A question about the equivalence principle.

[stevendaryl]

I posted it a few years back in this post, but the LaTeX has changed and so now it is all math processing errors:
https://www.physicsforums.com/showthread.php?t=236880&page=4

I will try to re-work it and re-post it here.

It seems like the Doppler derivation is the worst of all possible worlds, as far as difficulty; you have to take into account everything: the changing distances between the two clocks, the time dilations experienced by the clocks, and the transit time for light.

Of course, right after launch, you can compute the Doppler shift approximately by assuming that the clocks are still moving nonrelativistically.

 

[Dale]

I posted it a few years back in this post, but the LaTeX has changed and so now it is all math processing errors:
https://www.physicsforums.com/showthread.php?t=236880&page=4

I will try to re-work it and re-post it here.

Here it is with the math processing errors fixed:

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:
$$\begin{array}{l} x_d(t)=\frac{g t^2}{2} \\ x_p(t)=H-c t \end{array}$$ 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.
$$\begin{array}{l} x_d\left(t_d\right)=x_p\left(t_d\right) \\ \frac{g t_d^2}{2}=H-c t_d \\ t_d=\frac{\sqrt{c^2+2 g H}-c}{g} \end{array}$$ eq 0.2

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

 

[Dale]

It seems like the Doppler derivation is the worst of all possible worlds, as far as difficulty; you have to take into account everything: the changing distances between the two clocks, the time dilations experienced by the clocks, and the transit time for light.

That is why I called it "the brute force approach". However, it has the distinct benefit of giving the same formula as the standard weak field approximation to GR does. I doubt that the length contraction approach does.

 

[stevendaryl]

That is why I called it "the brute force approach". However, it has the distinct benefit of giving the same formula as the standard weak field approximation to GR does. I doubt that the length contraction approach does.

As I said, considering only length contraction gives approximately the correct answer long after the launch time, but not immediately after launch.

Let x be the location of the rear of the rocket at time t (t and x measured in the launch frame). Let dx/dt = v.

Let x₂ be the location of the front of the rocket at time t, where initially, the front is a distance L away from the rear. Using length contraction, we compute:

x₂ = x + L √(1-(v/c)²)

v₂ = v - vL/c² dv/dt 1/√(1-(v/c)²)

Now, we use (derivation skipped) the fact that if the acceleration is g in the rocket's instantaneous rest frame, then in the launch frame, it is g/γ³. Using the fact that γ = 1/√(1-(v/c)²), we get:

v₂ = v - vgL/c² (1-(v/c)²)

Computing 1-(v₂/c)², we get:

1-(v₂/c)² = (1-(v/c)²)(1+2(v/c)²gL/c²)
(ignoring higher powers of L)

√(1-(v₂/c)²) = √(1-(v/c)²)(1+(v/c)²gL/c²)

Letting T₂ be the time on the front clock, and T be the time on the rear clock, we find:

T₂ = ∫√(1-(v/c)²)(1+(v/c)²gL/c²) dt

In the limit v→c (long after launch), this becomes

T₂ = T(1 + gL/c²)

(where I used the formula T = ∫√(1-(v/c)²) dt

 

[yuiop]

In the limit v→c (long after launch), this becomes

T₂ = T(1 + gL/c²)

(where I used the formula T = ∫√(1-(v/c)²) dt

That looks like the binomial expansion approximation for

$$T_2 = \frac{T}{(1 - gL/c^2)}$$

By the way, is g measured at the front or back (or middle)? It makes a big difference.

 

[Austin0]

Well, I was trying to see the contributions of length contraction and relativity of simultaneity to the discrepancy between the front and rear clocks in the comoving frame. Obviously, if there is no length contraction, then it is all due to relativity of simultaneity.



