timedilation contradiction

[hu??]

When a rocket is standing on earth the time in the top is going faster that the time at ground/rocket-engine level due to gravitational time-dilation.

Using the equivalence principle with the same rocket accelerating in deep space, again an observer in the top will see the time passing by faster in the top then a traveler at rocket-engine level.

To me this looks like a contradiction because they both will experience the same acceleration and duration of acceleration!

I am confused: how is this possible? What am i missing?

[Bufofrog]

31 minutes ago, hu?? said:

Using the equivalence principle with the same rocket accelerating in deep space, again an observer in the top will see the time passing by faster in the top then a traveler at rocket-engine level.

Why do you think that?  There is no difference in gravity like on earth and both observers are experiencing the same acceleration.

[hu??]

2 minutes ago, Bufofrog said:

Why do you think that?  There is no difference in gravity like on earth and both observers are experiencing the same acceleration.

The gravitational time-dilation is not caused by the difference in gravitation gradient (caused by the distance from a gravitational attractor), but only by the difference in gravitational potential (i hope i use the correct terminology). Hence the equivalence principle.

If that is not convincing: tt is also possible to derive the time-dilation within the accelerating rocket simply using SR.

[Bufofrog]

Just now, hu?? said:

The gravitational time-dilation is not caused by the difference in gravitation gradient (caused by the distance from a gravitational attractor), but only by the difference in gravitational potential (i hope i use the correct terminology). Hence the equivalence principle.

So again, why do you think there is a difference for the 2 observers?  There is no difference in gravitational potential, velocity or acceleration between the observers.

[Ghideon]

1 hour ago, hu?? said:

Using the equivalence principle with the same rocket accelerating in deep space, again an observer in the top will see the time passing by faster in the top then a traveler at rocket-engine level.

I think the comparison is not completely right. I think you need two rockets, one accelerating equal to gravitation at ground level on earth. The second rocket accelerates slightly less, same as gravity at the top of the rocket standing on earth. Two observers, one in each accelerating rocket, will be affected by different acceleration as they are in the stationary rocked on earth. Does that makes sense? 

(This is the same as @Bufofrog said, just differently worded)

 

Edited January 26, 2020 by Ghideon

[swansont]

37 minutes ago, Bufofrog said:

So again, why do you think there is a difference for the 2 observers?  There is no difference in gravitational potential, velocity or acceleration between the observers.

But you’d be able to tell the difference between gravity and an acceleration if there was no difference.

The question to answer, I think, is how do you compare the clocks? Show what happens to the signal.

[MigL]

Swansont is correct.

If we look at gravitational time dilation as the loss of energy of an EM reference signal as it travels up a gravitational well, then an EM signal ( light clock ) would lose energy travelling in the direction of acceleration, and gain when travelling opposite the direction of acceleration.
By gain/loss of energy I mean increased/decreased frequency and/or shortened/lengthened wavelength.

IOW the equivalence principle holds.

Edited January 27, 2020 by MigL

[Markus Hanke]

9 hours ago, hu?? said:

I am confused: how is this possible? What am i missing?

The equivalence principle states that uniform acceleration is equivalent to the presence of a uniform gravitational field within the rocket. Any region of spacetime with uniform gravity is in fact flat, it has no curvature. This is why you can derive what happens in this rocket from SR, which is a model of flat Minkowski spacetime.

The same is not true for the surface of the Earth - the gravitational field of any central mass is not uniform, but tidal. If you had sensitive enough instruments, you could detect geodesic deviation within the rocket (though the effects would be small, so the field within such a small region is very nearly uniform). So these two scenarios are physically distinguishable, at least in principle, given sensitive enough instruments.

[md65536]

11 hours ago, hu?? said:

To me this looks like a contradiction because they both will experience the same acceleration and duration of acceleration!

MigL explains why it's not a contradiction and is expected.

If you're asking why there is a difference in times at the top and bottom of the rocket, when they should have identical experiences, it's because the difference in time is only relative, not something they experience locally. They actually both experience the exact same thing as each other regarding time: "My clock runs at 1 s/s, clocks above me run faster, clocks below me run slower."

[hu??]

2 hours ago, Markus Hanke said:

The same is not true for the surface of the Earth - the gravitational field of any central mass is not uniform, but tidal. If you had sensitive enough instruments, you could detect geodesic deviation within the rocket (though the effects would be small, so the field within such a small region is very nearly uniform). So these two scenarios are physically distinguishable, at least in principle, given sensitive enough instruments.

I know this. I did not want to complicate things by mentioning this.because it is not relevant here.

[hu??]

2 hours ago, md65536 said:

MigL explains why it's not a contradiction and is expected.

If you're asking why there is a difference in times at the top and bottom of the rocket, when they should have identical experiences, it's because the difference in time is only relative, not something they experience locally. They actually both experience the exact same thing as each other regarding time: "My clock runs at 1 s/s, clocks above me run faster, clocks below me run slower."

Correct, this is not a contradiction.

When the travelers synchronize clocks before acceleration, then accelerate with 1 in to top and one in the bottom of the rocket, then after acceleration they compare clocks again: they will not have experienced the same time!

 

Or do they and is my reasoning incorrect?

 

Edited January 27, 2020 by hu??

[MigL]

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.

