Different Clock Rates Throughout Accelerating Spaceship

1977ub

But this is what I'm not sure I can believe anymore. Specifically that t2 and t4 are both equal to the same amount of time in the RF. Once the hurlers start pushing the boulders, the vehicles star moving, and their clocks no longer appear to be synchronized with one another from RF. Frontward clock now appears to be set later than Rearward clock. Thus RF must conclude that boulders are finished being hurled at different RF times.

stevendaryl

You have to keep straight two different scenarios: (1) The front and rear accelerate so that the distance between them remains constant (as viewed by the people in the rockets). (2) The front and rear accelerate at exactly the same time (according to the initial rest frame) and in exactly the same way (according to the initial rest frame).

In scenario (1):

• The clock in the front runs faster than the clock in the rear, according to the original rest frame, and also according to those aboard the rockets.
• The distance between the front and rear contracts, according to the original rest frame.
• The distance between the front and rear remains constant, according to those aboard the rockets.
• The acceleration felt by those in the front is less than the acceleration felt by those in the rear.

In scenario (2):

• The clocks in the front and rear run at the same rate, according to the original rest frame.
• The front clock runs faster than the rear clock, according to the people in the rockets.
• The distance between the front and rear remains constant, according to the original rest frame.
• The distance between the front and rear expands, according to those aboard the rockets.
• The acceleration felt by those in the front is the same as the acceleration felt by those in the rear.

1977ub

In my later posts, I settled on a 3rd scenario I found more fundamental, in which I do not presume to know how things will turn out, but instead start with identical programmed equipment. We've got identical springs compressed holding identical boulders. Both parties in their identical but separate vehicles activate the release on their own copy of the equipment at t0 which both parties in their vehicles and an observer in the rest frame agree is simultaneous, since at t0 all are at rest wrt one another. So how does that pan out? Presumably to the rest observer, the boulders *complete* their release at different times, since throughout the spring release, the vehicles are moving, and for him their simultaneity is lost.

PAllen

If I understand it, your 3d scenario is identical to stevendaryl's scenario (2). In which case it doesn't correspond in a meaningful way to an idealization of real rocket or to a tall building sitting on the surface of large planet (these are scenario (1) as stevendaryl labels them).

You've got boulders expelled by identical springs at t0 per starting rest frame. This means they must be the same distance apart and moving at the same speed per this starting rest frame. If they each expel another boulder at t0+1 per their own watches (which are still in synch per the rest frame (but slow), but not per each other), again their speed and distance and clocks are in synch per the starting rest frame. Per each other, their clocks are out of synch and they have moved further apart.

1977ub

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

3) Therefore, the final moment of boulder release from the spring will be different between the vehicles as seen by the rest frame.

I am describing the case where the apparatus, the program, the intent of the two vehicles is the same, but I think that they end up out of synch at then end of the acceleration period, as seen in the rest frame, therefore we can't in a general way say that they have "accelerated in exactly the same way in the rest frame." The intended so, they began so, but because the acceleration takes time, and they are moving during that time, they have not ended up so. They *began* accelerating in exactly the same way in the rest frame. They entire acceleration process did not take place in exactly the same way in the rest frame.

PAllen

No. They would be out of synch with starting frame if they followed a synchronization procedure while they were in motion. Since they started synchronized in the initial frame, and follow identical physical process, they remain in synch in the initial frame, and out of synch with each other.
Nope. It will be in synch from the initial frame; it will appear out of synch to each vehicle, if they were to apply a synchronization procedure.

stevendaryl

That's not a 3rd scenario, that's scenario 2. If the front and rear follow exactly identical actions that are (initially) simultaneous in the initial rest frame, then they will always be synchronized in the initial rest frame, and the distance between them will remain constant, according to the initial rest frame.

No. If they are doing the same actions, starting at the same time, they will finish at the same time.

