Equivalence of Time Dilation in Different Gravitational and Accelerating Frames

[Chrisc]

Two clocks in a gravitational field separated by an altitude of x, exhibit constant time dilation.
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?

 

[lzkelley]

Two clocks in a gravitational field, separated by any height - will not experience the same acceleration - nor the same dilation.

 

[Chrisc]

That's true, but not my question.
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.

 

[Dale]

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

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.

 

[Chrisc]

I don't think so.
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.

 

[Dale]

Well, if the distance changed in the gravitational field then the time dilation would also be time varying in the gravitational field. If you want an equivalent situation then you need to have the distance as a function of time be the same in both the accelerating and gravitational cases. If the distance is changing in one and not in the other then obviously the results will be different.

 

[Chrisc]

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

 

[Dale]

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

As measured in the accelerating/gravitating frame.

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.

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?

 

[pmb_phy]

I don't think so.
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.

Your assertion above, i.e. Two clocks accelerating in free space separated by the length x of their ship, exhibit non-constant time dilation is not correct for a uniformly accelerating frame of reference. In such a frame the clock rate is a constant for both clocks as measured by an observer at rest in the accelerating frame. 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?

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

 

[Dale]

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?

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.

 

[pmb_phy]

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.

Those are measurements taken by different observers. No one observer in a static g-field (or a uniformly accelerating frame) will ever measure the frequency of light to change as or propagates through the spacetime. In any case, the same would happen for observers in a gravitational field. If an observer initially measures the frequency of light comming from a steady source (a source which is at rest in the g-field and which produces light with the same freuquency indpendant of time) then he'd measure that frequency to change if he moved to a place lower in the field. Same with the accelerating observer. No experiment can determine whether you're in a uniformly accelerating frame of reference or in a uniform gravitational field. That is what the principle states. If measurement shows otherwise then the equivalence principle would be wrong.

Pete

 

[Dale]

Oh, I misunderstood your point at first. I agree, I cannot think of any reason for the blueshift to change over time either.

 

[Chrisc]

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.

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

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.

 

[Dale]

As the ship approaches c the distance traveled by the light signal approaches 0.

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.

 

[Chrisc]

It approaches c in the frame of any observer that measures it in constant acceleration. In the ship the distance x, (length of the ship) and the constancy of the speed of light are as (x/c) what determines the clock at the front to be running fast relative to the clock at the back.

 

[pmb_phy]

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

Thanks for clarifying. If this is the case then your statement This results in the frequency of light signals increasing as measured by the clock at the back of the frame. was confusing.

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.

No. It is because the clock is actually ticking faster as reckoned by the accelerating observers.

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.

You're making coordinate dependant statements. As such please specify which observer
you are referring to. As measured by an observer at rest in the ship the distance is constant.

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.

Then you are mistaken. You are assuming that "we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system. is the equivalence principle. It is not in that neither the weak or strong equivalence principle make such an assertion.

Pete

 

[Dale]

It approaches c in the frame of any observer that measures it in constant acceleration.

Sure, but that frame is an inertial frame so it has nothing whatsoever to do with the equivalence principle.

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.

 

[Chrisc]

pmb_phy, DaleSpam, all of your comments make it appear you are not familiar with this thought experiment.
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.

 

[pmb_phy]

pmb_phy, DaleSpam, all of your comments make it appear you are not familiar with this thought experiment.

Ya never know!

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.

Yup. That is exactly how I understand gravitational time dilation to be.

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.

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.

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.

Okay so far (except for the speed of light in a g-field).

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.

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".

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.

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.

Pete

 

[Chrisc]

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.

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.

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".

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.
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?

 

[Dale]

all of your comments make it appear you are not familiar with this thought experiment.

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.

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

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.

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.

 

[Chrisc]

OK, you two are either arguing without thinking or not reading carefully, or assuming I am saying something I am not.
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.

 

[Dale]

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.

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?

 

[yuiop]

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.

 

[Chrisc]

DaleSpam, start with a frame of reference that is the rocket-ship containing the two clocks Ac and Bc at the front A, and back B, of the ship respectively.
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 kev, I don't know why you are suggesting the rate of acceleration of A and B differ with respect to C as they are the front and back of a single rocket ship accelerating with respect to C.
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.

