Does mass really increase with speed

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  • #51
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This is incorrect; there is no relativistic [itex]\gamma[/itex] factor in the formula for the Schwarzschild radius. The correct formula is

[tex]r_{s} = \frac{2 G m}{c^{2}}[/tex]

where [itex]G[/itex] is Newton's gravitational constant.
In my notation above [itex]\gamma[/itex] is the gravitational constant. Let me know if you have serious comments.
 
  • #52
PeterDonis
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The rest mass is related to total energy and total momentum
This part is OK, but you've left out something: how did the two objects come to have such high velocities in the first place? The energy that got them moving at those speeds had to come from somewhere. Put another way, in the center of mass frame the total energy of the system has *always* been what you are calculating as the "new" rest mass; it's just that before the two rockets were launched, the energy that launched them had to be stored somewhere else, in some other object from which the rockets originated.

What this means is that, if the combined total energy of the rockets is enough for them to be inside their common Schwarzschild radius now, then the original object that spawned them must have been a black hole. Which means they couldn't have escaped in the first place.
 
  • #53
PeterDonis
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In my notation above [itex]\gamma[/itex] is the gravitational constant. Let me know if you have serious comments.
See my post #52. That notation is not standard, which is why it confused me for a bit.
 
  • #54
PeterDonis
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As to the question "Does mass really increase with speed?" my idea is, that if true, then the creation of black holes would be observer dependent. Therefore it can't be true. Perhaps this was already mentioned in one of the posts, I didn' read all so far.
Your reasoning is correct; the creation of black holes is not observer dependent, therefore mass can't "really" increase with speed. The general point has been made before in this thread, but nobody has put it specifically the way you did, so yours was a good addition.

If it sometimes seems like mass is increasing with speed, that's because something has been left out of the analysis. See my post #52 in response to DrStupid for an example.
 
  • #55
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This part is OK, but you've left out something: how did the two objects come to have such high velocities in the first place?
That doesn't matter. But if you need an explanation you might assume that they have been accelerated by rockets.

Put another way, in the center of mass frame the total energy of the system has *always* been what you are calculating as the "new" rest mass;
There is no "new" rest mass and of course rest mass is conserved in isolated systems.

What this means is that, if the combined total energy of the rockets is enough for them to be inside their common Schwarzschild radius now, then the original object that spawned them must have been a black hole.
The objects don't have to be spawned by a single source. They could have been started from very distant positions. But as I mentioned above that doesn't matter. The simple question is: Does the acceleration of the bodies depend on their velocities? If no than two bodies that does not collapse at low velocity will never collapse no matter how fast they are shoot together. If yes than this is an example where gravitation increases with speed.
 
  • #56
atyy
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Hmm. We do talk about the independence of physical observables from the choice of coordinates in GR as being a kind of gauge invariance, by analogy with electromagnetism, but thinking about it I'm not sure if the analogy fully holds.

In electromagnetism there are certainly "gauge variant" quantities as distinct from "gauge invariant" ones: the potential [itex]A_{u}[/itex] is gauge variant, but the field tensor [itex]F_{uv}[/itex] is gauge invariant. The reason we say this is that we can change the potential by the gradient of a scalar, [itex]A'_{u} = A_{u} + \partial_{u} \phi[/itex], without changing any physical predictions; but the reason it doesn't change any physical predictions is that it doesn't change the field tensor [itex]F_{uv} = \partial_{u} A_{v} - \partial_{v} A_{u}[/itex] (because mixed partial derivatives commute), and the field tensor is what determines the physical predictions.

