How do inertial frames centered on a black hole's horizon work?

[Ben Niehoff]

I think now I actually agree with PAllen, that there really isn't a natural way to define "simultaneous" for spatially-separated events in curved spacetime. The prescription I gave is coordinate-free and unique, but I don't think it would naturally "feel" more simultaneous, when applied to distant events, than any other convention. After all, the precise location of any distant event (i.e., its coordinates in any chosen coordinate system) must be inferred from light signals received from that event.

Nearby events of course must agree with a LIF, which agrees with normal coordinates, etc.

So I will say, some simultaneity convention can be chosen, but it should be based on something that can be clearly defined without coordinates.

 

[Ben Niehoff]

your proposed 'unique' simultaneity will fail for moderately complex non-inertial paths event in SR (for example, a W shapped path for the traveling twin).

I am aware of this. By 'unique', I do not mean that distant events get assigned a unique line of simultaneity. I mean only that from a given observer in a given direction, a unique line of simultaneity can be defined. But it is fully possible for these lines to intersect at some distant event. I never proposed to define a global time coordinate this way.

 

[Dale]

Ack! "Simultaneity" is a real physical thing, not a coordinate artifact. ...

a line of simultaneity satisfies a simple, first-order ordinary differential equation with one initial condition. It is a geodesic which emanates from a given event along the direction of a vector, at that event, orthogonal to some observer's worldline.

In what sense is such a geodesic a "real physical thing"? It has no mass, no energy, no charge, none of the characteristics most people would talk about as being typical of "real physical thing"s. For non-inertial observer's you cannot write the laws of physics in terms of those "real physical thing"s and one of the postulates of relativity is that even for inertial observers the laws of physics are no different for the observer if they use them or others which are not orthogonal.

Personally, I don't worry too much about "real", but I just don't see any justification for your assertion.

 

[PAllen]

I am aware of this. By 'unique', I do not mean that distant events get assigned a unique line of simultaneity. I mean only that from a given observer in a given direction, a unique line of simultaneity can be defined. But it is fully possible for these lines to intersect at some distant event. I never proposed to define a global time coordinate this way.

It is obviously fine to say a particular simultaneity definition has limited scope. However, if a traveler is trying to correlate events on his home world with his own events, the proposition that home 7pm to 8pm corresponds to traveler 2 pm to 2:30 pm and also to traveler 2:31 pm to 3:01 pm is absurd. A sane physicist would insist they must choose a different simultaneity convention to correlate their times with home times.

 

[TrickyDicky]

Going back to the solution of the "puzzle", I see the original way it was put forward in the blog linked by Peter, which is slightly different than how it is described in the OP, it stresses that the discrepancy in separation of signals in an local inertial frame that contains the EH between what the Equivalence principle prescribes and what GR predicts should be enough to reject black holes and proof of the internal inconsistence of GR because the EP is at the theoretical core of GR.
My answer to that is that the author of this blog has a basic misunderstanding of the EP as it is used in GR, as happens often he is loosing sight of the 4-dimensionality of GR, this 4-dimensionality introduces an element of ambiguity in the EP that generates endless discussions when it is not acknowledged. In this case it is argued that relative positions of signals in a LIF can be used to show inconsistency of GR, that would be true if we where talkinkg only about the spatial hypersurfaces, but clearly since GR is a 4-dimensional theory there is margin to accommodate both perspectives if we are talking of different frames of reference or different motion states.

 

[PeterDonis]

I see the original way it was put forward in the blog linked by Peter, which is slightly different than how it is described in the OP

That's because the original blog post doesn't even do a good job of stating the puzzle to begin with. It's cluttered with irrelevant talk about "laws" that aren't stated properly anyway, and it misrepresents what GR (and SR) actually say. (For example, it claims that signals can be sent outward through a Rindler horizon.) I tried to extract the one thing that actually seemed worth thinking about.

the author of this blog has a basic misunderstanding of the EP

Yes, indeed. But I think it's even more basic than that: the author doesn't understand how to properly set up a local inertial frame centered on the horizon. The post implicitly assumes that the horizon is "a point in space", when in fact it's an outgoing light ray.

For extra fun, try looking at the author's other blog post about his new theory of gravity (it's linked to in the blog post I linked to in the OP).

 

[kugbol]

My answer to that is that the author of this blog has a basic misunderstanding of the EP as it is used in GR, as happens often he is loosing sight of the 4-dimensionality of GR, this 4-dimensionality introduces an element of ambiguity in the EP that generates endless discussions when it is not acknowledged.

Which statement in the blog's diagram is false, then? I'm not seeing an incorrect statement there. In particular:

When we let all of the particles above the horizon be escaping to infinity, GR demands that the cloud be splitting apart at the horizon. Nothing about a negligible tidal force mandates such splitting; this behavior is unique to frames straddling the horizon.

How can the cloud not be splitting into two pieces (at least), so that GR is compliant with its EP, or how can the cloud necessarily be splitting in two and yet GR somehow stays compliant with its EP even though the cloud wouldn't need to be splitting in two in every LIF? I'm not seeing the solution. It seems clear to me that the cloud must be splitting in two, because the particles below the horizon are destined to reach the singularity, whereas the escaping particles move ever further away from the hole. It also seems clear to me that no such behavior is required for an LIF; that is, in an LIF the cloud could possibly stay intact given the same velocity specified for only some of its particles, in this case the escaping ones. The author of this blog may well be off base but I'd sure like to know which statement in that diagram is false. To my knowledge there is no ambiguity about the EP to the extent where logically opposite behavior is predicted for an LIF vs. a true inertial frame.

 

[kugbol]

This is a red herring. The answer to the puzzle must lie in curvature effects, because if we could truly ignore all curvature effects, then the axioms of the local inertial frame would demand that the probes collide.

Be careful; the solution must work for all LIFs, because the EP implies that all LIFs are equivalent to one another.

