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

[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.

 

[Passionflower]

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".

Why do you think that is the case?

A coordinate can present different things at different regions of the chart.

 

[kugbol]

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.

The ruler wasn't launched. I didn't specify its velocity. I specified only its location, in that it straddles the horizon initially; we agree it then must be moving inward as required by GR. I didn't specify that its upper end is moving outward. I concluded that the particle must be moving toward a higher number on the ruler.

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.

Then quote something from my thought experiment and show what's wrong with it. It met the definition of a sound argument. Your finding here contradicts the conclusion of my experiment. In fact it contradicts the finding of a physical experiment, the gold standard in physics. It's easy for me to do the elevator / skyscraper experiment using a ruler, a coin, and a high chair. The EP says the result of my experiment is what's expected for the same experiment in the astronaut's frame, as long as the ruler is likewise moving downward, which is demanded by GR when I let the ruler be straddling the horizon.

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.

This logic doesn't solve the paradox I noted. The paradox involves the ruler and the particle (or probe). If the astronaut's frame were an LIF then the particle could in principle be moving toward a lower number on the ruler. But GR allows the particle to move only to a higher number on the ruler. I don't deny that GR says what you say here; nevertheless the paradox remains intact. It appears to be a contradiction within GR.

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

Moving inward means that all parts of the ruler (whose lower end is marked zero) move to a lower r coordinate. Meanwhile the escaping particle's r coordinate ever increases. Therefore it can move only toward a higher number on the ruler.

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?

Spacetime diagrams are overkill for such a simple thought experiment. When the elevator car free falls downward in the skyscraper the r coordinates are simply the floors passing by. The r coordinates have fixed spacing within any inertial frame. There's no way part of the ruler in the elevator car can be moving to a lower floor, while another part of the ruler moves to a higher floor, while the ruler is radially oriented and isn't being stretched. The same idea applies within an elevator car (absent the skyscraper) falling through the horizon of a black hole.

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.

But I didn't specify that, which is important. I simplified the paradox to make it easier to see. I concluded that when the ruler is falling downward (as specified in the skyscraper) it's impossible for it to be moving "in the positive x direction" faster than the particle, which is moving upward. This is easy to visualize in the skyscraper, and easy to confirm by physical experiment using a ruler, coin, and high chair. Then I used the EP to determine that the same result must occur in the astronaut's frame. There's nothing unique about the astronaut's frame in regards to fixed spacing of r coordinates passing by.

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.

Yes, it's obvious in the elevator car in the skyscraper. I intentionally made that obvious. I didn't specify that the particle is moving towards a higher number on the ruler; that was a conclusion. I did specify that the ruler is falling downward whereas the particle is moving upward. What you're missing is, when I let that same elevator car be falling through the horizon of a black hole, I didn't specify that the ruler in it is moving inward or downward. Instead I specified only its location, namely that it straddles the horizon. GR then demanded that it be moving inward or downward, so that the EP guarantees that the result is the same as in my skyscraper experiment: the particle can move only toward a higher number on the ruler. If the astronaut's frame were a valid LIF, the same as any other, specifying only the location of the ruler and not its velocity would not guarantee that same result. Instead the particle could in principle be moving toward a lower number on the ruler. Hence there is a paradox. GR demanded that the result of my experiments in each of two LIFs be the same even though I specified a different set of initial conditions.

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.

Sure, but I didn't do that.

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. ...

I read all this but don't think it makes the thought experiment simpler. It seems the only way mine could be simpler is to do the blog author's method. I've removed every extraneous feature from your puzzle. It now concludes within an arbitrarily short time as measured by the astronaut. Our whole disagreement boils down to whether part of the ruler can have a decreasing r coordinate while another part of it has an increasing r coordinate. I say it can't. It obviously can't when fully above the horizon, and I say it can't even when straddling the horizon.

(The blog author's method eliminates our disagreement. In that method, some of a cloud of particles are below the horizon while at least one particle above the horizon is escaping. It's then obvious that they can't all be moving in formation, all moving outward, as could be the case in an LIF when the velocity of only one of the particles is specified.)

... 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.

Yes, and yet my conclusion for the elevator car in the skyscraper remains both obvious and valid. As you say, of course the particle's going to outrun the ruler. It's fine to specify that the ruler be moving downward relative to the ground. That affects the results of experiments in the elevator car even when the observer doesn't know anything about "downward". The EP let's us do an obvious thought experiment in the skyscraper, or even a physical experiment at home, and apply that to the astronaut's frame falling through the horizon. When I let the ruler straddle the horizon the EP ensures the result of the experiments will be the same, leaving a paradox.

 

[mfb]

I think it is not so easy to say "tidal gravity can be ignored" - it can be made arbitrarily small, but then you have to make the velocity difference small as well! Small velocity differences lead to a longer typical timescale, and therefore more time for tidal effects to act.
Well, I like Kruskal coordinates, where no paradox appears as the event horizon is not a coordinate singularity.