In the case of gravity, the analogue of a "gauge transformation", a change of coordinates, can change the components of the things that actually determine physical predictions, such as the electromagnetic field tensor [itex]F_{uv}[/itex]; the reason it doesn't change the actual physical predictions is that those predictions are expressed as scalars, i.e., contractions of vectors and tensors, and those do not change with a coordinate transformation, even though the individual components do. So in this case, we could say that the vectors and tensors themselves are the "gauge variant" quantities, and only the scalars are "gauge invariant"; but "gauge variant" here has a different meaning than it did in the electromagnetism case, because the things we are calling "gauge variant" are directly used to make physical predictions, not indirectly as in the electromagnetism case.
So perhaps there isn't such a great distinction between the seemingly different answers that gravity does or does not change between observers, if we are able to associate a gauge (say Fermi normal coordinates) with each observer? Strictly speaking, this can't be done completely locally, and the "distant stars" are needed, but that will be true for observables such as the expansion scalar in the FRW case anyway (take the congruence of geodesics as "distant stars"). (There's also the complication that a single chart may not cover all of spacetime, but let's ignore that here.) So the loose answer would be gravity does change (we are allowed to associate a gauge with an observer), but the strict answer would be that gravity does not change (we insist that a spacetime is really the whole diffeomorphism equivalent class of metrics). To tie the the loose and strict answers together, could one say gravity does change between observers, but they agree on their disagreement, so there is no problem (just the same as simultaneity in SR)?
 
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  • #57
PeterDonis
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That doesn't matter. But if you need an explanation you might assume that they have been accelerated by rockets.
Which means the rockets had to expend energy by burning fuel--also it means there was rocket exhaust ejected. To properly compute the invariant mass of the entire system you have to take all these things into account.

There is no "new" rest mass and of course rest mass is conserved in isolated systems.
I didn't say there was "new" rest mass, nor did I think you said there was. I said that the invariant mass of the entire system, taken as a whole, is constant. That follows from conservation of energy. (I assume that you were imagining an isolated system, i.e., the rockets don't interact with anything else except, possibly, each other, and of course with their fuel supplies and rocket exhausts.) So if the rockets are inside the Schwarzschild radius of the system as a whole *after* they have accelerated to high speed, they must have been inside the Schwarzschild radius of the system as a whole *before* they accelerated to high speed.

The objects don't have to be spawned by a single source. They could have been started from very distant positions. But as I mentioned above that doesn't matter.
I agree, it doesn't matter for what I said above; what I said above is true no matter how the total invariant mass of the system as a whole is split up among the rest masses and kinetic energies of its parts.

The simple question is: Does the acceleration of the bodies depend on their velocities?
No, it doesn't. That is, if by "acceleration" you mean the acceleration the bodies actually feel, and which would be measured by accelerometers carried along with the bodies. If you mean something else, then please clarify what you mean.
 
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  • #58
PeterDonis
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To tie the the loose and strict answers together, could one say gravity does change between observers, but they agree on their disagreement, so there is no problem (just the same as simultaneity in SR)?
For a suitable definition of the term "gravity", yes, you could say this.
 
  • #59
timmdeeg
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So if the rockets are inside the Schwarzschild radius of the system as a whole *after* they have accelerated to high speed, they must have been inside the Schwarzschild radius of the system as a whole *before* they accelerated to high speed.
Just to understand this correctly: before acceleration the rockets have been freely falling and were expecting to reach the singularity in proper time x. After accelaration they can't even hover but are able to bend their worldlines such as to reach the singularity a little later that x.
If you say "they accelerated to high speed", do you mean in relation to a part p of the rocket, which was left freely falling before acceleration? However is the rocket after acceleration still within the light-cone of p?
 
  • #60
PeterDonis
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Just to understand this correctly: before acceleration the rockets have been freely falling and were expecting to reach the singularity in proper time x. After accelaration they can't even hover but are able to bend their worldlines such as to reach the singularity a little later that x.
Yes, that's more or less what I was imagining.

If you say "they accelerated to high speed", do you mean in relation to a part p of the rocket, which was left freely falling before acceleration?
Yes. However, I'm not sure that this is totally consistent, because the two rockets are supposed to move in opposite directions, spatially, but I'm not sure that's possible inside a black hole's horizon unless at least one rocket moves *towards* the singularity (relative to p), not away from it. I wasn't trying to construct a detailed scenario; I was just pointing out an obvious implication of the claim that the rockets were "inside each other's Schwarzschild radius".

However is the rocket after acceleration still within the light-cone of p?
It depends on what part of p's worldline you look at. Obviously, since the rockets move on timelike worldlines, they will remain in the future light cone of the event on p's worldline at which they left p and started accelerating. But there will be some point on p's worldline to the future of that event where the rockets will move outside the future light cone of p; in other words, there will be some point after which p can no longer send light signals to either rocket. From the rockets' point of view, this will be because p is closer to the singularity than they are, to the point that any light emitted by p will fall into the singularity before it reaches them.
 
