# Rindler Horizon calculation

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1. Nov 19, 2014

### Mentz114

Greetings. For my own edification I calculated a set of the congruence of uniformly accelerated observers in flat spacetime in the spherical polar chart. These observers accelerate radially outwards from some $r$ so that their horizons are at the same position $r_h$. This requires that $r_h=r-1/a$ or $a=1/(r-r_h)$.

To solve for $a(r)$ I boosted the comoving frame field by $\beta(r)$ and set the resulting proper acceleration $\beta \partial_r\beta/\gamma^3$ equal to $1/(r-r_h)$. The solution of that ode gives
\begin{align} \beta &= \frac{\sqrt{{\log\left( r-r_h\right) }^{2}-1}}{\log\left( r-r_h\right) } \end{align}
This is correct in that when substituted into the proper acceleration it gives $1/(r-r_h)$.

But this says that $r>r_h+e$ (!). I don't like that $e$, nor the fact that there is this region outside the horizon where the coords don't seem to work.

Any idea what's going on ?

(I can show the workings if required)

2. Nov 19, 2014

### Staff: Mentor

No, they're not. The Rindler horizon is not a point. It's a pair of null surfaces (more precisely, a pair of 3-surfaces with one null and two spacelike tangent vectors). Two different accelerating observers can only share the same Rindler horizon if they're both accelerating in the same direction (and with proper accelerations appropriately adjusted based on their respective distances from their common Rindler horizon).

First, there can't possibly be a "comoving" frame field, because these observers are not comoving (except possibly for a single instant when they are all at rest at the same $r$).

Second, which direction are you boosting in? You can't boost in all directions at once.

3. Nov 19, 2014

### Mentz114

Two different accelerating observers can only share the same Rindler horizon if they're both accelerating in the same direction (and with proper accelerations appropriately adjusted based on their respective distances from their common Rindler horizon).

That is exactly what I was describing. I even give the acceleration required for different points. They are all accelerating in the same direction - radially out.
So they all have the nearest point to the horizon at $r_h$.

The frame I boost is $f^0=\partial_t,f^1=\partial_r,f^2 = r\partial_\theta, f^3=r\sin(\phi)\partial_\phi$. The boost is in the $r$ direction.

4. Nov 19, 2014

### Staff: Mentor

Radially outward is not "the same direction" in the appropriate sense. "The same direction" in the sense required to have a common Rindler horizon means the angle between the two acceleration vectors is zero. In your scenario, an observer accelerating radially outward at, for example, $\theta = \pi / 2, \phi = 0$ is not accelerating "in the same direction" (in the sense required for a common Rindler horizon) as an observer accelerating radially outward at $\theta = \pi / 2, \phi = \pi$. They are accelerating in opposite directions as far as their Rindler horizons are concerned.

Which is a different direction for each distinct pair of $\theta, \phi$ values. Not only in the sense required to determine the Rindler horizon, but in the sense required to boost. You can't boost in the $r$ direction everywhere; that's not a Lorentz transformation.

5. Nov 19, 2014

### Mentz114

I understand what you're saying, but the calculation is done in a local frame where $x,y,z$ are just orthogonal axes whose orientation is irrelevant. I think your objection could apply to the LeMaitre frame field also. However, even ignoring this, my calculation works for observers with equal $\phi$ and $\theta$ coordinates. It is possible to set up a line of equal horizon observers in these coordinates.

I'm trying to see what constraints the symmetry puts on the horizons, and so far it is interesting except for $\beta$ being strange.

6. Nov 20, 2014

### Staff: Mentor

Yes, but it will be a different local frame for different observers moving radially outward at different $\theta, \phi$ coordinates.

If you mean that you think different Lemaitre observers should have different black hole horizons, no. First of all, I think you really mean "hovering" observers who are at different angular coordinates, since those are the ones that are accelerated; Lemaitre observers are freely falling into the black hole.

More importantly, Schwarzschild spacetime is curved, not flat, so "hovering" observers who are accelerating in different directions can still have the same black hole horizon. (So do all other observers, since in Schwarzschild spacetime the horizon is a unique global feature of the spacetime, not an observer-dependent feature as the Rindler horizon is in flat spacetime.)

Well, sure, that's just the standard family of Rindler observers in different coordinates. Was that what you meant to be describing in the OP?

7. Nov 20, 2014

### Mentz114

I don't think so. To maintain spherical symmetry they are all the same local frame. This is flat spacetime.