 

[PeterDonis]

I modified your puzzle to make it easier to visualize.

Ease of visualization may be misleading, because you're trying to visualize it in space instead of in spacetime. This encourages you to misinterpret the relationship between the global r coordinate and the LIF's x coordinate. The blog post author does the same thing; he even draws a spatial diagram that makes the mistake.

There is no problem with using extended free-falling objects in an SR puzzle; a particle is as good as a ruler.

As long as the ruler fits within the size of the LIF, sure. But if the ruler is almost as large as the LIF (as it is in your version of the puzzle), the whole experiment within the LIF can only last a little longer than a single instant.

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

No, this is not correct, and it is a key mistake you continue to make even though I have pointed it out. Once again, have you actually drawn a spacetime diagram of the LIF and drawn how lines of constant r look in it? They are *not* straight lines; and they are certainly not *vertical* straight lines (i.e., they are not lines of constant x coordinate in the LIF). Either do as I'm asking you or stop making this incorrect claim.

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"

No, this is not correct. The fact that it happens to be true in your skyscraper scenario does not mean it will be true in all scenarios. The r coordinate is a global coordinate, not a local one, and the EP says nothing about how a global coordinate has to relate to local coordinates within an LIF. At the horizon of a black hole, the relationship between the global r coordinate and the local coordinates within an LIF is very different from what you are used to. Here's a hint: the horizon itself, which is an outgoing null line (path of a light ray in the positive x direction) in the LIF of the puzzle, is a line of *constant* r.

The ruler has a constant velocity in the astronaut's frame

Within the LIF, yes.

and the spacing of the r coordinates are fixed in the frame

Wrong. They are not. (And the r coordinate is not the spatial coordinate in that frame anyway, as above.) Draw the spacetime diagram of the LIF, and draw the lines of constant r in it, and you will see.

The ruler wasn't launched. I didn't specify its velocity. I specified only its location, in that it straddles the horizon initially; we agree it then must be moving inward as required by GR. I didn't specify that its upper end is moving outward. I concluded that the particle must be moving toward a higher number on the ruler.

And it's perfectly possible to set up a scenario within the LIF straddling the horizon such that the particle *is* moving toward a higher number on the ruler. If that is your scenario, there's no puzzle at all. The only way there is even a puzzle to begin with is if the particle starts out moving towards a lower number on the ruler, even though it is moving outward with escape velocity. Wasn't that your intent? If not, then there's no puzzle at all, so what's the problem?

Then quote something from my thought experiment and show what's wrong with it.

I have, repeatedly, and I'll do it again, repeatedly, below: you are misunderstanding how the global r coordinate relates to local coordinates within the LIF. If you can't or won't fix this misunderstanding, there's not much to discuss.

If the astronaut's frame were an LIF then the particle could in principle be moving toward a lower number on the ruler.

And it can be.

But GR allows the particle to move only to a higher number on the ruler.

No, it doesn't. You are misunderstanding how the global r coordinate relates to the local coordinates within the LIF. It is perfectly possible for the particle to have strictly increasing r but still be moving, within the LIF, towards a lower number on the ruler. This implies, of course, that the upper end of the ruler itself also has strictly increasing r within the LIF, even though the ruler's center of mass has strictly decreasing r. That's also permitted by GR. If you correctly understood how the global r coordinate relates to the local coordinates within the LIF, you would see why there is no contradiction.

the paradox remains intact. It appears to be a contradiction within GR.

No, just an error in your understanding.

Moving inward means that all parts of the ruler (whose lower end is marked zero) move to a lower r coordinate.

No, it doesn't. See above.

the escaping particle's r coordinate ever increases. Therefore it can move only toward a higher number on the ruler.

Wrong. See above.

Spacetime diagrams are overkill for such a simple thought experiment.

In other words, you presume to say there is a contradiction within GR, when you won't even use the most basic tools of GR. That's like saying you've found an error in arithmetic while saying that writing down Arabic numerals is "overkill" for such a simple problem.

That said, drawing a spacetime diagram is not the only possible way for you to correct your misunderstanding about the r coordinate. You could also do it by writing out the math for the coordinate transformation between a global coordinate chart and the local coordinates within the LIF. If you prefer to do it that way, go ahead. I suggested the spacetime diagram because it seems easier to me to do it that way.

When the elevator car free falls downward in the skyscraper the r coordinates are simply the floors passing by.

Yes, but that's a special feature of the skyscraper scenario; it is *not* true in all scenarios. The relationship between the global r coordinate and the coordinates in an LIF is not always the same.

The r coordinates have fixed spacing within any inertial frame.

Wrong. See above.

But I didn't specify that

In which case there is no puzzle at all (see above), so what's the problem?

I concluded that when the ruler is falling downward (as specified in the skyscraper) it's impossible for it to be moving "in the positive x direction" faster than the particle, which is moving upward.