  • #61
timmdeeg
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But there will be some point on p's worldline to the future of that event where the rockets will move outside the future light cone of p; in other words, there will be some point after which p can no longer send light signals to either rocket. From the rockets' point of view, this will be because p is closer to the singularity than they are, to the point that any light emitted by p will fall into the singularity before it reaches them.
Yes.
Thank you for your comments, its very helpful.
 
  • #62
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Ok the penny dropped.
Referring back to the balance readout between the 2 rockets, seen from earth and from the rockets. See post # 13.

I stated something like: since we need to take account of force transformations, the 2 sets of observers might not see the same value on the readout. This is not true.

The solution is that the balance has its own internal spring. If we also apply the same force transformation on this spring, the readout will stay pointed to the exact same point. So, although there just might exist different forces, both observers must see the same value of readout.
 
  • #63
PeterDonis
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The solution is that the balance has its own internal spring. If we also apply the same force transformation on this spring, the readout will stay pointed to the exact same point. So, although there just might exist different forces, both observers must see the same value of readout.
Yes, you've got it.
 
  • #64
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PeterDonis: some great explanations...thank you!....
 
  • #65
timmdeeg
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It depends on what part of p's worldline you look at. Obviously, since the rockets move on timelike worldlines, they will remain in the future light cone of the event on p's worldline at which they left p and started accelerating.
Sorry, after overthinking that I started to be puzzled.

For simplicity A and B shall be in free fall together (no radial distance) inside the horizon. At event u B decides to accelerate in the direction opposite to the fall. Now, after the time interval dt A and B should be separated by a radial distance dr. Then however a lightpulse of A can't reach B, because A's light-cone is tipped towards the singularity.

This sounds strange however. How could B be outside of A's future lightcone having taken a time-like path? Your argument is convincing.

Any help what's wrong with my reasoning is appreciated. Further, this issue seems not within the context of this thread. Please give me a hint in case I should ask elsewhere.

Thanks
 
  • #66
PeterDonis
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timmdeeg, remember that light cones are "attached" to *events* in spacetime, not just to observers whose worldlines pass through those events. As an observer moves along his worldline, the light cone at the event he is passing through changes.

Also, remember that light rays move in spacetime, not just space. To determine whether one observer can reach another with a light ray, it's not enough just to look at their respective spatial locations. You have to consider time as well.

At event u B decides to accelerate in the direction opposite to the fall. Now, after the time interval dt A and B should be separated by a radial distance dr. Then however a lightpulse of A can't reach B, because A's light-cone is tipped towards the singularity.
Not necessarily. It's true that light can't move outward (i.e., can't increase its r coordinate) inside the horizon. But B is moving inward, so still may be possible for "outgoing" light from A to reach B; the light just has to move inward slower than A does, so B can catch up to it. When the light reaches B, that corresponds to B entering the future light cone of the event from which A *emitted* the light; but that event, of course, will *not* be in the future light cone of events further along A's worldline.

Viewed in spacetime, A's future light cone at the event where he emits the outgoing ray is indeed tipped towards the singularity, but that just means the "outgoing" side of the light cone no longer points in the direction of increasing r. It still points in the direction of increasing time, so it's still possible for B's worldline to cross it.

Btw, I should explain why I said "increasing time" just now instead of specifying a coordinate like t. First of all, Schwarzschild coordinates are singular at the horizon, and inside the horizon r and t switch roles: r is timelike and t is spacelike. So these coordinates are not good ones to use when trying to understand what's going on at or inside the horizon.

The picture of the light cones tilting inward towards the singularity comes from Eddington-Finkelstein coordinates:

http://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinates

The light cones also look similar to this in Painleve coordinates:

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

In both these coordinates, r is spacelike inside the horizon, which is a lot easier to deal with. However, in these coordinates, the "time" coordinate is *also* spacelike inside the horizon! (In other words, all four coordinates inside the horizon are spacelike.) So we can't really use their time coordinates either to indicate the direction of increasing time inside the horizon.

