Equivalence of Time Dilation in Different Gravitational and Accelerating Frames

[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.

 

[yuiop]

...
I'll have to brush up on LaTex first.

Hi Chrisc,
to get started click on the equation below. In the window that pops up, you will see an example of the text you need to embed in your post to create a latex image.

$$w = w_0\frac{(1+v/c)}{\sqrt{1-v^2/c^2}}$$

 

[Chrisc]

Thanks kev, very handy link. It's come a long way since I last looked at LaTex.

 

[Dale]

Hi Chrisc,

I appreciate the effort. I believe that your eq 2.0 is correct to a first order approximation for low velocities, although your derivation was not very clear. I would have derived it this way:

Consider the momentarily co-moving inertial reference frame where the detector is initially at rest at the origin and accelerates in the positive x direction with an acceleration of g and where the emitter emits a photon while at rest at a position of H on the positive x axis. We can express the worldlines of the photon and detector in this reference frame as:
$$\begin{array}{l} x_{\text{detector}}(t)=\frac{g t^2}{2} \\ x_{\text{photon}}(t)=H-c t \end{array}$$ eq 0.1

We can deterimne the time when the detector meets the photon by setting the two equations in 0.1 equal to each other and solving for t.
$$\begin{array}{l} x_{\text{detector}}\left(t_{\text{detection}}\right)=x_{\text{photon} }\left(t_{\text{detection}}\right) \\ \frac{g t_{\text{detection}}^2}{2}=H-c t_{\text{detection}} \\ t_{\text{detection}}=\frac{\sqrt{c^2+2 g H}-c}{g} \end{array}$$ eq 0.2

Multiplying the resulting detection time by g gives us the velocity of the detector relative to the velocity of the emitter.
$$v=\dot{x}_{\text{detector}}(t_{\text{detection}})=g t_{\text{detection}}=\sqrt{c^2+2 g H}-c$$ eq 0.3

Substituting eq 0.3 into the relativistic Doppler equation (eq 1.0) and simplifying gives us:
$$\frac{\omega }{\omega _0}=\frac{1}{\sqrt{\frac{2 c}{\sqrt{c^2+2 g H}}-1}}$$ eq 1.1

And finally, taking a Taylor series expansion about g=0 and simplifying gives.
$$\frac{\omega }{\omega _0}=1+\frac{H g}{c^2}+O\left(g^2\right)$$ eq 2.0

Please note that your eq 2.0 is constant wrt time since g H and c are all constant wrt time. Note also that eq 1.0, the relativistic Doppler equation, does not depend on the distance between emission and detection, only the relative velocity of the emitter and detector. The fact that this distance becomes smaller as the rocket's velocity increases is irrelevant, although you keep mentioning it.

Finally, note that your eq 3.0 is dimensionally inconsistent and doesn't follow from eq 2.0.

 

[Chrisc]

Thanks DaleSpam, you're equations all make sense and the Taylor series is a handy expression I was not familiar with.
But, you have shown me how to calculate relativistic doppler shift of wavelength between two comoving frames
accelerating with respect to an inertial frame and how to express the infinite function of the same.
This is not the point of this discussion and therefore the reason you disagree with me.
We are dealing with the frequency of detection not the frequency detected.
Which is to say the number of light signals detected at B per unit time (local time at B) not
the number of photons/wavelengths detected per unit time.
The reason I keep mentioning it is because the "frequency of detection" is the evidence B uses
to calculate the rate of the clock at A. The frequency of emission, (again not the frequency of light emitted
but the number of emissions at A per unit time [local time at A]) is not the same as the
frequency of detection at B. This evidence is what B uses to determine the clock at A is
running fast with respect to the clock at B.
I used the doppler shift equations because the "difference" in the frequency of emission and the
frequency of detection as calculated at B, vary "with" the "increase" of the instantaneous velocity
of the ship with respect to an inertial observer.

 

[Dale]

We are dealing with the frequency of detection not the frequency detected.

It is the same thing. Let's say that a transmitter transmits one thousand flashes every second and simultaneously broadcasts a pure sinusoidal 1 kHz radio signal. If the frequency detected of the pure sinusoidal signal is 1.001 kHz then the frequency of detection of the flashes must be 1001 flashes per second.

 

[Chrisc]

Yes DaleSpam, if you set the frequency of emission equal to the frequency of flashes then of course they will be and remain numerically equal as counted by observers in any frame the signal passes through. And although they will remain equal to each other, they will not remain the same over time as A accelerates with respect to any frame.
If A moves with constant velocity with respect to and away from M, the frequency detected at M will be less than the frequency emitted at A, but will remain constant as the motion between A and M is constant. This is "classical" evidence of the relative motion between A and M. In the case where this constant velocity approaches c, there will be an increasingly apparent discrepancy in the classical addition of velocities which is what the principle of relativity reconciles via the Lorentz transformation and the time dilation of A and M become important.

