A question about the equivalence principle.

harrylin

As I elaborated, that's roughly correct.
Hehe you can surely say that!

Cheers,
Harald

yuiop

The apparent field inside an accelerating rocket also gets weaker at greater distance from the apparent source. Accelerometers attached to the nose of the rocket indicate less proper acceleration than accelerometers attached to the tail of the rocket.
The snag is that the by calculating the relative time dilation by using the relative velocities at the front and back of the rocket one comes to the conclusion that the relative rates of clocks on board the rocket as measured by observers on board the rocket increases over time which is simply not true. The relative rates of clocks on board the rocket as measured by observers on board the rocket is constant over time.

Austin0

Hi yuiop you have returned ;-)

i agree with your assumption that it appears that the relative dilation due to coordinate velocity would increase. That was my intuition also but after thought i became less sure.
That was partly my interest in the analysis I described.

But if we are right and the dilation differential would increase up the velocity curve this would seem to present a major problem.

As I said, the two clocks in question could be unconnected.
Independently accelerating clocks that would maintain the proper contracted separation simply as a consequence of the acceleration differential.

It seems clear that the fundamental SR principles must pertain.I.e. The gamma relation between velocity and proper time rates and deltas. And the clock hypothesis.

So any short interval measurements of coordinate velocity and clock comparisons for those intervals must correspond to the fundamental relationship where ever they taken along the course of acceleration.

Simply connecting the two clocks into a single system should not have any effect on this relationship.

So if the results predicted by the Rindler coordinates do not agree with the results predicted by the fundamental principles of SR there would be a real question.

What possible physics would account for this ?

i.e. what would prevent the velocity dilation from occurring?

Which prediction should be considered valid?

Would you agree there would be a question?

It was actually this question which prompted my primitive attempt at calculating the dilation factor I posted earlier.

You missed this one. regarding Steve's derivative approach.

So have you calculated the differential due to velocity actually/ or is it just a logical guess? ;-0
thanks

harrylin

Good one! Yes, there is a very slight difference due to length contraction - and it's always the same for a given distance between clocks and given acceleration. However, if one allows the Earth sensor to free-fall inside or next to the rocket, it can detect no gradient at all, not even in principle.
And I now wonder if one would nevertheless interpret the (incredibly small) difference in accelerometer readings as due to a real field, with what distance to the gravitation source that would correspond, and with what mass - could one really be fooled?

yuiop

At the time it was a estimate, but I have now done the calculations just to check and can confirm that clock speed differential between front and back increases over time if we use the velocity time dilation approach. For example if we have a constant proper acceleration of 1 at the back of the rocket and 0.1 at the front and use units of c=1, then initially the time dilation ratio (back/front) is 1 at time t=0, 0.71 at t=1 and 0.456 at t=2. This is not what would be observed on board the rocket(s) because the time (t) here is the simultaneous coordinate time in the launch frame and the notion of what is simultaneous is different on board the rocket(s). This is the crux of the matter that I shall come back to.
After doing the calculations I am now sure that is what would be observed as coordinate velocity increases relative to any given MCIRF, as measured relative to the simultaneous coordinate time in that given MCIRF. As above, it is the notion of simultaneity that is important and this changes with reference frames and you have not been careful to specify one.
We are right but it does not present a problem ;)
Agree.
Agree, these fundamental SR principles do hold.
Nothing. Velocity time dilation still occurs, but the effect cancels out. Consider two signals emitted simultaneously from the front and back of a rocket at coordinate time t=0 as measured in a given MCIRF. To give this MCIRF a label I will call it the launch frame (LF), but there is nothing special about this MCIRF and I could do an identical analysis in any other MCIRF. When the signal from the front arrives at the back let us say the velocity in the LF is vb and when the signal from the back arrives at the front, the velocity in the LF is vf. It turns out that with Born rigid acceleration, vb = vf. This means that sqrt(1-(vb)^2) = sqrt(1-(vf)^2) so at the time of the reception of the signals the velocities and velocity based time dilation is identical at back and front and so there is no differential time dialtion between back and front due to velocity time dilation. All the apparent differential time dilation observed on board the rocket is due to classical Doppler shift according to the observer in the inertial LF.
They are both valid. One is the interpretation according to the simultaneity of an inertial reference frame and the other is the interpretation due to a different notion of simultaneity in an accelerating reference frame.
Not really. We are used to observer dependent quantities such as coordinate velocity, time, distance, time dilation, length contraction, simultaneity etc. being different in different inertial reference frame (by definition) in regular unaccelerated SR.
Well now we are back to the crux of the problem. I have been putting this off because while I feel I understand it intuitively in my head, putting it into words is not so easy :) The first difficulty is how to define a notion of simultaneity for the observers on board the rocket, when there own reference clocks appear to be running at different rates in different locations on board the rocket. If we ask the observers at the front and back of the rocket to send signals simultaneously in there own reference frame, how are they going to arrange that? One way to do this would be to agree a convention that defines "simultaneous" events at the front and back of the rocket as being events that are both simultaneous in a shared MCIRF. If they do this they will observe that the signals are received at the back and front "simultaneously" because the reception events will be simultaneous in a shared MCIRF, (but this shared MCIRF will not be the same shared MCIRF that the signals were emitted simultaneously in). For a practical example, lets say we an inertial rocket ir1 moving at 0.6c and another ir2 moving at 0.8c both relative to the LF. After launch we tell the observers at the front and back to send a signal at the moment they are at rest with ir1, then for a suitable length accelerating rocket, they will both receive signals when they are momentarily at rest with ir2. You might argue that I have cheated here, because I have simply defined, rather than derived a notion of simultaneous for the accelerating rocket observers.