The only coordinates I'm aware of where the "time" coordinate is timelike everywhere in a black hole spacetime are Kruskal coordinates:

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

In these coordinates, light rays are always 45 degree lines, so seeing the causal structure of the spacetime is easy. For instance, these coordinates make it easy to see why everything, including light, has to move in the direction of decreasing r inside the horizon: in that region, lines of constant r are spacelike lines (hyperbolas in the upper region of the diagram), since they're more horizontal than vertical. It's also easy to see why you can't avoid the singularity: it's one of those "constant r" hyperbolas, so it's to your future no matter where you are inside the horizon. But rather than get too deep into all this, I just said the direction of increasing "time" above.
 
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  • #67
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Which means the rockets had to expend energy by burning fuel--also it means there was rocket exhaust ejected. To properly compute the invariant mass of the entire system you have to take all these things into account.
I'm not interested in the entire system but in the two objects only. For the question discussed the ejected exhaust is completely negligible.

So if the rockets are inside the Schwarzschild radius of the system as a whole *after* they have accelerated to high speed, they must have been inside the Schwarzschild radius of the system as a whole *before* they accelerated to high speed.
Sorry, but that's rubbish. The Schwarzschild radius is limited but the starting distance between the rockets is not.

what I said above is true no matter how the total invariant mass of the system as a whole is split up among the rest masses and kinetic energies of its parts.
I am not talking about the gravity of the whole system. I am talking about the gravity of each body.

No, it doesn't. That is, if by "acceleration" you mean the acceleration the bodies actually feel, and which would be measured by accelerometers carried along with the bodies.
This "acceleration" is always zero because the bodies are in free fall and a free falling body doesn't "feel" any acceleration. That's not very helpful.

If you mean something else, then please clarify what you mean.
I mean the derivation of velocity with respect to time. That's the definition of acceleration.
 
  • #68
PeterDonis
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I'm not interested in the entire system but in the two objects only. For the question discussed the ejected exhaust is completely negligible.
It most certainly is not. You are postulating rockets that can reach high relativistic speeds. Take a look at the relativistic rocket equation to see the mass ratio required to achieve a given gamma factor:

http://www.desy.de/user/projects/Physics/Relativity/SR/rocket.html

For the kinds of speeds necessary to realize your scenario, the mass ratio will be huge; so by focusing on just the two objects and ignoring all the fuel and exhaust it took to get them to their speeds, you are ignoring by far the largest energies in the problem.

The Schwarzschild radius is limited but the starting distance between the rockets is not.
You postulated a scenario such that, if we calculate a "Schwarzschild radius" using the rockets' relativistic masses, they will be inside each other's Schwarzschild radius. That limits the starting distance between them.

I am not talking about the gravity of the whole system. I am talking about the gravity of each body.
As I said above, you are ignoring most of the energy in the system if you focus only on the two rockets. All that energy gravitates. You can't just ignore it.

I mean the derivation of velocity with respect to time. That's the definition of acceleration.
Velocity as in ordinary 3-velocity? That's frame dependent. Or velocity as in 4-velocity? That can be represented as a covariant 4-vector, but then what derivative do we take? The derivative with respect to coordinate time, or with respect to the object's proper time? It makes a difference.

The answer I gave you, that the acceleration that bodies actually feel does not depend on their velocities, is the only answer that IMO has any physical meaning, because the acceleration bodies actually feel has physical meaning. So that's the only kind I care about. That kind of acceleration is defined as the derivative of the object's 4-velocity with respect to its proper time. Since the derivative of the 4-velocity is independent of the 4-velocity itself (i.e., by applying appropriate forces to the object we can make the derivative of its 4-velocity anything we like, regardless of the 4-velocity itself), I answered that the object's acceleration is independent of its velocity.

If you care about some other kind of acceleration, or some other kind of velocity, you need to specify what kind. Just saying "velocity" or "the derivative of velocity with respect to time", as I noted above, won't do; it doesn't actually specify what you mean, because the terms "velocity" and "time" are ambiguous.
 
  • #69
PeterDonis
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The Schwarzschild radius is limited but the starting distance between the rockets is not.
I realized after making my first response to this that you may be thinking of a scenario where the rockets are moving towards each other, not away. I was assuming they were moving away from each other.

