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In a rocket accelerating at 1g, if you are standing in the top ( nose ) and let go of a ball, does it not drop to the bottom of the rocket ?
It gains kinetic from potential energy exactly the same as if you let go at the Earth's surface from the same height as the length of the rocket.

EM radiation climbing out of the gravitational well to that height on Earth, will be red-shifted, while EM radiation falling into the gravity well will be blue-shifted.
This effect is due to gravitational time dilation; the frequency of the EMR are the reference pulses ( light clock ) that measure the dilation.

We can now use the equivalence principle to make the exact same statements about the rocket accelerating at 1g.
EMR climbing up the rocket ( towards the nose ) will be red-shifted, while EMR falling to the bottom of the rocket ( towards the tail ) will be blue shifted.
This is due to equivalent gravitational time dilation, and the effect is exactly equivalent to the 1g , Earth case.

Please point out where you see an error in my thinking, Mordred.

Edited by MigL

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I don't, I pointed out a different examination that employs proper acceleration. One that the Bells spaceship paradox has been worked out in. The ships and string are just added descriptives.

You will find that your redshift relations can be shown through the same transformations.

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The top and bottom of the rocket sitting on Earth remain the same distance away from each other, and as specified feel the same gravitational forces.

If you want the rocket in space to have the top and bottom remain the same distance away from each other (in their frames), that's called Born rigidity. If you make the rocket Born rigid, the top and bottom will need to have different rates of acceleration, and different proper acceleration. Then the equivalence principle doesn't apply, at least not to say that the space rocket top and bottom are equivalent to the Earth rocket top and bottom.

If you want the equivalence principle to apply, just specify that the top and bottom have the same proper acceleration. Don't worry that the spaceship eventually pulls itself apart, you can't constrain everything how you want.

There's no contradictions... could the problem be that the two rockets are necessarily different in some way or another? I think the resolution is that the equivalence principle applies locally. it doesn't say that distant clocks and rulers will be equivalent. (Or I suppose it can be applied to the whole rocket if it were in freefall?)

Edited by md65536

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2 minutes ago, md65536 said:

The top and bottom of the rocket sitting on Earth remain the same distance away from each other, and as specified feel the same gravitational forces.

If you want the rocket in space to have the top and bottom remain the same distance away, that's called Born rigidity. If you make the rocket Born rigid, the top and bottom will need to have different rates of acceleration, and different proper acceleration. Then the equivalence principle doesn't apply, at least not to say that the space rocket top and bottom are equivalent to the Earth rocket top and bottom.

If you want the equivalence principle to apply, just specify that the top and bottom have the same proper acceleration. Don't worry that the spaceship eventually pulls itself apart, you can't constrain everything how you want.

There's no contradictions... could the problem be that the two rockets are necessarily different in some way or another? I think the resolution is that the equivalence principle applies locally. it doesn't say that distant clocks and rulers will be equivalent.

Bingo. +1...

Edited by Mordred

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5 hours ago, md65536 said:

The top and bottom of the rocket sitting on Earth remain the same distance away from each other, and as specified feel the same gravitational forces.

If you want the rocket in space to have the top and bottom remain the same distance away from each other (in their frames), that's called Born rigidity. If you make the rocket Born rigid, the top and bottom will need to have different rates of acceleration, and different proper acceleration.

I realized this last night myself as well: the proper local acceleration depends on the 'height' within the rocket because the observed differences in speed of time. It even does explain the contradiction i started with: the travelers will experience different travel-time although traveling the same.

I think i just have to accept relativity produces weird effects. 🤪

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Perhaps another way of looking at this is to consider what is seen as the two clocks pass a "stationary" observer. The clock at the front will show some time dilation because of its speed. By the time the second clock passes the same point it will have been accelerating for longer and so will be going faster and show greater time dilation.

The relative time dilation between the two clocks in this case must be the same as that seen because of the difference in potential in the rocket's frame of reference. (We had a similar thread a whole ago where it was shown that the time dilation in a centrifuge can be equally well described in terms of the relative speed or the acceleration caused by the circular path.)

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13 hours ago, hu?? said:

I think i just have to accept relativity produces weird effects. 🤪

When you start dealing with Acceleration and curvature you can often get counter intuitive results.

It's one of the reasons it's a good policy to be able to do the calculations where those calculations are viable.

Bells paradox is a good example in order to maintain the seperation distance one would think rocket a and b must have the same proper acceleration. However this isn't true as the lead ships proper acceleration must be lower than the trailing ship to maintain seperation distance

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Frankly, I hadn't considered that the equivalence principle wouldn't apply because,although the stresses might tear a string, a 1g acceleration wouldn't tear apart a rocket. Nor had I given Born rigidity much thought.
I had gone on the assumption that, in the given example, any differences in ( proper ) acceleration would be negligible and probably unmeasurable, and so could be considered equal for our purpose.

Then again, the time dilation for the height of a rocket, would also be extremely small.
( sometimes you make one assumption too many )

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In 1g, for a 1m height difference, it’s about a part in 10^16 frequency shift. Current cutting-edge optical clock technology is a couple orders of magnitude better than that.

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True, but could you measure acceleration that accurately without inferring/calculating it from the time dilation frequency shift ?

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On 1/27/2020 at 4:13 PM, MigL said:

To come together again to compare clocks, one or the other will have to move to the other's location.
Either climbing up, or down, the equivalent gravitational well.

Sorry to be absent for some time.