[Strange]

There is also the fact that the rocket is not perfectly rigid so the back end, near the engines, will start accelerating before the front. As a result there will always be a small difference in speed.. Which then introduces the problem of simultaneity: how do you define "when" you make measurements in the two positions (does this mean there is always a difference in acceleration, front and back? Dependent on the frame of reference.)

[Bufofrog]

9 hours ago, Markus Hanke said:

The equivalence principle states that uniform acceleration is equivalent to the presence of a uniform gravitational field within the rocket. Any region of spacetime with uniform gravity is in fact flat, it has no curvature. This is why you can derive what happens in this rocket from SR, which is a model of flat Minkowski spacetime.

The same is not true for the surface of the Earth - the gravitational field of any central mass is not uniform, but tidal. If you had sensitive enough instruments, you could detect geodesic deviation within the rocket (though the effects would be small, so the field within such a small region is very nearly uniform). So these two scenarios are physically distinguishable, at least in principle, given sensitive enough instruments.

 

7 hours ago, hu?? said:

I know this. I did not want to complicate things by mentioning this.because it is not relevant here.

What do mean it is not relevant?  That is the crux of the discussion!  That is the reason the 2 scenarios are not equivalent.

[swansont]

34 minutes ago, Bufofrog said:

 

What do mean it is not relevant?  That is the crux of the discussion!  That is the reason the 2 scenarios are not equivalent.

It's not relevant because that wasn't the scenario that was presented. The followup post made it clear that the gravity gradient was not the issue being brought up, and as several people have made clear, it is not the reason for the timing difference. What Markus said was true, and is also not being applied to the problem.

[DimaMazin]

Frame of escaping traveler and frame of arriving the same traveler have slower time relative each other than relative to home frame. Therefore traveled clock shows less time than home clock at meeting.

Edited January 27, 2020 by DimaMazin

[Bufofrog]

39 minutes ago, swansont said:

It's not relevant because that wasn't the scenario that was presented. The followup post made it clear that the gravity gradient was not the issue being brought up, and as several people have made clear, it is not the reason for the timing difference. What Markus said was true, and is also not being applied to the problem.

Geeze!  I complete misread the OP!  My bad.

[hu??]

I though this thought experiment would be clear. Appearently it is not. The time-dilation will be ridiculously small for any practical rocket unless you are using a start-trek like impulse drive for a very long lime, and we all know that is not available. So the though experiment is just that, and handles with an idealized 'rocket' or call it a space-hook if you like. We assume the rocket is idealized rigid. Clock synchronization is done while the rocket is not accelerating!

Edit: i inadvertently posted to early.

Edited January 27, 2020 by hu??

[Strange]

55 minutes ago, hu?? said:

We assume the rocket is idealized rigid.

That is not possible. Many apparent paradoxes are resolved by realising this.

[hu??]

5 minutes ago, Strange said:

That is not possible. Many apparent paradoxes are resolved by realising this.

Correct.

I can be wrong, but i think not this one. If so, please show me how.

[Mordred]

Well for one thing there are no rigid rods in relativity. Markus reply is accurate.

[hu??]

2 hours ago, Mordred said:

Well for one thing there are no rigid rods in relativity. Markus reply is accurate.

i know it is accurate, but in this case it is not relevant.

I you want to go into technical details: the deformation can also be compensated for with actuators so the rod will act as completely rigid up to measurement precision.

[Mordred]

Sure why don't you. Start with the standard Lorentz transforms. Then use instantaneous velocity for acceleration. Or if you prefer a commoving inertial frame.

 As Marcus mentioned the field front and back of spacecraft will be approximately uniform.

PS don't forget length contraction as well as time dilation.

Edited January 27, 2020 by Mordred

[Strange]

5 minutes ago, Mordred said:

As Marcus mentioned the field front and back of spacecraft will be approximately uniform

Yes, but because the spacecraft is accelerating, it will require energy to move an object from the rear to the front of the craft. Therefore the "gravitational" (due to acceleration) potential at the front must be higher than at the rear, so clocks must run faster. No?

I'm not sure why @hu?? thinks there is a contradiction here, though. @MigL has provided a couple of answers that seem to fully explain it, as far as I can see.

[Mordred]

If you placed an accelerometer front and rear of the same spacecraft in linear acceleration the acceleration would be the same. So at each time segment the instantaneous velocity must also be the same between the two points.

If you placed an accelerometer front and rear of the same spacecraft in linear acceleration the acceleration would be the same. So at each time segment the instantaneous velocity must also be the same between the two points. Albeit a slight delay rear to front assuming rear end thrust. 

This scenario reminds me of Bells spaceship paradox but in this case the craft is held together my electrostatic forces. In Bells paradox it's two spaceships with a string between them. (The string will break).

However depending on the observer relativity of simultaneity will be involved. However you will need a reference frame as in the lab frame there is no seperation but in the MCIF frame there is. (Momentary commoving inertial frame.)

In essence to break it down the answer will depend on if your in the lab frame or in the MCIF frame. In the lab frame there is no seperation between points a and b and the acceleration is simultaneous and of equal value.

In the MCIF frame you will have a seperation between points A and B due to different accelerations.

Anyways here is a link that discusses the effects of proper acceleration. The formulas in this article can be applied to this thought experiment.

https://arxiv.org/abs/physics/0601179

Edited January 27, 2020 by Mordred