Another approach is to speed up the rear clock (as mentioned in an earlier post) so that it appears to be running at the same rate as the front clock. Now we have an unequivocal method of defining simultaneous in the accelerating rocket reference frame and can synchronise clocks using the usual Einstein clock synchronisation convention. Now when we send signals from the front and back using the on board rocket reference clocks, the signals arrive simultaneously at the back and front respectively according to the rocket clocks and the elapsed time between sending and receiving is equal according to both the front and back accelerating rocket observers. At no point here have we had to refer to the MCIRF's but if we compare results, simultaneous emission and simultaneous reception of signals as measured by the accelerating rocket observers, agrees with simultaneous as measured in the MCIRF's. Now the rocket observers know they had to speed up the rear clock to make it run at the same rate as the front clock so they know the real proper time rate of the rear clock must be slower than the proper time rate of the front clock. If they consider themselves to be stationary in their own reference frame they would ascribe this differential clock rate to a gravitational field and this coincides with the proper acceleration they can feel and and measure. The observers in a given inertial reference frame outside the rocket would ascribe the differential clock rates observed by the rocket observers to classical Doppler shift.

The differences between the Born rigid accelerating rocket and a real gravitational field, due to tidal effects, are significant enough, that we would be only be fooled if limited to measurements in a very restricted local region.

stevendaryl

I'm not sure I understand your question. You're asking why the inertial coordinate system is relevant to the accelerated observer? In some ways, it's just a matter of convention for what the accelerated observer considers two "simultaneous" events. One convention is to use the same notion of simultaneity as a comoving inertial observer.

No, that's not correct. If you wait a short time δt after launch, the rear clock will be traveling at speed v_rear = g_rear δt, and the front clock will be traveling at speed v_front = g_front δt. So just based on that, you would expect a relative rate of the clocks to be
√(1- (g_rear δt)²/c²))/√(1- (g_front δt)²/c²))
but the actual ratio of rates is
g_front /g_rear
those aren't close at all. The first goes to 1 as δt → 0, but the second doesn't.

PeterDonis

I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, but this doesn't look right; the ratio of clock rates *should* be governed by the first formula, based on the two velocities. The ratio of clock rates is certainly *not* equal to the ratio of accelerations. And the ratio *should* go to 1 as delta t -> 0, because at launch, which is what delta t -> 0 means, the two clocks are at rest relative to each other, so their clock rates are the same.

Some web pages that seem to me to be relevant are Greg Egan's page on the Rindler horizon:

http://gregegan.customer.netspace.ne...erHorizon.html

And the Usenet Physics FAQ pages on SR and acceleration (which links to the page on the relativistic rocket equation, another good resource), and on the clock hypothesis:

http://math.ucr.edu/home/baez/physic...eleration.html

http://math.ucr.edu/home/baez/physic.../SR/clock.html

stevendaryl

But that doesn't give the correct answer, which is that, according to measurements performed by observers aboard the rocket, the front clock always runs faster than the rear clock, the ratio of their rates is always the same: 1 + gL/c².

PeterDonis

It's possible that I'm misunderstanding the scenario you are talking about. When you say that a short time after launch, each clock will be traveling at a velocity g delta t, where g is the acceleration of the clock, that tells me that exactly at time t = 0, i.e., before the time delta t elapses, each clock is at rest in the same "launch frame". If the two clocks are at rest relative to each other at some instant, then their clock rates, at that instant, are the same. See the web page I linked to on the clock postulate. My understanding is that that is the scenario you are talking about.

