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grav
2009-Jan-08, 11:24 PM
I have a dilemma, mentioned here (http://www.bautforum.com/questions-answers/83164-simple-relativity-question-2.html#post1404895). Here is the first part of it. Observers are positioned stationary every few hundred meters within a very, very long tunnel. One observer in the center of the tunnel then instantly accelerates to some relative speed toward one end. Due to the new relative speed, the length of the tunnel is now contracted to the observer and the distance between all of the stationary observers has contracted according to the moving observer as well. The moving observer also measures all of the stationary observers travelling past him at the same relative speed regardless of distance. Light coming from stationary observers behind him is all redshifted by the same amount regardless of the distance of the source and light coming from stationary observers in front of him is all blueshifted by the same amount regardless of distance.

Okay, so now, instead of an instant acceleration to v, let's just give our observer a constant acceleration toward one end of the tunnel. As the observer's relative speed to the rest frame increases, contraction of the tunnel takes place. Since the entire length steadily contracts, that means that our observer will see the stationary observers pull closer to him in front and in back at a rate that is proportional to the distance. He will see stationary observers at some distance to the front and back pulling closer to him at some rate due to the contraction while stationary observers at twice the distance pull closer at nearly twice the rate, only limited by the speed of light at the greatest distances. So from his point of view, minus the local relative speed between himself and the tunnel, all stationary observers are moving toward him with a relative speed which increases with distance from him as he accelerates.

First of all, is my scenario correct? If so, what happens when he simply turns off his engines and ceases to accelerate? Shouldn't all stationary observers which are now moving toward him from front and back with relative speeds that increase with distance continue to move toward him inertially at those same speeds? Or would turning off the engines somehow affect space-time since he is no longer accelerating and space-time contracting, so that the stationary observers suddenly stop travelling toward him except for the local relative speed to the tunnel which continues to persist, now all along the length?

gzhpcu
2009-Jan-09, 07:24 AM
Your scenario seems correct to me.

When he turns off his engines and ceases to acclerate, then, I would think, the stationary observers stop traveling toward him except for the local relative speed to the tunnel, etc. as you describe. However, they would not suddenly stop, but would start stopping when the contraction of space having ceased reaches them at the speed of light.

grant hutchison
2009-Jan-09, 04:38 PM
You need to bear in mind that the moving observer's acceleration continuously changes his location relative to the stationary observers. So the increasing Lorentz contraction astern is offset by the increasing distance of the stationary observers astern, while the increasing Lorentz contraction ahead compounds the velocity of approach of the observers ahead.
Measured in the accelerating observer's proper distance, his point of departure falls every more slowly behind, tending asymptotically to a limiting distance. Observers positioned ahead of the accelerating observer move in quickly, slow down, and then "pile up" in the space astern between the accelerating observer and his point of departure.
At any given instant for the accelerating observer, the stationary observers ahead and astern are distributed with equal spacing in his proper distance, the spacing dictated by Lorentz contraction. And that's the way they'll stay as soon the observer stops accelerating, although they'll now be moving relative to the coasting observer at a uniform, constant rate.

(All of this, of course, is a description of the coordinate behaviour under the Lorentz transformation. Visible effects will be different because of relativistic aberration.)

Grant Hutchison

grav
2009-Jan-10, 10:57 PM
Thank you very much, gzhpcu and Grant.

However, they would not suddenly stop, but would start stopping when the contraction of space having ceased reaches them at the speed of light.Gzhpcu, I'm wondering about your last statement now also. I can see that a material object doesn't react along its length faster than the speed of light due to the "scissor effect", and that if the sun "disappeared", the missing gravity wouldn't be felt at a distance faster than the speed of light also, but I'm not sure that a distortion in space-time as with something like gravity which affects all objects would be the same as the effect on a particular observer's perspective of space-time due to their own acceleration. It could be, though, I don't know. The Equivalence Principle or simultaneity might have something to say about that.

