Does mass really increase with speed

[atyy]

Do you mean why *does* one say this? I am saying the (rate of change of the) expansion *is* due to the gravity of stress-energy.

Well, I was asking why it seemed that the rate of change of expansion, rather than the expansion itself was called "gravity".

No, it isn't. The redshift of Galaxy X observed at Earth is a physical observable; anyone who calculates it will calculate the same answer. And the Earth is not at rest in the "comoving" frame, so it's not a preferred observer anyway. The point is that the *change with time* of a particular galaxy's redshift, if it is far enough away not to be gravitationally bound to our local cluster of galaxies, is an observable that tells us about the gravity of the universe as a whole, i.e., the combined gravity of all the objects in it.



I don't have to define this to make the argument I'm making. I don't have to interpret the redshift as actual "relative motion"; I only have to interpret its change over time as "deceleration".

Anyway, I wasn't referring to the redshifts, but to the idea that expansion is "gravity". The expansion is defined relative to a set of preferred observers.

My suggestion about relative motion not being defined was not related to any of those. It was trying to think of a completely different way of answering the OP question.

Not sure which of my "what ifs" you were responding to here. I am not aware of any particular solutions for masses moving on parallel worldlines but in opposite directions. If you mean the Chazy-Curzon vacuum, no, the two massive objects in it are not charged. I was hoping to find a good quick reference about it online, but I haven't.

I was asking about the one with two massive objects stationary relative to each other.

 

[PeterDonis]

Well, I was asking why it seemed that the rate of change of expansion, rather than the expansion itself was called "gravity".

Because the expansion itself is just due to an initial impulse that got applied to all the matter in the universe. (We're talking about the standard, zero cosmological constant FRW model, so we're leaving out stuff like inflation and dark energy.) Without gravity, the expansion would just keep on at the same rate forever; the limiting case of the FRW model with zero stress-energy tensor does exactly this. It's the *change* in the expansion, the fact that it slows down because there is nonzero stress-energy present, that indicates gravity.

The expansion is defined relative to a set of preferred observers.

This is true in the sense that the usual meaning of the phrase "the universe is expanding" is that the "comoving" observers are all moving away from each other (they all see each other as having a redshift). But those observers are not picked out arbitrarily: they are picked out by the fact that they see the universe as homogeneous and isotropic. In other words, their worldlines, and the 3-surfaces orthogonal to them, match up with a particular symmetry of the spacetime. So actually, the fact that the universe is expanding is a property of the spacetime, not just of the observers.

I was asking about the one with two massive objects stationary relative to each other.

Ah, ok. If I find any good online source on the Chazy-Curzon vacuum I'll post a link.

 

[atyy]

Because the expansion itself is just due to an initial impulse that got applied to all the matter in the universe. (We're talking about the standard, zero cosmological constant FRW model, so we're leaving out stuff like inflation and dark energy.) Without gravity, the expansion would just keep on at the same rate forever; the limiting case of the FRW model with zero stress-energy tensor does exactly this. It's the *change* in the expansion, the fact that it slows down because there is nonzero stress-energy present, that indicates gravity.

I see, so you count even the zero stress-energy tensor FRW case as expanding. But isn't that spacetime flat, so although it could be described as expanding, those observers are no longer reflecting such a particular symmetry of the spacetime, are they?

This is true in the sense that the usual meaning of the phrase "the universe is expanding" is that the "comoving" observers are all moving away from each other (they all see each other as having a redshift). But those observers are not picked out arbitrarily: they are picked out by the fact that they see the universe as homogeneous and isotropic. In other words, their worldlines, and the 3-surfaces orthogonal to them, match up with a particular symmetry of the spacetime. So actually, the fact that the universe is expanding is a property of the spacetime, not just of the observers.

Agreed. It's basically a matter of convention whether one says the expansion is observer-dependent or not. I was picking the former.

Ah, ok. If I find any good online source on the Chazy-Curzon vacuum I'll post a link.

Thanks. At least now I know what to google.

I guess the odd thing in GR is that one could try to say all questions are meaningless unless they talk about gauge invariant quantities. OTOH, we can make any gauge variant quantities gauge invariant by invariantly specifying events and worldlines and frames. The latter could be a bit restricted by saying that we only allow worldlines and frames that are not test particles, ie. their stress-energy must be accounted for in the stress-energy tensor in the Einstein field equations. But that's probably too strong, because even if we could do that, those observers would be left without light beams and test particles to probe their spacetime.

 

[PeterDonis]

I see, so you count even the zero stress-energy tensor FRW case as expanding. But isn't that spacetime flat, so although it could be described as expanding, those observers are no longer reflecting such a particular symmetry of the spacetime, are they?

