# Crossing the Event Horizon of a Black Hole

1. Dec 5, 2009

### DiamondGeezer

Reading "Exploring Black Holes" I find that a particle following a geodesic towards a black hole always reaches the speed of light c when crossing the "event horizon" regardless of the reduced circumference that the particle begins at (1mm or 10 million light years).

This would also apparently happen to a particle which has appeared out of the "quantum foam" where its anti-particle manages to escape (the basis of the so-called "Hawking Radiation")

Is this correct? (I have a supplementary question if this is so)

2. Dec 5, 2009

### Nabeshin

Where in the book do you find this claim? On its face, it seems patently false, and also there is no reference frame for the velocity given. Do you have more info? I have the book right here for reference.

3. Dec 5, 2009

### bcrowell

Staff Emeritus
It seems reasonable to me, if you take it as a statement about the Doppler shift of light emitted from the horizon and observed at infinity. The observer at infinity can take the infinite Doppler shift as indicating motion of the source at c.

4. Dec 5, 2009

### DiamondGeezer

From "Exploring Black Holes" by Taylor and Wheeler, as I said in the original post. Pages 3-14 onwards talk about the speed of the "particle" as approaching the speed of light as measured by shell observers outside of the event horizon and at the bottom of page 3-15 we find

At the reduced Circumference r=8M or four times the Schwarzschild radius (2M) the particle falling from rest at infinity is moving inward at half the speef of light, as witnessed by shell observers. As the particle crosses the event horizon at r=2M, nearby shell observers record it as moving at the speed of light...​

Last edited: Dec 5, 2009
5. Dec 5, 2009

### bcrowell

Staff Emeritus
I have the book in front of me. You say that the book's statement holds "regardless of the reduced circumference that the particle begins at (1mm or 10 million light years)," but I don't see anywhere in the book that it says that. Lots of the equations have labels like "[24. from rest at $r=\infty$]." This part of the book also goes into excruciating detail about different observers and their measuring procedures. Your statement that "a particle following a geodesic towards a black hole always reaches the speed of light" is expressed in a way that doesn't take these distinctions into account. I think the observer they refer to as the "shell observer" won't agree that all particles zip by at c, no matter where they're released from.

6. Dec 5, 2009

### Nabeshin

Hrm.. they derive:
$$v= -\sqrt{\frac{2M}{r}}$$

But I did a quick derivation and keep getting,
$$\frac{dr}{d \tau} = -\sqrt{\frac{2M}{r}} \left(1-\frac{2M}{r}\right)^{1/2}$$

For the infalling particle (assumed stationary at infinity), I get,
$$u^\alpha = \left(\left(1-\frac{2M}{r}\right)^{-1},-\sqrt{\frac{2M}{r}},0,0\right)$$
And for an observer stationary at some radius R,
$$u'^{\alpha}= \left(\left(1-\frac{2M}{R}\right)^{-1/2},0,0,0\right)$$
So,
$$\frac{dr}{dt} = \frac{\frac{dr}{d\tau}}{\frac{dt}{d\tau}}= \frac{u^r}{u^t} = -\sqrt{\frac{2M}{r}} \left(1-\frac{2M}{r}\right)$$
And then,
$$\frac{dr}{d\tau'} = \frac{dr}{dt} \frac{dt}{d\tau'} = \frac{dr}{dt} u'^{t} = -\sqrt{\frac{2M}{r}} \left(1-\frac{2M}{r}\right) \left(1-\frac{2M}{R}\right)^{-1/2}$$
And in the case where r=R,
$$v = -\sqrt{\frac{2M}{R}} \left(1-\frac{2M}{R}\right)^{1/2}$$

Did I mess something up or am I simply not calculating the same thing as them?

7. Dec 6, 2009

### DiamondGeezer

No, you're fine.

The result of all that is that when R=2M, the velocity v=-1 (ie the speed of light, c, in natural units in the opposite direction to R)

So does that mean we accept that infalling particles that follow geodesics get to the speed of light at the event horizon of a black hole?

8. Dec 6, 2009

### stevebd1

According to Wheeler and taylor, there are three types of in-falling radial plunger-

Drip (dropped from rest at ro)

Rain (dropped from rest at infinity)

Hail (hurled inward at speed vfar from a great distance).