And that's the way they'll stay as soon the observer stops accelerating, although they'll now be moving relative to the coasting observer at a uniform, constant rate.
Grant, by your last statement, do you mean just the contracted distances will remain the same or the distances and the apparent relative speeds as well? In other words, do you have an idea about whether the apparent relative speeds of the stationary observers should remain the same as previously observed upon ceasing deceleration, although now travelling inertially, or if they should all stop at once except for the local relative speed, or if they should immediately stop as soon as the change in local space-time reaches them at the speed of light, or perhaps even gradually?

grant hutchison
2009-Jan-10, 11:06 PM
Grant, by your last statement, do you mean just the contracted distances will remain the same or the distances and the apparent relative speeds as well? In other words, do you have an idea about whether the apparent relative speeds of the stationary observers should remain the same as previously observed upon ceasing deceleration, although now travelling inertially, or if they should all stop at once except for the local relative speed, or if they should stop as the change in space-time reaches them at the speed of light?There's a relative velocity due to the observer's velocity at the time he stops accelerating. That will we the same for all stationary observers, and is what I meant by "moving relative to the coasting observer at a uniform, constant rate".
Superimposed on that was a coordinate effect generated by the changing Lorentz contraction. The Lorentz contraction ceases to vary as soon as the acceleration stops. So that "velocity" disappears as soon as acceleration stops. There's no influence from that which needs to propagate at light-speed: it's a coordinate effect, not an effect on the distant objects.

Grant Hutchison

grav
2009-Jan-10, 11:20 PM
There's a relative velocity due to the observer's velocity at the time he stops accelerating. That will we the same for all stationary observers, and is what I meant by "moving relative to the coasting observer at a uniform, constant rate".
Superimposed on that was a coordinate effect generated by the changing Lorentz contraction. The Lorentz contraction ceases to vary as soon as the acceleration stops. So that "velocity" disappears as soon as acceleration stops. There's no influence from that which needs to propagate at light-speed: it's a coordinate effect, not an effect on the distant objects.

Grant HutchisonOkay, great. Thanks again, Grant. That's what I was thinking too, that it's just a local coordinate effect, so all the stationary observers will immediately stop except for the local relative speed, which would be the relative speed the moving observer has with the stationary observers as he passes them while accelerating, since the Lorentz contraction won't apply to the local zero distance between them as they coincide.

So on to the second part of my dilemma. :) Will the apparent relative speeds due to the contraction that takes place while accelerating produce a blueshift of the light that is received from the stationary observers in either direction, which becomes greater with distance, minus the immediate local relative speed? If so, would the light that is received from a large distance behind suddenly change from tremendously blueshifted to equally redshifted from all distances behind once the moving observer becomes inertial?

grant hutchison
2009-Jan-10, 11:29 PM
So on to the second part of my dilemma. :) Will the apparent relative speeds due to the contraction that takes place while accelerating produce a blueshift of the light that is received from the stationary observers in either direction, which becomes greater with distance, minus the immediate local relative speed? If so, would the light that is received from a large distance behind suddenly change from tremendously blueshifted to equally redshifted from all distances behind?No. The relativistic Doppler depends on the relative motion of the frames at the instant of emission and the instant of reception. It doesn't care about what the observers have been doing with their coordinatization in the meantime.

Grant Hutchison

grav
2009-Jan-10, 11:36 PM
No. The relativistic Doppler depends on the relative motion of the frames at the instant of emission and the instant of reception. It doesn't care about what the observers have been doing with their coordinatization in the meantime.

Grant HutchisonOkay, but wouldn't that affect the concept of the Hubble expansion? I mean, if a contraction of space-time doesn't produce a blueshift, then expansion shouldn't produce a redshift either, right? Would that mean galaxies are travelling away from each other through space, then, not with it?

grant hutchison
2009-Jan-10, 11:54 PM
Okay, but wouldn't that affect the concept of the Hubble expansion? I mean, if a contraction of space-time doesn't produce a blueshift, then expansion shouldn't produce a redshift either, right? Would that mean galaxies are travelling away from each other through space, then, not with it?Different things.
As the Universe expands, the photon traverses space which is expanding.
In your scenario, the photon simply passes through a spacetime block which different observers are coordinatizing in different ways.

Analogy: Paris and New York are moving apart slowly because of continental drift: that's a real effect. You'll also get a different numerical value if you measure the distance in kilometres or miles: that's a coordinate effect.

Grant Hutchison

grav
2009-Jan-11, 01:01 AM
Different things.
As the Universe expands, the photon traverses space which is expanding.
In your scenario, the photon simply passes through a spacetime block which different observers are coordinatizing in different ways.Yes, that makes sense. I was trying to determine what difference there might be between the two scenarios and what you just said nailed it. I'm still not completely convinced, however, because it seems to me they are both coordinate effects, but now at least I know what I would need to look at and compare, so I'll start with just the inertial observer.