As for what the observer actually does see, I should expect that initially, the distance between the probes is decreasing, but it should actually reach a turnaround point and start increasing. After all, the observer has tossed one ball up at escape velocity, and then a second ball at less than escape velocity. The second ball must return.

But this behavior isn't predicted for all LIFs, so it's not predicted by the solution. The solution must allow the distance between the probes (as measured in the observer's LIF) to decrease at a constant rate for as long as the LIF is valid (possibly years) or until the probes collide, as would happen in other LIFs. These probes each have a constant velocity as measured in the observer's LIF.

 

[kugbol]

Bear in mind also that the puzzle does not say the two probes collide in the local inertial frame; it only says they approach each other, even though the first probe is increasing its radial coordinate and the second is decreasing it.

There's an assumption being made here, worthy of investigation. The probes cannot be approaching each other and moving in opposite directions away from a radial coordinate between them. The probes are approaching each other according to SR, which the EP tells us we are free to employ. Then the radial coordinate of the second probe must be increasing faster than the radial coordinate of the first probe increases, so that they would eventually collide (be at the same radial coordinate) if the frame were large enough. The EP let's us reach this conclusion even though GR disagrees.

In this thread I've seen you and others focus on predictions specific to GR, while (in your solutions at least) ignoring the laws of physics for an LIF. In the OP you asked "What gives?" The answer can't be the laws of physics in an LIF; the EP doesn't allow those to give, of course. A proposed solution to your thought experiment is invalid if it doesn't agree with the laws of physics for all LIFs as the EP implies it must. I suggest verifying that the behavior predicted by a proposed solution is exactly what is expected in an LIF falling toward a speck of dust in an otherwise gravity-free universe.

 

[kugbol]

so $dr/d\tau$ must be negative, i.e., $r$ must decrease.

Agreed. But how is that compatible with the fact that in the thought experiment's LIF the second probe would (according to SR) approach the first probe, whose $r$ always increases? How can they be approaching each other in that LIF when the first probe moves away from the same thing the second probe moves toward?

 

[PAllen]

Agreed. But how is that compatible with the fact that in the thought experiment's LIF the second probe would (according to SR) approach the first probe, whose $r$ always increases? How can they be approaching each other in that LIF when the first probe moves away from the same thing the second probe moves toward?

An local inertial frame is limited in time as well as space. Note that the r coordinate does not correspond to a positional coordinate in any LIF. For a while, assuming it were possible to 'see' the horizon - a light like surface - you would see the second probe overtaking the first probe, and the light like surface overtaking it even faster. This is possible because free faller r is decreasing much faster than second probe r, so it is moving away, very rapidly, in LIF. However, before the horizon can overtake the outer probe, as the free faller moves farther from the horizon, the probes are far enough in space and time that an LIF model deviates from reality. Without doing a geometric optics calculation, the following is an educated guess: I would think that despite the probes being inertial, it would shift to look similar to the situation of an accelerating probe with a head start on light, and another accelerating probe (with faster apparent acceleration) emitted with a lag on light. Distance contraction would be so extreme that the infaller probably never sees divergence of the paths, but will soon determine that an LIF model starts to be inaccurate.

 

[PeterDonis]

How can the cloud not be splitting into two pieces (at least), so that GR is compliant with its EP, or how can the cloud necessarily be splitting in two and yet GR somehow stays compliant with its EP even though the cloud wouldn't need to be splitting in two in every LIF?

The cloud is *not* splitting within the LIF. That's the point. Within the LIF, the cloud behaves just like you'd expect it to, just like it behaves in any other LIF. It only starts splitting later, outside the LIF. The LIF only covers a small piece of spacetime; the splitting of the cloud occurs later, outside that piece.

 

[PeterDonis]

But how is that compatible with the fact that in the thought experiment's LIF the second probe would (according to SR) approach the first probe, whose $r$ always increases? How can they be approaching each other in that LIF when the first probe moves away from the same thing the second probe moves toward?

Have you read through this thread? In particular, have you read the multiple posts explicating the fact that "distance" relative to the coordinate r is not the same as distance relative to the LIF?

 

[bossman27]

I hope this isn't too elementary a question, but what exactly defines the validity of an LIF? Or rather, how does one know when a LIF is no longer a valid? I've read through this thread, as well as a few online sources, and all I can gather is that while in the most strict sense an LIF is only valid if made arbitrarily small, it's valid for "regions small enough that non-uniformities of the gravitational field are too small to measure." I guess a more specific question would be, at what point exactly does the astronaut's LIF no longer contain the two probes?

 

[PAllen]

I hope this isn't too elementary a question, but what exactly defines the validity of an LIF? Or rather, how does one know when a LIF is no longer a valid? I've read through this thread, as well as a few online sources, and all I can gather is that while in the most strict sense an LIF is only valid if made arbitrarily small, it's valid for "regions small enough that non-uniformities of the gravitational field are too small to measure." I guess a more specific question would be, at what point exactly does the astronaut's LIF no longer contain the two probes?

It is not a yes/no question. Given a proposed measurement, you can compute the error of assuming LIF. Then the question becomes what is your acceptable error. The error is only mathematically zero for a point. Beyond that, you get continuous growth of error. The growth of the error depends on the measurement. For curved spacetime, there are a few special types of measurements sensitive to second derivative of metric. For these, there is no size that will act as an LIF. For almost all measurements, the error is proportional to tidal effects at first, but then will become non-linear. Think of it as higher order terms of a Taylor expansion becoming dominant.

 

[PeterDonis]

I guess a more specific question would be, at what point exactly does the astronaut's LIF no longer contain the two probes?

Just to wrap up the "puzzle", I'm going to show the math on this. I'll start with the equation for "escape velocity" at a radius $r = 2M ( 1 + 2 \epsilon )$, where $\epsilon << 1$, i.e., just above the horizon. The reason for the factor of 2 in front of $\epsilon$ will be clear in a moment.

