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Two clocks accelerating in free space separated by the length x of their ship, exhibit non-constant time dilation.

How is the principle of equivalence sustained when this fundamental measure of acceleration differs?

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Two clocks accelerating in free space separated by the length x of their ship, exhibit non-constant time dilation.

How is the principle of equivalence sustained when this fundamental measure of acceleration differs?

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1) A clock at a height x above the Earth's surface will run faster than an identical clock directly below it on the Earth's surface as measured by the latter.

2) A clock at the front of an accelerating frame will run faster than an identical clock directly behind it (on the axis of motion) in the same frame as measured by the latter.

The time dilation in 1) is constant. i.e. the difference in "rate" of time measured between the clocks is the same no matter when it is measured.

The time dilation in 2) is non-constant. i.e. the difference in the "rate" of time measured between the clocks "increases" over time.

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I don't think this is correct. I am pretty sure that the time dilation in 2) is constant over time if the distance between the clocks is not changing as measured in the accelerating frame.The time dilation in 2) is non-constant. i.e. the difference in the "rate" of time measured between the clocks "increases" over time.

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The distance between the front clock at the time of emitting a light signal and the position of the back clock at the time the signal is observed, must continue to decrease in an accelerating frame. This results in the frequency of light signals increasing as measured by the clock at the back of the frame.

The clock at the back thus measures the clock in the front is running fast, but it also finds it is running faster with each measurement.

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The clocks in 1) and 2) can be thought of as in the front and back of a rocket-ship, let the distance between them be x,(the length of the ship).

In 1) the rocket-ship is standing upright on the Earth's surface.

In 2) it is accelerating in free space.

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As measured in the accelerating/gravitating frame.The distance between the two clocks does not change in 1) or 2).

Then I am pretty sure that the time dilation is not time varying in either 1) or 2). If you believe it is could you post a derivation?The clocks in 1) and 2) can be thought of as in the front and back of a rocket-ship, let the distance between them be x,(the length of the ship).

In 1) the rocket-ship is standing upright on the Earth's surface.

In 2) it is accelerating in free space.

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Your assertion above, i.e.

The distance between the front clock at the time of emitting a light signal and the position of the back clock at the time the signal is observed, must continue to decrease in an accelerating frame. This results in the frequency of light signals increasing as measured by the clock at the back of the frame.

The clock at the back thus measures the clock in the front is running fast, but it also finds it is running faster with each measurement.

Also there are two forms of the equivalence principle. The weak form states

A uniformly accelerating frame of reference is equivalent to (e.g. has the same metric tensor as) a uniform gravitational field.

The strong equivalence principle states

Any physical law which can be expressed in tensor notation in SR has exactly the same form in a locally inertial frame of reference, even in a curved spacetime.

Einstein never claimed that an arbitrary gravitational field (e.g. one corresponding to a curved spacetime) is equivalent to an arbitrarily accelerating frame of reference, in fact it they aren't.

Pete

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Are you sure about that? If you consider the momentarily co-moving inertial frame when a photon leaves the front then, by the time it reaches the back, the back clock has accelerated and there is some Doppler shift. My understanding is that this was the essence of the derivation for the gravitational blueshift.A clock at the back of the ship will not detect changes in the wavelength of a beam of light as measured by by a clock at the front of the shift. There is no increase in frequency as you claim there is. Why do you there would be such a change?

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Those are measurements taken byAre you sure about that? If you consider the momentarily co-moving inertial frame when a photon leaves the front then, by the time it reaches the back, the back clock has accelerated and there is some Doppler shift. My understanding is that this was the essence of the derivation for the gravitational blueshift.

Pete

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A clock at the back of the ship will not detect changes in the wavelength of a beam of light as measured by by a clock at the front of the shift. There is no increase in frequency as you claim there is. Why do you there would be such a change?

We are talking about the time of light signals as the frequency of emission and detection, not the wavelength of light.

If the rate of the clock at the front of the ship is measured to increase at all relative to the clock at the back of the ship, it is because the distance traveled by the light signal is less than the length of the ship. The distance traveled by the light signal continues to decrease with the increased instantaneous velocity of acceleration therefore the rate of the clock at the front of the ship will continue to increase relative to the clock at the back of the ship. As the ship approaches c the distance traveled by the light signal approaches 0.

I am not talking about arbitrary physical equivalence, I am talking about the fact that the principle of equivalence ("we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system." (Einstein 1907)) is statement about the "instantaneous" physical equivalence of the laws.Einstein never claimed that an arbitrary gravitational field (e.g. one corresponding to a curved spacetime) is equivalent to an arbitrarily accelerating frame of reference, in fact it they aren't.