I think that this scenario you have postulated needs to be nailed down more precisely. How about giving some actual numbers? You don't need to give many; just the following:

(1) The rest mass of the rockets. (Just one number, we'll assume it applies to both rockets.) This should be just the payload, i.e., just the part that is there when the rockets are inside each other's Schwarzschild radius according to you. As an example, the rest mass of the Apollo command module was approximately 20 metric tons (20,000 kg).

(2) The distance between the rockets whey they are supposedly inside each other's Schwarzschild radius. (This is distance as seen in the "lab" frame, the frame in which the rockets are moving at ultra-relativistic speeds.) This will fix the invariant mass of the total system.

(3) The relative directions the rockets are traveling in; this will fix the combined momentum of the rockets. (This is also to avoid the kind of confusion I mentioned above.)
 
  • #70
timmdeeg
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Thank you for very helpful explanations, Peter.

Not necessarily. It's true that light can't move outward (i.e., can't increase its r coordinate) inside the horizon. But B is moving inward, so still may be possible for "outgoing" light from A to reach B; the light just has to move inward slower than A does, so B can catch up to it.
Yes, understood, this is the point I haven't realized.

First of all, Schwarzschild coordinates are singular at the horizon, and inside the horizon r and t switch roles: r is timelike and t is spacelike.
This is hard to imagine. The only layman interpretation I am aware of sounds like this: r is timelike inside the horizon, as it has only one direction, like time flows only in one direction. But the weirdness seems "only" to be a matter of the choosen coordinates and can be transformed away, you mentioned the Kruskal coordinates already.
 
  • #71
PeterDonis
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This is hard to imagine. The only layman interpretation I am aware of sounds like this: r is timelike inside the horizon, as it has only one direction, like time flows only in one direction. But the weirdness seems "only" to be a matter of the choosen coordinates and can be transformed away, you mentioned the Kruskal coordinates already.
The terminology of calling a coordinate "timelike" or "spacelike" is unfortunate since it doesn't really convey what's going one, especially if what looks like the *same* coordinate (r in this case) is said to be timelike in one coordinate chart (the interior Schwarzschild chart) and spacelike in others (ingoing Eddington-Finkelstein and Painleve). Here's what I think is a better way of looking at it:

A "coordinate" like r is really a shorthand way of referring to two different things. One is a set of surfaces in the spacetime: each surface is labeled with a unique value of the coordinate, and every event in the spacetime lies on one and only one of the surfaces. For example, in Schwarzschild spacetime, there is a set of surfaces of constant r that satisfies the above properties.

The second thing a coordinate refers to is a directional derivative: for example, r corresponds to [itex]\partial / \partial r[/itex], the rate of change of something in the "r direction". The thing to remember about this is to avoid the "second fundamental confusion of calculus" (I learned this term from George Jones, one of the mentors here, who pointed me at a reference to it in one of Roger Penrose's books): partial derivatives can change depending on what other variables are being held constant. So a coordinate defined as a directional derivative will depend on what other coordinates it is combined with in a specific chart.

You can probably see what's coming next: when you change coordinate charts, the two things above do not necessarily change together. For example, in all three of the coordinate charts for Schwarzschild spacetime that I mentioned above, the first aspect of the "r" coordinate is the same: i.e., the "r" coordinate in all three charts refers to the *same* set of surfaces of constant r. What changes from chart to chart is the directional derivative. This seems to be the usual convention for coordinate nomenclature: a given coordinate name, such as "r", is applied to a given set of curves; then the changes in the directional derivative between charts are captured by calling the coordinate "timelike" or "spacelike" in different charts, according to the direction the derivative points in.

