Are all objects crossing the event horizon travelling at C?

In summary, if you fall from a few meters away, you'll have a faster velocity than if you fall from infinity, but the two will hit the horizon at the same moment.
  • #1
Stonius
23
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Im sure there must be something I'm missing here. Can someone please explain to me?

You're outside a black hole event horizon and you have two identical objects.

If one object were released from only 1 meter above the event horizon, and another were released from a million km away, shouldn't the further object hit the horizon going faster than the one that was only released from a meter? Both objects should cross the event horizon at very different velocities.

But then, when either of those two objects cross the horizon, time comes to a standstill for the object. Any motion therefore occurs in an infinitely small amount of time. So I then conclude that they must be traveling at C. When they hit the event horizon

But if they both cross at C, then the object that was only 1 meter above the event horizon must have been accelerated faster than the one that was a million km away as they both crossed the horizon at the same velocity.

How could one object have a faster rate of acceleration than another when they were both in the same gravitational field?
 
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  • #2
the gravitational field itself causes time to stop relative to the observer. Not the observers velocity.
 
  • #3
Stonius said:
If one object were released from only 1 meter above the event horizon, and another were released from a million km away, shouldn't the further object hit the horizon going faster than the one that was only released from a meter? Both objects should cross the event horizon at very different velocities.
Stonius said:
But then, when either of those two objects cross the horizon, time comes to a standstill for the object. Any motion therefore occurs in an infinitely small amount of time. So I then conclude that they must be traveling at C. When they hit the event horizon
To begin with, you're confusing two different "times": coordinate time or Schwarzschild time t as measured by a distant observer, and the object's proper time τ as measured by someone riding along on it. When you say they have different velocities, you're talking about the rate of change of distance with respect to the proper time, dr/dτ, and indeed this will be different for the two objects. When you say "time comes to a standstill" (which is really not a very good way to put it! :yuck:), what you mean is that the Schwarzschild time coordinate t → ∞ at the horizon, and a distant observer will see things happening slower and slower. Time does NOT slow to a standstill for someone riding on the object.
 
  • #4
Hi Pseudo Epslion,
I get that it's not velocity that causes time to stop for the dropped objects, and yet the effect is the same, isn't it? While gravity is distorting spacetimetime, the effect is the same as an accelerated frame, right?

Hey Bill_K,
I think I get it - there is no observer frame where you can compare the timings of events - is that what you mean by 'Schwarzschild time' (the time experienced by an observed close to the event hroizon)? From the outside observers point of view both objects take forever to hit the horizon although the 1m away one will hit it first. Yet from the point of view from the objects, the rest of the universe speeds up infinitely until the rest of time happens in a moment (forgive my lack of technical jargon). Or as you said, "Schwarzschild time coordinate t → ∞"

Lets say you wanted to drop the 1m ball so that it hit the horizon at the same moment (local object time) that the distant object would hit the horizon. In other words, say they both aim to hit the horizon at 5 oclock. Using einsteins clock synchronisation and lorenz calculations for the journey of the distant object this would be possible to achieve, no?

So would you drop the 1m ball as the distant object went past, or before it got there in order for them to hit the horizon at the same time? I know you'd have to take into account time dilation for the first part of the distant objects journey, but theoretically its possble, no?
 
  • #5
The very short version:

The observer who falls from infinity does have a different velocity than the one who falls from a few meters. However, their velocity relative to the event horizion is "c", because the event horizon is trapped light. And light always moves at "c" relative to any observer. So it doesn't matter that the two falling observers have "different" velocities, you expect that the relative velocity between the timelike worldline of the infalling observer, and the lightlike worldline of the event horizion, will be "c".

The not so short version.

The velocity "relative to the event horizon" isn't described except insofar as you compute the velocity at arbitrary r, and take the limit as r-> 2M.

Detailed calculations are in https://www.physicsforums.com/showpost.php?p=602558&postcount=29 and https://www.physicsforums.com/showpost.php?p=621802&postcount=30 for observers with different energy parameters. An observer who falls in from infinity will have E=1, an observer who falls in from a "short" distance will have E approaching (but not quite equal to) zero.

See the "short" version for why this result makes sense. The relative velocity between any observer following a timelike worldline and a surface following a light-like worldline will be "c", and the event horizion falls into the later class.
 
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  • #6
Hi Pervect,

Thanks for the detailed reply. I read it 4 times and am still trying to understand some bits. Hope you don't mind if I try to clarify?

pervect said:
The very short version:

The observer who falls from infinity does have a different velocity than the one who falls from a few meters. However, their velocity relative to the event horizion is "c", because the event horizon is trapped light.

Do you mean that their velocity relative to the event horizon is C in the instant when the objects cross it or during their entire freefall?

pervect said:
And light always moves at "c" relative to any observer. So it doesn't matter that the two falling observers have "different" velocities, you expect that the relative velocity between the timelike worldline of the infalling observer, and the lightlike worldline of the event horizion, will be "c".