The moving observer is travelling at a relative speed to the tunnel of v. The stationary observers are positioned every light-second along the tunnel's length. A stationary observer at the end of the tunnel emits pulses of light every second, so that the pulses coincide with all of the other stationary observers in the tunnel simultaneously every second as well. The moving observer then coincides with one of the stationary observers at the same time that a pulse of light also coincides. According to the moving observer, the next stationary observer is a distance of L d ahead, where d is the distance according to the stationary observers of one light-second and L is the length contraction of that distance according to the moving observer. The simultaneity shift puts the next stationary observer future-forward according to the moving observer's clock by t = (L d) v / (c^2 - v^2), so a pulse passed the next stationary observer that long ago. According to the moving observer, then, the next stationary observer would have been at L d + v t when the pulse passed her, and the pulse would have since travelled a distance of c t since then, so the next pulse would now be at a distance of L d + v t - c t from the pulse the moving observer currently coincides with.

The frequency ratio between that emitted and that observed, therefore, would become fo / fe = (c / wo) / (c / we) = [c / (L d + v t - c t)] / (c / d) = d / [L d - (c-v) t] = d / [L d - (c-v) (L d) v / (c^2 - v^2)] = (1 / L) / [1 - v / (c+v)] = (1 / L) [c / (c+v)] = (c+v) / L = sqrt[(1 + v/c) / (1 - v/c)]. That only achieves normal Relativistic Doppler, of course, but I will also attempt the same thing by applying the formulas for acceleration if I can and see what happens, although it might take me a little while to work it through.

grav
2009-Jan-11, 01:42 AM
Well, I've once again come up against the same problem I've had a few times before that requires a resolution. If I apply the formula for the distance travelled by an accelerating observer directly, which is d = (c^2 / a) * sqrt[(1 + (a t / c)^2) - 1], then the distance between the stationary observers according to the moving observer will stay the same, with no contraction observed. I've still got a couple of threads up in the air due to this. The discussion in this thread, however, gives me an idea about this. I'm thinking that the relative speed attained, which is v = a t / sqrt[1 + (a t / c)^2], is the local relative speed of the moving observer to the part of the tunnel he currently coincides with. In that case, so should be the distance travelled. For instance, if the distance travelled from rest with respect to the tunnel is found to be 3 light-seconds, then the moving observer will have passed exactly 3 stationary observers. If he is trying to discern his distance from, say, the 10th stationary observer, then, he would will be 7 light-seconds away according to the stationary frame, so the same for the moving observer, but that distance will be Lorentz contracted according to the moving observer due to the current local relative speed. Does that sound correct?

grav
2009-Jan-11, 02:20 AM
Ahah. I think I'm starting to get it now, even without the math. It seems there might still be a blueshift in both directions, but it would not vary with distance or with the apparent relative speed to the stationary observers. The wavelengths would vary in the same way as the distances between the stationary observers according to the moving observer. In other words, although stationary observers at a very large distance would have a very large apparent relative speed, the distance between one stationary observer and the next would contract in the same way as those that are closer, so the moving observer measures the same distance between consecutive stationary observers at any distance from him. In the same way, the distance between consecutive pulses of light might contract for an overall blueshift, but regardless of and independent to the apparent relative speed of the emitter, only due to the Lorentz contraction with the local relative speed, plus normal Relativistic Doppler for the local relative speed itself. So either there will be an constant blueshift in both directions due to the contraction or not at all, but it should not become greater with greater distance of the emitter. I will keep working on it.

grant hutchison
2009-Jan-11, 02:48 AM
So either there will be an constant blueshift in both directions due to the contraction or not at all, but it should not become greater with greater distance of the emitter."Not at all" is the answer you're looking for.

Grant Hutchison

grav
2009-Jan-11, 05:41 AM
"Not at all" is the answer you're looking for.