And this conclusion does not generalize, because it assumes that any object moving in the positive x direction must have an increasing r coordinate, which happens to be true in the skyscraper scenario, but is not always true.

This is easy to visualize in the skyscraper, and easy to confirm by physical experiment using a ruler, coin, and high chair. Then I used the EP to determine that the same result must occur in the astronaut's frame.

Which is incorrect, because the EP doesn't say anything about global coordinates; it only says something about the local coordinates within the LIF. And, as I've said many, many times now, your assumption that the two are always the same is incorrect; it happens to be true in the skyscraper scenario, but it's not always true.

There's nothing unique about the astronaut's frame in regards to fixed spacing of r coordinates passing by.

Yes, there is, because the r coordinate and the LIF's x coordinate are not always the same. See above.

I could go on, since there's still a fair portion of your most recent post left, but I don't see much point; it would just be repeating and repeating what I've already said.

 

[PeterDonis]

I think it is not so easy to say "tidal gravity can be ignored"

As I said in post #69 (and in previous posts, I think), it can't be ignored in the full explanation of why the horizon doesn't catch up to the probe moving outward at escape velocity; but it can be ignored in the explanation of how the probe can be moving slower than the horizon in the LIF even though the probe has a strictly increasing r coordinate.

Well, I like Kruskal coordinates, where no paradox appears as the event horizon is not a coordinate singularity.

Kruskal coordinates also make it a lot easier to visualize what's going on in a spacetime diagram of the LIF: you basically take a really small patch of the Kruskal diagram that's centered on a point on the horizon and enlarge it to a scale you can easily see.

 

[PeterDonis]

Well, kugbol, I thought there wasn't much point in responding to the rest of your post, but on re-reading the last part I found a few more things worth responding to.

Our whole disagreement boils down to whether part of the ruler can have a decreasing r coordinate while another part of it has an increasing r coordinate. I say it can't.

And GR says it can, so yes, this is a point of disagreement.

It obviously can't when fully above the horizon

Why not? Have you considered *all* possible LIFs above the horizon? Including ones that are moving towards the horizon at speeds very close to the speed of light? Have you actually computed what lines of constant r look like in such LIFs?

and I say it can't even when straddling the horizon.

Even if you were correct about LIFs above the horizon (which I don't think you are, see above), your conclusion would not extend to LIFs straddling the horizon, because curves of constant r are no longer timelike there, and your argument doesn't even make sense if curves of constant r are not timelike.

The blog author's method eliminates our disagreement. In that method, some of a cloud of particles are below the horizon while at least one particle above the horizon is escaping. It's then obvious that they can't all be moving in formation

No, it isn't. It is perfectly possible to set up an LIF straddling the horizon in which, within the LIF, all the cloud particles *are* moving in formation in the positive x direction, at any speed less than the speed of light. Once again, you are making an incorrect assumption about how the global r coordinate relates to the local x coordinate within the LIF.

 

[jwatts]

I don't know if this was already brought up so here it goes. In the problem you send out two probes away from the black hole (orthogonally I guess). The first one is sent out before crossing the horizon, the second one is sent out after. The "observation" is then that the second probe is moving near the speed of light away from you. However, inside the even horizon all paths converge to the singularity in space and time. Therefore the second probe could never be sent closer to the even horizon than where he is. The astronaut and the probe would be side by side, no matter what speed he tried to send it backwards at.

 

[PeterDonis]

I don't know if this was already brought up so here it goes.

It has, but not precisely in the words you give.

The "observation" is then that the second probe is moving near the speed of light away from you.

Correct. However, that does *not* mean the second probe is getting closer to the horizon. It isn't. See below.

However, inside the even horizon all paths converge to the singularity in space and time.

Eventually, yes.

Therefore the second probe could never be sent closer to the even horizon than where he is.

It can't get closer to the EH than the point at which it is launched. But the point at which it is launched is *not* "where he is", because the astronaut is falling inward towards the singularity too.

The astronaut and the probe would be side by side, no matter what speed he tried to send it backwards at.

No: relative to the astronaut, the second probe is indeed moving "outward" at almost the speed of light. But because the astronaut and the probe are inside the horizon, even an object moving "outward" at almost the speed of light still has a decreasing r coordinate. But the r coordinate of the probe is not decreasing as fast as the r coordinate of the astronaut; so relative to the astronaut the probe is still moving outward--its distance from the astronaut in the "outward" direction is increasing.

I should note that the above description might still be misleading, because it implies that the r coordinate is an ordinary "space" coordinate inside the horizon. It isn't, as I've already said several times in this thread in response to other posters. The direction of decreasing r inside the horizon is really the direction of future time; that's why I kept putting "outward" in quotes in the above. The direction in which the second probe is launched, relative to the astronaut, is *not* the direction of increasing r, because that direction points into the past.