So are you saying that from the point of view of either object, the event horizon will appear to approach at the speed of light? But this is really due to the effects of gravitational time dilation. Neither object would experience inertia, being in freefall. So while the near object would see itself as having rapidly accelerated to C (relative to the event horizon) with no inertia over a very short period of time, an independent observer hovering somewhere above the event horizon would see the two objects having very different velocities (and of course would say that neither ever reached the horizon in the end anyway). Have I got that right, or am I missing the point?

pervect said:
The not so short version.

The velocity "relative to the event horizon" isn't described except insofar as you compute the velocity at arbitrary r, and take the limit as r-> 2M.

Not sure what r and M denote. Radius? and Mass? and why is r-> 2M the limit? What do you mean 'isn't described'? Sorry, I'm not very mathematical.

pervect said:
Detailed calculations are in https://www.physicsforums.com/showpost.php?p=602558&postcount=29 and https://www.physicsforums.com/showpost.php?p=621802&postcount=30 for observers with different energy parameters. An observer who falls in from infinity will have E=1, an observer who falls in from a "short" distance will have E approaching (but not quite equal to) zero.

See the "short" version for why this result makes sense. The relative velocity between any observer following a timelike worldline and a surface following a light-like worldline will be "c", and the event horizion falls into the later class.
 
  • #7
Stonius said:
Do you mean that their velocity relative to the event horizon is C in the instant when the objects cross it or during their entire freefall?
A free faller (whatever their initial conditions) can only speak of the event horizon's motion relative to them, not the other way around. To speak of speed of an object relative to the horizon you would need to posit an observer on the horizon, and this is impossible because the horizon is a light like surface. You can no more talk about speed relative to the horizon than you can talk about your speed relative to light.

Yes, the horizon passing at c is at the moment of its passing. The horizon's speed relative to the free faller before crossing and after crossing would depend on coordinates chosen - there is no unique meaning that can be given to these statements. My guess is that for Fermi-normal coordinates (a natural choice for the free faller), the horizon would approach at less than c, pass at c, and recede at greater than c. I have not calculated this though (thus my uncertainty) - Fermi normal coordinates are complex to calculate.
Stonius said:
So are you saying that from the point of view of either object, the event horizon will appear to approach at the speed of light? But this is really due to the effects of gravitational time dilation.
It's got nothing to do with gravitational time dilation. It has to do with the simple fact that the horizon a light like surface, and light moves at c locally for all observers everywhere.
Stonius said:
Neither object would experience inertia, being in freefall. So while the near object would see itself as having rapidly accelerated to C (relative to the event horizon)
A free faller doesn't experience itself moving or accelerating at all, by definition. It sees the black hole approaching faster and faster, with the horizon passing at c - no matter where it started free fall from.
Stonius said:
with no inertia over a very short period of time, an independent observer hovering somewhere above the event horizon would see the two objects having very different velocities (and of course would say that neither ever reached the horizon in the end anyway). Have I got that right, or am I missing the point?
It all depends on what you man by 'see' versus 'say' (modeling based on theory). If you mean literally see, and interpret relative velocity of two objects at a distance from you based on their doppler, you would never see either object cross the horizon and you would see both their doppler's go to infinite red shift. You could interpret this as both objects slowing to asymptotically zero radial speed and asymptotically meeting and stopping at the horizon at your t=∞. However, what I would say, based on theory, is that what I am seeing is an image frozen by extreme gravity, and that both objects pass the horizon and both continue to have velocity relative to each other, and reach the singularity at presumably different times.
Stonius said:
Not sure what r and M denote. Radius? and Mass? and why is r-> 2M the limit? What do you mean 'isn't described'? Sorry, I'm not very mathematical.

r is radial coordinate in Schwarzschild coordinates. In natural units in which c=1, the event horizon radius is 2M, M being the mass of the BH.
 
  • #8
The horizon's speed relative to the free faller before crossing and after crossing would depend on coordinates chosen - there is no unique meaning that can be given to these statements. My guess is that for Fermi-normal coordinates (a natural choice for the free faller), the horizon would approach at less than c, pass at c, and recede at greater than c.
Since the horizon is a null surface, the relative velocity between faller and horizon can't be anything but c, ever.
 
  • #9
Stonius said:
Hi Pervect,

Thanks for the detailed reply. I read it 4 times and am still trying to understand some bits. Hope you don't mind if I try to clarify?



Do you mean that their velocity relative to the event horizon is C in the instant when the objects cross it or during their entire freefall?

The event horizon has no valid "frame of reference", any more than any other thing moving at "c" does. The solution I suggested previously was talking the limit of the velocity to a static (hovering) observer, as that observer got closer and closer to the event horizon.

If you don't like that solution, another solution is to use the frame of reference of the infalling observer (which does exist) and say that the event horizon is approaching that observer at "c" in their frame. That may be a simpler way to do it.

So are you saying that from the point of view of either object, the event horizon will appear to approach at the speed of light?

Yes. That's a good way of looking at it. If you get the two objects to cross the event horizon at the same time, each object will measure the other object as having a different velocity - and they will both measure the event horizon approaching them at "c".