The equation for "escape velocity" is (note that I will be freely using the binomial expansion and discarding higher-order terms in $\epsilon$ where necessary):

$$v_e = \sqrt{\frac{2M}{r}} = \sqrt{\frac{2M}{2M (1 + 2 \epsilon )}} = \left( 1 + 2 \epsilon \right)^{-1/2} = 1 - \epsilon$$

So the factor of 2 above makes the expression for $v_e$ cleaner.

But what does $v_e$ mean, physically? It is the "escape velocity" at radius r, relative to a "static" observer at radius r. That means the escaping object is moving outward at $v_e$, relative to the static observer; but the astronaut, who is free-falling inward from rest at infinity, is moving *inward* at $v_e$. So the relative velocity of the escaping object and the static observer is given by the relativistic velocity addition formula:

$$v = \frac{v_e + v_e}{1 + v_e^2} = \frac{2 (1 - \epsilon)}{1 + \left( 1 - \epsilon \right)^2} = \frac{2 (1 - \epsilon)}{2 - 2 \epsilon + \epsilon^2} = \frac{1}{1 + \epsilon^2 / [ 2 ( 1 - \epsilon ) ]} = \frac{1}{1 + \epsilon^2 / 2} = 1 - \frac{1}{2} \epsilon^2$$

Now, an interesting point that I didn't raise before: the "puzzle" can actually be stated even more simply, and sharply, than I did in the OP. Here's how: in the astronaut's local inertial frame, the horizon is moving outward at the speed of light. The probe launched outward at escape velocity just above the horizon is moving outward at *less* than the speed of light. But if it's moving slower than the horizon, how can it possibly escape?

The answer, of course, is found by computing the time it would take for the horizon to catch up with the probe, with respect to the local inertial frame. To do that, we need to first compute how far the probe is from the astronaut at the instant the astronaut crosses the horizon. So we need the time with respect to the LIF between the probe's launch and the astronaut crossing the horizon. This is simple since we have the radial coordinate at which the probe is launched; it's just the standard formula for proper time to fall for a Painleve observer from radius $2M (1 + 2 \epsilon )$ to radius $2M$:

$$\tau = \frac{2}{3} [ \sqrt{\frac{r}{2M}} ( r ) - 2M ] = \frac{2}{3} [ \sqrt{1 + 2 \epsilon} ( 2M ) ( 1 + 2 \epsilon ) - 2M ] = \frac{4}{3} M [ ( 1 + \epsilon ) ( 1 + 2 \epsilon ) - 1 ] = \frac{4}{3} M ( 3 \epsilon) = 4 M \epsilon$$

The distance the probe moves in this time is then just $v \tau$, but since $v$ differs from 1 only by a term quadratic in $\epsilon$, then to the order of approximation we are using, the distance is simply $D = \tau = 4 M \epsilon$ (the correction term is cubic in $\epsilon$).

The time it will take for the horizon to catch up to the probe is then $T = D / (1 - v)$, the distance divided by the "closure speed", the difference of the probe's velocity and the horizon's velocity (which is 1). This gives

$$T = \frac{\tau}{1 - v} = 4 M \epsilon \frac{2}{\epsilon^2} = \frac{8 M}{\epsilon}$$

Since the horizon starts at time t = 0 in the LIF, and moves outward at speed 1, this will also be the distance from the LIF's origin at which the horizon would catch the probe, according to the LIF. And since $\epsilon << 1$, we can see that $T >> 8M$.

Now we need to compare this to the size of the LIF; i.e., we need to answer the question, at what distance (or time) from the origin of the LIF will tidal gravity become non-negligible? The easiest way to quantify this roughly (which, as we will see, is more than enough) is to look at the components of the Riemann curvature tensor at the origin of the LIF; they are all of order $M / r^3 = M / (2M)^3 = 1 / 8 M^2$. If we write an expression for the metric in the LIF:

$$g_{ab} = \eta_{ab} + O(x^2)$$

then the quadratic correction terms will be proportional to the Riemann tensor components; and since they are quadratic, the "distance" (in space or time) at which the corrections are of order unity is given by the inverse square root of the Riemann tensor components. So the acceptable size of the LIF (in space and time), within which the corrections to the metric due to curvature are negligible, is given by

$$\delta << \sqrt{8} M$$

But we saw above that the time for the horizon to catch up to the probe was $T >> 8 M$; and this is *much* greater (by two factors, so to speak) than the size of the LIF. So tidal gravity/spacetime curvature will become non-negligible long before the horizon would be able to catch the probe.

A final note: what happens once tidal gravity becomes non-negligible? One way to approach this is to ask, what if there were no tidal gravity? What would happen? Well, the horizon would catch up to the probe, meaning it would fall below the horizon (and then meet up with the second probe, which was launched from below the horizon). Tidal gravity basically pulls the horizon down so it can't catch the probe. (More precisely, tidal gravity means the hole pulls the horizon down faster than it pulls the escaping probe down, because of their initial spatial separation.) So the horizon starts out catching up to the probe, but tidal gravity causes it to stop catching up and start receding from the probe.

(A good exercise, btw, is to re-do the above analysis in the LIF of the first probe.)

 

[kugbol]

An local inertial frame is limited in time as well as space. Note that the r coordinate does not correspond to a positional coordinate in any LIF. For a while, assuming it were possible to 'see' the horizon - a light like surface - you would see the second probe overtaking the first probe, and the light like surface overtaking it even faster. This is possible because free faller r is decreasing much faster than second probe r, so it is moving away, very rapidly, in LIF. However, before the horizon can overtake the outer probe, as the free faller moves farther from the horizon, the probes are far enough in space and time that an LIF model deviates from reality.