 

[Dale]

DaleSpam, start with a frame of reference that is the rocket-ship containing the two clocks Ac and Bc at the front A, and back B, of the ship respectively.
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.

sure, I accept the equivalence principle. That isn't at issue. I just don't accept your unsupported claim about the time dilation factor changing over time in the accelerating frame. You need to derive that rigorously and explicitly.

 

[pmb_phy]

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.

Actually all one needs to do is to state the rate at which the clock is ticking relative to a local observer.

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.

That is incorrect. Let S be the initial inertial frame of referance S. As the ship's speed increases there will be an increased blue shift in the frequency of the light according to observers which are in front of it and a decreased frequency in the direction behind it (assuming a source which emits light of the same frequency in all directions). The greater the the speed the greater the blue shift. However since the ship is moving away then by the time the light gets to it the clock Ac will have increased its own speed resulting in a frequency change. But this change will be constant since each observer is accelerating. However the clock at A is not accelerating with the same acceleration as the clock at B. This may seem odd since the ship is behaving like a rigid body in the ship's rest frame but it turns out that they must accelerate at different rates. In fact an observer rest at A will have a different weight than the same observer at rest at B! If the accelerations were the same then the distance between the ships as measured by observers in S will remain fixed. but we know that the faster an object goes the shorter it is. Therefore the acceleration of the front of th ship must be less than the acceleration of the back of the ship if the ship is to have a constant length.

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

On the contrary. I agree with both the experimental mechanics and am not questioning the mechanics. I'm explaining why your interpretation is wrong.

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?

The "means"?? I don't know what you mean by that. If you're asking for a mechanism then we've already agreed on the mechanism, i.e. light at B is emitted towards A. Since A's velocity has increased relative to S then A will detect a lower frequency. Likewise light emitted at A towards B is blueshifted.

You keep on describing some arbitrary inertial reference frame rather than the accelerating reference frame.

Actually he did state what frame he was referring to, i.e. the Earth's frame of reference (assuming one is far enough from the Earth so that the gravitational field of the Earth won't mess with the experiment's results).

OK, you two are either arguing without thinking or not reading carefully, or assuming I am saying something I am not.

Or you made a mistake and can't see it yet.

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

And herein lies the problem. When you use the term "change" do you mean "changes with time" or "changes with position"?

Dalespam and I both agree that the rate at which clocks run depends on position in the field. Observer's at A emit light light and the observer's at B receive the with an increased in frequency. The difference measured will not change with time however. I.e. if A emits a beam of light towards B then B will observer the light with a higher frequency than if measured locally but the difference in value will not change in time as observed by observers at rest at B.

Read kev's explanation. He is exactly right.

Hi kev, I don't know why you are suggesting the rate of acceleration of A and B differ with respect to C as they are the front and back of a single rocket ship accelerating with respect to C.

This is probably where you are making a mistake. The reason the rates of acceleration are different is to allow the ship to maintain a constant length in the ships frame of reference but to let it shrink in the Earth's frame of reference due to Lorentz contraction.

Best wishes

Pete

 

[Chrisc]

As the ship's speed increases there will be an increased blue shift in the frequency of the light according to observers which are in front of it

So you agree it continues to increase, does not stay shifted at a specific frequency.

However since the ship is moving away then by the time the light gets to it the clock Ac will have increased its own speed resulting in a frequency change. But this change will be constant since each observer is accelerating.

This "However" is immediately contradicted by this "However".

However the clock at A is not accelerating with the same acceleration as the clock at B.

Ac increases speed at a constant rate with respect to observers, but A is not accelerating with the same acceleration as B?

This may seem odd since the ship is behaving like a rigid body in the ship's rest frame but it turns out that they must accelerate at different rates.

"They must" by decree, by doctrine, by penalty of death, or by reason?

If the accelerations were the same then the distance between the ships as measured by observers in S will remain fixed.
but we know that the faster an object goes the shorter it is.

There is one ship. If it accelerates with respect to C, its instantaneous velocity with respect to C will increase over time.

Therefore the acceleration of the front of th ship must be less than the acceleration of the back of the ship if the ship is to have a constant length.

So let me get this straight. You think there "must" be a difference in the rate of acceleration between A and B "because" the ship "must" have constant length. In other words you are not reasoning the laws of physics, your are assuming they fit your beliefs.