Okay, let's pick a time t after the two clocks have been accelerating. Let the event e₁ be an event taking place at the rear clock at time t, and let e₂ be an event taking place at the front clock at time t. Let the coordinates of e₁ be (x₁,t) and let the coordinates of e₂ be (x₂,t), as measured in the launch frame. Let the corresponding coordinates in the momentary inertial rest frame be (x₁',t₁') and (x₂',t₂'). Letting δt' be the difference between t₁' and t₂', and letting δx be the difference between x₁ and x₂, the Lorentz transforms tell us that:

δt' = γ (δt - v/c² δx)

We chose the two events so that they are simultaneous in the launch frame, so δt = 0. So we have:

δt' = -γ v/c² δx

But δx is the distance between the front and the rear, as measured in the launch frame. So by length contraction, that is L/γ, where L is the length of the rocket in its comoving frame. So we have:

δt' = -γ v/c² L/γ
= - v/c² L

As time goes on, v→c, so this expression approaches

δt' = -L/c

So the desynchronization effect doesn't keep growing, it approaches a fixed constant (which happens to be the length of time required for light to travel from the rear to the front, in the comoving frame; hmm, not sure what the significance of that is).

In contrast, the discrepancy due to length contraction keeps getting bigger and bigger.

Hi I thought you were comparing the accelerating frames clocks to the momentarily comoving inertial frames.

In the case you are describing here as I understand it you are comparing the launch frame clocks to the CMIRF's correct??

In this case I am fairly certain that throughout the course of acceleration the relationship will be simply vL. Specifically the clock at the rear of any particular MCIRF will be running vL (current v) ahead of the front clock relative to the simultaneity of the launch frame. SO as v increases so will the desynchronization.
But that does not apply to the accelerating clocks without resynchronization because they will retain their original synchronization with the additional increase at the front due to the relatively faster rate of the front clock. SO relative to any MCIRF the front clock will be running ahead which is what I was talking about before when I misunderstood your approach.

 

[Dale]

Using length contraction, we compute:

x₂ = x + L √(1-(v/c)²)

I am not even sure this step is correct since the Length contraction formula is a feature of the Lorentz transform which relates two inertial frames. I think that you are going to get into some thorny issues about synchronization here.

 

[Austin0]

As I said, considering only length contraction gives approximately the correct answer long after the launch time, but not immediately after launch.

Let x be the location of the rear of the rocket at time t (t and x measured in the launch frame). Let dx/dt = v.

Let x₂ be the location of the front of the rocket at time t, where initially, the front is a distance L away from the rear. Using length contraction, we compute:

x₂ = x + L √(1-(v/c)²)

v₂ = v - vL/c² dv/dt 1/√(1-(v/c)²)

Could you explain what you are doing here. I am unclear on the reason for division by acceleration times gamma. Also is it v-vL divided by... or is it v - (vL divided by...)
Sorry my math is not up to your level.

Now, we use (derivation skipped) the fact that if the acceleration is g in the rocket's instantaneous rest frame, then in the launch frame, it is g/γ³. Using the fact that γ = 1/√(1-(v/c)²), we get:

v₂ = v - vgL/c² (1-(v/c)²)

Computing 1-(v₂/c)², we get:

1-(v₂/c)² = (1-(v/c)²)(1+2(v/c)²gL/c²)
(ignoring higher powers of L)

√(1-(v₂/c)²) = √(1-(v/c)²)(1+(v/c)²gL/c²)

Letting T₂ be the time on the front clock, and T be the time on the rear clock, we find:

T₂ = ∫√(1-(v/c)²)(1+(v/c)²gL/c²) dt

In the limit v→c (long after launch), this becomes

T₂ = T(1 + gL/c²)

(where I used the formula T = ∫√(1-(v/c)²) dt

Is this the other (non brute) method DaleSpam mentioned?

 

[Dale]

No, it seems even more brutish than the brute force method.

The elegant method is to take the transform and use it to compute the metric. Any time dilation will fall out automatically. Or, if you already know the metric, you can even skip the first step.

 

[Austin0]

Here it is with the math processing errors fixed:

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:
$$\begin{array}{l} x_d(t)=\frac{g t^2}{2} \\ x_p(t)=H-c t \end{array}$$ 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.
$$\begin{array}{l} x_d\left(t_d\right)=x_p\left(t_d\right) \\ \frac{g t_d^2}{2}=H-c t_d \\ t_d=\frac{\sqrt{c^2+2 g H}-c}{g} \end{array}$$ eq 0.2

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

Hi I can folow most of this but am unsure about the conceptual setup.