They don't have to climb: the rocket just has to stop accelerating. That is the whole thing. for instance, when the rocket stops accelerating half-way the trip, then turns around and starts accelerating again. Then stops accelerating when the velocity is zero relative to the starting point. The observers will disagree on the trip-time!

I think that is just weird.

Edited by hu??

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37 minutes ago, hu?? said:

The observers will disagree on the trip-time!

If so (and I don't know if that is correct or not) this is because they will disagree about exactly when the accelerations starts and stops (relativity of simultaneity).

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1 hour ago, Strange said:

If so (and I don't know if that is correct or not) this is because they will disagree about exactly when the accelerations starts and stops (relativity of simultaneity).

I don't think that is relevant, because that can be compensated for by a mechanic construction (e.g. with the rocker motor is positioned in the middle of the rocket).
But for the sake of the argument lets assume the rocket stays in the back, what would that entail?

Let's see:

The difference in start and stop is given by a constant time:
For a complete rigid rocket this only depending the height:

delta_t = h/c

where:
h: height difference between observers
c : speed of light.

The top observer will start later, but also stop later. The result is that it is 'pushed' as long as the bottom observer (in the reference frame of the bottom observer).

Now the difference in traveltime, can be calculated as follows.

The timedilation is given by (for relative small acceleration and 'height' difference):

v = a * h/c
T = t1/t0 = 1/sqrt( 1 - v^2/c^2 )

where:
a: acceleration of rocket as observed by observer in the bottom.

The result T is slightly lower than in reality, but the difference will only be become significant for very large a or h.

So, the total traveltime disagreement is:

dt = T * tb

where:
tb: total traveltime for observer in bottom of of rocket.

Clearly, since delta_t is constant (and canceled out in the end, and can be compensated for) while dt depends on the total traveltime of observer in the bottom, the observers will disagree on the total traveltime at the end of the trip.

Edited by hu??

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3 hours ago, hu?? said:

I don't think that is relevant, because that can be compensated for by a mechanic construction (e.g. with the rocker motor is positioned in the middle of the rocket).

I don't see why the position of the motor is relevant to relativity of simultaneity.

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On 2/6/2020 at 9:30 PM, Strange said:

I don't see why the position of the motor is relevant to relativity of simultaneity.

Indeed, it is completely irrelevant.

I only mentioned this to show that it is possible to nullify the difference in start of acceleration for both observers by positioning the rocket motor between both observers instead of one at the level of the motor and the other one a some distance.

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2 hours ago, hu?? said:

Indeed, it is completely irrelevant.

I only mentioned this to show that it is possible to nullify the difference in start of acceleration for both observers by positioning the rocket motor between both observers instead of one at the level of the motor and the other one a some distance.

Well then, I suppose what I really meant was: I don't see why the position of the motor changes the difference in acceleration of the two observers. (The point I made on Jan 28 still stands.)

I may be wrong, but I think there is only a difference in the clock rate while they are accelerating. If the spaceship stops accelerating then their clocks will tick at the same rate (but they will have accumulated a time difference). But if they now decelerate back to their initial (zero) speed then the situation will reverse and their clocks will be synchronised again.

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On 2/6/2020 at 10:08 AM, hu?? said:

I don't think that is relevant, because that can be compensated for by a mechanic construction (e.g. with the rocker motor is positioned in the middle of the rocket).
But for the sake of the argument lets assume the rocket stays in the back, what would that entail?

[...]

For a complete rigid rocket this only depending the height:

Instead of describing a physical rocket and figuring out how the different points of it accelerate, you can specify how you want the different points to accelerate. If all you're comparing is two points, you can have the two points move independently and then not even care about the physical aspects of a rocket. For example, if you want to see what happens when they perform the same maneuvers, specify that the two points have identical acceleration measured from some inertial frame (eg. Earth frame). Or if you want the rocket to be rigid, use Born rigidity equations.

It's especially pointless to describe the physical aspects of the rocket, then ignore the physics in some tiny details (like assuming it's completely rigid with one source of acceleration, which is impossible), and then try to figure out other tiny details of the physics.

4 hours ago, Strange said:

I may be wrong, but I think there is only a difference in the clock rate while they are accelerating. If the spaceship stops accelerating then their clocks will tick at the same rate (but they will have accumulated a time difference). But if they now decelerate back to their initial (zero) speed then the situation will reverse and their clocks will be synchronised again.

Sure. If the top and bottom accelerate at the same time and rate according to an observer on Earth, those clocks always read the same from Earth. While accelerating, the bottom clock ticks slower than the top (in their reference frames), this can be verified from the Earth frame just by considering the always increasing time it takes light signals to go from the bottom to the top (takes longer because the top is moving away during the time the light travels) vs top to bottom (takes less time).

If the rocket then coasts, Earth says their clocks still read the same. On the rocket, the clocks now tick at the same rate but the rear clock is behind, in agreement with relativity of simultaneity.

If the rocket reverses and returns to relative rest with the Earth, still with the same timing and rate of acceleration as measured by Earth, the clocks as always remain the same according to Earth, and now the rocket agrees with that.

This would describe the situation in Bell's paradox, where eventually the rocket (fixed length in the Earth frame) rips apart. If you change that, so the clocks don't always have the same velocity as each other as measured by Earth, they can end up still out of sync after returning to Earth's inertial frame.

Edited by md65536

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15 minutes ago, md65536 said:

Sure.

+1 for confirming what I thought.

And, of course, you would have also got +1 if you had explained that I was wrong!

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