relativity dilemma #2

[grav]

Okay, say a traveller accelerates away from Earth at 1g. Then he should never receive any light that lies further than one light-year behind him, right? This would be the Rindler horizon. Now after quite some time, another traveller accelerates away from Earth in the same direction at 10g, from behind the first's horizon. The relative proper acceleration between the first and the second would be 9g, correct? The relevant acceleration between them might be less than that, however, but certainly not zero or negative. So there should be some positive acceleration between the first and the second, and so the second should eventually catch up to the first within finite time.

But let's say the second emitted a pulse of light when he left, and according to relativity, no matter how fast he travels, he will never catch up to the light. So that light is forever in front of him. That means, then, that if he were to eventually catch up to the first traveller, the light will have already passed the first and now be somewhere in front of both of them. But the first traveller can never have observed the light in finite time because it originated from behind his horizon. That means the second traveller can never catch up to the first, regardless of his acceleration. But the relevant acceleration should always be positive, right?

I suddenly remember something publius was saying in the Interstellar Traveller thread about a negative acceleration behind the horizon, but I think I thought that was for the apparent acceleration for a pulse of light or something. I'll look back through it. I suppose it must follow for another accelerating traveller that originates from behind the horizon as well, regardless of the original acceleration, right?

[publius]

Grav,

The difference in proper acceleration, which is just the "force you feel in your frame", or more generally, the deviation from your geodesic as measured by your own proper ruler and clock, is actually irrelevant for comparisons like you're making.

The first observer never even sees the second observe leave home if that second observer leaves after the horizon "wave" (as seen in the frame of the earth) passes by them. The first observer never even sees that proper time happen.

The second observer goes into his own Rindler frame, but that frame is "out of sync" with the first so to speak. That second observer can see the first obsever accelerating away (and that will transform to something very different in his own, second, Rindler frame), but the first observer nevers see any of it happen, as it is behind is own horizon.

You've already got the formulas for the coordinate accelerations, velocity, and distance travelled in a Lorentz frame watching someone accelerate. Use them for accelerating travellers, where one starts out behind the horizon of the other. What is the distance between them (in your frame) vs time? Let the 'a', the proper acceleration of the second, chasing observer get as large as you like. What is that distance vs time?

-Richard

[grav]

I'll work on that, but I'm really just wondering about what the equivalence principle says about this. If the first traveller is stationary and a second begins to accelerate toward him at relevant acceleration of 9g from one light-year away, then that should be exactly the same as the first acclerating at 1g and the second at 10g, with some slight adjustment depending on whether it is a relevant or proper acceleration. Otherwise, what are these accelerations relative to? Only to each other at this point, right? The only difference here is that the first has been travelling for a while and has aquired a relevant velocity as well. But we could then just say that both begin accelerating at the same time with one light-year distance originally between them when they were stationary to each other, and these accelerations are then relative to another point that remains stationary. Wouldn't a stationary observer eventually see both travellers arrive at the same relevant distance?

[01101001]

Okay, say a traveller accelerates away from Earth at 1g. Then he should never receive any light that lies further than one light-year behind him, right?

Question: where does the one lightyear distance come from? Because the traveller happened to have left a planet whose orbit took 1 year?

[publius]

Question: where does the one lightyear distance come from? Because the traveller happened to have left a planet whose orbit took 1 year?


No, it comes from that fact that c^2/a, the distance to the Rindler horizon, where 'a' is the proper acceleration, is just about equal to 1 light year when a is 1 earth g. Or put another way, it's a cute coincidence that we live on a planet whose orbital period is about the same time as c divided by its surface gravity.


-Richard

[grant hutchison]

Question: where does the one lightyear distance come from? Because the traveller happened to have left a planet whose orbit took 1 year?In a roundabout, entirely coincidental way, yes: it so happens that the length of our year makes the distance light travels in a year similar to the distance that falls out of the relativistic acceleration.
The actual distance from the equation is c2/a, which at 1g acceleration is ~(3x108 x 3x108)/10 = 9x1015 metres. And, since there are ~3x107 seconds in a year, one light year is ~3x108 x 3x107 = 9x1015 metres.
So it's a coincidence.