Grant HutchisonYes, I'm not done working through the math yet (it's too much fun to pass up the opportunity - yeah, right :) ), but I'm thinking that would indeed make the most sense, so I would have to agree. As Ken G would also say, what happens at a distance for an observer seems more of a mental and mathematical construct of convience for the observer, not so much reality itself per say. So it makes more sense that what happens locally for the observer should occur smoothly all the way up to the point of shutting off the engines, whereas the frequency one measures at cutting them off would be the same frequency one measures when inertial with the same local relative speed.

gzhpcu
2009-Jan-11, 07:02 AM
There's a relative velocity due to the observer's velocity at the time he stops accelerating. That will we the same for all stationary observers, and is what I meant by "moving relative to the coasting observer at a uniform, constant rate".
Superimposed on that was a coordinate effect generated by the changing Lorentz contraction. The Lorentz contraction ceases to vary as soon as the acceleration stops. So that "velocity" disappears as soon as acceleration stops. There's no influence from that which needs to propagate at light-speed: it's a coordinate effect, not an effect on the distant objects.

Grant Hutchison
You are right, of course. Yet, even with the Lorentz contraction disappearing instantaneously, it will become apparent to the travelling observer only when the light from the stationary observers reaches him.

grav
2009-Jan-11, 06:27 PM
Okay, I've been trying to work it out on paper, but it got to be too involved and too much to keep up with so I wrote a program for it instead where I can just plug in the values for the acceleration, time of acceleration, and the original distances. But here's my problem now. When I find for the new relative speed for the stationary observers toward the moving observer, say close to c, and then find for the distance travelled by using the distance formula accordingly with the new distance Lorentz contracted on top of that, the apparent relative speed becomes greater than c. This is impossible because if two consecutive stationary observers send light pulses toward each other in the stationary frame, the pulse will coincide between them, but according to the moving observer the furthest stationary observer is moving faster than light toward him when one includes the contraction, so travels faster than the pulse according to him, and the two pulses would coincide somewhere behind the furthest stationary observer according to the moving observer instead, then, but this can't be since all observers must agree whether that event coincides between the two stationary observers or behind them. What do I do? What is the correct formula for distance travelled in a contracted frame?

grant hutchison
2009-Jan-11, 06:42 PM
There isn't a meaningful "distant travelled" in this scenario, for the accelerating observer. His coordinates keep changing. All you have is the instantaneous proper distance between him and his departure point, which approaches an asymptotic limit as acceleration continues. The stationary observers measure a distance travelled against time that asymptotically converges on lightspeed.

Grant Hutchison

grav
2009-Jan-11, 06:55 PM
There isn't a meaningful "distant travelled" in this scenario, for the accelerating observer. His coordinates keep changing. All you have is the instantaneous proper distance between him and his departure point, which approaches an asymptotic limit as acceleration continues. The stationary observers measure a distance travelled against time that asymptotically converges on lightspeed.

Grant HutchisonI can see now that even the apparent relative speed due to contraction cannot exceed light speed. What I was doing before was to find the distance travelled according to the stationary frame and then contract that for the accelerating observer, but I also realize now that by using the distance formula for that, I was using the stationary frame's clock as well. So what I guess I'll need to do is to find the distance travelled and the time on the accelerating observer's clock, both according to the stationary frame's clock, and then contract the distance and convert to the time on the accelerating observer's clock instead. I'll know I will most likely have the correct distance formula according to an accelerating observer when the result of the program matches that of normal Relativistic Doppler according to the current local relative speed. I did it before somewhere on here, but would you happen to know what the time on the accelerating observer's clock would read according to the stationary observer's clock offhand, or vice versa?

grav
2009-Jan-11, 08:16 PM
I did it before somewhere on here, but would you happen to know what the time on the accelerating observer's clock would read according to the stationary observer's clock offhand, or vice versa?Okay, I've got it. All I had to do was to integrate dt sqrt[1 - (v/c)^2], replacing v with the formula for the relative speed of an accelerating observer, to get t' = (c / a) * asinh-1(a t / c) = (c / a) * ln[a t / c + sqrt(1 + (a t / c)^2)].

grant hutchison
2009-Jan-11, 08:49 PM
t' = (c / a) * asinh-1(a t / c)That's the fella. :)

Grant Hutchison

grav
2009-Jan-11, 09:28 PM
Okay, solving for (a t / c), I get

a t' / c = ln[(a t / c) + sqrt(1 + (a t / c)^2)]

e^(a t' / c) = (a t / c) + sqrt[1 + (a t / c)^2]

