component with what? Because I'm using t,r,theta and phi?! Since this is stationary the 4-velocity is (1,0,0,0)

PeterDonis said:
Do you mean in Schwarzschild spacetime?
Yes, which I then contrasted to Kerr.

PAllen said:
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.

Vick said:
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.

PeterDonis said:
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.

PAllen said:
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##.

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!

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