This doesn't seem to pass a logic test. First we eliminate any issues related to the free faller being too far from the probes. Use a technique common to SR texts to simplify such thought experiments. In the LIF employ a string of observers at rest with respect to the astronaut's frame, each with their own measuring equipment. Let all measurements be taken by these observers when the probes pass right by them, so that speed of light delays and other issues with taking measurements from afar are eliminated. The observers record their measurements, and when the experiment is over the measurements are collected and analyzed to draw conclusions. (The LIF can possibly remain valid for years as measured on the observers' wristwatches, given a sufficiently massive black hole.)

Let the second probe be launched right at the horizon (so it straddles the horizon initially). Let the upper end (the end furthest away from the black hole) of the second probe be further above the horizon than the upper end of the first probe. (We let the second probe be longer than the first probe.) Now, according to any of the string of observers taking measurements directly, the second probe cannot possibly be overtaking the first probe, or else the second probe would be passing outward through the horizon and escaping, which of course GR doesn't allow. GR only allows the second probe to fall inward toward the center of the black hole. Whereas the first probe is moving outward, away from the black hole. GR allows only one way in which these probes can be moving relative to each other. Whereas in an LIF there are 3 ways allowed: second probe overtaking (in the outward direction) first probe, first probe overtaking second probe, probes at rest with respect to each other.

The practicalities of launching the probes can be ignored. We are free to imagine the probes already in free fall at the start of the thought experiment. Let the first probe be a particle and the second probe a rod, with the first probe above the horizon and escaping, alongside the second probe which is straddling the horizon. We are free to imagine this because it's all valid in an LIF. Nothing that's valid in an LIF can be denied by GR.

The consensus in this thread as to the solution to the OP's thought experiment seems to be that the LIF becomes invalid before a paradox can arise. We see above, however, that the thought experiment concludes instantly, in a single moment as measured in the LIF, with everything measurable by observers (who are at rest in the astronaut's frame) right next to the probes. What am I missing? (Using worded logic only please, since that can be both scientifically valid and simpler.)

 

[PeterDonis]

Let the second probe be launched right at the horizon (so it straddles the horizon initially).

But it cannot be "at rest" relative to the horizon; the horizon is moving outward at the speed of light in the LIF. So no matter what velocity you give this probe in the LIF, the horizon will be moving faster.

Let the upper end (the end furthest away from the black hole) of the second probe be further above the horizon than the upper end of the first probe. (We let the second probe be longer than the first probe.)

When is this supposed to be the case? The first probe has already been launched outward. Do you mean that at the instant, in the LIF, that the second probe gets launched, its upper end is still further above the horizon than the upper end of the first probe, even though the latter is already moving outward (at close to the speed of light, since it has to be moving at escape velocity)?

Now, according to any of the string of observers taking measurements directly, the second probe cannot possibly be overtaking the first probe, or else the second probe would be passing outward through the horizon and escaping, which of course GR doesn't allow.

Nope, this is not correct. I assume you intend the second probe to be launched outward with a greater velocity, relative to the LIF, than the first, right? If that is the case, then the second probe's upper end will move *away* from the first probe's upper end (assuming, as I did above, that the first probe's upper end is still below the second probe's upper end when the second probe is launched). And all of your string of observers will make observations that show that.

However, the above will only be true within the LIF, and the LIF is pretty small; under the conditions you have given, the size of the LIF can't be much larger than the length of the second probe. So within a very short time, the probes will leave the LIF, and the laws of SR can no longer be used after that point to predict what happens to them. The laws of GR say that at some point after the two probes leave the LIF, the second probe's upper end will stop moving away from the first probe's upper end; it will fall back, and the first probe's upper end will pass it. Soon after that, the lower end of the first probe will pass the upper end of the second; and soon after *that*, the upper end of the second probe will fall below the horizon.

GR only allows the second probe to fall inward toward the center of the black hole. Whereas the first probe is moving outward, away from the black hole.

This is true if the words "inward" and "outward" are defined relative to the global radial coordinate r. But the spatial coordinate in the LIF is *not* r. So it's perfectly possible for everything I said above to be true, *and* for what you said in the above quote to be true as well.

However, this does bring up an important point which wasn't raised in my original version of the puzzle. The probes in my original version were idealized as point particles, so they only have a single "spatial position" at a given time in any frame. Your probes are extended objects, so they occupy a range of spatial positions; and the relationship between spatial position in the LIF and the global radial coordinate r *changes* within the LIF. What you stated in the quote above is, strictly speaking, true only for the centers of mass of the probes: the second probe's CoM has decreasing r, while the first probe's CoM has increasing r.

However, because of the probes' spatial extension, the second probe's upper end starts out with *increasing* r, not decreasing r; that's how it can move away from the first probe's upper end. This only lasts while the second probe remains within the LIF, though. Once it exits the LIF, as I said above, its upper end will at some point reverse direction relative to the first probe's upper end.

Note also that, once the second probe exits the LIF, tidal gravity is no longer negligible. That means the second probe will be stretched by this process. The stretching will not be detectable within the LIF, but it will be once the probe exits the LIF.

One final note: a key point that was identified in the discussion in this thread is that radial "distance" with respect to the r coordinate is not a well-defined "proper distance". This is another reason why your assumptions about what is implied by the behavior of the r coordinates of the probes' upper ends are not valid.

GR allows only one way in which these probes can be moving relative to each other.

This is incorrect. GR allows all the ways that are allowed in the LIF. But the relationship between the LIF's spatial coordinate and the global radial coordinate r is not as simple as you think it is; see above.

 

[Passionflower]

So the relative velocity of the escaping object and the static observer is given by the relativistic velocity addition formula:

$$v = \frac{v_e + v_e}{1 + v_e^2} = \frac{2 (1 - \epsilon)}{1 + \left( 1 - \epsilon \right)^2} = \frac{2 (1 - \epsilon)}{2 - 2 \epsilon + \epsilon^2} = \frac{1}{1 + \epsilon^2 / [ 2 ( 1 - \epsilon ) ]} = \frac{1}{1 + \epsilon^2 / 2} = 1 - \frac{1}{2} \epsilon^2$$

I suppose you mean 'the astronaut' instead of the static observer?

in the astronaut's local inertial frame, the horizon is moving outward at the speed of light.