[e^(a t' / c) - (a t / c)]^2 = 1 + (a t / c)^2

e^(2 a t' / c) - 2 e^(a t' / c) (a t / c) = 1

(a t / c) = [e^(2 a t' / c) - 1] / [2 e^(a t' / c)]

Then solving for the formula for the contracted distance using the time according to the accelerating observer produces

d' = L d

where L = 1 / sqrt[1 + (a t / c)^2] and d = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1],

d' = (c^2 / a) [1 - 1 / sqrt(1 + (a t / c)^2)]

= (c^2 / a) [1 - 1 / sqrt(1 + (e^(2 a t' / c) - 1)^2 / (4 e^(2 a t' / c)))]

= (c^2 / a) [1 - 1 / sqrt(1 + [(e^(a t' / c) - e^(-a t' / c)) / 2]^2)]

= (c^2 / a) [1 - 1 / sqrt(1 + [sinh(a t' / c)]^2)]

so this should be the distance travelled by the accelerating observer according to his own clock and measure of distance to the point of origin.

grant hutchison
2009-Jan-11, 10:30 PM
(c^2 / a) [1 - 1 / sqrt(1 + [sinh(a t' / c)]^2)]Yes.
You can tighten it up a little by using your hyperbolic trig identities to simplifying that sqrt(1 + sinh2(at'/c)). Makes it easier to see what happens at the limit.

Grant Hutchison

grav
2009-Jan-11, 11:41 PM
Yes.
You can tighten it up a little by using your hyperbolic trig identities to simplifying that sqrt(1 + sinh2(at'/c)). Makes it easier to see what happens at the limit.

Grant HutchisonOh okay, cool, thanks. :) I found that cos2(x) - sinh2(x) = 1, so cosh(x) = sqrt[1 + sinh2(x)], and I found tanh(x) = sinh(x) / cosh(x). So let's see. We had (a t / c) = sinh(a t' / c), so substituting that into each of the formulas for Lorentz contraction, distance travelled, and the relative speed, all according to the accelerating observer's clock, then, that gives

L = 1 / sqrt[1 + (a t / c)^2]
= 1 / sqrt[1 + sinh2(a t' / c)]
= 1 / cosh(a t' /c)
= sech(a t' / c)

d' = L d
= [1 / cosh(a t' / c)] [(c^2 / a) (sqrt(1 + (a t / c)^2) - 1)]
= [1 / cosh(a t' / c)] [(c^2 / a) (cosh(a t' / c) - 1]
= (c^2 / a) [1 - 1 / cosh(a t' / c)]
= (c^2 / a) [1 - sech(a t' / c)]

v / c = (a t / c) / sqrt[1 + (a t / c)^2]
= sinh(a t' / c) / cosh(a t' / c)
= tanh(a t' / c)

grant hutchison
2009-Jan-12, 12:09 AM
Well, the hyperbolic secant is a step too far for my delicate constitution.
But otherwise yes.

Grant Hutchison

mugaliens
2009-Jan-12, 07:29 PM
Grav, does this mean you've finally reached the end of your relativistic pursuits you've been at for the last, oh, year?

Nice work!

grav
2009-Jan-13, 02:34 AM
Grav, does this mean you've finally reached the end of your relativistic pursuits you've been at for the last, oh, year?Well, more like over the last couple of years :shifty:, but I do indeed feel that I have finally gone as far as I can with SR, though, yes, but mostly with the tremendous help from members here on BAUT through discussion and debate (I tend to question every detail :rolleyes:), like with my main mentor Richard, as well as Ken, Grant, and Hh. So that's it for the "easy" stuff, I guess. Now begins a whole new chapter with accelerating observers and GR, as I've been struggling to do over the last month or two. I've picked up some new methods along the way that might make things a little easier, but it will still be slow going for a while, I fear, at least until I can get the ball rolling, as I'm attempting to do with this thread.

Nice work!Thanks. :) It definitely helps having great teachers. BAUT rocks! :dance:

grav
2009-Jan-15, 10:53 PM
Okay, well, I don't know what I'm doing wrong, but instead of normal Relativistic Doppler, my program is giving me sqrt[(1+v/c) / (1-v/c)] / [1 - (v/c)^2] for the result of the blueshift in the forward direction. I will need that same type of configuration to run future scenarios, so I will have to figure out what I did wrong. If I can't find the error after a few more reviews, I will go ahead and post it to see if anybody else can spot it with a fresh perspective.