Grant Hutchison

[Kwalish Kid]

Otherwise, what are these accelerations relative to? Only to each other at this point, right? The only difference here is that the first has been travelling for a while and has aquired a relevant velocity as well. But we could then just say that both begin accelerating at the same time with one light-year distance originally between them when they were stationary to each other, and these accelerations are then relative to another point that remains stationary. Wouldn't a stationary observer eventually see both travellers arrive at the same relevant distance?
I think, if you pick a reference frame in which to found the actual accelerations, you will see that most of your difficulties disappear. Often apparent problems in SR are due to forgetting that times and distances vary between frames.

[grant hutchison]

Or put another way, it's a cute coincidence that we live on a planet whose orbital period is about the same time as c divided by its surface gravity.Which I first encountered in a novel by TJ Bass: either The Godwhale or its sequel. Bass's characters discover that the human species emerged at a latitude and geological epoch which gave precise equality for yr = c/g, as if this were some cosmic law in operation behind the evolution of intelligence. When I first read the story, in the 70s, I recall being particularly struck by the fact that the dimensional analysis was right - velocity over acceleration gives time, so the "equality" works in any set of units.
I spent an hour or so wondering if Bass might not have stumbled on some stunning truth ...

Grant Hutchison

[01101001]

In a roundabout, entirely coincidental way, yes: it so happens that the length of our year makes the distance light travels in a year similar to the distance that falls out of the relativistic acceleration.
[...]
So it's a coincidence.

So after as little as 1 second of travel away from earth, at 1g, then I cannot receive light from anything, say Alpha Centauri, more than about 1 lightyear behind me? That doesn't seem right. Is it that light leaving a distant object at the same time I start accelerating cannot reach me?

I didn't follow all the Rindler talk, but does it really imply what grav stated:


Okay, say a traveller accelerates away from Earth at 1g. Then he should never receive any light that lies further than one light-year behind him, right?

Never? Not even right away? Not even light that was already on its way to me? Or does that "light that lies further" mean the photons themselves from one lightyear cannot reach me.

[grav]

So after as little as 1 second of travel away from earth, at 1g, then I cannot receive light from anything, say Alpha Centauri, more than about 1 lightyear behind me? That doesn't seem right. Is it that light leaving a distant object at the same time I start accelerating cannot reach me?

I didn't follow all the Rindler talk, but does it really imply what grav stated:



Never? Not even right away? Not even light that was already on its way to me? Or does that "light that lies further" mean the photons themselves from one lightyear cannot reach me.The pulses that were originally beyond the distance to the Rindler horizon behind us when we begin to accelerate will never catch up to us. Those that were in front of it (already on their way) will reach us, but we will receive them with less and less frequency and they will become more and more redshifted and time dilated. All distant objects (behind us) will eventually just fade to oblivion, and we will never see more than one year of time (t=c/a) pass for anything behind us along our line of motion.

[EDIT-Distant objects behind us will eventually appear to fade to oblivion, but it will take an eternity for this to happen, so they will never completely "disappear" within finite time.]

[Ken G]

So after as little as 1 second of travel away from earth, at 1g, then I cannot receive light from anything, say Alpha Centauri, more than about 1 lightyear behind me? That doesn't seem right.What isn't making sense is that you're forgetting to include that you have to maintain the 1 g acceleration forever, and nothing that happened before you started to accelerate "counts". That's implicit in the global Rindler metric. Another way to say that is, metrics are really local animals, and have to be globally integrated including everything that is happening, but if you keep the conditions constant forever, then you can more easily extend local metrics globally.

Never? Not even right away? Not even light that was already on its way to me? Or does that "light that lies further" mean the photons themselves from one lightyear cannot reach me.

The latter. The way I think of it, the Rindler metric is a device that is useful for constant acceleration, but it isn't "real" because no nonlocal metric is "real". Reality is determined by local observers, and nonlocal ones can only conceptualize reality. The best way to do that conceptualization is just to rely on the local observers, and then transform their results into whatever global coordinate system you like. But don't be surprised is strange things happen when that global coordinate system is poorly chosen (as would be a north/south/east/west system extrapolated to the North pole). Rindler is poorly chosen unless you throw out anything that happened before you accelerated, and you maintain constant acceleration. It's not a "real" GR metric, which are always local and just need to know what true gravity is doing. But it can be used to show that light emitted from 1 LY away when you accelerate won't ever reach you, so it has its uses.