The formula 1 + gL/c describes something a bit different than what I would call the "ratio of instantaneous clock rates", but I'd rather not get into that until I'm sure I correctly understand the scenario you're talking about.

stevendaryl

To talk about clock "rates", you have to refer to two different points in time. To talk about the relative rates of two clocks, you need 4 events, and you need a notion of simultaneity.

e₁ = the rear clock shows time 0.
Let F = the instantaneous rest frame of the rear clock at this event.

e₂ = some event at the front clock that is simultaneous with e₁ in frame F . Let T₂ = the time showing on the front clock at event e₂.

e₃ = the rear clock shows time δT
Let F' be the instantaneous rest frame of the rear clock at this event.

e₄ = some event at the front clock that is simultaneous with e₃ in frame F'. Let T₄ = the time showing on the front clock at event e₄.

So from the point of view of an observer in the rear of the rocket, the rear clock advanced by δT, and the front clock advanced by: T₄ - T₂

Then the ratio of clock rates, as measured by the rear clock is:

R_front/R_rear
= (T₄ - T₂)/δT

As δT → 0, this ratio does not go to 1, but goes to 1 + gL/c², where L is the length of the rocket, and g is the acceleration of the rear.

PeterDonis

I agree with the second sentence given the first. I take the first as your definition of how you are using the term "clock rate"; it is not the only possible definition, but I have no problem using yours for this discussion.

You've left out a key piece of information here: is the front clock at rest, in frame F, at event e2? (By hypothesis the rear clock is at rest in frame F at event e1.)

Same question here. I ask these, again, to make sure I understand the scenario you are talking about. I think the answer to both is "yes", but I would really like confirmation before going into details, since there's no point in my talking about a different scenario than you are talking about.

stevendaryl

That's not really relevant to the question of what is the ratio of clock rates, as measured by an observer in the rear of the rocket. But in the way I've set things up, the answer is "yes".

Whether the front clock is at rest at events e₂ and e₄ doesn't seem relevant to me. But if you work out the details for the constant proper acceleration, it turns out to be true, that in frame F, the front clock is at rest at event e₂, and in frame F', the front clock is at rest at event e₄.

PeterDonis

Thanks for the confirmation. The reason it's relevant (other than giving me confirmation that I was correctly understanding your scenario) is the clock postulate, which I referred to before. Consider what you have here: you have two pairs of events, at each of which two objects are mutually at rest. By the clock postulate, then, their "instantaneous clock rates" should be the same, because their velocities, relative to any inertial frame, are the same, therefore their "time dilation factors", relative to any inertial frame, are the same. Yet between their respective events, one object (the front clock) experiences more elapsed time than the other (the rear clock).

In fact it's even weirder than that. Consider the entire worldlines of the two clocks; I'll write their equations in the "launch frame" as follows:

Rear clock: x^2 - t^2 = R^2

Front clock: x^2 - t^2 = (R + L)^2

Now draw any straight line through the origin, in the "launch frame", t = vx, where v >= 0. This is either a horizontal line (for v = 0) or a line sloping up and to the right at less than 45 degrees (for v > 0). Call the event where the line intersects the rear clock's worldline Rv, and the event where the line intersects the front clock's worldline Fv. Then all of the following are true:

(1) For any v, the line t = vx is a "line of simultaneity" in the instantaneous rest frame of the rear clock at Rv *and* of the front clock at Fv. Therefore, Rv and Fv are simultaneous as seen by both the front and the rear clocks.

(2) For any v, relative to the launch frame, the rear clock at Rv and the front clock at Fv are both moving at velocity v. Therefore, the front and rear clocks both see each other as at mutual rest at these events, and they both have the same "time dilation" factor at these events (because of the clock postulate).

(3) For any v, the proper time experienced by the front clock between F0 and Fv is greater, by the ratio (R+L)/R, than the proper time experienced by the rear clock between R0 and Rv.

So we have two clocks, which remain a constant distance apart at mutual rest, and have the same "time dilation factor" at any pair of corresponding events, and yet experience different proper times between corresponding events. I realize all this is true; I just point it out to explain why one has to be careful using the term "clock rate" without more explanation. Many people will think "time dilation factor", as in 1/sqrt(1 - v^2), when they see "clock rate" (after all, that's what I initially thought when I saw it), but that's not what you mean by the term. In particular, I think this may be one source of confusion in some of Austin0's posts.

stevendaryl

For problems involving acceleration, time dilation is not the only consideration. Let's consider a variant of the old twin paradox. We have two twins that are the same age, but live on different planets, light-years apart. One twin on his 20th birthday hops into a rocket and accelerates rapidly to nearly the speed of light and gets to his other twin in just one year, according to the clock in his rocket. But the other twin has aged 20 years during the trip. How did that happen? From the point of view of the traveling twin, the trip only lasted a year. How could the other twin age 20 years?