Do you say that is true only when the astronaut exactly crosses the horizon or at all times?

 

[PeterDonis]

I suppose you mean 'the astronaut' instead of the static observer?

Oops, yes, the relative velocity with the $\epsilon^2$ in it is the relative velocity of the astronaut (free-falling into the hole) and the first probe (shot outward at escape velocity just above the horizon). Thanks for catching that!

Do you say that is true only when the astronaut exactly crosses the horizon or at all times?

Strictly speaking, we can only assign an invariant meaning to "the velocity of the horizon relative to the astronaut" at the event where the two curves, astronaut worldline and horizon, cross. For practical purposes, though, we can say that the horizon is moving outward at c in any local inertial frame that includes the horizon, within the extent of that LIF.

 

[Passionflower]

It seems to me that from the perspective of the astronaut after he crosses the horizon the absolute horizon retreats faster than light. The effect is similar to the metric expansion of space in inflation models but the difference is that the effect is much larger closer to the astronaut because the curvature is larger here. It seems logical that the two probes seem to converge from the astronaut's perspective.

Consider: How does the whole universe look when the astronaut looks back after he passed the physical horizon?

 

[PeterDonis]

It seems to me that from the perspective of the astronaut after he crosses the horizon the absolute horizon retreats faster than light.

There may be a coordinate chart that would indicate this, but there's no invariant way to specify it that I can see.

The effect is similar to the metric expansion of space in inflation models

I'm not sure I see the similarity. It's true that Schwarzschild spacetime below the horizon is not static, but what family of observers would correspond to the "comoving" observers in an FRW spacetime, who see the universe as expanding?

the effect is much larger closer to the astronaut because the curvature is larger here.

Not necessarily; you can make the curvature at the horizon as small as you like by making the black hole's mass large enough.

Consider: How does the whole universe look when the astronaut looks back after he passed the physical horizon?

If the astronaut is a Painleve observer, IIRC the universe would look somewhat redshifted.

 

[Passionflower]

I'm not sure I see the similarity. It's true that Schwarzschild spacetime below the horizon is not static, but what family of observers would correspond to the "comoving" observers in an FRW spacetime, who see the universe as expanding?

It is obviously not exactly the same as an FRW spacetime however would you disagree that the event horizon for the astronaut acts like a cosmic event horizon? After all the astronaut once passed can never reach the event horizon again let alone pass it. And if you would agree with this notion how but attribute it to curvature of space. The curvature is obviously not identical to the curvature of a FRW spacetime.

Not necessarily; you can make the curvature at the horizon as small as you like by making the black hole's mass large enough.

True.

If the astronaut is a Painleve observer, IIRC the universe would look somewhat redshifted.

It would gradually get redder and redder, also similar to the metric expansion of space where stars 'slipping' beyond the cosmic event horizon leave a faint red print behind.

Interestingly if one decomposes the Doppler effect into a velocity and gravitational component for the astronaut then outside of the event horizon the gravitational and velocity shift effects increase (resp. blue shift and red shift), outside the event horizon the velocity based redshift is stronger and eventually at the event horizon the resulting shift becomes exactly 0.5.

 

[PeterDonis]

It is obviously not exactly the same as an FRW spacetime however would you disagree that the event horizon for the astronaut acts like a cosmic event horizon. After all the astronaut once passed can never reach the event horizon again let alone pass it.

Usually the cosmic event horizon is described according to how observers "outside" it see things (galaxies that pass beyond the horizon are no longer visible); but yes, this way of looking at it applies in both cases.

 

[kugbol]

The laws of GR say that at some point after the two probes leave the LIF, the second probe's upper end will stop moving away from the first probe's upper end; it will fall back, and the first probe's upper end will pass it. Soon after that, the lower end of the first probe will pass the upper end of the second; and soon after *that*, the upper end of the second probe will fall below the horizon.

... the words "inward" and "outward" are defined relative to the global radial coordinate r. But the spatial coordinate in the LIF is *not* r.

I say your first point is false even though your second is true. What's happening in terms of inward or outward relative to the global radial coordinate r affects what can possibly be measured in the LIF.

Usage of black holes, launching at relativistic velocities, etc. make this harder to see. A more mundane experiment can have all the important qualities to prove my point.

You're in a skyscraper's elevator car with snapped cabling, free-falling toward the ground. Within your car is a free-falling radially-oriented ruler, its downward end marked zero. There's a free-falling particle initially right next to the ruler. The ruler is moving directly toward the ground, that's a given. It's also given that the particle is moving directly toward the building's rooftop. The parameters I've specified enforce that there is only one way in which you could measure the particle moving relative to the ruler: toward a higher number on it. That's the conclusion no matter the ruler's velocity as you measure it. It also doesn't matter that you don't know which direction is upward or downward. What you can possibly measure in your LIF is restricted by the parameters given relative to that global system.

Now keep the experiment the same except: Let the elevator car be falling through the horizon of a black hole. Unlike in the skyscraper version the ruler's movement relative to the global radial coordinate r (like downward or inward) isn't specified. Let the ruler be straddling the horizon initially. Let the particle be above the horizon and escaping to infinity. Given the parameters I've specified, there is only one way in which which you could measure the particle moving relative to the ruler: toward a higher number on it. Again the velocity of the ruler as you measure it doesn't matter, nor does your inability to discern upward or downward. When we let the ruler be straddling the horizon initially, GR demands that it be falling downward, like the ruler in the skyscraper's car. Merely specifying the ruler's location within your LIF restricted its movement relative to the global radial coordinate r, in turn restricting what you can possibly measure in your LIF.