Yes, of course I meant a hovering observer. I realize there are problems interpreting a black hole horizon as a Rindler horizon. That's not what I wanted to do at this point. Can the Schwarzschild freely-falling observer receive receive information from inside the horizon ? ( I guess not but I'm not certain )

Yes. With the constraint that they all share the closest point to the horizon. I think my calculation is correct but I could be misinterpreting something.

A similar case is the Langevin congruence of rotating observers in axially symmetric coordinates. There is proper (centripetal) acceleration of $-\omega^2r/(1-\omega^2r^2)$. Where is the horizon ? I guess at inwards of the observer.

8. Nov 20, 2014

### Staff: Mentor

Either you're missing my point, or we're mixing up different scenarios. In flat spacetime, in spherical coordinates, if there are two observers, observer A at $r = R, \theta = \pi / 2, \phi = 0$, and observer B at $r = R, \theta = \pi / 2, \phi = \pi$, and both are accelerating radially outward (i.e., in their local $\partial / \partial r$ direction), then a boost in observer A's $\partial / \partial r$ direction is in the opposite direction from a boost in observer B's $\partial / \partial r$ direction. You can arrange things so that A and B are both at rest in the same inertial frame at one instant, but that's all.

If we have two other observers, C at $r = R_1, \theta = \pi / 2, \phi = 0$, and D at $r = R_2 > R_1, \theta = \pi / 2, \phi = 0$, then a boost in the $\partial / \partial r$ direction is the same for both of them, and at any event on either one's worldline, you can find an inertial frame in which both of them are momentarily at rest.

I had thought you were describing the A and B scenario, which was what my comment referred to. It now appears that you meant to describe the C and D scenario; if so, my comment doesn't apply to what you were trying to describe.

Once he falls inside it, yes. Not while he's still outside it.

I'm not sure what you mean by this. If they are all members of the same family of Rindler observers, they all share the same Rindler horizon, by definition. It's not an additional constraint.

Observers in the Langevin congruence do not all share the same horizon; they can't, because their accelerations point in different directions and vary with position in the "wrong" way (as compared with the Rindler congruence). A given Langevin observer's horizon will be outward of him, not inward (and, as seen in an inertial frame, it will move as the observer rotates).

9. Nov 20, 2014

### pervect

Staff Emeritus
It might help some if you write down the actual congruence you're interested in - I suppose theta and phi are constant, so you just need to specify t(tau) and r(tau)? And - hmmm, that won't fill all of space-time, I suppose you need another coordinate, a coordinate time at which the worldline starts to accelerate? I'm not sure about that last part, but I'll add it in to hopefully make my thoughts a bit clearer.

If I am understanding the congruence correctly, any individual observer on the congruence is going to have a Rindler horizon on a plane tangent to the sphere at r = $r_h$ (coordinates in the inertial frame).

If we ask what events P an individual observer can receive signal from ,having $r < r_h$ will insure that P is not visible, but having $r > r_h$ will not guarantee that P is visible, if my visualization is correct.

10. Nov 20, 2014

### Mentz114

I don't think it matters for the reasons I gave earlier ( we can align our Fermi-Walker spatial coords so the 'x'' points in the new 'r' direction and boost in the local 'x direction'). But we can go with C,D scenario and calculate the radius dependent acceleration which is required to give the common Rindler horizon.

Thanks. As I suspected.

You are referring to my 'closest point' expression.

The thing is this. We can write the 2D locus of the horizon in local polar-coords as $r=1/a\cos(\theta)$ with the origin at the initial point and $-\pi/2<\theta<\pi/2$. The curvature is $k=-a\,{cos\left( t\right) }^{2}\,\left( cos\left( t\right) \,sin\left( t\right) +{cos\left( t\right) }^{2}-2\right)$ which obviously depends on $a$ which is different for each observer. So their horizons share common points but have different spatial curvatures. Is this meaningful ?
( $k$ has values $a,\ a/2,\ 0$ at $\theta=0,\ \pm\pi/4,\ \pm\pi/2$ respectively.)

Ok, but we can freeze and look at one in particular as representative of any. Now the position of the horizon is $r_h=r\pm\frac{1-{r}^{2}\,{\omega}^{2}}{r\,{\omega}^{2}}$ where only one sign here is correct but which one ? With the '+' sign we get $1/(\omega^2r)$, otherwise $\frac{2\,{r}^{2}\,{\omega}^{2}-1}{r\,{\omega}^{2}}$.

