Crossing the Event Horizon of a Black Hole

[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)

 

[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.

 

[bcrowell]

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.

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.

 

[DiamondGeezer]

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.

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...​

 

[bcrowell]

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.

 

[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?

 

[DiamondGeezer]

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?

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?

 

[stevebd1]

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

Drip (dropped from rest at r_o)

Rain (dropped from rest at infinity)

Hail (hurled inward at speed v_far from a great distance).Velocity of in-falling object relative to shell frame (v_shell) (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.

 

[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?

 

[bcrowell]

Thanks, stevebd1, for that very informative post.

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)

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.

 

[George Jones]

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?

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.

 

[Nabeshin]

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.

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

 

[George Jones]

Very nice.

Nearest I can reckon, the best way to proceed now is to take r=R and..:

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,

https://www.physicsforums.com/showthread.php?p=621802#post621802.

So the minus sign is a little bit troubling here, but proceeding anyways...

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.

v=\sqrt{\frac{2M}{R}}
Which is the same up to a sign, but I'm content with that. Pretty cool, George

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:

https://www.physicsforums.com/showthread.php?p=848684#post848684.

 

[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.

 

[DiamondGeezer]

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,

https://www.physicsforums.com/showthread.php?p=621802#post621802.


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:

https://www.physicsforums.com/showthread.php?p=848684#post848684.

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

 

[George Jones]

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.

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.

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

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.

 

[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.

 

[DiamondGeezer]

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.

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.

 

[George Jones]

Why does that sound so evasive, George?

I'm not trying to be evasive, I'm trying to be careful .

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.

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.

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.

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).

 

[bcrowell]

Why does that sound so evasive, George?

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.

 

[DiamondGeezer]

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.

I accept that. But it does lead to a problem. Let's take the shell observers to the limit as r \rightarrow 2M

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).

The reasoning is not so difficult but the implications of any particle actually reaching the event horizon of a black hole are very clear. From the perspective of any particle (or astronaut), the black hole appears to approach with ever increasing speed, until at the event horizon, that speed is c.

The shell observers do not see a particle (or astronaut) reach c, but they infer it from measurements as the infalling particle accelerates through the shells.

But that brings a problem:

Q: If an infalling particle asymtotically reaches c, the kinetic energy of that particle increases without limit. Yet the kinetic energy comes from the gravitational field of the black hole, so what happens when the kinetic energy of a particle exceeds that of the total energy of the black hole? Of the Universe?

 

[Dmitry67]

Q: If an infalling particle asymtotically reaches c, the kinetic energy of that particle increases without limit. Yet the kinetic energy comes from the gravitational field of the black hole, so what happens when the kinetic energy of a particle exceeds that of the total energy of the black hole? Of the Universe?

It exceeds speed of light only in the coordinate system of a distant observer. GR has no problems with superluminal speed of far objects in curved spacetime.

So the superluminal speed has no physical consequences: you can not observe objects passing you faster then light, and all relative speeds, even inside the BH, are <c

 

[Dmitry67]

The reasoning is not so difficult but the implications of any particle actually reaching the event horizon of a black hole are very clear. From the perspective of any particle (or astronaut), the black hole appears to approach with ever increasing speed, until at the event horizon, that speed is c.

Event horizon is not in object. The position of apparent event horizon is observer-dependent, so different observers do not agree on the position of the horizon. No observer (including freely falling) can reach horizon in his own frame. So, for the freely falling observer, the event horizon recedes in front of him, so he never reaches it.

Regarding the signularity, in non-rotating BH singularity is space-like, so it is in the future, so we can not say that 'it approaches at some speed'

 

[atyy]

Q: If an infalling particle asymtotically reaches c, the kinetic energy of that particle increases without limit. Yet the kinetic energy comes from the gravitational field of the black hole, so what happens when the kinetic energy of a particle exceeds that of the total energy of the black hole? Of the Universe?

This is the relative velocity as seen in a succession of local inertial frames. Even in SR a particle that is traveling with constant velocity relative to any particular inertial frame will increase its velocity as one changes inertial frame.