The reason the direction is still called "outward", even though it's not the direction of increasing r inside the horizon, is that, if the astronaut carries a gyroscope with him, and points it directly radially outward when he is far away from the hole, for example by pointing it at some distant star that is directly overhead, then even after he crosses the horizon, the gyroscope will still be pointing directly at the same distant star, so the astronaut can use the gyroscope to determine the direction in which to launch the probe.

 

[Lino]

All, Apologies for "coming late to the party" - this was a very interesting thread and discussion. That been said, is the approach just over-kill? (PeterDonis, as the quote in your signature says, "Nature cannot be fooled", so) is there must be a simple solution / description, or am I missing the point of the puzzle?

How about, given that we are only considering the LIF of the free-faller, start by ignoring the BH and horizon. You get a static observer in empty space, launching two probes at different times and with different velocities - of course the second (faster) probe is going to start to close on the first (slower) probe. Re-introducing the BH and horizon (an accelerator and a tipping point), the first probe will escape to infinity so it's velocity (relative to the free-faller) will likewise tend toward infinity. On the other hand, the second probe cannot escape and its velocity (relative to the free-faller) will approach zero, where upon it starts to "fall" with the free-faller. Hence, (in the free-fallers LIF) the probes first appear to approach each other and then separate, so no paradox. (At this point, the answer could be enhanced by the introduction of calculations and diagrams to determine times / points / velocities at various stages.)

Regards,

Noel.

 

[PeterDonis]

That been said, is the approach just over-kill?

Apparently not, since when the scenario is stated in a simple, straightforward way, people raise all sorts of objections based on misconceptions about how the LIF of a free-faller crossing the horizon works.

That said, your "simple solution" has some errors in it too. See below.

You get a static observer in empty space, launching two probes at different times and with different velocities - of course the second (faster) probe is going to start to close on the first (slower) probe.

OK so far.

Re-introducing the BH and horizon (an accelerator and a tipping point)

I'm not sure this is a good way of viewing the BH and horizon. The important point about the BH is that it produces tidal gravity; the important point about the horizon is that, in the LIF of a free-faller crossing the horizon, the horizon itself is an outgoing light ray. The "acceleration due to gravity" produced by the BH (btw, it's questionable whether that concept even makes sense at or inside the horizon) doesn't come into the problem at all, and I'm not sure what "tipping point" is supposed to mean in reference to the horizon anyway.

the first probe will escape to infinity so it's velocity (relative to the free-faller) will likewise tend toward infinity.

This is true, but I don't see how it follows from the fact that the BH is an "accelerator" and the horizon is a "tipping point". The key fact is that the first probe is moving at escape velocity for its altitude above the horizon.

On the other hand, the second probe cannot escape

True--but again, I don't see how it follows from your premises. The key fact is that it is impossible for any object to escape from inside the horizon, because to do so the object would have to move faster than light.

and its velocity (relative to the free-faller) will approach zero

This is not correct; the second probe's velocity relative to the free-faller continually increases.

Hence, (in the free-fallers LIF) the probes first appear to approach each other and then separate, so no paradox.

This is not correct either; within the free-faller's LIF the two probes approach each other, period. There is no detectable change in their relative velocity within the LIF; that's the point. The two probes only start to separate *after* they have left the free-faller's LIF. (The reason they stop approaching each other and start to separate is the BH's tidal gravity, which is not detectable within the LIF.)

 

[Lino]

Thanks PeterDonis. I appreciate the feedback - but I would, if you don't mind, like to check a couple points.

I'm not sure this is a good way of viewing the BH and horizon. The important point about the BH is that it produces tidal gravity; the important point about the horizon is that, in the LIF of a free-faller crossing the horizon, the horizon itself is an outgoing light ray. The "acceleration due to gravity" produced by the BH (btw, it's questionable whether that concept even makes sense at or inside the horizon) doesn't come into the problem at all, and I'm not sure what "tipping point" is supposed to mean in reference to the horizon anyway.

Is the problem with the wording "accelerator and tipping point"? If so, apologies for the clumsy language - I agree. In my own mind, I was only looking for a short set of words that reflect the different accelerations at different distances from the free-faller. However, if the problem relates to the tidal gravity, I thought that you had specifically excluded this from consideration (i.e. pick a large enough BH and it effectively eliminates it). Are you saying that you want to limit the size of the BH so that the tidal forces are not negligible?

This is not correct; the second probe's velocity relative to the free-faller continually increases.

I have two questions in relation to this: (i) based on your previous comment (re "it's questionable whether that concept [of acceleration] even makes sense at or inside the horizon") how does it make sense to say that the relative velocity is increasing or decreasing? and (ii) I appreciate how the relative velocity would APPROACH zero (but never actually reach zero) and so the distance and velocity between them will always increase, but are saying that the increase would be more than this "tending to almost zero" rate?