Velocity of in-falling object relative to shell frame (vshell) (i.e. as observed from a specific radius)-

Drip

$$v_{shell}=-\left(1-\frac{2M}{r_o}\right)^{-1/2}\left(\frac{2M}{r}-\frac{2M}{r_o}\right)^{1/2}$$

Rain

$$v_{shell}=-\left(\frac{2M}{r}\right)^{1/2}$$

Hail

$$v_{shell}=-\left[\frac{2M}{r}+v_{far}^2\left(1-\frac{2M}{r}\right)\right]^{1/2}$$

which all equal c at the EH. In all three cases, multiply by $(1-2M/r)$ for the velocity of the in-falling object as observed from infinity- dr/dt.

Hypothetically, if you were to hover very close to a BH's event horizon, you would see an in-falling object approach close to c, then as it passed you and approached the EH, it would slow down and appear to freeze at the EH relative to you. This would always be the case, regardless of how close you hovered to the EH, technically you would have to be exactly on the EH to see the object cross at c.

Last edited: Dec 6, 2009
9. Dec 6, 2009

### DiamondGeezer

I think that the answer I would give is that there is a big difference (which is covered at length in EBH) between "seeing" and "inferring from measurements at various distances". It is clear from the text that an infalling astronaut would experience the black hole rushing towards him/her at an ever increasing speed and could infer that the speed at the event horizon would be c

I think this refers to the bookkeeper metric rather than what is actually "seen" by shell observers.

Can we take this as read and move on?

10. Dec 6, 2009

### bcrowell

Staff Emeritus
Thanks, stevebd1, for that very informative post.

I think the answer to your question may require some delicate handling of limits. For example, let's take the case of a drip from $r_o/2M=1+2\epsilon$, observed by a shell observer at $r/2M=1+\epsilon$. So for example, an observer 1 mm above the event horizon could be observing something infalling from 2 mm above the event horizon. In this situation, I get $v_{shell}=-1/\sqrt{2}$, not -1.

If I'm understanding your original question correctly, you're saying we have two particles, A and B, emitted near the event horizon. A is going to escape, and B is going to fall in. Here we have an $\epsilon$ that's on the order of the Planck length divided by the mass of the black hole. I think the quantum-mechanical uncertainty in the initial position of each particle is also on the order of the Planck length.

Do you have material particles in mind, or photons? If they're photons, then of course their velocity is c according to any local observer, but it isn't necessarily c as measured in Schwarzschild coordinates.

Last edited: Dec 6, 2009
11. Dec 6, 2009

### George Jones

Staff Emeritus
You didn't calculate the same thing. You calculated (change in radial coordinate)/(change in shell observer proper time), but speed measured by a shell observer is given by (change in shell observer proper distance)/(change in shell observer proper time). Think about how to do this, and, if you need any hints, just ask.

Here is another (slick) way to calculate relative physical (not coordinate) speed between two observers who are coincident at an event. Suppose the two 4-velocities are $u$ and $u'$. Then, $\gamma = \left( 1 - v^2 \right)^{-1/2} = g \left( u , u' \right) = g_{\alpha \beta}u^\alpha u'^\beta.$ This is an invariant quantity, and, consequently, can be calculated using any cordinate system/basis. What do you get for speed when you use your 4-velocities above?

This works in all coordinate systems in both special and general relativity.

Last edited: Dec 6, 2009
12. Dec 6, 2009

### Nabeshin

Thanks for the response. I'll try and stick with the way I was doing it originally, and then look at your "slick" trick afterwards. If I understand you correctly, you're saying I need something like this:
$$\frac{dr}{d\tau'} \frac{dr'}{dr} = \frac{dr'}{d\tau'}$$
Nearest I can reckon, the best way to proceed now is to take r=R and..:
$$\frac{dr'}{d\tau'} = \frac{dr}{d\tau'} \gamma = \frac{dr}{d\tau'} \left(1-\left(\frac{dr}{d\tau}\right)^2\right)^{-1/2} = -\sqrt{\frac{2M}{R}} \left(1-\frac{2M}{R}\right) \left(1-\frac{2M}{R}\right)^{-1/2} \left(1-\frac{2M}{R}\right)^{-1/2}=-\sqrt{\frac{2M}{R}}$$

Now let's see about what you were talking about...
$$\gamma= g_{\alpha \beta} u^{\alpha} u'^{\beta} = g_{t t} u^{t} u'^{t}$$
Since u'^r is equal to zero...
$$\gamma=-\left(1-\frac{2M}{R}\right) \left(1-\frac{2M}{R}\right)^{-1} \left(1-\frac{2M}{R}\right)^{-1/2}=-\left(1-\frac{2M}{R}\right)^{-1/2}$$
$$\left(1-v^2\right)^{-1/2}=-\left(1-\frac{2M}{R}\right)^{-1/2}$$
So the minus sign is a little bit troubling here, but proceeding anyways...
$$v=\sqrt{\frac{2M}{R}}$$
Which is the same up to a sign, but I'm content with that. Pretty cool, George

13. Dec 6, 2009

### George Jones

Staff Emeritus
Very nice.