Unlike the "constant" example above, the "continuous" emission of an EM wave from A will, with respect to an inertial observer M, "continue" to shift toward the lower end of EM spectrum with the "continued" acceleration of A with respect to and away from M.
In this case we have the "classical" evidence of acceleration between A and M. This relative motion is also subject to the principle of relativity and will with sufficient velocity between A and M, require the principle of relativity to reconcile the limit of the instantaneous velocity between A and M.

Now we must consider the measurements of B which is "classically" at rest with respect to A and "classically" under the same acceleration as A with respect to M. This is where the difference between "flashes" and "wavelength" becomes apparent and important to the considerations of an observer at B. In the classical limit, the observer can measure a discrepancy in the rate of flashes at a very low number of flashes. If the rate of flashes emitted at A is agreed to be 1 per second, the observer at B only has to mark a single, finite, discrete event - the tick of his clock and the detection of the flash. If they do not agree, he immediately knows there is a reconciliation required to explain the event. If on the other hand he attempts to measure wavelength he must measure the incident wave from trough to trough during which time his position has "potentially" changed. If we allow the wavelength to be very short so as to "nearly" negate the potential motion of B, we approach the classical limits of measurements made by B and the relativistic effects must be considered to significantly affect his measurements.
It is because he does not know if his frame is accelerating or in a gravitational field that he cannot "assume" his motion except with respect to A. Based on the constancy of the speed of light he has no choice but to reconcile the increased rate of flashes is proof of the increased rate of the clock at A. This is true for gravitation and acceleration.

Now consider the rate of flashes and the increasing rate of motion due to acceleration with respect to M. B will as mentioned above, detect each flash as a discrete event. The number of these events will increase per unit time as marked by the clock at B. B has no choice but to reconcile the increased and increasing rate of flashes as proof of the increased and increasing rate of the clock at A. This is NOT true of similar events marked in a gravitational field. B now knows the ship is accelerating in free space, i.e. the force on the ship is inertial.

 

[Dale]

And although they will remain equal to each other, they will not remain the same over time as A accelerates with respect to any frame.

Yes they will remain the same over time. I refer you back to our eq 2.0

This is where the difference between "flashes" and "wavelength" becomes apparent and important to the considerations of an observer at B. In the classical limit, the observer can measure a discrepancy in the rate of flashes at a very low number of flashes. If the rate of flashes emitted at A is agreed to be 1 per second, the observer at B only has to mark a single, finite, discrete event - the tick of his clock and the detection of the flash. If they do not agree, he immediately knows there is a reconciliation required to explain the event. If on the other hand he attempts to measure wavelength he must measure the incident wave from trough to trough during which time his position has "potentially" changed.

I'm sorry, but this is just silly. You cannot measure the frequency of flashes by measuring a single event. You have to measure the time between a minimum of two flashes to get a "frequency of detection" just like you need to measure a minimum of two peaks on a sinusoidal signal to get a "detected frequency". A frequency is, by definition, the inverse of a period. A period is a finite duration of time, not a single event. Your assertions here are absurd.

There is no difference between "frequency of detection" and "detected frequency". If one shifts or dilates so will the other.

Now consider the rate of flashes and the increasing rate of motion due to acceleration with respect to M. B will as mentioned above, detect each flash as a discrete event. The number of these events will increase per unit time as marked by the clock at B.

So derive it already! Despite all of your talk the only thing you have managed to demonstrate so far is that the Doppler shift detected at B is constant over time, which directly contradicts your claims.

 

[Chrisc]

Yes they will remain the same over time. I refer you back to our eq 2.0

I'm sorry, but this is just silly. You cannot measure the frequency of flashes by measuring a single event. You have to measure the time between a minimum of two flashes to get a "frequency of detection" just like you need to measure a minimum of two peaks on a sinusoidal signal to get a "detected frequency". A frequency is, by definition, the inverse of a period. A period is a finite duration of time, not a single event. Your assertions here are absurd.

I'm sorry, I did not mean to imply one signal was in itself all that was necessary. I meant that the coincidence of detection and the tick of the clock is all that was needed. It does take at least two signals or one signal at a prearranged time according to H/c after the clocks are synchronized. What you're missing or ignoring is that B does not know the nature of his acceleration, so the measuring of an event of A's clock against his clock is the most accurate, "finite" measure of A's clock he can make.
If A and B (the ship) are comoving under inertial acceleration with respect to C, the redshift of an EM wavelength emitted at A with respect to C, will be almost canceled by the (blue shift) of B's motion with respect to M. The difference in the velocity of A with respect to C at time of emission and the velocity of B with respect to C at time of detection will remain constant. This, as far as I can tell is your perception of the "frequency" or "wavelength" proof of constant time dilation. Please let me know if I've misunderstood you.