In the meantime, I figured out another somewhat simpler way to determine the blueshift observed. According to the stationary frame, the frequency is one pulse per second, so one pulse passes a given point per second in the same frame. Also, if an observer were moving toward the emitter, then that observer will encounter additional pulses equal to the distance travelled in the stationary frame divided by the distance between pulses. So for instance, if pulses are emitted every second in the stationary frame, then they pass a stationary point every second. So if an observer were to travel toward the emitter for 11 seconds, covering a distance of 7 light-seconds while doing so, time and distance according to the stationary frame, then the observer will have encountered 11 + 7 = 18 pulses along the way.

So all we have to do now is to find the number of pulses that encounter an observer between a time of t and t + dt, and between a distance of d_t and d_{t + dt}, according to the stationary frame, add them together and then divide that by the time that has passed for the observer from t to t + dt to get the frequency as it observed by the observer during dt', or the number of pulses that pass per observer's time. The number of pulses encountered during a known time t and t + dt is just f t and f (t + dt), and over the distance the observer travels is d_t f / c and d_{t + dt} f / c, so the observed frequency becomes fo = [((t + dt) - t) f + (d_{t + dt} - d_t) f / c] / [t'_{t + dt} - t'_t]. The ratio of the frequency observed to that which was emitted, where fe = f, becomes

fo / fe

= [((t + dt) - t) + (d_{t + dt} - d_t) / c] / [t'_{t + dt} - t'_t]

= [dt + (c / a) [(sqrt(1 + (a (t + dt) / c)^2) - 1) - (sqrt(1 + (a t / c)^2) - 1)]] / [(c / a) (log(a (t + dt) / c + sqrt(1 + (a (t + dt) / c)^2)) - log(a t / c + sqrt(1 + (a t / c)^2)))]

Now, I could extend sqrt and log to a few terms in each series in order to cancel dt and reduce that way, but I took the easy way out and wrote a program in UBasic for it instead, as seen below.

10 point 30
20 cls
30 C=3*10^8
40 F=1
50 A=10
60 T=1000000000
70 Dt=0.00000000000000001
100 D=(C^2/A)*(sqrt(1+(A*T/C)^2)-1)
110 N=T*F+D*F/C
150 D2=(C^2/A)*(sqrt(1+(A*(T+Dt)/C)^2)-1)
160 N2=(T+Dt)*F+D2*F/C
200 Tt=(C/A)*log(A*(T+Dt)/C+sqrt(1+(A*(T+Dt)/C)^2))-(C/A)*log(A*T/C+sqrt(1+(
A*T/C)^2))
210 Fo=(N2-N)/Tt
220 Rf=Fo/F
230 print Rf
240 print
250 V=A*T/sqrt(1+(A*T/C)^2)
260 Vv=sqrt((1+V/C)/(1-V/C))
270 print Vv

The results of the program give the blueshift in terms of the formula above and also for that according to normal Relativistic Doppler for the instantaneous local relative speed. The two results are nearly exactly the same for any arbitrary values of a and t, to 25 digits for the values in the example shown, and become more precisely the same with increasingly small dt.

grav
2009-Jan-16, 04:08 AM
Okay, I think I figured out what the problem with the first program is, but now there is a new dilemma involved. Let's say that the acceleration and time that passes in the stationary frame are known, a and t. d_t is the distance travelled by an accelerating observer during t according to the stationary frame and d_{t + dt) is the distance travelled an infinitesimal amount of time later. Here, d_t = (c^2 / a) [sqrt(1 + (a t / c )^2) - 1] and t is replaced with t + dt for d_{t + dt}. The instantaneous relative speed of the accelerating observer according to the stationary frame, then, is v = [d_{t + dt} - d_t] / dt.

The times that pass for the accelerating observer when t and t + dt pass for the stationary frame are t'_t and t'_{t + dt}, where t'_t = (c/a) log[a t / c + sqrt(1 + (a t /c )^2)]. The distances travelled are d'_t and d'_{t + dt}, where d'_t = (c^2 / a) [1 - 1 / sqrt(1 + (a t / c)^2]. The instantaneous relative speed according to the accelerating observer, then, is v' = [d'_{t + dt} - d'_t] / [t'_{t + dt} - t'_t].