[01101001]

Thanks. I'm not quibbling with Rindler. I'm not even much interested in it. MEGO (http://www.answers.com/topic/mego-computer-jargon-1). The language grav used just threw me. You have corrected my interpretation.

Please go on about your business. Thanks, again.

[grav]

In looking through some material about the Lorentz contraction, I have come across something important that refers to the Rindler horizon. When I first attempted to find the values for the Rindler horizon, I used Euclidean values to compare to and get an idea of what was going on before jumping into relativity. When I used relativistic terms, I found the same thing, but twice as great, and found it exactly as Rindler describes. The thing is, though, I did it wrong, by still continuing to use a Euclidean form, and then so probably did Rindler. That might make this ATM, though, so if so, please move it to there.

What I was doing was finding the values for the relevant velocites, and moving the accelerating observer ahead accordingly, but keeping c constant the whole time. The thing is, though, that by keeping c constant, I was actually finding for a stationary observer, like freezing the accelerating observer momentarily while the light caught up over that normal distance while stationary. What I should have been doing was adding the two velocities together for the speed of light. Otherwise, the light actually slows down in respect to the observer over time instead of remaining constant for the particular frame of reference we are considering. After accelerating indefinitely this way, and by keeping the speed of light at c as if he were stationary, the speed of light would eventually reach zero. But taking the SR approach to this, if the speed of light is constant for any instantaneous velocity, then it is constant during the entire acceleration. And that means it would be c when considering the observer as stationary, but c+v when the observer is travelling at v, which is allowed when considering the effects from the same frame of reference, so that (c+v)-v=c. What that would say about the Rindler horizon, then, is that light always catches up to an accelerating observer from any distance in finite time and so does another observer accelerating a little faster behind him to catch up. I would have to rework the formulas to find out the details for this, though. But that would basically mean that the Rindler horizon does not exist, except for cases where an acceleration takes place between two points that really are stationary, as with a black hole, but that would just be the event horizon. I still have to think about this some more, though.

[publius]

Grav,

Something you may not exactly be aware of is the constancy of the speed of light holds globally only for inertial observers (and inertial observers in flat space-time, zero Riemann curvature).

The Rindler observer sees light going past him, where he is, his origin, at c, but the speed of light is not constant in his coordinates far away. This is the expression for the coordinate speed of light in Rindler coordinates:

c(x) = c*sqrt(1 + ax/c^2), where a is his proper acceleration.

It increases without limit ahead of him, and slows to zero at the horizon, x = -c^2/a.

THe Rindler observer never sees light he fired behind him cross the horizon at all, just like everything else on a horizon crossing path.

-Richard

[grav]

Grav,

Something you may not exactly be aware of is the constancy of the speed of light holds globally only for inertial observers (and inertial observers in flat space-time, zero Riemann curvature).

The Rindler observer sees light going past him, where he is, his origin, at c, but the speed of light is not constant in his coordinates far away. This is the expression for the coordinate speed of light in Rindler coordinates:

c(x) = c*sqrt(1 + ax/c^2), where a is his proper acceleration.

It increases without limit ahead of him, and slows to zero at the horizon, x = -c^2/a.

THe Rindler observer never sees light he fired behind him cross the horizon at all, just like everything else on a horizon crossing path.

-RichardActually, that would not be the speed of light at that distance currently, but the speed when the light actually catches up to the observer from that original distance. Notice that the speed of light coming from a distance greater than -c^2/a does not become negative, but complex, simply meaning that it never catches up and so represents a physical impossibility, and so gives a complex value. Therefore, that would be the speed for when the observer actually receives and measures it, according to this equation, and anything beyond -c^2/a will never be received in order to be measured at any "real" speed.