Well, what you have to take into account is that simultaneity is relative. Let e₁ be the traveling twin celebrated his 20th birthday. Let e₂ be the event at which the distant twin celebrated his 20th birthday. Let F be the "launch" frame of the traveling twin, and let F' be the "traveling" frame. Events e₁ and e₂ are simultaneous in frame F, but not in frame F'; in frame F', e₂ took place long, long before e₁. So in jumping from frame F to frame F', the traveling twin changed his notion of what time "now" is on the distant planet, and so he changed his notion of how old the distant twin is.

So for an accelerated observer, the time on a distant clock changes both because that clock advances, and also because the observer's notion of what is "now" for the distant clock changes.

PeterDonis

Yes, the "simultaneity lines" t = vx that I described illustrate this: they change slope as v increases, meaning that the direction between the rear clock's "now" and the front clock's "now" changes. That means a given segment of the rear clock's worldline includes a set of simultaneity lines that sweep over a larger segment of the front clock's worldline. It's the hyperbolic equivalent of concentric circles, where the same angle sweeps out a longer arc on the circle with a larger radius.

yuiop

Just to summarise, the various formulas written down by Dalespam, stevendaryl and myself for the relative red shift of clock rates on board the accelerating rocket, as measured by observers on board the accelerating rocket are all equivalent, i.e.:

[tex]\frac{\Delta\tau_2}{\Delta\tau_1} = (1+g_1L/c^2) = \frac{R_2}{R_1} = \frac{g_1}{g_2}[/tex]

To the accelerated observers on board the rocket, the clocks remain at constant distance from each other, so from their point of view, none of the observed red shift is due to classical Doppler shift caused by relative velocities.

Austin0

Only one. The launch frame.

i am not sure if this actually tracks. If the signals are simultaneous in the emission MCIRF it appears unlikely the reception ,in a frame which would be many frames and spatial distance removed, would be simultaneous. I will have to work on this.

even if your assumption is correct and the reception is simultaneous wrt the MCIRF how does this imply no velocity dilation?? by definition there is no motion relative to the MCIRF's
so any calculation based on the MCIRF couldn't reveal relative velocity between front and back.

When you say here the observer in the LF do you mean the initial launch frame or the current mCIRF ,,,,earlier you were refering to MCIrfs as LF
In any case Doppler shift is as you say ,apparent dilation, so not really relevant

I think you misunderstood my question. I am aware of the problems implementing simultaneity in this circumstance , i mentioned a few earlier.
the question was why take an approach which had these problems and the inherent ambiguity of the result due to these problems.??

well I have to disagree here. Simply scaling the clocks does not make it an inertial frame.
It is still an accelerating system.
Even disregarding the acceleration/velocity differential, a synchronization which works for one velocity cannot work for other different velocities. Yeah?? How could it?

I think you have misunderstood both my approach and point.

regarding measurement of relative clock rates at the front and back from the launch frame. LF accelerating system AF
Simultaneity is not an issue. measurement of clock rates of course requires some interval of time a single event doesn't work.

so in LF at (bx_o,t₀) (fx₀ ,t₀) we get

observations bT'₀ and fT'₀ of AF ..

at some later point at (bx₁, t₁), (fx₁,t₁)

we get observations bT'₁ and fT'₁ of AF

(bx) t₁ - (bx)t₀ = (bx)dt

(fx)t₁ -(fx)t₀ = (fx)dt

bT₁-bT₀ =dbT'

fT₁ - fT₀= dfT

then (fx)dt/[itex]\gamma[/itex] =dfT' and (bx)dt/[itex]\gamma[/itex]=dbT' or not

Would you agree that in this circumstance the simultaneity of the LF clocks at either measurement is not important because there is no direct comparison between the observations between them?
The comparison is between an observation of the back clock with a later observation of the back clock.Etc.

Of course it would be necessary to calculate the proper times for these events using the Rindler coordinates to make these comparisons.

Also the measurement points could be widely separated say 0.7c and 0.8c

More and more i suspect that the velocity dilation would be insignificant and might possibly agree with the Rindler predictions. I.e. would not increase with greater velocities. The acceleration magnitudes you used were totally unrealistic. The back of the rocket quickly passing the front and leaving it in the dust ;-) so I am still unsure.
thanks