My assumption is that the emitter at H is at the front of the accelerating system
so at eq 3 the relative velocity is reated to the MCIRF at the time of emission . Is this correct?
If this is the case then H is constant wrt the MCIRF's but not to the launch frame?

What is the Taylor series expansion doing in this case??

Does this caculation result in a constant gamma between front and back??

Thanks

 

[Austin0]

No, it seems even more brutish than the brute force method.

The elegant method is to take the transform and use it to compute the metric. Any time dilation will fall out automatically. Or, if you already know the metric, you can even skip the first step.

Well your description is certainly elegant: Minimal, refined and completely enigmatic.

I can think of possible interpretations for the first sentance but nary a glimpse of falling dilations. Just kidding!. Could you perhaps be a little more exspansive in explaining what may be to you, obvious?

Thanks

 

[Dale]

My assumption is that the emitter at H is at the front of the accelerating system

Sorry about the confusion. It is always clearer in my head than on paper. Yes, the emitter is assumed to be at the front.

so at eq 3 the relative velocity is reated to the MCIRF at the time of emission . Is this correct?

Yes.

If this is the case then H is constant wrt the MCIRF's but not to the launch frame?

Yes, that is an assumption that I never stated explicitly. I was assuming a rocket under continuous acceleration where any transient strains had long since died out.

What is the Taylor series expansion doing in this case??

Approximating the time dilation to first order.

Does this caculation result in a constant gamma between front and back??

Yes.

 

[Dale]

Could you perhaps be a little more exspansive in explaining what may be to you, obvious?

It will probably be easier to just work a couple of examples. I will try to do that tomorrow.

 

[stevendaryl]

I am not even sure this step is correct since the Length contraction formula is a feature of the Lorentz transform which relates two inertial frames. I think that you are going to get into some thorny issues about synchronization here.

The formula I gave was for the length of the rocket as viewed in the launch frame, under the assumption that the original length L is very small compared with the characteristic length c²/g, where g is the acceleration of the rear of the rocket. In this case, the length of the rocket in the comoving frame will be L, and you can transform back to get the length in the original frame as L/gamma. Of course, the various parts of the rocket are not at rest relative to each other if the rocket is accelerating, so the "comoving frame" is slightly ambiguous, but if the rocket is small enough compared with c²/g, that doesn't make much difference.

The exact result for a Born-rigid rocket undergoing constant proper acceleration is this:

1. x_rear = √((ct)² + R²)
2. x_front = √((ct)² + (R+L)²)
   Taking the derivative of equation 1 to find the velocity, and expanding equation 2 in powers of L gives us the following:
3. x_front = x_rear + L √(1-(v_rear/c)²)

where R is the characteristic length c²/g.

With the given form for x_rear, we can take a derivative to get v_rear, and we can prove that
√(1-(v_rear/c)²) = R/x_rear
which is what I used for equation 3.

I suspect that my derivation is actually circular, in a sense; the way that you can derive that the front of the rocket must have the form in equation 2 is by demanding that each infinitesimal segment dL of the rocket has length dL/γ in the moving frame, where γ is computed for that segment.

 

[stevendaryl]

Hi I thought you were comparing the accelerating frames clocks to the momentarily comoving inertial frames.

No. There actually is not very much difference between those two, if the length of the rocket is sufficiently small. I was comparing the comoving inertial frame to the launch frame.

In this case I am fairly certain that throughout the course of acceleration the relationship will be simply vL. Specifically the clock at the rear of any particular MCIRF will be running vL (current v) ahead of the front clock relative to the simultaneity of the launch frame. SO as v increases so will the desynchronization.

I'm not sure that we are disagreeing. I said that long after launching, the velocity approaches c, and so the desynchronization approaches the constant L/c.

But that does not apply to the accelerating clocks without resynchronization because they will retain their original synchronization with the additional increase at the front due to the relatively faster rate of the front clock. SO relative to any MCIRF the front clock will be running ahead which is what I was talking about before when I misunderstood your approach.