I wrote a program to compare the relative speeds according to each frame, shown below. It turned out that the relative speed according to the accelerating observer was smaller by a factor of sqrt[1 - (v/c)^2]. That is the same amount the distance is contracted to the accelerating observer, so then I tried the same program but using non-contracted distances for d'_t and d'_{t + dt}, where d'_t = (c^2 / a) [sqrt(1 + (a t / c)^2) - 1] instead, same as for the stationary frame, but then, surprisingly, the relative speed according to the accelerating observer becomes greater than that measured by the stationary frame by 1 / sqrt[1 - (v/c)^2]. So it ended up becoming greater than the first result for the accelerating observer by 1 / [1 - (v/c)^2]. Either way, the local relative speeds don't match for each of the frames. What is going on here?

10 point 30
20 cls
30 C=3*10^8
40 A=10
50 T=1000000
60 Dt=0.000000001
100 D1=(C^2/A)*(sqrt(1+(A*T/C)^2)-1)
110 D2=(C^2/A)*(sqrt(1+(A*(T+Dt)/C)^2)-1)
120 T3=(C/A)*log(A*T/C+sqrt(1+(A*T/C)^2))
130 T4=(C/A)*log(A*(T+Dt)/C+sqrt(1+(A*(T+Dt)/C)^2))
140 D3=(C^2/A)*(1-1/sqrt(1+(A*T/C)^2))
150 D4=(C^2/A)*(1-1/sqrt(1+(A*(T+Dt)/C)^2))
160 V=(D2-D1)/Dt
170 Vv=(D4-D3)/(T4-T3)
180 print V
190 print
200 print Vv
210 print
220 print V*sqrt(1-(V/C)^2)

grav
2009-Jan-16, 02:00 PM
I see what's happening now. The relative speed I'm finding for in the accelerating frame is not the local relative speed at all, but the speed relative to the point of origin of acceleration starting at v=0 in the stationary frame, so it is the culmination of the local relative speed plus the contraction at a distance. It is the apparent relative speed to the point of origin of acceleration over the distance travelled while that distance is being contracted, so the apparent speed is smaller. While the local relative speed is just v = v' for both frames, then, the apparent relative speed to the point of origin of acceleration, according to the accelerating observer, becomes v' = v * L = v * sqrt[1 - (v/c)^2] = v / sqrt[1 + (a t / c)^2] = (a t) / [1 + (a t / c)^2].

grav
2009-Jan-17, 02:59 AM
The relative speed to the point of origin of acceleration in the stationary frame according to the accelerating observer was found to be v' = dd' / dt' = (a t) / [1 + (a t /c)^2]. Now I want to know what the relative speed would be to some point originally separated from the point of origin by a distance of x in the stationary frame. So that would be

v' = [(d'_{t + dt} + x'_{t + dt}) - (d'_t + x'_t)] / dt'

= [dd' + dx'] / dt'

= [(c^2 / a) [(1 - 1 / sqrt(1 + (a (t + dt) / c)^2)) - (1 - 1 / sqrt(1 + (a t / c)^2))] + x [(1 / sqrt(1 + a (t + dt) / c)^2)) - (1 / sqrt(1 + a t / c)^2))] ] / dt'

= [(c^2 / a) [1 / sqrt(1 + (a t / c)^2) - 1 / sqrt(1 + a (t + dt) / c)^2)] - x [1 / sqrt(1 + (a t / c)^2) - 1 / sqrt(1 + (a (t + dt) / c)^2)] ] / dt'

Now things simplify immensely since we can easily see here that dx' = - dd' x / (c^2 / a), so

v' = [dd' - dd' x / (c^2 / a)] / dt'

= (dd' / dt') [1 - x / (c^2 / a)]

Also, we already know from before that dd' / dt' = a t / [1 + (a t / c)^2], therefore

v' = a t [1 - x / (c^2 / a)] / [1 + (a t /c )^2]

So let's use this to verify that v' = v locally. To do this, we just find x as the distance the accelerating observer travels according to the stationary frame in order to coincide with it locally after a time of t, hence x = - (c^2 / a) [sqrt(1 + (a t / c)^2) - 1], negative due to the convention I'm using of considering x to be behind the point of origin along the direction of travel. So now we get

v' = a t [1 - x / (c^2 / a)] / [1 + (a t / c)^2]