[grav]

Ironically, although the Rindler horizon is meant for relativistic effects, it is really found only by Euclidean means. In flat space, it is half that of Rindler. By using relativistic effects for the instantaneous relevant velocities, but still using the speed of light as c, instead of that relative to the observer, then we get the same distance as that of Rindler. If we do it correctly, however, and consider the light to be travelling at c+v in respect to the observer travelling at each instantaneous velocity of v, for (c+v)-v=c, then we actually get no horizon at all, but only that same time that the light would have taken to travel to the observer from the original distance, as if he had never begun accelerating. The reason is that in Euclidean, we are considering that the observer can obtain any speed over time, but the speed of light remains at c, so the observer can actually catch up to the light in front of him and pass it. With relativity, however, he can never catch the light that moves away in front of him, and therefore, by the same token, he can never escape the light coming from behind him, and so he can never outrun it, and will always receive light from any finite distance in some finite time, regardless of how fast he travels.

[grav]

c(x) = c*sqrt(1 + ax/c^2), where a is his proper acceleration.
publius,

I have been trying to reproduce that formula, trying different methods that might have been used for this, even working backwards from it, but to no avail. I want to see where any error might lie in its structure, if it does. Would you happen to know how this can be found, built up from the other relativistic formulas, such as for relevant velocity and acceleration and such?

[publius]

publius,

I have been trying to reproduce that formula, trying different methods that might have been used for this, even working backwards from it, but to no avail. I want to see where any error might lie in its structure, if it does. Would you happen to know how this can be found, built up from the other relativistic formulas, such as for relevant velocity and acceleration and such?
Grav,

The easiest way to see that is from the Rindler metric itself:

ds^2 = (1 + gx/c^2) (c dt)^2 - dx^2.

Light always follows a "null path", ds = 0, so just set ds = 0, and get the coordinate speed (as a function of x) in the Rindler frame:

0 = (1 + gx/c^2) (c dt)^2 - dx^2 -->

(dx/cdt)^2 = (1 + gx/c^2) -->

dx/dt = c * sqrt(1 + gx/c^2)

Now, the Rindler metric "simply" comes from the transform from Minkowski to the local ruler and clock of the accelerating observer. One can get the same thing by "simply" transforming a null geodesic in Minkowski,
x = x0 +/- ct

into the Rindler coordinates using the transform relations directly.


-Richard

[grav]

Grav,

The easiest way to see that is from the Rindler metric itself:

ds^2 = (1 + gx/c^2) (c dt)^2 - dx^2.

Light always follows a "null path", ds = 0, so just set ds = 0, and get the coordinate speed (as a function of x) in the Rindler frame:

0 = (1 + gx/c^2) (c dt)^2 - dx^2 -->

(dx/cdt)^2 = (1 + gx/c^2) -->

dx/dt = c * sqrt(1 + gx/c^2)

Now, the Rindler metric "simply" comes from the transform from Minkowski to the local ruler and clock of the accelerating observer. One can get the same thing by "simply" transforming a null geodesic in Minkowski,
x = x0 +/- ct

into the Rindler coordinates using the transform relations directly.


-RichardThanks, publius. :)

[grav]

Before relativity, the apparent motion of bodies depended on how they each moved in respect to a medium. The formulas related them as such, and what an observer at K watching K' might see would be different than what K' saw at K. By removing that medium, as relativity does, then all motions are only seen in respect to each other, and so there is always a reciprocity between them. That is what Einstein did with SR. He did have trouble with GR, however, and could never completely let loose of the idea of an ether. But either there is an ether or there isn't. We can't have it both ways. If there is not, which is what the whole idea of relativity is based upon, then there is necessarily reciprocity for accelerating observers as well, since there is nothing else to relate to except each other. In that case, what an observer "stationary" at K sees of an accelerating observer at K' will be exactly the same as what the accelerating observer sees at K. If we invoke the Mach principle here as Einstein attempts to do with this, then we go right back to relating to a stationary frame of reference, as with an ether, and we must consider velocities in relation to this frame, just as with the ether. But by considering it from the frame of reference of the stationary observer, we eliminate the potential mishap of misinterpreting for the speed of light within the accelerating observer's frame, by unintentionally relating it to some absolute frame in space (stationary). And so what the stationary observer sees of the Rindler horizon in relation to the accelerating observer (or if everything else accelerated away together), which is none (but an increasing redshift and time dilation still occur), is exactly as it would be for the accelerating observer for everything else that remains stationary.