... within the free-faller's LIF the two probes approach each other, period. There is no detectable change in their relative velocity within the LIF; that's the point. The two probes only start to separate *after* they have left the free-faller's LIF. (The reason they stop approaching each other and start to separate is the BH's tidal gravity, which is not detectable within the LIF.)

In my mind, this really relates to (i) above, but if one of the statements (either increasing or decreasing) is acceptable, then within the LIF the rate that either probe is "pulled back" (again, I might need to apologise for this wording) will vary with distance from the free-faller. Or, are you saying that, in the LIF, there is no variation in the gravitational field?

Thanks again for your response.

Regards,

Noel.

 

[PeterDonis]

Is the problem with the wording "accelerator and tipping point"?

Yes, but mainly because I wasn't sure exactly what you meant by them. You clarify it somewhat with this:

I was only looking for a short set of words that reflect the different accelerations at different distances from the free-faller.

This is really another way of saying the BH produces tidal gravity, which, as I said, is not detectable within the LIF, but certainly does come into play once the objects leave the LIF.

However, if the problem relates to the tidal gravity, I thought that you had specifically excluded this from consideration (i.e. pick a large enough BH and it effectively eliminates it).

More precisely, I said that *within the LIF*, tidal gravity is negligible, and for a large enough BH, the initial conditions of the problem can be realized within the LIF. The key initial condition is that differences in "escape velocity" *are* detectable within the LIF, even though tidal gravity is not; whether or not this is true depends on the size of the BH--the larger the better.

(i) based on your previous comment (re "it's questionable whether that concept [of acceleration] even makes sense at or inside the horizon")

That's not quite what I said; what I said was that it's questionable whether the concept of "acceleration due to gravity produced by the BH" makes sense at or inside the horizon. There are other concepts associated with the term "acceleration" that are perfectly well-defined at or inside the horizon.

how does it make sense to say that the relative velocity is increasing or decreasing?

"Relative velocity" has a direct physical interpretation as the observed Doppler shift of light signals from one object to the other. So defining relative velocity doesn't depend on defining "acceleration due to gravity". (Changes in this observed Doppler shift are one of the concepts of "acceleration" that is well-defined at or inside the horizon.)

(ii) I appreciate how the relative velocity would APPROACH zero

No, that's not correct--read what I said carefully. I said the second probe's velocity relative to the free-faller *continually increases*. It does not approach zero; it starts at some positive value, the value at launch, and gets larger and larger.

within the LIF the rate that either probe is "pulled back"

Once again, read what I said carefully. Within the LIF, the probes' relative velocity does not change; the second probe does not get "pulled back". The "pulling back" is an effect of tidal gravity and only comes into play once the probes leave the LIF.

are you saying that, in the LIF, there is no variation in the gravitational field?

There is no detectable tidal gravity, yes. But as noted above, there *is* a detectable change in escape velocity. Having to make these sorts of careful distinctions is one reason the term "gravitational field" is not usually used without qualification.

 

[Lino]

Thanks PeterDonis. I'm afraid (glad?) that this means lots more reading for me! As always, your insights are greatly appreciated.

Regards,

Noel.

 

[PeterDonis]

To help clarify what I've been describing in this thread, here is a spacetime diagram of a local inertial frame that is falling through the horizon of a black hole:

The time and space axes of this diagram have units of $2 M \epsilon$, where $M$ is the mass of the black hole and $\epsilon$ is a small parameter (in the diagram I have used $\epsilon = 0.1$, which is not really very small, but makes the differences between the various curves easier to see).

The time (vertical) axis of this diagram is the worldline of an astronaut falling through the horizon of a black hole; the origin of the diagram is the event at which he crosses the horizon.

The blue line is the horizon.

The green line starting at (t, x) = (-1, 0) is the worldline of the first probe, which is launched at escape velocity from just outside the horizon.

The red line starting at (t, x) = (1, 0) is the worldline of the second probe, which is launched at a speed closer to the speed of light than that of the first probe, from just inside the horizon.

The light gray lines are lines of constant radial coordinate $r$ (other than the horizon, which is the curve $r = 2M$); the increment of $r$ between adjacent lines is the same as the scaling of the axes, i.e., $2 M \epsilon$. Note that these lines converge as you go from the lower left to the upper right of the diagram; the lines outside the horizon are timelike (though just barely) and the lines inside the horizon are spacelike (just barely). Note also that the green line has (slowly) increasing $r$, while the red line has (slowly) decreasing $r$ (this can be seen from the relationship between these lines and the gray lines that start out at the same events on the time axis), even though the red and green lines are converging (because the red line has a higher velocity than the green).

I will post a follow-up with more details on the math behind this diagram (it differs in some details from the math I posted earlier in the thread, to make things easier to plot).

 

[PeterDonis]

Here are more details on the math behind the spacetime diagram I just posted. Some details are changed from post #51 in this thread, where I went through similar calculations; this doesn't affect the physics, it just happened to make it easier for me to generate the plot. For brevity, though, I have left out steps in the computation that are substantially the same as the ones I gave in post #51, so it may help to read that post as well.