In GR energy conservation is generally weird, but a conserved energy for a freely falling particle in the Schwarzschild solution is given by Eq 7.38 and 7.41 of http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html

 

[DiamondGeezer]

What's interesting is that no-one has pointed out that if a particle travels at the speed of light at the EH in one frame of reference then it travels at the speed of light in all of the others.

 

[atyy]

What's interesting is that no-one has pointed out that if a particle travels at the speed of light at the EH in one frame of reference then it travels at the speed of light in all of the others.

The particle does not travel at the speed of light at the EH relative to any local inertial frame at the event horizon.

 

[DiamondGeezer]

The particle does not travel at the speed of light at the EH relative to any local inertial frame at the event horizon.

There is no "local inertial frame" at the event horizon.

 

[DiamondGeezer]

Event horizon is not in object. The position of apparent event horizon is observer-dependent, so different observers do not agree on the position of the horizon. No observer (including freely falling) can reach horizon in his own frame. So, for the freely falling observer, the event horizon recedes in front of him, so he never reaches it.

You should ring up Edwin Taylor and tell him, because he's under the strange delusion that the event horizon is reachable in the finite time of an infalling particle or astronaut. Especially as he devotes an entire chapter of "Exporing Black Holes" to describing just that.

Regarding the signularity, in non-rotating BH singularity is space-like, so it is in the future, so we can not say that 'it approaches at some speed'

This makes absolutely no sense at all.

 

[George Jones]

There is no "local inertial frame" at the event horizon.

There most certainly are local inertial reference frames at the event horizon

The particle does not travel at the speed of light at the EH relative to any local inertial frame at the event horizon.

and what atty states is quite correct. This is very much related to

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).

It exceeds speed of light only in the coordinate system of a distant observer. GR has no problems with superluminal speed of far objects in curved spacetime.

So the superluminal speed has no physical consequences: you can not observe objects passing you faster then light, and all relative speeds, even inside the BH, are <c

 

[stevebd1]

Might it be fair to say that as an object crosses the event horizon, the time dilation is zero at exactly 2M and according to Gullstrand-Painleve coordinates, time outside the EH increases to infinity (i.e., the universe outside the EH comes to an end) so the only frames of reference the in-falling object relate to once at/past 2M are those inside the EH.

 

[Dmitry67]

What's interesting is that no-one has pointed out that if a particle travels at the speed of light at the EH in one frame of reference then it travels at the speed of light in all of the others.

This is true only in SR, not in GR.

 

[Dmitry67]

i.e., the universe outside the EH comes to an end

No, infalling observer will NOT see all future history of the Universe.

 

[stevebd1]

No, infalling observer will NOT see all future history of the Universe.

I didn't say that Dmitry, I simply said that the universe outside the EH ceases to exist for the in-falling observer.

 

[Dmitry67]

I didn't say that Dmitry, I simply said that the universe outside the EH ceases to exist for the in-falling observer.

In what sense?
Some of the light from the outside still catches him, so he can see a history of what happened outside few seconds after he crossed the horizon. Even more, he continues to receive that light until he desitegrates in the singularity.

 

[George Jones]

Might it be fair to say that as an object crosses the event horizon, the time dilation is zero at exactly 2M and according to Gullstrand-Painleve coordinates, time outside the EH increases to infinity (i.e., the universe outside the EH comes to an end) so the only frames of reference the in-falling object relate to once at/past 2M are those inside the EH.

Let event A be the event for which an in-falling observer is on the event horizon, i.e., A is part of both the observer's worldline and the event horizon. The observer can use the radar method to establish a local inertial coordinate system that has A as the origin. Part of this local system will label events outside the horizon, and part of the system will label events inside the horizon. The radar method requires the observer to make wristwatch recordings both before and after A, but this is the same as for special relativity.

 

[DrGreg]

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&#039;. Then, \gamma = \left( 1 - v^2 \right)^{-1/2} = g \left( u , u&#039; \right) = g_{\alpha \beta}u^\alpha u&#039;^\beta. This is an invariant quantity, and, consequently, can be calculated using any cordinate system/basis.