The "means"?? I don't know what you mean by that. If you're asking for a mechanism then we've already agreed on the mechanism, i.e. light at B is emitted towards A. Since A's velocity has increased relative to S then A will detect a lower frequency. Likewise light emitted at A towards B is blueshifted.

Perhaps you've just digressed to the wavelength argument again.

Or you made a mistake and can't see it yet.

That is quite possible. I have made many mistakes in my life. But I tend to consider mistakes those things I recognize as mistakes not those labeled by others, so I hope you will persist in your attempts to show me my mistake, not just state that I have made one.

And herein lies the problem. When you use the term "change" do you mean "changes with time" or "changes with position"?

I mean change in the number per second, or frequency of detection.

Dalespam and I both agree that the rate at which clocks run depends on position in the field. Observer's at A emit light light and the observer's at B receive the with an increased in frequency. The difference measured will not change with time however. I.e. if A emits a beam of light towards B then B will observer the light with a higher frequency than if measured locally but the difference in value will not change in time as observed by observers at rest at B.

Again, I suspect you are mistaking frequency of detection for wavelength.

This is probably where you are making a mistake. The reason the rates of acceleration are different is to allow the ship to maintain a constant length in the ships frame of reference but to let it shrink in the Earth's frame of reference due to Lorentz contraction.

Let me restate what I read in this.
"The reason" of your position "is to allow" a measurement you cannot reason, so that you may then "let" this unreasoned notion agree with something you can reason.

I am going to attempt to post an image that should make my point clear. Please let me know if we are still talking about the same experiment.

 

[pmb_phy]

So you agree it continues to increase, does not stay shifted at a specific frequency.

If you are referring to measurements made by observers in the initial inertial frame S then yes. That has never been in question. When you make coordinate dependant statements, as you are in here, please state which coordinate system you are referring to.

This "However" is immediately contradicted by this "However".

I'm not even sure what that means but let me emphasize that it is of the utmost importance who is making these measurements. Please be sure you understand which observer is being spoken of. In this case I was referring to the observer at rest relative to clock Ac. If I was unclear about that I appologize.

Ac increases speed at a constant rate with respect to observers, but A is not accelerating with the same acceleration as B?

That is correct.

"They must" by decree, by doctrine, by penalty of death, or by reason?

The term "must" as I used it above means that if it is otherwise a paradox/contradiction will arise. Its a matter of calculation based on the equivalence principle.

There is one ship. If it accelerates with respect to C, its instantaneous velocity with respect to C will increase over time.

The speed of "the ship" is ill-defined since different parts of the ship accelerate at different rates.

So let me get this straight. You think there "must" be a difference in the rate of acceleration between A and B "because" the ship "must" have constant length.

As measured in the ship's frame of reference, yes.

In other words you are not reasoning the laws of physics, your are assuming they fit your beliefs.

Hardly. It is a deduction arrived at by the equivalence principle. I.e. if the ship is at rest but sitting on the Earth's surface then the height of the ship will remain constant. According to the principle of equivalence the same thing must happen in the accelerating frame of reference.

Perhaps you've just digressed to the wavelength argument again.

I was asking you what you meant by "means". What did you mean by that?

..so I hope you will persist in your attempts to show me my mistake, not just state that I have made one.

Nah. Of course not. In fact I encourage you to keep going until you have convinced yourself of what you seek.

"The reason" of your position "is to allow" a measurement you cannot reason, ...

Nope. If I can't reason it, or find a description of a reason, then I won't make any such assertion .. or at least try not to.

...so that you may then "let" this unreasoned notion agree with something you can reason.

I recall nothing here in any of my comments in which I said I couldn't reason something.

I am going to attempt to post an image that should make my point clear. Please let me know if we are still talking about the same experiment.

Okay. It looks like we're talking about the same thing.

Best regards

Pete

 

[Dale]

I am going to attempt to post an image that should make my point clear. Please let me know if we are still talking about the same experiment.

That is a very nice figure!

You still haven't derived anything. None of your verbal arguments nor this excellent diagram prove that the time dilation is non-constant as measured by B.

I also think you still fail to understand that the equivalence principle relates a gravitational field to an accelerating reference frame, not an inertial one. Your frame C is inertial and so it is completely irrelevant to the discussion about the equivalence principle.