I'm not sure whether we disagree, or not. In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c²/g (where g is the acceleration of the rear). In the launch frame, the rates for both clocks start out the same, but the ratio gradually approaches the same value.

 

[stevendaryl]

Is this the other (non brute) method DaleSpam mentioned?

I don't know.

Let me do the derivation a little more slowly.

I'm assuming that when the rear of the rocket is traveling at speed v_rear, then the length of the rocket will be approximately L√(1-(v_rear/c)²) by length contraction (where L is the original length of the rocket). That means that if we let x_rear be the location of the rear, and x_front be the location of the front, then, approximately:

x_front = x_rear + L√(1-(v_rear/c)²)

Now, to compute v_front, the velocity of the front of the rocket, we take a derivative of x_front with respect to t.

1. d/dt x_front = v_front (by definition)
2. d/dt x_rear = v_rear (by definition)
3. d/dt L√(1-(v_rear/c)²)
   = -Lv_rear/c² (d/dt v_rear)/√(1-(v_rear/c)²)

Are you asking how I derived equation 3? It's just calculus, using the fact that
d/dt f(v_rear ) = (d/dv_rear f(v_rear )) dv_rear /dt

In this particular case f(v_rear ) = L√(1-(v_rear/c)²)
d/dv_rear L√(1-(v_rear/c)²)
= -Lv_rear/c²/√(1-(v_rear/c)²)

Putting equations 1,2, and 3 together gives
v_front = v_rear - Lv_rear/c² (d/dt v_rear)/√(1-(v_rear/c)²)

 

[yuiop]

The formula I gave was for the length of the rocket as viewed in the launch frame, under the assumption that the original length L is very small compared with the characteristic length c²/g, where g is the acceleration of the rear of the rocket. In this case, the length of the rocket in the comoving frame will be L, and you can transform back to get the length in the original frame as L/gamma. Of course, the various parts of the rocket are not at rest relative to each other if the rocket is accelerating, so the "comoving frame" is slightly ambiguous, but if the rocket is small enough compared with c²/g, that doesn't make much difference.

They are not at rest wrt each other in a given inertial reference frame, when the MCIRF of accelerating rocket has significant relative velocity to that given inertial reference frame, but momentarily in any MCIRF they are at rest wrt each other and in the accelerating reference frame of the rocket they are permanently at rest wrt each other. On board the rocket, any part of the rocket has a constant radar (and ruler) distance from any other part of the rocket so in that sense they are at rest wrt each other. The ambiguity only arises if you fail to specify which reference frame the measurements are made from.

 

[yuiop]

I'm not sure that we are disagreeing. I said that long after launching, the velocity approaches c, and so the desynchronization approaches the constant L/c.

I am not sure what you are getting at here. Let's say the clocks are initially synchronised in the launch frame. After launch let's say that when 1 second elapses on the rear clock, 4 seconds elapses on the front clock in the MCIRF. It then follows that after 2 seconds elapses on the rear clock, 8 seconds will have elapsed on the front clock in the new MCIRF in which the rocket is momentarily at rest. The ratio between the time elapsed on the front clock versus the rear clock is always constant in any given MCIRF. If we stay with the measurements made in the initial inertial launch frame, then the desynchronisation is always increasing.

I'm not sure whether we disagree, or not. In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c²/g (where g is the acceleration of the rear).

This approximation is good if R is large and L is small, but becomes very inaccurate if R is small and L is large.

 

[stevendaryl]

They are not at rest wrt each other in a given inertial reference frame, when the MCIRF of accelerating rocket has significant relative velocity to that given inertial reference frame, but momentarily in any MCIRF they are at rest wrt each other

You are exactly right! Somehow, either I never did the calculation, or forgot it, but I just convinced myself of that this morning. The description of the rocket's position and velocity as a function of time in the launch frame I always knew was this:

1. x_rear = √((ct)² + R²)
2. v_rear = c² t/√((ct)² + R²)
3. x_front = √((ct)² + (R+L)²)
4. v_front = c² t/√((ct)² + (R+L)²)

(So at time t=0, both the front and the rear have speed 0.) But what's amazing is that equations 1-4 aren't just true in the launch frame, it's true in every inertial frame in which the rocket is momentarily at rest. That's pretty amazing; the rocket launches at time t=0 in frame F. At some time later, the rocket is at rest in some frame F' moving at speed v relative to F'. That time is t'=0, according to frame F', and equations 1-4 still hold, with x replaced by x' and t replaced by t'.