= a t [1 + (c^2 / a) [sqrt(1 + (a t / c)^2) - 1] / (c^2 / a)] / [1 + (a t /c )^2]

= a t [1 + sqrt(1 + (a t / c)^2) - 1] / [1 + (a t / c)^2]

= a t / sqrt[1 + (a t / c)^2]

where v = a t / sqrt[1 + (a t /c )^2], so v' = v

Let's also find the point behind the horizon where v' = 0. That is where

v' = a t [1 - x / (c^2 /a)] / [1 + (a t /c )^2] = 0

1 - x / (c^2 / a) = 0

x / (c^2 / a) = 1

x = c^2 / a

That is the Rindler horizon. Anything that originally lies at that distance behind the accelerating observer in the stationary frame will continue to lie at that distance according to the accelerating observer while accelerating, independent of the time that passes since t is not in the solution. Any points further behind or forward of it will be seen by the accelerating observer to move toward the horizon due to the contraction while accelerating as well.

Somewhere in this forum, a long time ago, it was explained to me, not sure if by Richard himself or not, that the Rindler horizon was the point where light coming from beyond that distance cannot catch up to the accelerating observer. I ran a few scenarios for that and found it could be the case, but only if one assumes a partially Euclidean geometry or something, I can't remember the details so I'll have to look them up, but thinking in the back of my head that according to the stationary frame, the light travels at c while the accelerating observer never exceeds it, so the light should always catch up. Later, based upon the definition of the Rindler horizon as light struggling to catch up to the accelerating observer, I then refuted Richard's use of the Rindler horizon as anything other than that of a pure flight of light effect. I can see now it does directly apply to real space-time according to the accelerating observer. Sorry about that, Richard.

grav
2009-Jan-18, 03:58 AM
Now let's do a little bit with simultaneity. If space-time is the same from the perspective of the accelerating observer a moment before cutting off the engines as it is a moment after the observer becomes inertial, then the times that the observer reads on the clocks in the stationary frame are the same as if the observer were inertial at that instantaneous local relative speed the whole time. So the time lag at a distance is tl = L d v / (c^2 - v^2), same as between inertial frames, d being the distance as measured in the stationary frame. From that we get

tl = L d v / (c^2 - v^2), L = 1 / sqrt[1 + (a t / c)^2], v = a t / sqrt[1 + (a t / c)^2]

tl = d a t / ( [1 + (a t / c)^2] [c^2 - (a t)^2 / [1 + (a t / c)^2]] )

= d a t / (c^2 [1 + (a t / c)^2] [1 - (a t / c)^2 / [1 + (a t / c)^2]] )

= d a t / (c^2 [1 + (a t / c)^2] [1 / [1 + (a t / c)^2] )

= d a t / c^2

An emitter at an original distance of d = (c^2 / a) behind the observer in the stationary frame when the observer begins accelerating will always lie at that distance during acceleration, so d' = d always for that point only, at the Rindler horizon. The time the accelerating observer reads locally in the stationary frame is t while his own clock reads t'. At some distance away while accelerating, however, the observer will read t_d = t + tl_d on the stationary frame's clock at that distance. So at the Rindler horizon, the observer reads t_d = t + tl_d = t + d a t / c^2, where d = x = - (c^2 / a) using positive convention in the direction of travel, therefore t_d = t + x a t / c^2 = t + (- c^2 / a) a t / c^2 = t - t = 0. So a clock at the Rindler horizon reads the same time as when the observer began accelerating and will always read the same as long as the observer accelerates steadily at the same rate. Anywhere beyond the Rindler horizon will begin moving back in time with further distance.

Of course, that doesn't include flight of light effects, so I probably shouldn't have said the observer will read those times now that I think about it, but that is what happens to space-time, while what he actually reads depends upon the time the light takes to reach him. For instance, at the Rindler horizon, what the observer actually sees upon beginning to accelerate will not suddenly freeze or anything, but will begin redshifting more and more, the pulses reaching him more slowly, and the rate of time the observer sees passing upon the stationary clock at that distance when the observer begins to accelerate at T = 0 will steadily decrease, although still moving forward from an original reading of T = - d / c = - x / c = - c / a , ever more slowly toward a limit of T = 0.