But this is really due to the effects of gravitational time dilation. Neither object would experience inertia, being in freefall.

I don't know what you mean by "experience inertia" at all. I suspect that if I did know what you mean I wouldn't like it though :-).

I don't suggest using the gravitational time dilation at all, especially not using the explanation with the "gravitational time dilation" as expressed in Schwarzschild coordiates. You didn't specify which coordinate system you were using, perhaps you believe that gravitational time dilation is indepenent of coordinates. It is NOT. Gravitational time dilation is just the rate of coordinate time (which requires a coordinate system) to proper time. So the term has no meaning unless you define a coordinate system. Since you didn't specify a coordinate system, I have to guess which one you are using - from your remarks, I infer that it is probably the Schwarzschild coordinates.

The problem with using Schwarzschild coordinates is the usual one - they are singular at the event horizon. This makes them confusing at best. Fortunately, there are other, non-singular coordinates one can use. Rather than beating your head against the wall by using coordinates ill-suited to the problem, you'd be better off learning about coordiates that are NOT singular. Such as Kruskal coordinates. Or Painleve coordinates. Or just not using coordinates at all.

I've given this advice a lot, but it seems it's rarely listened to. Still, I can hope you'll be the exception.

So while the near object would see itself as having rapidly accelerated to C (relative to the event horizon)

The near object would see itself as being in free fall. It would see the event horizon as moving towards it as "c". It would see various static observers accelerating towards it at increasingly high rates.

It would never see a static observer at the event horizon. Sufficiently close to the event horizon, it would see the accelerations of the static observers being unbounded (assuming that they exist in this thought experiment). A million gravities - a billiion - a trillion - if you get close enough to the event horizon, the static observer would have to be accelerating that hard.

As the object gets closer and closer to the event horizon, these static observers would approach the speed of light. Even if they only had an inch to do it in. Infinite acceleration can do that. But the object isn't accelerating that hard. It's accelerometer reads zero. What is really accelerating are the (hypothetical) static observers.

Not sure what r and M denote. Radius? and Mass? and why is r-> 2M the limit? What do you mean 'isn't described'? Sorry, I'm not very mathematical.

r is the radius of the black hole
M is the mass of the black hole.

Units are chosen so that G=c=1 (probably - I'd have to look). So r=2M is the same as r = 2GM/c^2 in conventional units, thus r=2M is the Schwarzschild radius in geometric units. Sorry for the abberviations, if the math is too advanced sticking to the easy explanation might be best.
 
  • #10
Bill_K said:
Since the horizon is a null surface, the relative velocity between faller and horizon can't be anything but c, ever.

Relative velocity can only be defined locally. At a distance, there is no unique definition. Coordinate velocity of a null path can easily differ from c. So it is not obvious to me what the coordinate velocity of the horizon would be in Fermi-Normal coordinates of a free faller where such coordinates cover its history form well before to well after horizon crossing. Such coordinates have Minkowski metric along the line (t,x,y,z)=(t,0,0,0), and also connection components vanish (for a free faller) along this line. However, away from this line, both metric and connection become non-trivial, so I would be somewhat surprised if the horizon's coordinate speed was c throughout these coordinates.
 
  • #11
PAllen said:
Such coordinates have Minkowski metric along the line (t,x,y,z)=(t,0,0,0), and also connection components vanish (for a free faller) along this line. However, away from this line, both metric and connection become non-trivial, so I would be somewhat surprised if the horizon's coordinate speed was c throughout these coordinates.

That's good enough to support the claim that the free-falling observer at rest in those coordinates observes the horizon approaching at the speed of light, is it not? That observer is following the (t,x,y,z)=(t,0,0,0) line so will never encounter anything but the Minkowski metric on his way down.

Of course it doesn't tell us anything about what any observer not following the same world line, especially one at a constant Schwarzschild r distance, is going to think.
 
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  • #12
Welcome to PF!

Hi Stonius! Welcome to PF! :smile:
Stonius said:
If one object were released from only 1 meter above the event horizon, and another were released from a million km away, shouldn't the further object hit the horizon going faster than the one that was only released from a meter? Both objects should cross the event horizon at very different velocities.

But if they both cross at C, then the object that was only 1 meter above the event horizon must have been accelerated faster than the one that was a million km away as they both crossed the horizon at the same velocity.

No.

The momentum is mv/√(c2 - v2), or m sinhα, where α is the rapidity, defined by v/c = tanhα

Then the momentum keeps increasing, but no matter how fast or slowly it does so, it will always end up with momentum msinh∞, ie ∞, and speed ctanh∞, = c :wink:
 
  • #13
Nugatory said:
That's good enough to support the claim that the free-falling observer at rest in those coordinates observes the horizon approaching at the speed of light, is it not? That observer is following the (t,x,y,z)=(t,0,0,0) line so will never encounter anything but the Minkowski metric on his way down.

Of course it doesn't tell us anything about what any observer not following the same world line, especially one at a constant Schwarzschild r distance, is going to think.