 

[yuiop]

... In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c²/g (where g is the acceleration of the rear).

This approximation is good if R is large and L is small, but becomes very inaccurate if R is small and L is large.

I may have to retract this last claim and you may be right here! The analysis I did with graphical software may have had limitations due to the accuracy of the software.

Here is a small "derivation" of the ratio of the clock rates being proportional to the ratio (R+L)/R.

The velocity of a clock on the rocket after proper time T is given as:

$$v = c - tanh(aT/c)$$

This can be rearranged to:

$$T = atanh(c-v)*c/a$$

Since the acceleration is equal to c^2/r this becomes:

$$T = atanh(c-v)*r/c$$

The ratio of the clock rates of two clocks at r₁ and r₂ respectively is then given by:

$$\frac{T_2}{T_1} = \frac{ atanh(c-v_2)*r_2/c}{atanh(c-v_1)*r_1/c}$$

In any given MCIRF, v is equal for all clocks at rest in that MCIRF, from the point of view of any other inertial reference frame, so we can assume v₁ = v₂ and can conclude:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{r_1+L}{r_1}$$

This is how more time elapses (from launch) on the front clock relative to the rear clock as seen in any MCIRF and is also a measure of the constant redshift seen by an observer at r₂ of a signal at r₁.

 

[stevendaryl]

I am not sure what you are getting at here. Let's say the clocks are initially synchronised in the launch frame. After launch let's say that when 1 second elapses on the rear clock, 4 seconds elapses on the front clock in the MCIRF. It then follows that after 2 seconds elapses on the rear clock, 8 seconds will have elapsed on the front clock in the new MCIRF in which the rocket is momentarily at rest. The ratio between the time elapsed on the front clock versus the rear clock is always constant in any given MCIRF. If we stay with the measurements made in the initial inertial launch frame, then the desynchronisation is always increasing. This approximation is good if R is large and L is small, but becomes very inaccurate if R is small and L is large.

I misunderstood how you were using the word "desynchronization". I was meaning "relativity of simultaneity", the fact that clocks that are synchronized in one frame are not synchronized in another frame. You are using to mean the discrepancy between the two clocks (regardless of the cause).

Let me try to explain what I meant by the two causes for the discrepancy between the two clocks.

Let e₁ be some event taking place at the rear clock. Let T₁ be the time on that clock at that event.

Let e₂ be some event taking place at the front clock that is simultaneous with e₁, according to the launch frame. Let T₂ be the time on that clock at that event.

Let e₃ be some event taking place at the front clock that is simultaneous with e₁, according to the instantaneous inertial reference frame of the rocket. Let T₃ be the time on that clock at that event.

Let δT₁ = T₂ - T₁
Let δT₂ = T₃ - T₂

δT₁ is completely due to length contraction; the front travels less than the rear, and so experiences less time dilation.

δT₂ is an additional effect due to relativity of simultaneity.

In the instantaneous rest frame of the rocket, the discrepancy between the times on the two clocks is the sum of the two δs. What I've been saying is that soon after launch, δT₂ is much bigger than δT₁. Long after launch, δT₁ eventually becomes much bigger than δT₂. (The latter approaches a maximum value of L/c). In the accelerated reference frame of the rocket itself, the only relevant number is the sum of the two, which grows steadily at a constant rate.

 

[Dale]

Could you perhaps be a little more exspansive in explaining what may be to you, obvious?