Is the terminology "physical speed" fairly standard amongst authors? Are there any other names for \sqrt{1 - |u_\alpha u&#039;^\alpha|^{-2}}?

 

[George Jones]

Is the terminology "physical speed" fairly standard amongst authors?

I don't know; probably not.

Are there any other names for \sqrt{1 - |u_\alpha u&#039;^\alpha|^{-2}}?

Relative speed.

For a short (maybe too short to be understandable) derivation of the result, see

https://www.physicsforums.com/showthread.php?p=1111069#post1111069.

Again, for coincident observers, this derivation works in both special and general relativity. In this thread, I do not agree with Oxymoron's interpretation of the relative velocity between two non-coincident observers in general relativity.

 

[DiamondGeezer]

I don't know; probably not.


Relative speed.

Is there any other kind?

For a short (maybe too short to be understandable) derivation of the result, see

https://www.physicsforums.com/showthread.php?p=1111069#post1111069.

Again, for coincident observers, this derivation works in both special and general relativity. In this thread, I do not agree with Oxymoron's interpretation of the relative velocity between two non-coincident observers in general relativity.

Its striking to me that this thread is short on explanation and really long on proof by assertion.

At the event horizon, there is no space separation between two events (or come to think of it, thereafter) so how can there be an inertial frame of reference?

Also, and you've yet to explain it, any free-falling particle getting asymptotically close to an event horizon will gain kinetic energy which eventually will be larger than the gravitational energy of the black hole. What happens then?

 

[DiamondGeezer]

It exceeds speed of light only in the coordinate system of a distant observer. GR has no problems with superluminal speed of far objects in curved spacetime.

So the superluminal speed has no physical consequences: you can not observe objects passing you faster then light, and all relative speeds, even inside the BH, are <c

An assertion without any proof. I haven't spoken about superluminal speed at all.

But just for the record, observing anything at all inside a black hole is going to be difficult, since light cones inside a BH all point away from you, so how will you see anything?

Secondly since there is no space-like separations, only time-like, how will you breathe when every atom in your lungs has no spatial separation?

Thirdly, how will you think when the electric signals in your brain can only move forward along the time axis?

 

[Dmitry67]

Also, and you've yet to explain it, any free-falling particle getting asymptotically close to an event horizon will gain kinetic energy which eventually will be larger than the gravitational energy of the black hole. What happens then?

No
As explained, you are mixing coordinate speed and physical speed.
free falling particle gains 'c' only the the refernce frame of external observer.
It is coordinate speed, not physical speed.

 

[Dmitry67]

Thirdly, how will you think when the electric signals in your brain can only move forward along the time axis?

For a freely falling observer, nothing interesting happens when he crosses the event horizon. Even more, if you fall into supermassive black hole, you can even miss the 'point of no return'. Ooops, I am already inside the horizon! Huston, we have a pro... aaaa!

Inside the BH time is flowing towards the center of the BH, but the observer's mframe rotates this way too, so for the observer time and space do not chnage their places - it happens only in the coordinate system of external observer.

http://www.valdostamuseum.org/hamsmith/DFblackIn.gif

 

[stevebd1]

Also, and you've yet to explain it, any free-falling particle getting asymptotically close to an event horizon will gain kinetic energy which eventually will be larger than the gravitational energy of the black hole. What happens then?

Unless I'm mistaken, gravitational potential is synonymous with the the time dilation, so as an object increases in speed in free fall towards an object of mass, the kinetic energy is only relative to the inertial frame and observers outside the gravitational field (i.e. at infinity) would not perceive/be aware of the the total increase in energy. It's a bit like an object that fell to Earth would have kinetic energy as it struck the ground but observers outside the gravitational field will observe that the activity of the object has slowed down (due to the time dilation). In order for the object to share the same frame as the observers outside the gravity field, work would have to be done to remove it from the gravity field, meaning the kinetic energy gained would be paid back, complying with the first law of thermodynamics. Hence a predicted free fall velocity of c results in a time dilation of zero.