The difference between "frequency" or "wavelength" and "rate of detection" is that by virtue of the constancy of the speed of light, there is NO cancellation of "rate of emission" because there is no "KNOWN" change in the displacement of A and C. All that is measured by B is an increased rate of detection that predicts an increased rate of the clock at A. Because there is no known displacement of A and C, and because the distance between A and B is constant, in the case of inertial acceleration, the rate of detection at B will continue to increase over time.

So derive it already! Despite all of your talk the only thing you have managed to demonstrate so far is that the Doppler shift detected at B is constant over time, which directly contradicts your claims.

I have derived it already. In fact I offered Feynman's derivation and you not only didn't accept it, you insisted on putting relativistic measures into what an observer at B cannot, unless and until they know the nature of their motion and even then it will not affect the key point of this discussion.

 

[Dale]

I'm sorry, I did not mean to imply one signal was in itself all that was necessary. I meant that the coincidence of detection and the tick of the clock is all that was needed. It does take at least two signals or one signal at a prearranged time according to H/c after the clocks are synchronized.

Same with the sine wave.

If A and B (the ship) are comoving under inertial acceleration with respect to C, the redshift of an EM wavelength emitted at A with respect to C, will be almost canceled by the (blue shift) of B's motion with respect to M. The difference in the velocity of A with respect to C at time of emission and the velocity of B with respect to C at time of detection will remain constant. This, as far as I can tell is your perception of the "frequency" or "wavelength" proof of constant time dilation. Please let me know if I've misunderstood you.

Pretty much. The only slight changes I would make to what you said here are that the blueshift more than cancels the redshift resulting in a net blueshift from A at the front of the rocket to B at the rear. That and the difference in velocity would almost certainly not be constant in C, it would use relativistic velocity addition. Also, "inertial acceleration" is a contradiction in terms.

The difference between "frequency" or "wavelength" and "rate of detection" is that by virtue of the constancy of the speed of light, there is NO cancellation of "rate of emission" because there is no "KNOWN" change in the displacement of A and C. All that is measured by B is an increased rate of detection that predicts an increased rate of the clock at A. Because there is no known displacement of A and C, and because the distance between A and B is constant, in the case of inertial acceleration, the rate of detection at B will continue to increase over time.

You keep saying the same words with no math to back them up.

I have derived it already. In fact I offered Feynman's derivation and you not only didn't accept it, you insisted on putting relativistic measures into what an observer at B cannot, unless and until they know the nature of their motion and even then it will not affect the key point of this discussion.

You have not derived it. I just looked back through the whole thread and the only post where you even came close to deriving anything was https://www.physicsforums.com/showpost.php?p=1748095&postcount=48". Not only was your derivation there extremely sloppy, but it directly contradicted your claim! I don't know how you could possibly think that what you have presented so far is even remotely convincing.

I'm sorry to be so critical, but honestly I am getting a little bored with the discussion. It has become quite repetitive. You really need to learn how to derive things clearly and methodically from first principles. This will give you a much stronger understanding of the physics and allow you to communicate your ideas more clearly. Frankly, the main thing demonstrated by this thread is that English is too vague to accurately convey the real concepts here.

 

[Chrisc]

...Also, "inertial acceleration" is a contradiction in terms.

It's a common term used to distinguish between "inertial" and "gravitational" acceleration.

You keep saying the same words with no math to back them up.

Please look at the chart posted below and let me know if you agree or disagree.
I think this will put it to rest.

 

[stillwonder]

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?

gravitational field is unfortunately not uniform. that's the reason of time difference of two clocks at different altitudes. IF you assume the g. filed is uniform, then both redshifts should match and stay constant (gravitational and accelerated).

Edit: there will be a redshift, but no time dilation

I was wondering about similar scenario since the consideration of uniform g. field goes against how gravitational redshift is explained.

 

[Dale]

Please look at the chart posted below and let me know if you agree or disagree.
I think this will put it to rest.

I agree with the drawing, it is exactly what you have been saying over and over and over. It doesn't support your claim.

It supports the claim that the time between A0 and B0 as measured by M is longer than the time between A1 and B1 as measured by M. That is not your claim. Your claim is that the time between B0 and B1 as measured by B is longer than the time between B1 and B2 as measured by B. This drawing doesn't demonstrate that, nor does anything you have written anywhere in this thread in either English or math.

 

[Mentz114]

Chrisc,

from your diagram I would say that the relative velocity (v) between the emitter and detector ( separated by length L) at some time t is -

$$v = at - a\left(t + \frac{L}{c}\right) = -\frac{aL}{c}$$ ( assuming at t=0, a=0 )

which is constant if acceleration a is constant. So the red-shift would not change over time.