No, it only tells us what I said from the beginning: the free faller sees the horizon pass at c. It does not say what speed they 'consider' the horizon to be approaching (using Fermi-Normal coordinates) well before they reach it; or what speed they consider it receding, well after it is passed. My guess remains the former would be < c, the latter > c. No arguments have been presented why e.g. z=ct should be a null path (of the horizon) in coordinates where the metric and connections deviate from Minkowski (as they do for any part of the horizon path away from the t axis).
 
  • #14
Nugatory said:
That's good enough to support the claim that the free-falling observer at rest in those coordinates observes the horizon approaching at the speed of light, is it not? That observer is following the (t,x,y,z)=(t,0,0,0) line so will never encounter anything but the Minkowski metric on his way down.

I don't think this is right. In a thread some time back there was quite a bit of discussion about what the local inertial frame of a free-faller crossing the horizon looks like. More precisely, I mean the LIF whose origin is the event at which the free-faller crosses the horizon, and whose t axis is the integral curve of the free-faller's 4-velocity at that event.

One thing that came out of that discussion was that the worldline of that free-faller is *not* a straight line (t,x,y,z)=(t,0,0,0) in such an inertial frame. (At least, that's the case for a Painleve observer, who is falling in from rest at infinity; that's the only case I worked out in that other thread. I don't see why it wouldn't also be the case for a free-faller with any other intial velocity, though.) Note that this is true even if the local inertial frame is made small enough that tidal gravity is negligible.

The reason for the above is that the "acceleration" of the Painleve observer's worldline (meaning the change in its 4-velocity with the change in radial coordinate r) is detectable on a length scale much smaller than the scale on which tidal gravity is detectable (at least, it is given a sufficiently large black hole). The other thread went into this in more detail; I can try to dig up a link to it if needed.
 
  • #15
PeterDonis said:
I don't think this is right. In a thread some time back there was quite a bit of discussion about what the local inertial frame of a free-faller crossing the horizon looks like. More precisely, I mean the LIF whose origin is the event at which the free-faller crosses the horizon, and whose t axis is the integral curve of the free-faller's 4-velocity at that event.

One thing that came out of that discussion was that the worldline of that free-faller is *not* a straight line (t,x,y,z)=(t,0,0,0) in such an inertial frame. (At least, that's the case for a Painleve observer, who is falling in from rest at infinity; that's the only case I worked out in that other thread. I don't see why it wouldn't also be the case for a free-faller with any other intial velocity, though.) Note that this is true even if the local inertial frame is made small enough that tidal gravity is negligible.

The reason for the above is that the "acceleration" of the Painleve observer's worldline (meaning the change in its 4-velocity with the change in radial coordinate r) is detectable on a length scale much smaller than the scale on which tidal gravity is detectable (at least, it is given a sufficiently large black hole). The other thread went into this in more detail; I can try to dig up a link to it if needed.

To further complicate, I proposed Fermi-Normal coordinates as one natural coordinates for a free faller to use. There are not the LIF Peter refers to. These have, as defining feature, that a chosen world line is given the coordinates (t,x,y,z)=(t,0,0,0). Thus, for a free faller, this would be the world line equation in these coordinates by definition. Further, their construction (even for a non-inertial t axis) achieve that the metric is Minkowski all along the t axis. However, for a non-inertial t axis, Fermi-Normal coordinates have non-vanishing connection along the t axis. For an inertial t axis, they achieve vanishing connection components all along the t axis. Both the metric and the connection deviate from minkowski away from the t axis.
 
  • #16
PAllen said:
I proposed Fermi-Normal coordinates as one natural coordinates for a free faller to use. There are not the LIF Peter refers to.

Yes, agreed. What's more, if I'm right in what I said about the worldline of a Painleve observer in the LIF (see below for further comment on that), Fermi Normal coordinates along the Painleve observer's worldline will not match up with the LIF at all except at the origin of the LIF; i.e., if we draw the usual coordinate grid of the LIF, the coordinate grid lines of the Fermi Normal coordinates will in general be curved, but the F-N t axis will pass through the origin of the LIF and will be vertical there (same slope as the LIF t axis), though not elsewhere, and the F-N x-axis passing through the LIF origin will be horizontal there (same slope as the LIF axis), though not elsewhere.

I agree these are "natural" coordinates for the free-faller to use if he wants to describe events over an extended region; however, it's good to bear in mind that there will not, in general, be a direct correspondence between his F-N coordinate values and actual measurements that he makes, even along his worldline. The only thing I think he can ensure a match for is coordinate time along his worldline matching proper time along his worldline (since he always has the freedom to scale coordinate time along the coordinate t axis however he wants).
 
  • #17
PAllen said:
Relative velocity can only be defined locally. At a distance, there is no unique definition. Coordinate velocity of a null path can easily differ from c. So it is not obvious to me what the coordinate velocity of the horizon would be in Fermi-Normal coordinates of a free faller where such coordinates cover its history form well before to well after horizon crossing. Such coordinates have Minkowski metric along the line (t,x,y,z)=(t,0,0,0), and also connection components vanish (for a free faller) along this line. However, away from this line, both metric and connection become non-trivial, so I would be somewhat surprised if the horizon's coordinate speed was c throughout these coordinates.