I still don't know which one is right. Help.

Last edited: Nov 20, 2014
11. Nov 20, 2014

### Mentz114

I gave the frame basis and the boost, I thought that was enough.
You've visualized my original scenario correctly I think. And you are right that I have picked a non-space filling subset of the boosted rest frame congruence which has a particular acceleration.

Sorry, I don't follow. The horizon by definition precludes receiving signals from $r<r_h$, surely ?

After examining the local metric in Fermi-Walker coords I think that the idea of a Rindler horizon in a local frame is flawed because the axes cannot extend to the required distances without becoming curved. The explicit calculation is from Poisson et al. in the famous point particles paper (arXiv:1102.0529v2 [gr-qc] 8 Feb 2011) from page 49, equations 9.14-9.16.
\begin{align} g_{tt} &= -\left[ 1 + 2a_a\hat{x}^a +\left(a_a\hat{x}^a \right)^2 + O(s^3) \right]\\ g_{ta} &= O(s^3) \\ g_{ab} &= \delta_b^a + O(s^3) \end{align}
The $\hat{x}^a$ are the distances from the curve and $s^2=\hat{x}^a\hat{x}_a$. The distance of the horizon from the curve is just $-1/a$ we have $a_a\hat{x}^a=1$ which means that $g_{tt}$ deviates from $-1$ by a large amount at the horizon. The other metric terms deviate by $O(s^2) \approx O(1/a^3)$ at the horizon.

Last edited: Nov 21, 2014
12. Nov 21, 2014

### Mentz114

I'm sorry I have to make three posts in succession but I just worked out that the method I described in the first post is wrong. I started with the wrong congruence. Using the proper Rindler congruence** for an observer with constant acceleration $a$, all we need is to set $a=1/(r_0-r_h)$ which means the acceleration is constant and the entire congruence has an horizon at the same place $r_h$. I think the spherical symmetry introduces some constraints but I haven't worked them out yet.

**I used coframe basis $e_0= \sqrt{{a}^{2}\,{t}^{2}+1}\ dt + at\ dx,\ \ e_1= at\ dx + \sqrt{{a}^{2}\,{t}^{2}+1}\ dt,\ \ e_2=r\ d\theta,\ \ e_3 = r\sin(\theta)\ d\phi$

So the problem that brought me here was a weird artefact and not even wrong.

Thanks for the stimulating discussion (so far).

Last edited: Nov 21, 2014
13. Nov 21, 2014

### pervect

Staff Emeritus
The region where $g_{00}$ goes to zero for the accelerating observer in Rindler coordinates is definitely a flat plane - at least in any version of the coordinates I've seen. I haven't actually worked out the details of the signal-catching-up problem to determine if the region where $g_{00}$ goes to zero is in fact the boundary of the region that can't send signals to the accelerating observer - I always thought it was, but I haven't worked it out in depth, and perhaps it's curved as you suggest.

14. Nov 21, 2014

### PAllen

I think there is a difference. Rindler coordinates are defined by YZ planes of observers each with the same proper acceleration (different planes have different proper acceleration). The plane of g00=0 represents the boundary of plane light waves able to catch up with a plane of Rindler observers. However, for a given accelerating observer, the causal horizon is curved surface only one point of which is on the g00=0 plane. This surface has axial symmetry in the x direction (conventional coordinate naming).

[Edit: While the above is true from symmetry conditions, the axially symmetric surface actually turns out to be a plane, when I compute it. Unless my computation is in error, the horizon for a uniformly accelerating observer is a flat plane orthogonal to their direction of acceleration, propagating at light speed.]

Last edited: Nov 21, 2014
15. Nov 21, 2014

### Mentz114

If you're using the Rindler chart in Minkowski coords this is true. But I think the surface for the accelerating frame is curved and the local polar equation is $r=1/a\cos(\theta)$. I think the curvature calculation above is right.

16. Nov 21, 2014

### PAllen

I disagree. I'll write up my calculation to show this, as above. The horizon, defined by signal propagation, for a single uniformly accelerating observer, is a flat plane. I do find this counter-intuitive, but it is what the math shows.

17. Nov 22, 2014

### Mentz114

Thank you, I look forward to that.