We want to launch the first probe at escape velocity at a value of $r$ just above the horizon; i.e., the launch point is at $r = 2M \left( 1 + \epsilon \right)$, where $\epsilon << 1$. The escape velocity formula is $v_e = \sqrt{2M / r}$, which gives $v_e =\sqrt{1 / \left( 1 + \epsilon \right)} \approx 1 - \epsilon / 2$. However, as I noted, this is the escape velocity relative to a static observer; the astronaut who is at rest in our LIF is actually falling radially inward at this same velocity, relative to a static observer. So the velocity of the first probe relative to the astronaut is $v_1 = 2 v_e / \left( 1 + v_e^2 \right) \approx 1 - \epsilon^2 / 8$.

We want the second probe to be moving faster than the first probe, relative to the astronaut, so that within the LIF, the two probes are converging. Any value closer to the speed of light than $v_1$ above will do; for simplicity I chose $v_2 = 1 - \epsilon^2 / 16$, i.e., $1 - v_1 = 2 \left( 1 - v_2 \right)$.

To establish the relationship between the global $r$ coordinate and the $t$, $x$ coordinates in our LIF, we need a global coordinate chart. The best one to use for this scenario is the Kruskal chart; basically, our LIF is just a small square patch of the Kruskal chart, centered on a point on the horizon $U = V$ in that chart, and rescaled appropriately. The line element in the Kruskal chart (omitting the angular coordinates) is

$$
ds^2 = \frac{32 M^3}{r} e^{- r / 2M} \left( - dV^2 + dU^2 \right)
$$

Since we are working very close to $r = 2M$, the scaling factor becomes simply $C^2 = 16M^2 / e$, and we can rewrite the line element as simply $ds^2 = - dT^2 + dX^2$, where $T = V / C$ and $X = U / C$. However, we want to make two further refinements to this. First, we want to translate the chart to some point well along the horizon, i.e., we want to move our little square patch to be centered on $(V, U) = (K, K)$, where $K$ is some positive constant. So we actually have $T + K = C V$ and $X + K = C U$.

Second, we want to rescale the axes of our diagram in units of $2 M \epsilon$, because that is the proper time for the astronaut to fall from the launch point of the first probe to the horizon:

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

So our final coordinates in the LIF are $t + k = C v$ and $x + k = C u$, where the lower-case $t$, $x$, $k$, $u$, $v$ are equal to their upper-case counterparts divided by $2 M \epsilon$.

Now we need to plug all of this into the equation for $r$ in terms of the Kruskal coordinates $U$, $V$:

$$
V^2 - U^2 = \pm \left( 1 - \frac{r}{2M} \right) e^{r / 2M}
$$

The $\pm$ in front on the RHS is so our formula covers both inside (plus sign) and outside (minus sign) the horizon. Plugging in and rearranging some factors, we have

$$
4 M^2 \epsilon^2 \left[ \left( t + k \right)^2 - \left( x + k \right)^2 \right] = \pm q \epsilon e C^2
$$

where $r / 2M = 1 + q \epsilon$, i.e., $q$ tells us which $r$ curve we are on. Plugging in the formula for $C^2$, making the obvious cancellations and rearranging again so that we have an expression for $t$ in terms of $x$ gives us:

$$
t = \sqrt{\left( x + k \right)^2 \pm \frac{4}{\epsilon} q} - k
$$

The only thing left to do now is to obtain a formula for $k$. We do this by noting that when $q = 1$ and $x = 0$, we should have $t = \pm 1$. Plugging these values into the above formula and solving for $k$ gives

$$
k = \frac{2}{\epsilon} \pm \frac{1}{2}
$$

where now the plus sign holds outside the horizon and the minus sign holds inside the horizon (i.e., the reverse of the signs for the $q$ term in the formula above).

For the actual diagram, as I noted, I chose $\epsilon = 0.1$, which is a rather large value but makes it easier to see the differences in the curves. The other key parameters are then $k = 20.5$ outside the horizon, $k = 19.5$ inside the horizon, $v_1 = 0.99875$, and $v_2 = 0.999375$.

 

[PeterDonis]

To complement the previous spacetime diagram I posted, here is a spacetime diagram of the "skyscraper" or "skydiver" LIF: i.e., the LIF of a "skydiver" who is freely falling in the gravitational field of a planet or star, say off the top of a tall building, and launches two probes outward with the same velocities, relative to him, as the astronaut launches his probes in the LIF falling through a black hole's horizon.

Note that everything looks the same as the previous diagram except for the lines of constant radial coordinate $r$ (the dark gray lines).