Yes, you're right, I was speaking loosely. The case where the stress-energy tensor is exactly zero is just Minkowski spacetime with a weird coordinate chart. But if we look at a case with non-zero stress-energy tensor, and gradually let the density approach zero, we find that the deceleration goes to zero. So viewed as a limiting case, the "zero energy, zero deceleration" case just shows that it's the deceleration that indicates gravity, not the expansion itself.

Agreed. It's basically a matter of convention whether one says the expansion is observer-dependent or not. I was picking the former.

No real argument, but it's worth noting that the fact that the set of comoving observers exists and that they all see the universe as expanding--i.e., the fact that it is possible to find a frame in which the universe is expanding everywhere--places limits on what observers in other states of motion can see as well. For instance, there is no observer who will see the universe contracting everywhere (assuming that all observers have the same direction of time). In fact, I believe there's no observer who will even see a "preponderance" of contraction over expansion (but I'll have to think some more to formulate that intuitive guess more precisely).

I guess the odd thing in GR is that one could try to say all questions are meaningless unless they talk about gauge invariant quantities. OTOH, we can make any gauge variant quantities gauge invariant by invariantly specifying events and worldlines and frames.

I agree, in the sense that specifying worldlines and frames specifies *which* particular invariants you are talking about. However, I would *not* say that this counts as making a variant quantity into an invariant quantity. Or at least, I would not word it that way. This may be a matter of choice of words rather than physics, but I think it's important. I would say that what a particular observer actually observes, in the sense of observable numbers (such as redshifts) can always be specified in an invariant way--that is, it can always be specified in terms of *only* invariant quantities, i.e., only scalars (which may be formed by contracting covariant objects like vectors and tensors). What we sometimes call "frame-dependent" quantities can actually be specified in frame-independent terms; for example, the energy you observe an object as having is the contraction of its 4-momentum with your 4-velocity.

The latter could be a bit restricted by saying that we only allow worldlines and frames that are not test particles, ie. their stress-energy must be accounted for in the stress-energy tensor in the Einstein field equations. But that's probably too strong, because even if we could do that, those observers would be left without light beams and test particles to probe their spacetime.

Yes, I agree, this restriction would be way too strong.

 

[PeterDonis]

That means this 4-acceleration is independent from the velocity of the body - even if it is so fast that we are inside our common Schwarzschild radius?

On re-reading I realized I may not have made clear the scenario I was describing, with an observer "hovering" above a gravitating body. By "hovering" I mean maintaining a constant height above the gravitating body--in other words, the relative velocity of the "hoverer" and the body is zero, and stays that way. So there is no "velocity" involved.

 

[atyy]

I agree, in the sense that specifying worldlines and frames specifies *which* particular invariants you are talking about. However, I would *not* say that this counts as making a variant quantity into an invariant quantity. Or at least, I would not word it that way. This may be a matter of choice of words rather than physics, but I think it's important. I would say that what a particular observer actually observes, in the sense of observable numbers (such as redshifts) can always be specified in an invariant way--that is, it can always be specified in terms of *only* invariant quantities, i.e., only scalars (which may be formed by contracting covariant objects like vectors and tensors). What we sometimes call "frame-dependent" quantities can actually be specified in frame-independent terms; for example, the energy you observe an object as having is the contraction of its 4-momentum with your 4-velocity.

So is there such a thing as a gauge variant quantity in GR if we are always allowed to put test events in spacetime?

The only thing that comes to mind is Shapiro delay, but that's not really even gauge variant since it's a delay compared to a non-existent Newtonian trajectory. As I understand, the real observable in Shapiro delay is the logarithmic form.

 

[PeterDonis]

So is there such a thing as a gauge variant quantity in GR if we are always allowed to put test events in spacetime?

Not sure what you mean here. I was trying to say that you don't even need the concept of "gauge variant" quantities at all; you can express everything in terms of invariants, even things that are often taken to be "frame dependent".

 

[atyy]

Not sure what you mean here. I was trying to say that you don't even need the concept of "gauge variant" quantities at all; you can express everything in terms of invariants, even things that are often taken to be "frame dependent".

Yes. What I'm asking is whether we can even define "gauge variant" as something distinct from "gauge invariant".

 

[PeterDonis]

Yes. What I'm asking is whether we can even define "gauge variant" as something distinct from "gauge invariant".

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 $A_{u}$ is gauge variant, but the field tensor $F_{uv}$ is gauge invariant. The reason we say this is that we can change the potential by the gradient of a scalar, $A'_{u} = A_{u} + \partial_{u} \phi$, without changing any physical predictions; but the reason it doesn't change any physical predictions is that it doesn't change the field tensor $F_{uv} = \partial_{u} A_{v} - \partial_{v} A_{u}$ (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 $F_{uv}$; 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.