You can approximate it reasonably well, because the fermi-normal metric for a free-falling observer (no proper accleration, zero angular momentum) is known from the Riemann.

See MTW, pg 332.

g_00 , for example, is ## \left( -1-R_{\hat{0}\hat{l}\hat{0}\hat{m}} \, x^\hat{l}x^\hat{m} \right)##

The whole metric (all of the metric coefficients) are given, but I'm too lazy to to type them all in. I think you'll find, though, that the spatial components of the metric will contribute at most equally to the speed of light changes as g_00, so you can get an idea of how fast the speed of light changes just from the above.

Basically when your tidal force ##R_{\hat{t}\hat{x}\hat{t}\hat{x}} * x^2 ## becomes a significant fraction of 1 (or c^2, depending on your unit choice), the curvature effects will start to be important and you'll see signficant time dilation in the metric, resulting in significant changes in the coordinate speeds of light.

The precision of the measurement and calculation determine what's "significant", as always.

Comparing the above expression for g_00 with the Newtonian approximation (1-2U) dt^2, we can see that the integral of the tidal force which is ## \int R_{\hat{t}\hat{x}\hat{t}\hat{x}} x dx ## has the same effect on time dilation as the usual Newtonian potential U (the integral of force * distance) in the approximation for g_00, as one might expect.

[add]
So, for the benefit of the lurkers, I'm saying that if one has a large black hole, with a weak tidal force, I believe that one can reasonably treat the rate of approach of the event horizon as "c", even before you get there, over quite a large distance. It's true that curvature effects cause issues with velocity comparisons non-locally, the above formula and detailed discussion gives one some idea of what non-locally means.
 
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  • #18
pervect said:
if one has a large black hole, with a weak tidal force, I believe that one can reasonably treat the rate of approach of the event horizon as "c", even before you get there, over quite a large distance.

I'm not sure that's all there is to it, because in the other thread I referred to (it was the "black hole horizon puzzle" thread), I showed that the "escape velocity" can vary detectably within an LIF centered on the event where a Painleve observer crosses the horizon, even though curvature is negligible within the LIF. That seems to indicate that the F-N coordinates centered on the Painleve observer's worldline must be detectably different from the LIF coordinates within the LIF, which would mean that the coordinate speed of light in the F-N coordinates would most likely no longer be c (since it is exactly c in the LIF coordinates).
 
  • #19
PeterDonis said:
I'm not sure that's all there is to it, because in the other thread I referred to (it was the "black hole horizon puzzle" thread), I showed that the "escape velocity" can vary detectably within an LIF centered on the event where a Painleve observer crosses the horizon, even though curvature is negligible within the LIF. That seems to indicate that the F-N coordinates centered on the Painleve observer's worldline must be detectably different from the LIF coordinates within the LIF, which would mean that the coordinate speed of light in the F-N coordinates would most likely no longer be c (since it is exactly c in the LIF coordinates).

I think this is right. The way I look at this in coordinate terms is that an LIF is best approximated by Riemann-Normal coordinates which have Minkowski metric and vanishing connection at a point, and deviations from flat geometry as slow as possible in all directions of spacetime. Meanwhile, Fermi-Normal for a free fall world line, while achieving Minkowski metric and vanishing connection along the t axis, have deviations from Minkowski metric and connection, that is relatively more rapid than for Riemann-Normal, and the t axis picks a preferred spacetime direction. Thus, deviation from coordinate speed of c for light would occur closer to the t axis in Fermi-normal coordinates than to the origin in Riemann-Normal coordinates.
 
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  • #20
PeterDonis said:
I'm not sure that's all there is to it, because in the other thread I referred to (it was the "black hole horizon puzzle" thread), I showed that the "escape velocity" can vary detectably within an LIF centered on the event where a Painleve observer crosses the horizon, even though curvature is negligible within the LIF. That seems to indicate that the F-N coordinates centered on the Painleve observer's worldline must be detectably different from the LIF coordinates within the LIF, which would mean that the coordinate speed of light in the F-N coordinates would most likely no longer be c (since it is exactly c in the LIF coordinates).

That's a rather long thread - could you give the post # where you presented your argument?

If we posit that the Riemann is varying "slowly", I don't really see where the approximation I mentioned would be breaking down, but I'd like to review your detailed argument.
 
  • #21
pervect said:
could you give the post # where you presented your argument?

There isn't really a single post in that thread where that argument is given, since it wasn't the primary topic of that thread. But the argument basically goes like this:

(1) A LIF centered on the event where a free-faller (specifically, a Painleve observer) crosses the horizon will have negligible tidal gravity within it.

(2) Within such an LIF, an object launched radially outward by the free-faller slightly above the horizon, just at escape velocity, will be moving at slightly less than the speed of light. (This means, of course, that within the LIF, that object will be getting *closer* to the horizon--or perhaps it's better to say the horizon will be catching up to it--even though the object's radial coordinate r is strictly increasing. That was the puzzle--at least, part of it--that prompted me to start that thread, but it's not really the primary question for this thread.)