The time and space axes here are scaled in units of $0.01 / a_0$, where $a_0$ is the proper acceleration of the point (top of the skyscraper, plane or ship hovering over the planet or star, or whatever it is) from which the skydiver jumps (this point is assumed to remain at constant radial coordinate $r$). Note that this means $1 / a_0 = 100$ in the units of the diagram, which if the diagram is assumed to cover a reasonable size scale in human terms (such as meters or even kilometers), equates to an *extremely* large proper acceleration. If 1 time/space unit in the diagram equates to 1 kilometer, for example, then $a_0$ is approximately 100 billion Earth gravities. I picked this value for $a_0$ so that the behavior of the lines of constant $r$ (slowly accelerating away from the free-falling skydiver at rest in the LIF) would be easier to see.

Most of the features of the diagram are exactly analogous to those of the previous one. The time axis of the diagram is the worldline of the skydiver. The blue line is an outgoing light ray that passes the skydiver halfway between his launching of the first and second probes; this is the analogue of the horizon (although of course it is emphatically *not* a line of constant $r$ in this LIF, nor is it an event horizon in this spacetime--see below). The green line is the first probe's worldline; the red line is the second probe's worldline. All of these lines are exactly identical, in terms of their coordinates within the LIF, as their analogues in the other diagram.

The dark gray lines, as noted above, are the lines of constant $r$; of course these are very different from those in the previous diagram. That is because this LIF has a very different relationship to the global coordinates in its spacetime than the first LIF has to the global coordinates in *its* spacetime. It is important to understand that that relationship has *nothing* to do with the equivalence principle: the EP is local, not global, and the EP does *not* say anything about what relationship local coordinates within an LIF must have to global coordinates in the spacetime in which the LIF is embedded.

Also, if we are going to compare "experiments" done in two different LIFs in order to check the EP, we have to compare experiments with identical initial conditions. That means, for example, that the two probes in the skydiver LIF must be launched with the same velocities *relative to the skydiver* as the two probes in the black hole LIF are launched relative to the infalling astronaut. It does *not* mean that the first probe in this LIF must be launched at escape velocity for its spacetime, at the point where it is launched. "Escape velocity" is a global concept, not a local one; and comparing experiments between two LIFs does *not* mean comparing their global coordinates or their global properties. It means comparing how they look *relative to the LIF*, and that is *all* it means. If we launched the first probe in this LIF at escape velocity for its launch point, we would be giving it a very different (and much slower) initial velocity relative to the skydiver; and that would mean we were running a different experiment and should not expect to get the same results.

Similar remarks apply to the outgoing light ray in this LIF that is analogous to the horizon in the black hole LIF. As noted above, that light ray's worldline in this LIF is certainly *not* an event horizon. But that doesn't matter; "event horizon", like escape velocity, is a global concept, not a local one. For the two LIFs to be the same for purposes of checking the validity of the EP, the light ray's worldline *relative to the skydiver* must be the same as the horizon worldline relative to the astronaut in the black hole LIF. That's what the diagram above shows.

One final note: comparing results between the two LIFs, for purposes of checking the validity of the EP, is also restricted to within the LIFs; more precisely, it is restricted to within the size of the *smaller* of the two LIFs. If we look at the ultimate fates of the two probes in the two experiments, they are obviously different: in the black hole case, the first probe just barely escapes to infinity while the second falls into the singularity; but in the skydiver case, both probes escape to infinity easily. However, there's a more subtle point here as well.

In the black hole case, the second probe never catches the first, even though we would predict that it would if we just extrapolated worldlines from within the LIF, taking no account of spacetime curvature. In the skydiver case, the second probe *will* catch the first eventually; at least, we are assuming that the global curvature of the spacetime is such as to allow this to be the case--which is possible for a wide range of assumptions about that curvature. It is even possible, given the right assumptions about the global curvature of the skydiver spacetime, that the second probe will catch the first *within the range of the skydiver LIF*. (This is because the proof I gave in post #51, showing that the predicted "catch-up distance" would always be much larger than the LIF size for the black hole case, does not hold for the skydiver case, since the assumption that the LIF was centered on the horizon was necessary for the proof.)

Doesn't this invalidate the EP, since we proved that the "catch-up" can never occur within the LIF for the black hole case? No, it doesn't, because if the "catch-up" does occur within the LIF for the skydiver case, it must be because the skydiver LIF can cover a much larger piece of spacetime than the black hole LIF (because the spacetime curvature in the skydiver case is much smaller), so that the "catch-up" can occur within the skydiver LIF while still being far outside the black hole LIF. But we are only allowed to compare the two LIFs over the range covered by the smaller one; otherwise we could always find a violation of the EP by simply calling globally flat Minkowski spacetime an "LIF", since there will always be some difference between a flat and a curved spacetime.

 

[PeterDonis]

Here's a quick summary of the math behind the skydiver LIF spacetime diagram that I just posted.

A curve of constant $r$ has constant proper acceleration, and we can use that to determine what it looks like in the LIF. The key thing we need is that, in an inertial frame, a curve of constant proper acceleration is a hyperbola. In the simplest case, where the asymptotes of the hyperbola cross at the origin, the equation for it is $X^2 - T^2 = 1 / a^2$, where $a$ is the proper acceleration.