 

[timmdeeg]

Yes, you're right, I was speaking loosely. The case where the stress-energy tensor is exactly zero is just Minkowski spacetime with a weird coordinate chart.

Talking about the empty universe there are two cases, the empty FRW universe (hyperbolic) and the Milne universe (flat Minkowski spacetime), which can be converted to each other by coordinate transformation. Interestingly the kinematic explosion model (Milne) is equivalent to the expansion model (FRW) in this case.

 

[PeterDonis]

timmdeeg, I see this is your first post. Welcome to PhysicsForums!

which can be converted to each other by coordinate transformation.

This is another way of saying they are the *same* spacetime, just described by different coordinate charts. That is, they both describe exactly the same physics.

Interestingly the kinematic explosion model (Milne) is equivalent to the expansion model (FRW) in this case.

Yes, because they both describe the same spacetime and the same physics.

 

[timmdeeg]

Thanks for your welcome, Peter.

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.

 

[DrStupid]

$r < \frac{{2 \cdot \gamma }}{{c^2 }}\sqrt {\left( {\frac{{m_1 \cdot c}}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot c}}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 - \left( {\frac{{m_1 \cdot v_1 }}{{\sqrt {c^2 - v_1^2 } }} + \frac{{m_2 \cdot v_2 }}{{\sqrt {c^2 - v_2^2 } }}} \right)^2 }$

Do you have a reference for this formula?

The distance of the bodies shall be below their common Schwarzschild radius:

$r < r_s$

The Schwarzschildradius depends on the rest mass according to

$r_s = \frac{{2 \cdot \gamma \cdot m}}{{c^2 }}$

The rest mass is related to total energy and total momentum according to

$E^2 = m^2 \cdot c^4 + p^2 \cdot c^2$

and total energy and total momentum depends on the rest masses and velocities of the two bodies according to

$E = \frac{{m_1 \cdot c^2 }}{{\sqrt {1 - \frac{{v_1^2 }}{{c^2 }}} }} + \frac{{m_2 \cdot c^2 }}{{\sqrt {1 - \frac{{v_2^2 }}{{c^2 }}} }}$

and

$p = \frac{{m_1 \cdot v_1 }}{{\sqrt {1 - \frac{{v_1^2 }}{{c^2 }}} }} + \frac{{m_2 \cdot v_2 }}{{\sqrt {1 - \frac{{v_2^2 }}{{c^2 }}} }}$

All together results in the formula above.

 

[PeterDonis]

The Schwarzschild radius depends on the rest mass according to

$r_s = \frac{{2 \cdot \gamma \cdot m}}{{c^2 }}$

This is incorrect; there is no relativistic $\gamma$ factor in the formula for the Schwarzschild radius. The correct formula is

$$r_{s} = \frac{2 G m}{c^{2}}$$

where $G$ is Newton's gravitational constant.

 

[DrStupid]

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.

You are right, but that doesn't mean that gravitational mass does not increase with speed. It might depend on the circumstances.

 

[DrStupid]

This is incorrect; there is no relativistic $\gamma$ factor in the formula for the Schwarzschild radius. The correct formula is

$$r_{s} = \frac{2 G m}{c^{2}}$$

where $G$ is Newton's gravitational constant.

In my notation above $\gamma$ is the gravitational constant. Let me know if you have serious comments.

 

[PeterDonis]

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.

 

[PeterDonis]

In my notation above $\gamma$ 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.

 

[PeterDonis]

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.

 

[DrStupid]

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.

 

[atyy]

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 $A_{u}$ is gauge variant, but the field tensor $F_{uv}$ is gauge invariant. The reason we say this is that we can change the potential by the gradient of a scalar, $A'_{u} = A_{u} + \partial_{u} \phi$, without changing any physical predictions; but the reason it doesn't change any physical predictions is that it doesn't change the field tensor $F_{uv} = \partial_{u} A_{v} - \partial_{v} A_{u}$ (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 $F_{uv}$; 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)?

 

[PeterDonis]

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.

 

[PeterDonis]

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.

 

[timmdeeg]

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?

 

[PeterDonis]

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.

 

[timmdeeg]

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.

 

[Per Oni]

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.

 

[PeterDonis]

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.

 

[Naty1]

PeterDonis: some great explanations...thank you!...

 

[timmdeeg]

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

 

[PeterDonis]

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.

 

[DrStupid]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

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.)

 

[timmdeeg]

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.