 

[pmb_phy]

I also think you still fail to understand that the equivalence principle relates a gravitational field to an accelerating reference frame, not an inertial one. Your frame C is inertial and so it is completely irrelevant to the discussion about the equivalence principle.

It should be noted here that a frame of reference which is in free-fall in a uniform gravitational field will be a globally inertial frame of reference, the spacetime being flat (in all frames/coordinate systems of course).

Pete

 

[Dale]

Yes, but there has been no discussion of dropping clocks

 

[pmb_phy]

Yes, but there has been no discussion of dropping clocks

But there has been talk of inertial frames in conjunction with gravitational fields, right? I mentioned it because I thought there was. If not then ... never mind.

Pete

 

[Chrisc]

The instantaneous velocity of the "accelerating" ship is:
t1=v1, at t2=v2, at t3=v3.
The distance traversed by each light signal is from emission at A to detectin at B is then:
t1=c-v1, at t2=c-v2, at t3=c-v3.
Each second marked by A is then determined by B according to the frequency of signals from A as
t1= c - v1/c, t2= c - v2/c, t3= c - v3/c

This means second 3 is less than second 2, and second 2 is less than second 1.
This is a "constant" change in the rate of time between A and B, and a "non-constant" time dilation
of A with respect to B.

In a gravitational field the time dilation of A with respect to B is "constant".

 

[Dale]

Hi Chrisc,

Your equations are not dimensionally correct, you haven't considered time dilation, you haven't considered length contraction, and your analysis is still from the point of view of an irrelevant inertial observer C instead of the accelerating observer B.

 

[pmb_phy]

The instantaneous velocity of the "accelerating" ship is:
t1=v1, at t2=v2, at t3=v3.
The distance traversed by each light signal is from emission at A to detectin at B is then:
t1=c-v1, at t2=c-v2, at t3=c-v3.

Please define your varialbles. I don't understand their meaning. First you use, say, t1 to refer to an acceleration and then later you use it to refer to the distance traveled by the light. It also appears as if you believe that all parts of an accelerating ship have the same velocities. That is incorrect. Recall that Lorentz contraction implies that the distance between tip and tail will be decreasing with time and therefore the tip and tail cannot be accelerating with the same value as measured in S (S = the initial inertial frame of reference).

We are speaking of an uniformly accelerating frame of reference, correct? If so then I think that you're a bit confused on what a uniformly accelerating frame of reference is.

Let me explain: If the origin of a coordinate system S' is undergoing uniform acceleration as measured in S then the origin of that coordinate system would eventually accelerate to and beyond the speed of light. If at T = 0 you measured the acceleration of a particle dropped from rest at the origin of the accelerating frame S' it would have the value of, say, a, as measured from observers at rest in S'. However it you were to later drop another particle from the origin the origing from rest then it would not have the same value of acceleration. Thus a coordinate system such as that you defined will not represent a uniformly accelerating frame of reference. Only in the non-relativistic limit would this be true.

This means second 3 is less than second 2, and second 2 is less than second 1.
This is a "constant" change in the rate of time between A and B, and a "non-constant" time dilation
of A with respect to B.

In a gravitational field the time dilation of A with respect to B is "constant".

It once again appears that you're confusing the coordinate systems. It appears as if you are using measurements taken from the inertial frame of reference S and not from the accelerating frame S'. If a rate is constant one frame of reference then that doesn't imply that a rate is constant in another frame of reference.

 

[Dale]

Hey Pete,

I have been thinking about this a little bit. Is it really possible to make an equivalent situation? If you have a uniform gravitational field then two vertically-separated points can maintain the same separation and undergo the same acceleration. But in an accelerated reference frame if they maintain the same separation then they must have different accelerations or if they have the same acceleration then their separation changes.

(all accelerations and separations are proper)

 

[MeJennifer]

If you have a uniform gravitational field then two vertically-separated points can maintain the same separation and undergo the same acceleration.

I do not think that is true.
Can you demonstrate what you say is true? For instance what is your definition of an uniform gravitational field?

 

[Dale]

My definition of a uniform gravitational field would be one where g is constant (not a function of time or position).

If A and B are (constant) positions in the field then, by symmetry, the separation between them must also be constant, right?

 

[MeJennifer]

Could you explain how you reason that when A and B are accelerating the same way their distance remains constant in a uniform gravitational field (a field where g is constant)?