(3) Thus, the change in escape velocity between the horizon (where it's c) and slightly above the horizon (where it's less than c) is detectable within the LIF. (More precisely, it is for a sufficiently large hole, such that tidal gravity at the horizon is small enough that the scenario can be set up in the first place.)

(4) Since a Painleve observer falls inward at escape velocity, and since changes in escape velocity are detectable within the LIF, changes in the Painleve observer's 4-velocity should also be detectable within the LIF. That means the Painleve observer's worldline within the LIF should not look like a straight line. In other words, the size scale over which the Painleve observer's 4-velocity changes detectably is smaller than the size scale over which tidal gravity is detectable.
 
  • #22
PeterDonis said:
There isn't really a single post in that thread where that argument is given, since it wasn't the primary topic of that thread. But the argument basically goes like this:

(1) A LIF centered on the event where a free-faller (specifically, a Painleve observer) crosses the horizon will have negligible tidal gravity within it.

(2) Within such an LIF, an object launched radially outward by the free-faller slightly above the horizon, just at escape velocity, will be moving at slightly less than the speed of light. (This means, of course, that within the LIF, that object will be getting *closer* to the horizon--or perhaps it's better to say the horizon will be catching up to it--even though the object's radial coordinate r is strictly increasing. That was the puzzle--at least, part of it--that prompted me to start that thread, but it's not really the primary question for this thread.)

(3) Thus, the change in escape velocity between the horizon (where it's c) and slightly above the horizon (where it's less than c) is detectable within the LIF. (More precisely, it is for a sufficiently large hole, such that tidal gravity at the horizon is small enough that the scenario can be set up in the first place.)

(4) Since a Painleve observer falls inward at escape velocity, and since changes in escape velocity are detectable within the LIF, changes in the Painleve observer's 4-velocity should also be detectable within the LIF. That means the Painleve observer's worldline within the LIF should not look like a straight line. In other words, the size scale over which the Painleve observer's 4-velocity changes detectably is smaller than the size scale over which tidal gravity is detectable.

I agree with 1) and 2)

I don't really agree with 3), about detectability of the escape velocity in the LIF. It seems to me that the issue is that no matter how large your LIF is, the LIF won't include infinity. Therefore, even if you have a good approximation for a large region of space-time, you can't answer the question of where the absolute horizion is, or whether the light beam from the horizon has "caught up". You can say that the light beam emitted from the horizon hasn't "caught up" within the confines in which your LIF is valid, but that's all. You can't make any statements about infinity, because infinity is outside the confines of your LIF.

I don't really see anything that suggests to me that eq (13.73) on pg 332 of MTW shouldn't provide a good estimate of the coordinate speed of light in a fermi-normal frame, either - though I would perhaps withdraw the observations about the "tidal force U potential-equivalence". If you stick to a single chart, I think this chart should give you a good estimate of how accurate the LIF approximation is, by including low order error terms.
 
  • #23
PAllen said:
I think this is right. The way I look at this in coordinate terms is that an LIF is best approximated by Riemann-Normal coordinates which have Minkowski metric and vanishing connection at a point, and deviations from flat geometry as slow as possible in all directions of spacetime. Meanwhile, Fermi-Normal for a free fall world line, while achieving Minkowski metric and vanishing connection along the t axis, have deviations from Minkowski metric and connection, that is relatively more rapid than for Riemann-Normal, and the t axis picks a preferred spacetime direction. Thus, deviation from coordinate speed of c for light would occur closer to the t axis in Fermi-normal coordinates than to the origin in Riemann-Normal coordinates.

If you only consider things that happen "now", and choose to orient the time axis of your Riemann normal coordinates parallel to the time axis of the Fermi normal coordinates, shouldn't the two systems give the same answer to questions about what's happening "now"?

For example: how far away is the event horizon "now"? What is the coordinate velocity of the event horizon "now"?
 
  • #24
If an observer stays very close to the event horizon, straight under the falling object, then the observer must move towards the object when the event horizon is bulging towards the object.

If the object is dropped at lower altitude, then it has less kinetic energy, and the event horizon does not bulge as much towards the approaching mass-energy, and an observer close to the event horizon does not move so much towards the falling object.
 
  • #25
pervect said:
If you only consider things that happen "now", and choose to orient the time axis of your Riemann normal coordinates parallel to the time axis of the Fermi normal coordinates, shouldn't the two systems give the same answer to questions about what's happening "now"?

For example: how far away is the event horizon "now"? What is the coordinate velocity of the event horizon "now"?

No, I don't think so. If you examine the metric along different inertial world lines through origin if Riemann-Normal coordinates, none is preferred, and the metric deviates from Minkowski (as slowly as collectively possible) along all such world lines. For Fermi-Normal coordinates based on a chosen inertial world line, that world line is preferred, and the metric remains exactly Minkowski along it.