Note that both experiments need last only an arbitrarily short time elapsed on your wristwatch, so that your LIF remains valid. To tie back to your point, my elevator car experiment shows that the second probe's upper end can never be measured to be moving away from the first probe's upper end.

It seems there's indeed a paradox within GR. (To be clear, the paradox is that GR also implies by way of its principle of equivalence that in the black hole example it must be possible in principle for you to measure the particle moving toward a lower number on the ruler. So GR seems to contradict itself.) The thought experiment you originally proposed, with a few simplifications, is highly intriguing! Where did I go wrong?

 

[PeterDonis]

I say your first point is false even though your second is true. What's happening in terms of inward or outward relative to the global radial coordinate r affects what can possibly be measured in the LIF.

No it doesn't, at least not the way you mean. You are ignoring a key point: curves of constant r (and constant theta, phi, but we're leaving out the angular coordinates for this discussion) are timelike outside the horizon, but they are spacelike inside the horizon (and the horizon itself is a null curve of constant r which marks the boundary between these two regions).

All of your intuitive arguments assume that curves of constant r are timelike, so they don't apply at or inside the horizon. The LIF we are discussing includes a portion that is inside the horizon, so your arguments don't apply to it.

The parameters I've specified enforce that there is only one way in which you could measure the particle moving relative to the ruler: toward a higher number on it. That's the conclusion no matter the ruler's velocity as you measure it.

This argument implicitly assumes that curves of constant height are timelike. You may not realize that, but it's true. Think about what would happen if a curve of constant height were null instead--for example, the path of a light beam.

Now keep the experiment the same except: Let the elevator car be falling through the horizon of a black hole. Unlike in the skyscraper version the ruler's movement relative to the global radial coordinate r (like downward or inward) isn't specified.

Yes it is, because you specify that the ruler is straddling the horizon. That means it *must* be moving inward. It's impossible for an object straddling the horizon to be at a constant or increasing r, even for an instant; that would require the object to be moving faster than light. Its r *must* be decreasing.

It seems there's indeed a paradox within GR.

Nope, just an error in your reasoning.

Where did I go wrong?

See above.

 

[PeterDonis]

All of your intuitive arguments assume that curves of constant r are timelike

Actually, on thinking it over some more, they assume even more than that. They assume that curves of constant r are "parallel", so to speak, to curves of constant spatial coordinate x in the LIF. That's not true either for the case of an LIF that straddles the horizon.

I think I've suggested before that to really understand what's going on in the LIF straddling the horizon, one needs to draw a spacetime diagram of it, with all the curves of interest (the horizon itself, curves of constant r, and the worldlines of all objects in the scenario) drawn correctly in it. If you do so, and then draw a spacetime diagram of your "skyscraper" scenario as well, you will see that they differ in some key respects.

 

[PeterDonis]

GR also implies by way of its principle of equivalence that in the black hole example it must be possible in principle for you to measure the particle moving toward a lower number on the ruler.

This is correct; but I should point out that the "paradox" you are pointing out (which isn't really a paradox, just a puzzle) actually has nothing to do with the equivalence principle. The puzzle has to do with the relationship between the LIF and the global r coordinate; but the EP doesn't say anything about how an LIF has to match up with a global coordinate, whether it's r or anything else. The EP only says that if I run an experiment with the same initial conditions in any LIF, I will get the same results; and that is true for the puzzle scenario. The difference between the LIF straddling the horizon and an LIF such as the one in your skyscraper example is purely in the relationship between the LIF and a global coordinate, which has nothing to do with the EP.

 

[someGorilla]

I had missed this one. I wanted to try myself at an answer but I read the first few answers on the first page of the thread and as far as I understand the answer is supposed to be limited to a local zone small enough to be approximately flat. So it's not an answer like "the projected collision point it outside of the validity of the Minkowski approximation". In this case I don't understand what exactly is being asked...

 

[someGorilla]

Having read till the beginning of the second page (yep I know I shouldn't have ) I understand the question to be "how can the two paths be diverging in one frame and converging in the other?"

It's a trick question!
I hand't got it at first since it's not so strange that two lines on two different coordinate charts in a curved spacetime form different angles between them. I say "in a curved spacetime" because if one insists in seeing the difference, one is deliberately considering the curvature.
In the locally flat reference frame of the falling astronaut, the two paths converge very slightly (the angle between the worldlines on a spacetime chart is very small).
In the other coordinates (Schwarzschild?) the two paths diverge very slightly (the angle between the worldlines is very small).
This very small difference is ascertainable only if the approximation is not so loose as to really make the local spacetime flat. With "local" I mean any zone large enough to contain both launch events.
So the trick is that the "paradox" statement pretends it's dealing with a flat spacetime, and then poses a question that rests on such a precise distinction as to reveal that it's not exactly flat.

Am I failing to see something else?

 

[kugbol]

I understand the question to be "how can the two paths be diverging in one frame and converging in the other?"

Yes.

Am I failing to see something else?

As the OP says, tidal gravity is *not* the answer. The spacetime can be treated as perfectly flat; the curvature is completely ignorable.

I say "in a curved spacetime" because if one insists in seeing the difference, one is deliberately considering the curvature.

GR implies that the curvature can indeed be ignored. No matter how flat we make the spacetime in the astronaut's frame by making the black hole more massive, the paradox remains.

Above I simplified the paradox to be "how can a particle move toward a higher number on a ruler in one frame and move toward a lower number on the same ruler in the other?" This shows that the solution to the paradox cannot be that the curvature cannot be ignored. The paradox would remain even if you considered the curvature.

 

[PeterDonis]

"how can a particle move toward a higher number on a ruler in one frame and move toward a lower number on the same ruler in the other?"

I responded to this already, but just to be clear, the original puzzle dealt only with point-like objects that occupy a single spatial coordinate. Your modified "ruler" version uses extended objects, and that brings in some issues that don't arise in the original puzzle.