I'm concerned about what I wrote about the local metric and F-W coords. I think in flat spacetime the terms in $O(s^3)$ should not be there. This gives the local the metric around the worldline as
\begin{align}
g_{tt} &= -\left[ 1 + 2a_a\hat{x}^a +\left(a_a\hat{x}^a \right)^2 \right]\\
g_{ta} &= 0 \\
g_{ab} &= \delta_b^a 0
\end{align}
which is flat everywhere but with an increased time-dilation due to the acceleration. The local coordinates cannot extend beyond the horizon, I imagine.
The acceleration vector $a_b=(0,a,0,0)$ and at the horizon the displacement from the curve is $\hat{x}^b=(0,1/a,0,0)$ so $g_{tt} = 4$, always.

Last edited: Nov 22, 2014
18. Nov 22, 2014

### PAllen

Ok, here is my derivation.

First, note that there are two horizons: an ingoing horizon of the boundary of events that can receive a signal from the eternally accelerating world line, and an outgoing horizon of the boundary of events that can send signals to the world line. For a given position in an inertial frame, the events two way causally connected to the accelerating world line are those after the ingoing horizon has passed, until the outgoing horizon has passed.

By symmetry, we can focus only on the outgoing horizon (the ingoing must be the same, moving in the other direction). Similarly, by symmetry, we can dispense with one spatial coordinate, since there must be axial symmetry about the direction of acceleration. Thus we deal with an accelerating world line with y=0, and look for the boundary of events starting from some t (take it zero) that can catch the world line. We get some curve curve in the XY plane. Rotating this around x, gives the spatial slice of the horizon for t=0. The surface propagates at c, and a later t slice is just the null propagation of the t=0 slice.

The equation of the eternally uniformly accelerating world line is well known to be given by:

x2 = a2+t2 (1)

Next, we note that outgoing light sphere from some event (0,x0,y0) is given by

t2 = (x - x0)2 + (y - y0)2 (2)

If this intersects the world line, we must have y=0, and x given as in (1). Substituting, and doing a little algebra gives:

√(a2+t2) = (a2+x02+y02)/(2 x0) (3)

Focus on 0 < x0 < a/2. Then the right hand side of (3) is > a, for any y0. Thus it has a solution. Thus, for any y0, only as x0 → 0, do we fail to find a solution. Thus the boundary of solutions (the horizon) is x0=0, any y0. Thus a flat plane, when rotated.

[edit: A more general argument the RHS of (3) > a for any x0 > 0 is to rearrange algebraically by subracting a and adding a, doing some algebra, getting (for the RHS of (3)):

(((a - x0)2 + y02) / (2 * x0) ) + a
]

Last edited: Nov 22, 2014
19. Nov 22, 2014

### Mentz114

Thanks. Looks convincing but I'm thinking about it.

I also want to correct an error (again) in the metric around the accelerating world line. The value of $g_{tt}$ is $-\left[1+2a_a\hat{x}^a +(a_a\hat{x}^a)^2 \right]$. For reasons given in post#17, and with $a<0$, $g_{tt}=0$ at the horizon, as it should,

20. Nov 22, 2014

### Staff: Mentor

Shouldn't this be $x^2 = ( 1 / a^2 ) + t^2$? Acceleration has units of inverse length. (I don't think this affects your argument, I just wanted to point it out.)

21. Nov 22, 2014

### PAllen

True, I used the reciprocal of acceleration as 'a', which is confusing, just to make the algebra slightly easier.

22. Nov 22, 2014

### Mentz114

I wrote this slightly differently and get the same conclusion.
Solving the equations for the accelerating observer (1) and the light circle centered at $x_0,y_0$ (2)
\begin{align} x^2 &= t^2 + \frac{1}{a}, \ y=0\\ t^2 &= (x-x_0)^2+(y-y_0)^2\\ \Rightarrow {y_0}^{2} &=\frac{2\,a\,\sqrt{{a}^{2}\,{t}^{2}+1}\,x_0-{a}^{2}\,{x_0}^{2}-1}{{a}^{2}} \end{align}
Equation (3) has a real non-zero solution iff $2a\sqrt{{a}^{2}\,{t}^{2}+1}\,x_0 \gt {x_0}^{2}+1$. The term on the left grows indefinately and so for some $t$ the curves will intersect if $x_0>0$.

Excellent. The metric constructed from the F-W transport and the above agree. Not only that but one can try to find the time taken for a signal to get to the observer from close to the horizon which I find interesting.

Thanks to all, my questions are answered ( and some).

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