It looks like you're using Lorentaz contraction. I'll have to think a little about this. Another way is to use the metric to compute proper distance directly, as is done on page 2-21, 2-22 of Exploring Black holes, and in the middle (just after "End of Motivation") of a previous post by me,

Which signature did you use for the metric? The expression(s) that I gave are valid for a + - - - metric. For a - + + + metric, the expression is $\gamma = - g \left( u , u' \right)$. With the - + + + metric, the "inner product" of any two future-directed timelike 4-vectors (as the two 4-velocities are) is negative.
To move from v^2 to v, you took a square root, which introduces +/-. If v is treated purely as a speed (no direction involved), then use the + sign. If more information is given, then a choice between signs can be made. In our example, where one observer is falling down, if the shell observer chooses a coordinate system that: increases in the direction away from the black hole (up), a negative sign is needed; increases in the direction towards the black hole (down), a positive sign is needed.

Another old post (slightly more abstract mathematics) of mine that vaguely is related to this thread:

14. Dec 6, 2009

### Nabeshin

Ah yeah, I'm used to the - + + + convention so that's what I've been using throughout, and your comments clear up the choice of sign with the square root.

Lorentz contraction should be legitimate, but with the caveat that r=R since we can treat that part of spacetime as locally minkowski, right? At least that's the way it seems to me.

15. Dec 6, 2009

### DiamondGeezer

So can you confirm George that the speed of an infalling particle at the EH is the speed of light?

16. Dec 6, 2009

### George Jones

Staff Emeritus
Lorentz contraction (used correctly) is always legitimate in GR. Lorentz contraction and time dilation often confuse me, and, personally, I almost always use the interval (metric) to compute times and distances, even in special relativity.
Very loosely, and I wouldn't say that. There is no shell observer at "r = 2M". A more precise statement (echoing bcrowell) is that as r approaches 2M, the relative speed between the infalling observer and the shell observer *approaches* the speed of light.

Note that it is possible to have two different observers coincident at an event on the event horizon, but, even in this case, the relative speed between the observers is always less than the speed of light.

17. Dec 6, 2009

### Jolb

Though I'm no expert on general relativity, I have heard on good authority that if you observe something falling into a (nonrotating uncharged) black hole, you actually never see it fall in. A clock falling into a black hole would appear to continually slow down (approaching 0 velocity) while getting more and more redshifted and ticking slower and slower, forever. I guess once the photons are redshifted enough, they just blend in with the background/vacuum radiation.

18. Dec 7, 2009

### DiamondGeezer

Why does that sound so evasive, George? The equations say that the particle *does* reach the speed of light at the EH, regardless of whether the particle is thrown at the black hole or simply falls in, and regardless of the starting position above the EH.

You might say that say that it asymptotically reaches the speed of light, but then it also asymptotically reaches the EH as well. Unless the EH represents some kind of hard barrier which causes everything to bounce off and never cross the EH, then that's what the equations suggest.

19. Dec 7, 2009

### George Jones

Staff Emeritus
I'm not trying to be evasive, I'm trying to be careful .
By "equations", I think you mean equations [42] and [47] from the chapter 3. Again, I have to point out that there only shell observers for (strictly) $r > 2M$, so these equations are not valid on or inside the event horizon.
There is a distinction between coordinate speed and physical speed. Physical speed can only be measured locally. Let any observer measure the speed of a photon that whizzes by in his local neighbourhood. The answer is the same (the speed of light) for all observers, above, at, or below the event horizon. Also, the measured speed (again by an observer above, at, or below the event horizon) of any matter in a local neighbourhood will also be less than this speed.

This shows that there is a fundamental physical difference between massive and massless particles.

Even in special relativity, there are (somewhat natural) coordinate-based definitions of "speed" according to which massive particles can have speeds greater than c (e.g., the Milne universe).

20. Dec 7, 2009

### bcrowell

Staff Emeritus
I think George has been very generous with his time in his thread, and has very carefully tried to cooperate with the other people in this thread in working out the issues in detail. DiamondGeezer, it seems like you keep insisting on a yes/no answer, but if you look back over the discussion, the entire thrust of the discussion has been that the original question was posed in a way that ignored some important distinctions. The question doesn't have a yes/no answer.