 

[DiamondGeezer]

No
As explained, you are mixing coordinate speed and physical speed.
free falling particle gains 'c' only the the refernce frame of external observer.
It is coordinate speed, not physical speed.

No I'm not. Physical speed is measured relative to a coordinate system using light as a measuring stick. If there is no spatial dimension to create a coordinate system, then "physical speed" means nothing. Quite how George manages to create an inertial coordinate system without a space dimension at the EH is one of those things I'm hoping to discover.

Since we're talking about invariants here, the 4-energy of a free-falling particle increases without limit as the particle reaches the event horizon, regardless of who is doing the measurement.

 

[DiamondGeezer]

For a freely falling observer, nothing interesting happens when he crosses the event horizon. Even more, if you fall into supermassive black hole, you can even miss the 'point of no return'. Ooops, I am already inside the horizon! Huston, we have a pro... aaaa!

Inside the BH time is flowing towards the center of the BH, but the observer's mframe rotates this way too, so for the observer time and space do not chnage their places - it happens only in the coordinate system of external observer.

This is what annoys me in this thread. "Proof by assertion" run riot.

"Inside the BH" whatever that means, the Schwarzschild metric for both space and time shows that spacelike and timelike separations are IMAGINARY, and no changing of coordinate systems changes that salient (and for some reason completely unexplained) fact.

Change the coordinate system after Kruskal and Szekeres and the same thing happens. Change the coordinate system to Taylor and Wheeler's "rain frame" and the same thing happens.

I note that Exploring Black Holes makes the same statement given without proof. But what does imaginary time and imaginary space (at right angles to our spacetime) actually mean?

http://www.valdostamuseum.org/hamsmith/DFblackIn.gif [/QUOTE][/URL]

Ooh look a picture!

Did you show how you can see anything inside a black hole yet?

 

[Dmitry67]

Look at the trajectory of infalling particle. Look at the point where it crosses the horizon. look at the lighcone. Now rotate the picture clockwise so the lightcone becomes vertical. This is falling observer perspective.

"the Schwarzschild metric for both space and time shows that spacelike and timelike separations are IMAGINARY"

This is wrong! It is just a ROTATION! So time becomes spacelike (in the coordinates of the distant observer) and one of the space coordinates becomes timelike (this part you are probably missing) but for the observer itself everything looks normal, again, just rotate the picture.

 

[Dmitry67]

Look at the infalling particle. I added red line - a path of another particle, say, another side of a spaceship. Initially they are spacially separated. Both particles are mutually inside the lightcones of each other, so they can 'see' each other (green lines).

Always - except the very final moments close to the singularity. Note that lightcones becomes narrower and narrower, it is a result of increasing tidal forces inside. At some point they look causial contact.

Timelike is NOT vertical, and spatial IS NOT horizontal. Time (for the freefalling observer) is in the middle of the lightcone.

 

[George Jones]

This is wrong! It is just a ROTATION! So time becomes spacelike (in the coordinates of the distant observer) and one of the space coordinates becomes timelike (this part you are probably missing) but for the observer itself everything looks normal, again, just rotate the picture.

I wouldn't say this. I would say this that Schwarzschild coordinates really make up two different coordinates systems. One system is valid outside the black hole, but not on the event horizon or inside the black hole. The other system is valid inside the black hole, but not on the event horizon or outside the black hole. The metric unambiguously determines the nature of each coordinate in each coordinate system.

So, inside the event horizon, space doesn't get rotated into a time direction, it's just that human-chosen labels (coordinates) have names (chosen by humans, not by nature) that are very misleading. See

https://www.physicsforums.com/showthread.php?p=1146536#post1146536.