Obviously if they are not accelerating their distance remains constant in such a field.

 

[Dale]

If A and B are (constant) positions in the field then, by symmetry, the separation between them must also be constant, right?

Could you explain how you reason that when A and B are accelerating the same way their distance remains constant in a uniform gravitational field (a field where g is constant)?

Obviously if they are not accelerating their distance remains constant in such a field.

If the position is constant then by definition the first and second derivatives are 0 so they are not accelerating in the field.

 

[MeJennifer]

Yes but you are claiming that when A and B who are spatially removed from each other accelerate the same way their spatial distance remains the same. Do you think there is any difference between flat spacetime and a uniform gravitational field with constant g?

 

[Dale]

I am not making any claims, I am trying to figure out how to make a uniformly accelerating field equivalent to a uniform gravitational field. From my understanding I should be able to do this without GR since a uniform gravity field is a flat spacetime. I am just having a mental block somewhere.

Two bodies at rest in a uniform gravity field have the same proper acceleration (the field is uniform) and constant proper distance (they are at rest).

Two bodies at rest in a uniformly accelerating reference frame have the same proper acceleration (the frame is uniformly accelerating) and constant proper distance (they are at rest).

Two bodies accelerating in an inertial reference frame with the same proper acceleration will have a non-constant proper distance.

So how do I translate the gratitational field into an equivalent accelerated reference frame and then into an inertial reference frame?

 

[pmb_phy]

Hey Pete,

I have been thinking about this a little bit. Is it really possible to make an equivalent situation? If you have a uniform gravitational field then two vertically-separated points can maintain the same separation and undergo the same acceleration.

Actually they don't have the same acceleration. The accelerations are different due to time dilation. I know that this sounds anti-intuitive .. but one has to expect that in relativity.

My definition of a uniform gravitational field would be one where g is constant (not a function of time or position).

This is the problem. The definition of a uniform gravitational field is a gravitational field with zero tidal gradients. This means zero spacetime curvature. If the acceleration due to gravity was the same for all values of, say, z then the spacetime would be curved.

Pete

 

[Dale]

Ah, that makes sense. So what is the expression for g that maintains zero curvature? I assume that the g is the same as the proper acceleration measured by an accelerometer at each point. And how is this then related to a "uniformly accelerating" reference frame of equivalence-principle thought experiments?

 

[Chrisc]

Once again you are both (DaleSpam and pmb_phy)trying to prime the observations of B. It seems you are both familiar with the principle of relativity and the transformations of the observations of an event from frame to frame. I suspect you are both far more capable mathematicians than I, but judging by your remarks, neither of you appear to be familiar with the "principle" of the principle of equivalence.
In "principle" the laws of mechanics predict identical observational evidence in both.
If you do not know whether your frame is accelerating or gravitating, you cannot define any of the relativistic attributes you are both asking me to define.
You are both asking me to make a definitive statement of acceleration so that you will have a reference frame from which to critique it. You are essentially arguing "against" the principle of equivalence by asking me to define what the principle states "cannot" be defined.

Consider the following without inserting any previous knowledge of the situation.
You are in a rocket-ship. We will call the back of the ship B, the front A.
You know there is a force acting on the ship and its contents by the measured mechanics of test bodies.
You decide to test the laws of mechanics to see if they will distinguish the force as acceleration or gravitation. (This cannot be done, in principle, but the exercise is necessary to my point)
You have two test bodies of different mass.
You make the following observations :
The test bodies both fall to the floor at the same rate.
The rate at which they fall is constant with every drop over time. ******(remember this one)*******
You place a light at the front (A) of the ship of a known and specific wavelength.
The wavelength of the light from A as measured at B is shortened.
You flash a beam across the ship and note that it strikes the wall lower than its position of emission.

This is the scenario of the original question. You are in this ship described above. You do not know if it is accelerating in free space , or if it is in the gravitational field of a large mass.
The point is you "do not know" because you "DO" know the mechanics of both define all of the above situations to be indistinguishable or "equivalent".
But because you know the mechanics of both you also know that light signals from A will indicate the clock is running faster than the clock at B. You know why this is true in both acceleration and gravitation. But because you know why this is true in acceleration, you also know that the rate at which these signals arrive at B will increase over time.
******Unlike the scenario I asked you to remember above, (the rate of falling objects will "always" remain constant as measured by B in an accelerating frame), light is not carried with the frame as masses are. Light will not reflect the continued increase in the instantaneous velocity of the ship. It will therefore be detected at B to arrive at increasingly shorter intervals.