The person in the front and the rear will not view their clocks as synchronized, but they will continue to be synchronized according to the initial rest frame.

stevendaryl

Why do you think that? If the front and rear are accelerating identically, then they will always be traveling at the same velocity, according to the initial rest frame.

1977ub

ok. thanks.

Austin0

How do they determine these effects within the frame???
Measure relative clock rates and distance??

1977ub

ok hope this is ok for this same thread. I'm moving in a more basic direction for understanding acceleration+SR:

For two intertial frames in relative motion, we can use gamma to describe how each observer measures the other's clock speed. Complete parity. Closely related to relativity of simultaneity. Fine.

Given an inertial frame RF,
and someone moving in a circle AF at velocity v with associated Lorentz gamma,

as far as I understand, RF still use gamma to determine rate of AF's clock?

what will AF use for RF's clock - 1/gamma ?

PAllen

They can determine (in principle) that the distance front to back of the rocket expanded by attaching a string that can't stretch; it will break. This is just Bell's spaceship paradox.

They can detect time difference between front and back clocks by exchanging signals.

PAllen

[yes, your RF can use gamma for the accelerated object.]

There are different philosophy's on this. To understand anything accelerating observers will measure (including see or photograph), it is simplest just to use any convenient inertial frame. The results of observations are invariant.

To try to come up with a frame for the accelerating observer, you run into the same issues as in GR: there is well defined local accelerated frame, just as there are well defined local frames in GR. However, there is no global frame for an accelerated observer in SR, just as there are no global frames in GR. What you can do, if you insist, is set up a coordinate system in which the accelerated observer remains at fixed coordinate position. Such a coordinate system may not be able to cover all of spacetime. Unfortunately, there are many ways to do this, none preferred. Once you have defined such coordinates (via transform from inertial frame), you can compute the metric in them. Then, using the metric, you can compute time dilation etc. per this coordinate system. It won't be as simple as a constant in place of gamma. The constant gamma results from the fact that the metric in the inertial SR frame is diag(1,-1,-1,-1). With a metric that varies by position and time, you need to integrate contraction of metric with path tangent vectors, instead of having a simple constant.

The up shot of all this is that there is no (preferred) answer to your question (what does the accelerated observer use in place of gamma?). It depends on what coordinate system you set up. On the other hand, let me stress again, if you want to know anything about what the accelerated observer measures or sees, just compute this in any convenient inertial frame.

1977ub

I'm asking a simpler question than how to define gamma if one is AF.

What RF calls gamma can be used to determine the click speed on AF and its inverse can be used to determine RF's click speed measured from AF.

I mean if RF measures one second while AF is making a revolution, and RF find AF's clock to have moved forward by .5 second, then it's a given that AF will find RFs clock to have clicked twice as fast as his own. I just want to verify it is this simple. I don't see how this can't be true.

PAllen

I use AO (accelerated observer) rather than AF, below, because there is no such thing as an accelerated frame (only choices of many possible coordinate systems).

For AO, the behavior they see on clocks in the inertial frame depend on where in the inertial frame they are, and the visual rates vary in time. That is, the observed behavior of inertial clocks will be both position and time dependent. The rates on these clocks averaged over time will show them (per the AO) to fast compared to AO clock. It is true that for pure circular motion at constant speed, the averaged rate seen on the inertial clocks will be gamma (as determined by RF) times faster than AO's clock.

1977ub

And inescapably, if we simply use a single clock in RF, and AO measures it once per revolution (passing right by it, say) then we can use 1/RF's-gamma to determine the time anti-dilation of RF as perceived by AO. I do understand that this will not apply to all of RF's clocks throughout a revolution. For one thing, AO is moving away from some of them while moving toward others, etc. "On average" AO must be able to use 1/RF's-gamma to determine the average speed of RF's clocks. Every time AO scrapes by clock-RF-0, RF finds AO's clock to have ticked slower by gamma, thus AO must find clock-RF-0 to have ticked faster by 1/RF's-gamma.

PAllen

Agreed. Just don't try to generalize this to other situations without understanding the complexities I described.