 

[Dmitry67]

Yes, I meant - rotation when you look at the diagram. SO we agree that when 2 sides of a spaceship are spacially separated, it is still valid inside the BH, so I still see everything inside normally (until it is ripped apart by tidal forces)

 

[DiamondGeezer]

I wouldn't say this. I would say this that Schwarzschild coordinates really make up two different coordinates systems. One system is valid outside the black hole, but not on the event horizon or inside the black hole. The other system is valid inside the black hole, but not on the event horizon or outside the black hole. The metric unambiguously determines the nature of each coordinate in each coordinate system.

It's this "one is valid for outside the black hole but not on the event horizon or inside the black hole" that interests me. Either the metric describes a continuous line integral that constitutes the worldline of an infalling particle or it doesn't.

In whatever case, the fact that the spacelike and timelike separations become imaginary beyond the event horizon means something, and it ain't rotation.

The classical form of the Schwarzschild Metric shows a singularity in the future of every spherically symmetric mass, not simply black holes.

So, inside the event horizon, space doesn't get rotated into a time direction, it's just that human-chosen labels (coordinates) have names (chosen by humans, not by nature) that are very misleading. See

https://www.physicsforums.com/showthread.php?p=1146536#post1146536.

Its abundantly clear that the Schwarzschild Metric unambiguously claims that inside of r=2M, whatever is there is at right angles to space and time in this Universe. However the coordinate system is transformed, that's what happens.

Another possibility is that the classical form of the Schwarzschild Metric which gives rise to the unphysical event horizon (that is, infalling particles gain infinite energy and reach the speed of light at the EH) is wrong.

I think I might know what that is.

 

[DrGreg]

To better understand a black hole's event horizon, and what happens at the horizon and the difference between above and below, it is well worth the effort to consider a different example of horizon.

If you are in an accelerating rocket in deep space, far from any gravitation, nevertheless an "apparent horizon" forms behind you. There are some minor technical differences between this "apparent horizon" and the "absolute horizon" of a black hole, but for the purposes of this discussion, the differences are irrelevant.

If the rocket has a constant proper acceleration of a, then the apparent horizon forms at a distance of c²/a behind you (as measured by yourself). Nothing, not even light, can cross the horizon towards you. If you drop an apple out of your rocket, you'll see the (red-shifted) apple slow down as it approaches the horizon and it never crosses it. That's what you see with your eyes and it's also what you calculate in your own coordinate system.

What makes this scenario easier to understand than a black hole is that you can examine what is happening in inertial Minkowski coordinates, or in your own accelerated "Rindler" coordinates. In inertial coordinates the apple reaches the horizon in a finite time and crosses it without incident.

We can ignore two space dimensions and consider inertial (t,x) coordinates. These are related to the rocket's Rindler coordinates (T,X) by

x = X \cosh \frac{aT}{c}
ct = X \sinh \frac{aT}{c}​

(for X > 0). The rocket is located at a constant X = c²/a in these coordinates, and the horizon is the limit as X \rightarrow 0. In inertial coordinates, the horizon is located at x = ct. There is a space-time diagram to illustrate tbis in post #9 of the "about the Rindler metric" thread[/color].

Post #15[/color] of that same thread shows an extra change of space-coordinate you can introduce

R = \frac{1}{2} \left[1 + \left(\frac{aX}{c^2}\right)^2 \right]​

which (in the simplified case where c = 1 = a) gives you a metric equation

ds^2 = (2R-1) dT^2 - \frac{dR^2}{2R-1}​

somewhat similar to the Schwarzschild metric. The post also shows how you can define a different coordinate transformation behind the horizon which gives rise to the same metric equation as above. And behind the horizon (which is at R=½), T measures distorted distance (not imaginary time) and R measures distorted time (not imaginary distance).

The point of all this is to show that almost all the strange things about horizons are due to the use of accelerating coordinate systems, and in particular using a coordinate system that fails at the horizon itself. In the accelerating rocket example, we can use ordinary inertial coordinates to show there's nothing weird "really" happening at all, only the accelerated coordinates making it seem weird.

Spend some time working through the maths of the accelerating Rindler rocket and its "Rindler horizon", and you should find a black hole's horizon easier to understand after that.