 

[Dale]

If you do not know whether your frame is accelerating or gravitating, you cannot define any of the relativistic attributes you are both asking me to define.

I don't know what you think I am asking you to define that is undefinable, but if you cannot even clearly define your terms then it is going to be difficult to convince anyone that your idea makes any sense.

You have made a specific claim: "The time dilation between two observers at rest in a uniformly accelerating reference frame is time varying whereas the equivalent time dilation between two observers at rest in a uniform gravitational field is not time varying". I am simply asking you to derive your claim rigorously.

At the end of your derivation you should wind up with something like the following:

tB = k tA

where tB is the time rate measured at B, tA is the time rate generated at A, and k is the time dilation factor. If your claim is correct then you should be able to show that k is constant in the case of a uniform gravity field and that k is a function of time in an equivalent accelerating reference frame. (All of which are clearly measurable so your "undefinable relativistic attributes" excuse doesn't apply)

So far you have not even made a decent attempt at a derivation.

... It will therefore be detected at B to arrive at increasingly shorter intervals.

If your claim were logical it could be backed up with math.

 

[Chrisc]

If it will make a difference, I'll give it a try.
I cannot embed the equations as I do not use a LaTex program. I will use the html as best I can and post an image to clarify the equations at the end of this post.
The following notation is used:
"^" denotes exponent.
"• "denotes multiplication
"sqrt" denotes square-root

The relation of the frequency of emission and detection for a frame in constant linear motion with respect to an inertial observer (respectively), includes the familiar Lorentz factor[(sqrt 1-v^2/c^2)] that gives the time dilation resulting from the relative motion between the frames.

w = w0• (1+v/c) / (sqrt 1-v^2/c^2) _______________ eq. 1.0
where w is the frequency detected by the inertial observer and w0 is the frequency emitted by the frame in constant linear motion with respect to the inertial observer, v is their relative velocity and c the speed of light.

In the example being discussed in this thread, the distance between emission at A and detection at B is a function of the velocity (instantaneous) of the ship. Thus the relation of the frequency of emission and detection is a doppler effect relation. If the velocity of the ship with respect to a base observer was significant, the Lorentz factor used above would show a noticeable change in the magnitude of the relationship of the two frequencies considered. This factor would not however alter the point of the discussion, it would only server to increase or decrease the total effect depending on the velocity of the ship with respect to the base observer.
The point of this discussion is whether or not there is "change" in the time dilation measured between the clocks at A and B in the ship accelerating with respect to a base observer.

For reasons of simplicity we can therefore omit the Lorentz factor without fear of affecting the principle except in terms of the total magnitude of the relationship.
We must also change the equation from velocity to acceleration as it affects the distance between emission and detection. We will use g to indicate the acceleration and H to indicate the distance from A to B. From each second to the next the velocity of B toward emission at A increases by gH/c.
B will always have this relation or "instantaneous velocity" with respect to the emission of A. Bearing in mind this relationship includes g, an expression of acceleration, it is understood that as a function of time, or second to second, the "relationship" remains the same, but g increases the instantaneous velocity of the ship with respect to a base observer, thereby increasing the doppler shift over time as expressed in the doppler equation
w=w0•(1+(gH)/(c^2) ____________________________eq. 2.0
Taking this relationship over successive emissions which amounts to successive instantaneous velocities with respect to the base observer we get
∆w=w0•(1+(vt2-vt1)H)/(c^2) _____________________eq. 3.0
where vt2 and vt1 are the instantaneous velocities of the ship with respect to a base observer at time 2 and time 1 respectively.

This relationship of the detection of signals at B with respect to the rate of emission at A is evidence of an ever increasing rate of time (emission) marked by the clock at A as detected at B.
A non-constant time dilation of acceleration.

 

[MeJennifer]

I cannot embed the equations as I do not use a LaTex program.

This forum has built-in Latex capabilities you do not need a separate program.

 

[Chrisc]

Thanks MeJennifer, I'll give a try.
I'll have to brush up on LaTex first.