As a concrete example, here's how things work out for all of the charts I have mentioned for Schwarzschild spacetime:

(1) The Schwarzschild chart. (Technically, there are actually two of these, exterior and interior, because the coordinates are singular on the horizon.) Outside the horizon, the directional derivatives look like this: [itex]\partial / \partial t[/itex] timelike; [itex]\partial / \partial r[/itex] spacelike; [itex]\partial / \partial \theta[/itex] spacelike; [itex]\partial / \partial \phi[/itex] spacelike. So a surface of constant t is a spacelike 3-surface; but a surface of constant r has one timelike and two spacelike dimensions. (I won't talk about surfaces of constant theta, phi here; angular coordinates work a little differently. The usual way of talking about them is just to say that, since the spacetime is spherically symmetric, we can think of it as a set of coordinate pairs (t, r), where each unique pair labels a 2-sphere, which is a spacelike 2-surface covering all possible values of theta, phi. So what I said above can be condensed to: outside the horizon, lines of constant t are spacelike, and lines of constant r are timelike, where each "line" is really a series of 2-spheres. The only exception is r = 0, which is a single point, and is not technically part of the spacetime because the curvature is infinite there--but that's a whole other post :smile:.)

Inside the horizon, the r and t derivatives switch directions: [itex]\partial / \partial t[/itex] is spacelike and [itex]\partial / \partial r[/itex] is timelike. This is what the common statements that "r is timelike inside the horizon" or "t is spacelike inside the horizon" refer to. You can also see that, inside the horizon, lines of constant *r* are now spacelike, and lines of constant *t* are now timelike. So the labeling of coordinates as "timelike" or "spacelike" will look backwards if you are looking at the lines of constant coordinate value instead of the directional derivatives.

(2) Ingoing Eddington-Finkelstein & Painleve charts. (I lump these together because they are the same in the aspects we're discussing; also I specify "ingoing" because there are also "outgoing" versions of these charts. I won't go into the difference here.) Outside the horizon, these are the same as the Schwarzschild exterior chart; [itex]\partial / \partial T[/itex] is timelike and the other three coordinate derivatives are spacelike. So (leaving out theta, phi again as above) lines of constant T are spacelike and lines of constant r are timelike. Note that we are using a different label, T, for the "time" coordinate because it refers to a different set of lines (or surfaces if we include the angular coordinates) than the Schwarzschild "t" coordinate does.

*On* the horizon (these charts are nonsingular at the horizon, so this is meaningful here), [itex]\partial / \partial T[/itex] is *null* in both charts. ("Null" means it points in the same direction in spacetime as a light ray--an outgoing light ray, in this case.) However, the other three coordinate derivatives stay spacelike in this chart. So on the horizon, lines of constant T are still spacelike, but lines of constant r are null. In fact, that is one way of stating the *definition* of the horizon: it is a null line (of 2-spheres) of constant r.

Inside the horizon, [itex]\partial / \partial T[/itex] is spacelike; this means that lines of constant r are spacelike. This is why it's impossible to "hover" at a constant r inside the horizon: you would have to move on a spacelike line, i.e., faster than light. But [itex]\partial / \partial r[/itex] is *also* spacelike inside the horizon in this chart; in other words, *all four* coordinates are spacelike inside the horizon! This seems very weird, but that's how it is; what it is really telling you is that, to get a timelike vector at all inside the horizon, you have to combine [itex]\partial / \partial T[/itex] and [itex]\partial / \partial r[/itex] with opposite signs; for example, a future-directed timelike curve will have positive [itex]\partial / \partial T[/itex] and negative [itex]\partial / \partial r[/itex]. This is just another way of saying that everything inside the horizon is forced to fall into the singularity. In Painleve coordinates, for example, an observer freely falling into the black hole from rest "at infinity" is described by the vector [itex]\partial / \partial T - \sqrt{2M / r} \partial / \partial r[/itex], where M is the mass of the hole in units where G = c = 1.

(3) The Kruskal chart. Here what we normally think of as "r" and "t" (or "T" in the Eddington or Painleve charts) are not coordinates at all: they are functions of the coordinates that are used to label curves. The actual coordinates T, X in the Kruskal chart don't have a straightforward physical interpretation, but they do have a key property that makes the chart nice for seeing the global structure of the spacetime: their directional derivatives work just like the ones for the standard Minkowski coordinates of special relativity. In other words, [itex]\partial / \partial T[/itex] is timelike everywhere, and [itex]\partial / \partial X[/itex] is spacelike everywhere, and their relationship is such that null curves (light rays) are always 45 degree lines in the chart.