This shows that the solution to the paradox cannot be that the curvature cannot be ignored.

I think it's also worth clarifying this point. First, with respect to the original puzzle: the "cleanest" way of posing the original puzzle, I think, is the way I did in post #51: in the LIF of an astronaut falling through the horizon, a probe launched radially outward at escape velocity at some r just slightly above the horizon is moving slower than the horizon (since the horizon is moving outward at the speed of light): so how can the probe possibly escape?

The answer to this has two parts, and we can only really ignore curvature in the first part. The first part, which is what I had in mind when I first posed the puzzle, is to understand that the r coordinate is not the same as "radial distance", so the fact that the probe launched at escape velocity has a strictly increasing r coordinate from the instant it's launched is not in any way inconsistent with the fact that, within the LIF of the infalling astronaut, the probe is moving outward slower than the horizon is. (The best way I know of to visualize this is to draw lines of constant r in a spacetime diagram of the LIF: note that they are *not* straight lines.)

The second part of the answer is the calculation I did in post #51, which shows that the LIF of the infalling astronaut is much too small to allow the horizon to actually catch up with the outgoing probe. But as soon as we consider a region of spacetime larger than a single LIF, we can't ignore curvature; and of course the full answer to why the probe escapes requires us to consider curvature, since it's curvature that makes the horizon stop catching up with the outgoing probe, and then start falling away from it.

With respect to the "ruler" version of the puzzle, as I noted in post #53, if the "ruler" is to be long enough that the outgoing (escaping) probe is still below the ruler's upper end when the ruler is launched, the ruler must be about the same size as the LIF. That means the ruler will leave the LIF almost immediately. For the very short time that the ruler is within the LIF, as I noted in post #53, it is perfectly possible for the ruler's upper end to be moving outward faster than the escaping probe, so that the escaping probe moves "downward" on the ruler.

(Yes, this implies that the second ruler's upper end has an increasing r coordinate, even though its center of mass has a decreasing r coordinate, and even though, within the LIF, tidal gravity is negligible so the ruler is not being stretched. Once again, the r coordinate is *not* the same as radial distance--indeed, at and inside the horizon the r coordinate is not even timelike. But once the ruler exits the LIF, it can be stretched by tidal gravity, and probably will be; its exact behavior will depend on its material constitution and the internal forces it can sustain. But in any case, the full explanation of how the ruler's motion relative to the escaping probe changes, so the probe stops moving "downward" on the ruler and starts moving "upward" until it is completely above the ruler, once again requires taking curvature into account.)

 

[kugbol]

All of your intuitive arguments assume that curves of constant r are timelike, so they don't apply at or inside the horizon. The LIF we are discussing includes a portion that is inside the horizon, so your arguments don't apply to it.

I relied only on GR's prediction that the ruler must be falling downward, and the EP. We agree on both of those. My arguments apply.

Yes it is, because you specify that the ruler is straddling the horizon. That means it *must* be moving inward. It's impossible for an object straddling the horizon to be at a constant or increasing r, even for an instant; that would require the object to be moving faster than light. Its r *must* be decreasing.

I specified that the ruler is straddling the horizon, and hence GR required it to be moving inward. It's an important point that my thought experiment doesn't specify that it's moving inward, for it means that if a paradox arises it wasn't a problem with my experiment. If GR restricts the movement of the ruler such that a paradox arises it's not my problem.

I think I've suggested before that to really understand what's going on in the LIF straddling the horizon, one needs to draw a spacetime diagram of it, with all the curves of interest (the horizon itself, curves of constant r, and the worldlines of all objects in the scenario) drawn correctly in it.

The EP tells me that I don't need to do this. I can treat this paradox as an SR problem only, with one requirement demanded by GR: the second probe or ruler must be moving inward.

We can't have it both ways. It can't be, as you say in the OP, that tidal gravity isn't the answer, and yet I must go outside of SR (beyond accepting that one requirement) to see the answer. We can't both ignore and not ignore the curvature.

The EP only says that if I run an experiment with the same initial conditions in any LIF, I will get the same results; and that is true for the puzzle scenario. The difference between the LIF straddling the horizon and an LIF such as the one in your skyscraper example is purely in the relationship between the LIF and a global coordinate, which has nothing to do with the EP.

We agree on what the EP says. I depended on that to determine that the result of my experiment in the skyscraper must be the same result as in the LIF straddling the horizon. The finding in both situations was this: there is only one way in which which you could measure the particle moving relative to the ruler: toward a higher number on it. Which left the paradox I noted. The paradox isn't resolved by your second statement here; I validly relied on what the EP says.

 

[PeterDonis]

I specified that the ruler is straddling the horizon, and hence GR required it to be moving inward. It's an important point that my thought experiment doesn't specify that it's moving inward, for it means that if a paradox arises it wasn't a problem with my experiment. If GR restricts the movement of the ruler such that a paradox arises it's not my problem.

No, you've got it backwards. You specified a LIF in which the ruler is launched with its center of mass at the origin, in such a way that its upper end is moving outward (i.e., in the positive x direction in the LIF) faster than a probe that was launched outward a very short time before. GR does not restrict you at all; you can specify *any* motion you like that works in an inertial frame. The fact that the ruler is "moving inward" relative to the global r coordinate (more precisely, that its center of mass is) has nothing to do with the LIF, and does not affect any physical predictions that are made using the LIF.

The only thing GR tells you that applies to the LIF is that, since you specified that the second probe "straddles the horizon" (for simplicity I interpret this to mean that the ruler center of mass is exactly on the horizon at the instant it is launched), the horizon appears in the LIF as an outgoing light ray that passes through the horizon. Both the ruler and the probe move relative to the horizon, within the LIF, exactly as they would in any inertial frame relative to a light ray passing through the origin in the positive x direction.

The EP tells me that I don't need to do this. I can treat this paradox as an SR problem only, with one requirement demanded by GR: the second probe or ruler must be moving inward.