I did this during my lunch break - could be wrong, or not relevant even.

M

 

[Dale]

Edit: there will be a redshift, but no time dilation

That is a logical contradiction. If the distance is not changing then the only way to get a redshift is to have time dilation. I refer you back to https://www.physicsforums.com/showpost.php?p=1749296&postcount=55".

 

[Chrisc]

Thanks DaleSpam and Mentz114, it is now quite clear why you're disagreeing, and thanks for putting the event IDs on my image DaleSpam, it helps make the problem much more obvious. The issue is not math it is the justification of B's measurement.
My claim has always been that the time between A0 and B0 as measured by B, is less than B's calculation of H/c and larger than the time between A1 and B1 measured by B. Where H is the distance between B and A at rest with respect to each other. Before you take issue with the claim that B is even capable of measuring A0 with respect to his own clock, remember all the events of emission are pre-designated by A and B to occur at regular intervals according to the clocks they synchronized before moving to the front and back of the ship.
Since you agree the time between A0 and B0 as measured by M is longer than the time between A1 and B1 as measured by M, then you must agree that the time between A0 and B0 as measured by B is less than H/c as measured by B and longer than the time between A1 and B1. If you do not agree with this you must explain how B measures the speed of light between A and B to decrease over time.
Once again, before you begin to calculate the relativistic addition of velocities, the point is not how "you" will explain away the speed of light, with the knowledge you have of the situation, it is how B explains away the speed of light, as we are discussing the principle of equivalence not the principle of a God's Eye frame.
The only options B has to reconcile the time of detection with the agreed rate of emissions are:
1) reject the constancy of the speed of light.
2) consider A's clock to be running faster than his
He will likely pick 2. and conclude:
the force he measures on the ship is a) gravitational or b) inertial acceleration
The testing of light signals as described above will, over time, tell him:
the ship is in inertial acceleration in free space.

 

[Mentz114]

Chris,

If you do not agree with this you must explain how B measures the speed of light between A and B to decrease over time.

I don't think this is what B will experience. At constant acceleration only external quantities like relative velocity and distance will change. The observer in the ship will experience constant ( unchanging with time) effects. Like a red-shift between the front and back of the ship.

M

 

[Chrisc]

Chris,

I don't think this is what B will experience.

Hi Mentz144
I am not suggesting it is.
I said if you do not agree with the previous sentence, you will be making the claim that the speed of light changes as measured by B.

 

[yuiop]

Could someone bring me up to speed here as this thread has got rather big to read through it all.

Are we specifically analysing:

1) Born rigid acceleration
The case of an accelerating rocket that has greater acceleration at the rear than at the tail at any given instant according to a non accelerating observer. That same obeserver would measure the rocket to be length contracting progressively more over time. An observer on the ship would measure the proper length of the ship to remain constant over time. Observers on the rocket would notice that an accelerometer indicates greater proper acceleration at the tail than at the nose.

2) Bell type acceleration.
The case of accelerating rocket that has equal acceleration at the nose and at the tail at any given instant, according to a non accelerating observer. That same observer would say the length of the rocket remains constant over time while an observer onboard the rocket would consider the rocket to be getting longer over time. Observers on the rocket would notice that an accelerometer indicates the same proper acceleration at the tail than at the nose.

 

[Dale]

Could someone bring me up to speed here as this thread has got rather big to read through it all.

Are we specifically analysing:

1) Born rigid acceleration
The case of an accelerating rocket that has greater acceleration at the rear than at the tail at any given instant according to a non accelerating observer. That same obeserver would measure the rocket to be length contracting progressively more over time. An observer on the ship would measure the proper length of the ship to remain constant over time. Observers on the rocket would notice that an accelerometer indicates greater proper acceleration at the tail than at the nose.

2) Bell type acceleration.
The case of accelerating rocket that has equal acceleration at the nose and at the tail at any given instant, according to a non accelerating observer. That same observer would say the length of the rocket remains constant over time while an observer onboard the rocket would consider the rocket to be getting longer over time. Observers on the rocket would notice that an accelerometer indicates the same proper acceleration at the tail than at the nose.

The proper distance (in the accelerating frame) was stipulated to be constant in https://www.physicsforums.com/showpost.php?p=1743187&postcount=7". So it is Born rigid acceleration.

 

[Mentz114]

Hi Mentz144
I am not suggesting it is.
I said if you do not agree with the previous sentence, you will be making the claim that the speed of light changes as measured by B.

Your logic is too tortuous for me.

It's a strange way to communicate, asserting that if I disgree with proposition A then I'll be claiming that proposition B is true etc. How do you know what I'll be claiming ? Maybe your logic is faulty and I'll claim something completely different.

Its all hand waving if you can't state whatever your problem is in equations.