# Are all objects crossing the event horizon travelling at C?

1. May 9, 2013

### Stonius

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 travelling 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?

2. May 9, 2013

### Pseudo Epsilon

the gravitational field itself causes time to stop relative to the observer. Not the observers velocity.

3. May 9, 2013

### Bill_K

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 Schwarzchild 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. May 9, 2013

### Stonius

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, isnt 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, "Schwarzchild 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. May 10, 2013

### pervect

Staff Emeritus
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.

Last edited: May 10, 2013
6. May 10, 2013

### Stonius

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?

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?

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.

7. May 10, 2013

### PAllen

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.
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.
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.
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.
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. May 10, 2013

### Bill_K

Since the horizon is a null surface, the relative velocity between faller and horizon can't be anything but c, ever.

9. May 10, 2013

### pervect

Staff Emeritus
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.

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".

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.

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.

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. May 10, 2013

### PAllen

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. May 10, 2013

### Staff: Mentor

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 Schwarzchild r distance, is going to think.

Last edited: May 10, 2013
12. May 10, 2013

### tiny-tim

Welcome to PF!

Hi Stonius! Welcome to PF!
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

13. May 10, 2013

### PAllen

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. May 10, 2013

### Staff: Mentor

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. May 10, 2013

### PAllen

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. May 10, 2013

### Staff: Mentor

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. May 10, 2013

### pervect

Staff Emeritus
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.

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.

Last edited: May 10, 2013
18. May 10, 2013

### Staff: Mentor

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. May 10, 2013

### PAllen

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.

Last edited: May 10, 2013
20. May 10, 2013

### pervect

Staff Emeritus
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. May 11, 2013

### Staff: Mentor

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. May 11, 2013

### pervect

Staff Emeritus
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. May 11, 2013

### pervect

Staff Emeritus
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. May 11, 2013

### jartsa

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. May 11, 2013

### PAllen

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. ]

Last edited: May 11, 2013