Pete

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I don't understand this at all. The ship doesn't approach c in the accelerating reference frame, it is always at rest in the accelerating reference frame. That is the whole point of the accelerating reference frame.As the ship approaches c the distance traveled by the light signal approaches 0.

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Thanks for clarifying. If this is the case then your statementWe are talking about the time of light signals as the frequency of emission and detection, not the wavelength of light.

No. It is because the clock isIf the rate of the clock at the front of the ship is measured to increase at all relative to the clock at the back of the ship, it is because the distance traveled by the light signal is less than the length of the ship.

You're making coordinate dependant statements. As such please specify which observerThe distance traveled by the light signal continues to decrease with the increased instantaneous velocity of acceleration therefore the rate of the clock at the front of the ship will continue to increase relative to the clock at the back of the ship. As the ship approaches c the distance traveled by the light signal approaches 0.

you are referring to. As measured by an observer at rest in the ship the distance is constant.

Then you are mistaken. You are assuming thatI am not talking about arbitrary physical equivalence, I am talking about the fact that the principle of equivalence ("we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system." (Einstein 1907)) is statement about the "instantaneous" physical equivalence of the laws.

Pete

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Sure, but that frame is an inertial frame so it has nothing whatsoever to do with the equivalence principle.It approaches c in the frame of any observer that measures it in constant acceleration.

The equivalence principle relates an accelerating reference frame to a gravitational field, not an inertial reference frame to a gravitational field. Of course you don't get equivalence in your comparison.

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Let me make it clear.

A rocket-ship stands upright on the Earth. We will call the front of the ship A and the back B.

We synchronize clocks, call them Ac and Bc and then place them at A and B respectively.

A clock "Ac" at the front of the ship "A" is designed to emit a flash of light at regular intervals toward B (let's say once per second).

An observer at the back of the ship (B) is marking the frequency of the signals (i.e. the time intervals between each flash)

against their clock Bc.

In a gravitational field, the observer at B will measure the light signals to arrive at a greater frequency than once per second according

to their clock Bc. This is gravitational time dilation.

When the ship is accelerating in free space, (let's say relative to the Earth) the observer at B will again measure an increased frequency

of the signals which they can only reconcile as follows:

The speed of light is constant, the length of the ship (x) is constant, therefore x/c is constant, therefore the clock at A is running faster than

the clock at B. This is the equivalence of gravitational time dilation

In a gravitational field, such as when it is standing on Earth, the clock at A, will remain at a constant rate of time that is faster than the

clock at B.

When accelerating, the clock at A will continue to increase its rate of time with respect to B as the ship's instantaneous velocity

relative to the Earth, increases with its acceleration. Why, because relative to the Earth and for the same reason Ac is fast in the

first place, the distance between the emission at A and detection at B decreases over time with the constant acceleration of A,B.

The constancy of the speed of light (i.e. independent of A and B) will result in a constantly decreasing time of detection at B between each signal.

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Ya never know! :tongue:pmb_phy, DaleSpam,all of your comments make it appear you are not familiar with this thought experiment.

Yup. That isA rocket-ship stands upright on the Earth. We will call the front of the ship A and the back B. We synchronize clocks, call them Ac and Bc and then place them at A and B respectively. A clock "Ac" at the front of the ship "A" is designed to emit a flash of light at regular intervals toward B (let's say once per second). An observer at the back of the ship (B) is marking the frequency of the signals (i.e. the time intervals between each flash) against their clock Bc. In a gravitational field, the observer at B will measure the light signals to arrive at a greater frequency than once per second according to their clock Bc. This is gravitational time dilation.

Woa there! Where did you get the idea that the speed of light is constant in a gravitational field. In fact it isn't. It is a function of position.When the ship is accelerating in free space, (let's say relative to the Earth) the observer at B will again measure an increased frequency of the signals which they can only reconcile as follows: The speed of light is constant, the length of the ship (x) is constant, therefore x/c is constant, therefore the clock at A is running faster than the clock at B.

Okay so far (except for the speed of light in a g-field).This is the equivalence of gravitational time dilation In a gravitational field, such as when it is standing on Earth, the clock at A, will remain at a constant rate of time that is faster than the clock at B.

I disagree. The rate at which the clocks tick will remain constant as measured in the ship's frame of reference. Why do you think that the rate will increase. Please state with what frame of reference you are referring to when you use the term "increase".When accelerating, the clock at A will continue to increase its rate of time with respect to B as the ship's instantaneous velocity relative to the Earth, increases with its acceleration.