[Edit: Compare discussion in MTW pp. 286 with pp. 332. The timelike basis vector preference of Fermi-Normal (332), versus no basis preference in Riemann-Normal (286) becomes explicit. Further, analysis in both sections supports my statements: the two normal coordinates differ in 3d order terms, with Riemann-Normal having smallest possible 3-order deviations from flatness.

Further, it is clear that my overall point is true for both such coordinates: at some distance from the horizon, there will be some deviation from c of the horizon. The only thing I remain uncertain of (because I would only believe a computation which I have not done so far) is whether the deviation from c for a FN coordinates for a radial free faller would have the character I propose: < c some distance before crossing, > c some distance after crossing. ]
 
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  • #26
PAllen said:
No, I don't think so. If you examine the metric along different inertial world lines through origin if Riemann-Normal coordinates, none is preferred, and the metric deviates from Minkowski (as slowly as collectively possible) along all such world lines. For Fermi-Normal coordinates based on a chosen inertial world line, that world line is preferred, and the metric remains exactly Minkowski along it.

[Edit: Compare discussion in MTW pp. 286 with pp. 332. The timelike basis vector preference of Fermi-Normal (332), versus no basis preference in Riemann-Normal (286) becomes explicit. Further, analysis in both sections supports my statements: the two normal coordinates differ in 3d order terms, with Riemann-Normal having smallest possible 3-order deviations from flatness.

Further, it is clear that my overall point is true for both such coordinates: at some distance from the horizon, there will be some deviation from c of the horizon. The only thing I remain uncertain of (because I would only believe a computation which I have not done so far) is whether the deviation from c for a FN coordinates for a radial free faller would have the character I propose: < c some distance before crossing, > c some distance after crossing. ]

If you choose your basis vectors at point P to generate the Riemann normal coordinates to be the same set of basis vectors that your Fermi normal observer uses, then, by construction, the set of space-like geodesics at point P will be the same set as used by the Fermi normal observer (in fig 13.4). And the technique of finding the distances from the geodesics will be the same too (pick a spacelike direction, and find the distance from the affine parameter along the geodesic).

So the set of points that are simultaneous with P will be the same set in both coordinate systems, and have the same distance coordinates.

As far as velocities go, I've come to realize I'm not sure what velocities we're talking about.

We mentioned "the velocity of light in Fermi Normal coordinates", as I recall. (I think MTW gives us a good idea of how this changes). Then we got off into a segue "but maybe Riemann normal coordinates would be beter...". (I think MTW answers this, too, but I'm not as familiar, I don't have any intuition about RN coordinates).

At this point, I want to step back and ask - isn't there some coordinate independent way of saying what we're trying to say? I've got some intuition about FN coordinates, not very much intuition about RN coordinates, but what I'd really find convincing is an argument that didn't depend on coordinates.

As to the light beam catching up - I think it should have more to do with relative acceleration than velocity. Basically, the only way to outrun a light beam is to accelerate away from it, the light beam will always be faster than any physical object. At any given point, the light beam will always be faster than the object. But the object can still outrun the light beam, given enough of a head start, if the object accelerates.

What's needed to make the argument convincing is some definition of what "relative acceleration" is in a context when we can't even compare velocities unambiguously. This goes back to formulating the problem in a coordinate independent manner.

At the moment I don't have anything coherent to say on the topic, though some ideas are trying to form.
 
  • #27
pervect said:
As to the light beam catching up - I think it should have more to do with relative acceleration than velocity. Basically, the only way to outrun a light beam is to accelerate away from it, the light beam will always be faster than any physical object.

Within a LIF this is true. But globally in a curved spacetime it isn't. An object launched outward from just outside the horizon, at escape velocity, can be moving on a geodesic and still "outrun" the horizon, because the spacetime is globally curved. But within the LIF I described, the object will be moving towards the horizon--or, equivalently, the horizon will be catching up with it.

pervect said:
What's needed to make the argument convincing is some definition of what "relative acceleration" is in a context when we can't even compare velocities unambiguously.

Change in observed Doppler shift of light signals from one object to the other. For example, the original puzzle scenario in that other thread went like this: a Painleve observer falling through the horizon launches two probes. Probe #1 is launched just outside the horizon, at escape velocity; call that velocity ##1 - \epsilon## (meaning coordinate velocity in the LIF of the Painleve observer, centered on the event where he crosses the horizon). Probe #2 is launched just inside the horizon, at velocity ##1 - \epsilon / 2## (again, this is coordinate velocity in the LIF). Both probes move geodesically once launched.

Now suppose Probe #1 emits light signals radially inward, at a constant known frequency (as measured by the emitter on Probe #1). Then these signals will initially show a Doppler blueshift when measured by Probe #2, indicating that the two probes are initially approaching each other. However, the Doppler blueshift of the signals, as observed by Probe #2, will get smaller and smaller, then go to zero, and then become a Doppler redshift that gets larger and larger.

Note that this change in Doppler shift is due to tidal gravity, for which the term "relative acceleration" can be problematic, since both probes are moving geodesically--no proper acceleration anywhere. However, I believe this usage of the term "relative acceleration" is fairly standard; MTW seems to use it often, for example.
 