We want to adjust this in several ways. First, we want to move the origin of the inertial frame so that the curve of constant $r$ that marks the altitude from which the skydiver initially jumps meets the $t$ axis (the skydiver's worldline) at some point $t = - k$, where $k$ is some constant. (In the diagram as I've shown it, $k = 5$, i.e., the skydiver jumps at $t = - 5$.) So we have $t = T - k$.

Second, we want to shift the origin along the $x$ axis by a distance $1 / a_0$, where $a_0$ is the proper acceleration of the skydiver's jump point (the top of the skyscraper, plane or spacecraft hovering, or whatever). So we have $x = X - 1 / a_0$.

Third, we want to label different curves of constant $r$ with a distance parameter $q$, instead of by their proper acceleration; in other words, we want $1 /a = 1 / a_0 + q$, where $a$ is the proper acceleration of an arbitrary curve of constant $r$. (This means, of course, that $q = 0$ gives the worldline of the skydiver's jump point.)

Putting all this together, and rearranging terms, gives us a formula for $t$ in terms of $x$ for a curve of constant $r$:

$$
t = \sqrt{\left( x + \frac{1}{a_0} \right)^2 - \left( q + \frac{1}{a_0} \right)^2} - k
$$

As I noted in the previous post, for the diagram as I've shown it, $1 / a_0 = 100$.

 

[craigi]

I think there's a simple solution this.

The first probe moves away from the astronaut and the event horizon. For the second probe, all forward time paths go towards the event horizon, so both probes move away from the astronaut and seperate.

There's two interesting things about it, firstly, to fire these probes would require a near infinite amount of energy and the shorter the time between firing the probes, the higher the inevitable impulse on the astronaut, which accelerates her towards the singularity. Secondly, it wouldn't look all that weird since looking away from the event horizon is looking along a backward time path. As in the non-relativistic case, she can receive light from the past, but is powerless to go or send anything there.

 

[PeterDonis]

The first probe moves away from the astronaut

Yes, that's part of the specification of the problem.

and the event horizon.

Not within the LIF. Within the LIF, the horizon is moving in the positive x direction at the speed of light, i.e., faster than the first probe.

For the second probe, all forward time paths go towards the event horizon

The second probe is launched *after* the astronaut has passed the horizon. Please read carefully.

so both probes move away from the astronaut

Yes; again, that's part of the specification of the problem.

and seperate.

No, they don't. The second probe is launched in the positive x direction with a higher speed than the first, so within the LIF, the distance between the two probes decreases.

Bear in mind that this is an old thread to which I added some diagrams recently just for information. Please read through the entire thread before commenting; it will help to show you what arguments have already been made and don't need to be re-made.

There's two interesting things about it, firstly, to fire these probes would require a near infinite amount of energy

Yes, but this is a thought experiment so we don't care about that.

and the shorter the time between firing the probes, the higher the inevitable impulse on the astronaut

Actually, I was assuming that the probes were fired using rockets, so they don't exert any impulse on the astronaut; the astronaut releases the rocket with essentially zero impulse, then it fires without exerting any force on him (the rocket exhaust takes up the impulse as it is ejected in the opposite direction from the probe). So the astronaut remains in free fall the whole time, as shown in the spacetime diagrams I posted.

 

[craigi]

The second probe is launched *after* the astronaut has passed the horizon. Please read carefully.



Yes; again, that's part of the specification of the problem.

Actually, I was assuming that the probes were fired using rockets, so they don't exert any impulse on the astronaut; the astronaut releases the rocket with essentially zero impulse, then it fires without exerting any force on him (the rocket exhaust takes up the imupulse as it is ejected in the opposite direction from the probe). So the astronaut remains in free fall the whole time, as shown in the spacetime diagrams I posted.

Bad terminology on my behalf. I was taking about the apparent horizon, but we can just replace that with the singularity to make my argument clearer.

Your rockets just shift the problem. How heavy is a rocket capable of imparting near infinite impulse?

 

[PeterDonis]

Your rockets just shift the problem. How heavy is a rocket capable of imparting near infinite impulse?

If it bothers you that much, then assume that the two probes free-fall in with the astronaut but not connected to him in any way. Since free-fall trajectories are independent of the mass of the object, all three objects (the astronaut, the first probe rocket, and the second probe rocket) fall along the same trajectory (at least until each probe's rocket fires to launch it), which is all that's required for the thought experiment.

(Also, there's nothing in the scenario that requires the probes themselves--the payloads of the probe rockets--to be heavy objects. They could be nanotechnology probes made of a few tens of atoms. Then the total mass of each probe rocket could still be much, much smaller than the mass of the astronaut with his life support system, even after allowing for all the rocket fuel necessary to launch the probes at speeds close to that of light.)

 

[Dale]

Several off-topic posts have been removed and the thread closed.