With what frame of reference are you referring to when you say that the distance increases. Also recall that the speed of light is not constant in a g-field but varies with position.Why, because relative to the Earth and for the same reason Ac is fast in the first place, the distance between the emission at A and detection at B decreases over time with the constant acceleration of A,B. The constancy of the speed of light (i.e. independent of A and B) will result in a constantly decreasing time of detection at B between each signal.

Pete

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You're right the "speed" of light is not the "same" in all positions, but it is constant at each position. It's constancy at B is all that's needed to calculate the rate of signals from A.Woa there! Where did you get the idea that the speed of light is constant in a gravitational field. In fact it isn't. It is a function of position.

The rate at which A ticks will remain constant as measured by an observer "at" A and the rate at which B ticks will remain constant as measured by an observer "at" B.I disagree. The rate at which the clocks tick will remain constant as measured in the ship's frame of reference. Why do you think that the rate will increase. Please state with what frame of reference you are referring to when you use the term "increase".

That is not the issue here.

The rate at which A ticks according to the measured rate of signals detected at B must increase with the increased velocity of the ship with respect to the Earth.

You seem to be agreeing with the experimental mechanics in principle and then questioning the mechanics you agree with.

If you agree with the principle of equivalence of acceleration and gravitation what do you think is the means by which a clock at A is detected to run faster than a clock at B in an accelerating frame?

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I am aware of the rocket thought experiment, but you seem to be unaware of what an accelerating reference frame is. You keep on describing some arbitrary inertial reference frame rather than the accelerating reference frame.all of your comments make it appear you are not familiar with this thought experiment.

The distance between the two clocks does not change in 1) or 2).

You also need to be a little more precise and consistent with your descriptions. If you want to have an equivalent situation then you cannot have the proper distance change in the case of the rocket and not in the gravitational case.the length of the ship (x) is constant ... the distance between the emission at A and detection at B decreases over time with the constant acceleration of A,B.

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So, I will refer you to the original version of the thought experiment.

"Six Not-So-Easy Pieces" Richard Feynman, ISBN 0-201-32842-9. Page 131

And I will ask again - if you agree with the principle of equivalence of acceleration and gravitation,

as you both claim to, how do you explain the change in the rate of the clock at A as measured by

the clock at B in an accelerating frame.

If you say it does not change, there is no principle of equivalence. If you say it does change

please tell me how it is measured to change.

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I'm sorry, but nothing you have said so far has established this. Could you derive your claims rigorously? With clearly defined terms and coordinates in both the accelerating and gravitating cases?And I will ask again - if you agree with the principle of equivalence of acceleration and gravitation,

as you both claim to, how do you explain the change in the rate of the clock at A as measured by

the clock at B in an accelerating frame.

If you say it does not change, there is no principle of equivalence. If you say it does change

please tell me how it is measured to change.

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When accelerating, the clock at A will continue to increase its rate of time with respect to B as the ship's instantaneous velocity relative to the Earth, increases with its acceleration.

Why, because relative to the Earth and for the same reason Ac is fast in the first place, the distance between the emission at A and detection at B decreases over time with the constant acceleration of A,B.

The constancy of the speed of light (i.e. independent of A and B) will result in a constantly decreasing time of detection at B between each signal.

You are forgetting that B is accelerating faster than A according to an observer at rest with the Earth frame. Call this observer C. (Consider C to be far enough away from Earth that C can be considered an inertial observer for practiccal purposes.) According to C the rate at which A's clock ticks is slowing down as A accelerates, but B is accelerating faster than A and B's clock rate is slowing down even faster than A's clock at any "instant" according to C. B's velocity at any instant is greater than A's at any instant according to C and the time dilation of B's clock is disproportionally greater than that of A's clock because time dilation is not linearly proportional to velocity. The additional time dilation of B's clock relative to A's clock according to C makes up for the shorter flight path of the time signals from A to B. In fact it exactly cancels out and B will always see A's clock as running faster than his own clock by a constant factor, just as in the gravitational case.

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A emits light signal once per second according to Ac.

You are at B measuring the frequency of the light signals emitted from A.

You measure the rate of light signals to be greater than once per second according to Bc.

Are you accelerating in free space, or are you in the gravitational field of a large mass, at rest on its surface?

According to the principle of equivalence, the laws of mechanics will be equally upheld in both.

Without reference to some outside measurement, you will not know which is the case.

If you agree with this so far, I can continue. If not, we will never resolve this discussion.

Hi

It sounds as if you are saying that since B's clock is slower with respect to A's as measured by B, then with respect to C it must be moving faster to account for the greater time dilation. I hope that is not the case.

More importantly, the time dilation of A and/or B with respect to C has no bearing on the rate of signals detected by B as emitted from A.

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