  • #28
PeterDonis said:
Change in observed Doppler shift of light signals from one object to the other. For example, the original puzzle scenario in that other thread went like this: a Painleve observer falling through the horizon launches two probes. Probe #1 is launched just outside the horizon, at escape velocity; call that velocity ##1 - \epsilon## (meaning coordinate velocity in the LIF of the Painleve observer, centered on the event where he crosses the horizon). Probe #2 is launched just inside the horizon, at velocity ##1 - \epsilon / 2## (again, this is coordinate velocity in the LIF). Both probes move geodesically once launched.

Now suppose Probe #1 emits light signals radially inward, at a constant known frequency (as measured by the emitter on Probe #1). Then these signals will initially show a Doppler blueshift when measured by Probe #2, indicating that the two probes are initially approaching each other. However, the Doppler blueshift of the signals, as observed by Probe #2, will get smaller and smaller, then go to zero, and then become a Doppler redshift that gets larger and larger.

Note that this change in Doppler shift is due to tidal gravity, for which the term "relative acceleration" can be problematic, since both probes are moving geodesically--no proper acceleration anywhere. However, I believe this usage of the term "relative acceleration" is fairly standard; MTW seems to use it often, for example.

I agree with this approach.

Basically you have some observed doppler shift ##d(\tau)## that represents the ratio of the transmitted frequency at one end of a lightlike geodesic to the received frequency at the other end. In this case I've chosen to paramaterize ##d(\tau)## to be this ratio parameterized by the time of reception.

There then exists a mapping ##M(\tau) = \int_{t_0}^{t} d(\tau) d \tau + C ## that maps the proper time of reception, ##\tau##, to the proper time of emission, ##M(\tau)##. If ##M(\infty)## is finite, ie. if ## \int_{t_0}^{\infty} d(\tau) d \tau ## is finite, then M maps an infinite time of reception to a finite time of emission, thus one will never be able to see events emitted after ## M(\tau)## on the emitters worldline.

The exponential doppler shift ##e^{-a \tau} ## of an observer undergoing constant proper acceleration is more than sufficient to make the above integral finite. But a constant proper acceleration (in flat space) is not necessary to make the integral finite. Clearly, though, in flat space, some sort of acceleration is necessary.

In any LIF, the doppler shift function d for geodesic motion will be a constant, and this integral will diverge, so one will never see this effect in flat space-time without some sort of proper acceleration (though it doesn't have to be constant).

Therefore to study this sort of horizon, one always needs to use more than a LIF.

One might try to make a case that this conflicts with the idea that any small enough region of space can be modeled with a LIF. But I don't think there is any real conflict with the basic principle, I would summarize the situation as being that it's not too surprising that no LIF can be large enough to cover events (such as the catching up of the light beam with the particle moving at escape velocity emitted just before the horizon) that are expected to only happen at spatial infinity if they happen at all.
 
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  • #29
pervect said:
I don't really agree with 3), about detectability of the escape velocity in the LIF.

This may already have been resolved in our discussion, but on re-reading I think it's worth pointing out that I didn't say escape velocity itself was detectable within the LIF; I said that *changes* in escape velocity were detectable. But then I clarified that what I really meant was that changes in the Painleve observer's 4-velocity are detectable. That 4-velocity is locally measurable, even though, as you say, the fact that it happens to be numerically equal to the escape velocity at the point of measurement is not (since it requires knowledge of global properties of the spacetime).
 

1. Are all objects traveling at the speed of light when crossing the event horizon?

No, not all objects are traveling at the speed of light when crossing the event horizon. The speed at which an object crosses the event horizon depends on its initial velocity and the gravitational pull of the black hole. In general, objects with greater initial velocities will cross the event horizon at speeds closer to the speed of light.

2. Why is the speed of light significant when discussing the event horizon?

The speed of light is significant when discussing the event horizon because it is the maximum speed at which anything can travel in the universe. As an object approaches the event horizon, the gravitational pull of the black hole becomes stronger, causing the object to accelerate. If the object were to reach the speed of light, it would not be able to escape the black hole's gravitational pull and would be pulled into the singularity.

3. Can objects ever escape the event horizon?

No, once an object crosses the event horizon, it cannot escape. This is because the escape velocity (the speed an object needs to travel in order to escape the gravitational pull of a body) at the event horizon is equal to the speed of light. As previously mentioned, nothing can travel faster than the speed of light, so once an object crosses the event horizon, it is pulled towards the singularity at the center of the black hole.

4. Is the speed of light constant when crossing the event horizon?

Yes, the speed of light is constant when crossing the event horizon. This is because the event horizon is defined as the point at which the escape velocity is equal to the speed of light. Therefore, no matter how fast an object is traveling, it will always cross the event horizon at the speed of light.

5. Do all black holes have the same event horizon?

No, the size of a black hole's event horizon depends on its mass. The more massive a black hole is, the larger its event horizon will be. This means that the escape velocity at the event horizon will also vary depending on the black hole's mass, but for all black holes, the escape velocity will always be equal to the speed of light at the event horizon.

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