In this chart, lines of constant r are hyperbolas outside and inside the horizon; and the horizon itself, r = 2M, is a null line, i.e., a 45-degree line. Actually, it is a *pair* of 45 degree lines in the "maximally extended" Kruskal chart, which is mathematically well defined but is not physically realistic (again, that's a whole other post); these lines are the asymptotes of the hyperbolas for r > 2M and r < 2M. For r > 2M, the hyperbolas are more vertical than horizontal, and for r < 2M, they are more horizontal than vertical, so it's easy to see how the nature of the r coordinate changes.

Lines of constant Schwarzschild t in the Kruskal chart are straight lines radiating from the origin (T = 0, X = 0, which corresponds to the point where the two horizon lines for r = 2M, the asymptotes of the r hyperbolas, cross). The exterior lines radiate to the left and right, and the interior lines radiate up and down. So again, it's easy to see how the nature of the Schwarzschild t coordinate changes from exterior to interior: the lines of constant t are obviously spacelike in the exterior and timelike in the interior.

Unfortunately, I don't know a simple way to describe how the lines of constant Painleve time or Eddington-Finkelstein time T (technically they aren't quite the same set of lines, but they're close) look on the Kruskal chart. But they are spacelike lines in both the exterior and interior regions.
 
  • #72
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The two rocket do have some gravitational attraction. Let us say that they start out separated by a distance L1 side by side. After 50 years Earth time we see through our telescope (at year 100 on Earth because the light had to return 50LYs) the distance is now L2 (a number smaller than L1). Likewise on the rockets the people (old people) see a distance L2. But they have aged only 1 year (pick gamma so this is so). The people on Earth conclude the gravitational force is low due to changing the separation by L1-L2 in 50 years where as the folks on the rocket conclude the force is larger due to a L1-L2 change in only 1 year. It seems like the rocket folks will think there is more force???
 
  • #73
PeterDonis
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30,718
9,701
Let us say that they start out separated by a distance L1 side by side. After 50 years Earth time we see through our telescope (at year 100 on Earth because the light had to return 50LYs) the distance is now L2 (a number smaller than L1).
Is the distance "side by side" perpendicular to the direction of the rockets' motion? If so, it won't appear to change as viewed from Earth (or from the rockets, of course) due to the rockets' motion.
 
  • #74
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3
Yes side by side. There is a small gravitational pull. Both rocket have some mass, rest mass or rest mass time gamma we can argue about but either way they have mass and gravitational attraction and over 50 years even a small attraction adds up.
 
  • #75
1,895
332
I think that this scenario you have postulated needs to be nailed down more precisely. How about giving some actual numbers? You don't need to give many; just the following:

(1) The rest mass of the rockets. (Just one number, we'll assume it applies to both rockets.)
To simplify the scenario let's assume two hypothetical spherical symmetric mass distributions with identical radius r and identical rest mass m that can be superposed without any interaction except gravity. If both objects are at rest they shouldn't collapse to a black hole. That means

[itex]r > \frac{{4 \cdot G \cdot m}}{{c^2 }}[/itex]

If you need actual numbers let's take r = 1 m and m = 3.354·1026 kg (59% the mass of Saturn).

Deformations due to the tidal forces shall be neglected.

(2) The distance between the rockets whey they are supposedly inside each other's Schwarzschild radius.
Due to the assumption above they will never be inside each other's Schwarzschild radius and if they are at rest they even can not be inside the Schwarzschild radius of the entire system. But if the bodies are moving with an identical absolute value v of the velocities this will happen for

[itex]d < 2 \cdot \left( {r_s - r} \right)[/itex]

with the common Schwarzschild radius

[itex]r_S = \frac{{4 \cdot G \cdot m}}{{c^2 \cdot \sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}
[/itex]

As the distance d (measured from center to center) can not be negative this is possible for

[itex]v > c \cdot \sqrt {1 - \left( {\frac{{4 \cdot G \cdot m}}{{r \cdot c^2 }}} \right)^2 }[/itex]

With the above mentioned values this minimum velocity would be 2,602·107 m/s (8,68% the speed of light).

(3) The relative directions the rockets are traveling in; this will fix the combined momentum of the rockets.
To avoid a resulting momentum or angular momentum it should be a head-on collision.
 

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