"Moving inward" doesn't mean what you think it means. See above.

We can't have it both ways. It can't be, as you say in the OP, that tidal gravity isn't the answer, and yet I must go outside of SR (beyond accepting that one requirement) to see the answer. We can't both ignore and not ignore the curvature.

Read my posts more carefully. I explained in detail that there are two parts to the full answer to the puzzle, and that you can ignore curvature only in the first part. The second part doesn't have anything to do with what happens within the LIF, so it's irrelevant to the argument you're making anyway. I've also said repeatedly that, as far as anything that happens within the LIF is concerned, you *can* ignore curvature. So I agree with you that you do not have to "go outside SR" to explain what happens within the LIF.

The only thing you are mistaken about, within the LIF, is what "moving inward" relative to the global r coordinate means. That has nothing to do with curvature; it has to do, as I said in what you quoted, with correctly drawing the lines of constant global r coordinate in a spacetime diagram of the LIF. Have you tried to do that? What do you think they look like?

The finding in both situations was this: there is only one way in which which you could measure the particle moving relative to the ruler: toward a higher number on it.

No, that's not what I understood you to be saying. You specified (at least, I think you did--certainly what I'm about to say is what you would need to specify if you want your ruler scenario to be relevant to my puzzle) that the ruler, within the LIF, is moving in the positive x direction *faster* than the particle. That means that, initially, within the LIF, the particle is moving towards a *lower* number on the ruler. Otherwise there's no puzzle: if you specify initially, within the LIF, that the particle is moving towards a higher number on the ruler, then of course it's going to outrun the ruler. GR let's you specify things either way, but you have to pick one way or the other: you can't have both be true at the same time, and GR doesn't claim you can. Nor does it claim that moving towards a lower number on the ruler must correspond to moving towards a lower global r coordinate, or that moving towards a higher global r coordinate must correspond to moving towards a higher number on the ruler; see above.

Here's what you should, IMO, be specifying in the skyscraper scenario: a ruler free-falling with its center of mass, at some instant, at the origin of a local inertial frame. A particle at the same instant at some small positive x-coordinate e; the length of the ruler is a bit larger than 2e, so the particle is closer to the origin than the ruler's "upper" end (meaning the end in the positive x direction). The particle is initially moving towards a lower number on the ruler, because the ruler is moving in the positive x direction faster than the particle is. Question: is it ever possible for the particle to change direction relative to the ruler (to start moving towards a higher rather than a lower number)? Answer: not within the LIF; but that's also true of the LIF straddling the horizon with the same initial conditions.

However, over a region of spacetime larger than a single LIF, where tidal gravity can come into play, it is perfectly possible for a "skyscraper" scenario to be set up such that the ruler's center of mass will eventually be falling faster than the particle is, and once that's the case, the particle will eventually have to start moving towards a larger number on the ruler. Of course, in such a scenario, the relationship between the ruler and particle positions relative to each other, and some global coordinate (whether it's r or anything else) will probably be different than they are in the horizon-straddling scenario: but that has nothing to do with the equivalence principle, since the EP says nothing about how the coordinates in a LIF are related to global coordinates. How can it? The whole point is that LIFs look the same even in spacetimes that globally look very different.

 

[kugbol]

I responded to this already, but just to be clear, the original puzzle dealt only with point-like objects that occupy a single spatial coordinate. Your modified "ruler" version uses extended objects, and that brings in some issues that don't arise in the original puzzle.

I modified your puzzle to make it easier to visualize. We're able to ignore tidal gravity here as you say in the OP, so it's an SR puzzle, the EP tells us. There is no problem with using extended free-falling objects in an SR puzzle; a particle is as good as a ruler. The question as I restated it remains valid.

(That said, I think the blog author's experiment is the cleanest and easiest to visualize of all. Either the particles are all moving in formation or they're not. No extended objects are employed.)

(Yes, this implies that the second ruler's upper end has an increasing r coordinate, even though its center of mass has a decreasing r coordinate, and even though, within the LIF, tidal gravity is negligible so the ruler is not being stretched. ...

If part of a radially-oriented ruler has a decreasing r coordinate then all of it does, when it is not being stretched.

Once again, the r coordinate is *not* the same as radial distance--indeed, at and inside the horizon the r coordinate is not even timelike.

Even though an r coordinate is not the same as radial distance, still an increasing r coordinate always means "moving further from" the center of the object in question, while a decreasing r coordinate always means "moving closer to". Regardless of its location relative to the horizon, part of the ruler cannot be moving closer to the center of the black hole, while another part of the ruler moves away from the black hole, all while the ruler is not being stretched.

The ruler has a constant velocity in the astronaut's frame, and the spacing of the r coordinates are fixed in the frame. Given these facts it's impossible to set up the situation you describe. If you were right you could take two rulers and slide them alongside each other such that a marking on ruler A moves to a lower number on ruler B, while another marking on ruler A moves to a higher number on ruler B, with neither ruler being stretched.

But once the ruler exits the LIF, ...

That's after my thought experiment concludes. My experiments lasts an arbitrarily short time. The paradox arises before the ruler exits the LIF. Nothing that happens after the experiment concludes need be considered.

I simplified your puzzle to remove the extraneous launching part. I simplified it to show the paradox in an arbitrarily short time interval as measured in the astronaut's frame. I pointed out that local observers (at rest in the astronaut's frame) can be employed to make measurements right next to the particle and ruler. A single observer local to the particle need only make a single measurement as to how the particle moves relative to the local part of the ruler. There is no need to consider the ends of the ruler, or what happens before or after that single measurement. I do understand why you mention the afterwards part. It's because you think part of the ruler can have an increasing r coordinate, when it's obvious the whole ruler must eventually dip below the horizon or else there's no denying it's stretching. But the increasing r coordinate claim is impossible for you to prove.