OK, so what I was talking about was using the metric to get an expression for time dilation as follows:

Time dilation in Minkowski spacetime:
$$c^2 d\tau^2=c^2 dt^2-dx^2-dy^2-dz^2$$$$\frac{d\tau^2}{dt^2}=1-\frac{1}{c^2}\left(\frac{dx^2}{dt^2}+\frac{dy^2}{dt^2}+\frac{dz^2}{dt^2}\right)$$$$\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2}}=1/\gamma$$

Time dilation in Schwarzschild spacetime:
$$c^2 d\tau^2 = \left(1-\frac{R}{r}\right) c^2 dt^2 - \left(1-\frac{R}{r}\right)^{-1} dr^2 - r^2 \left( d\theta^2 + sin^2\theta d\phi^2 \right)$$$$\frac{d\tau^2}{dt^2} = \left(1-\frac{R}{r}\right) - \frac{1}{c^2}\left(1-\frac{R}{r}\right)^{-1} \frac{dr^2}{dt^2} - \frac{r^2}{c^2} \left( \frac{d\theta^2}{dt^2} + sin^2\theta \frac{d\phi^2}{dt^2} \right)$$for an object at rest in the Schwarzschild coordinates all of the spatial derivatives are 0 leaving$$\frac{d\tau}{dt} = \sqrt{\left(1-\frac{R}{r}\right) }$$

Time dilation in Rindler coordinates:
$$c^2 d\tau^2 = \frac{g^2 x^2}{c^2}dt^2 - dx^2 - dy^2 - dz^2$$$$\frac{d\tau^2}{dt^2} = \frac{g^2 x^2}{c^4} -\frac{1}{c^2}\left( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2} \right)$$$$\frac{d\tau}{dt} = \sqrt{\frac{g^2 x^2}{c^4} -\frac{v^2}{c^2}}$$for an object at rest in the Rindler coordinates v=0 leaving$$\frac{d\tau}{dt} = \frac{g x}{c^2}$$

 

[stevendaryl]

I may have to retract this last claim and you may be right here! The analysis I did with graphical software may have had limitations due to the accuracy of the software.

Here is a small "derivation" of the ratio of the clock rates being proportional to the ratio (R+L)/R.

The velocity of a clock on the rocket after proper time T is given as:

$$v = c - tanh(aT/c)$$

Are you sure about that? The formulas that I use are:
v = c tanh(aT/c)
not
v = c - tanh(aT/c)

 

[yuiop]

... so we can assume v₁ = v₂ and can conclude:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{r_1+L}{r_1}$$

This is how more time elapses (from launch) on the front clock relative to the rear clock as seen in any MCIRF and is also a measure of the constant redshift seen by an observer at r₂ of a signal at r₁.

I think it is interesting to expand on this relation T2/T1 = R2/R1 (which appears to be an exact result) by comparing it to the gravitational time dilation in the Schwarzschild metric.

First the time dilation ratio in the Born rigid acceleration case, as a function of constant proper acceleration is given by:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{a_1}{a_2}$$

In the Schwarzschild metric the time dilation ratio of two stationary clocks in the field is given by:

$$\frac{T_2}{T_1} = \frac{\sqrt{1-2GM/(r_2c^2)}}{\sqrt{1-2GM/(r_1c^2)}}$$

and when we express this in terms of proper acceleration it becomes:

$$\frac{T_2}{T_1} = \sqrt{\frac{(c^2-2a_2r_2)}{(c^2-2a_1 r_1)}}$$

This appears to be very different to the Born rigid acceleration case with possibly a poor equivalence principle correlation, but maybe I am missing a gamma cubed factor in there somewhere between proper acceleration and coordinate acceleration. Any ideas?

 

[yuiop]

Are you sure about that? The formulas that I use are:
v = c tanh(aT/c)
not
v = c - tanh(aT/c)

Oops, I slipped up in quoting the given formula. Here is the derivation again using your correct expression. (The end result is still the same).

===============================

Here is a small "derivation" of the ratio of the clock rates being proportional to the ratio (R+L)/R.

The velocity of a clock on the rocket after proper time T is given as:

$$v = c * tanh(aT/c)$$

This can be rearranged to:

$$T = atanh(v/c)*c/a$$

Since the acceleration is equal to c^2/r this becomes:

$$T = atanh(v/c)*r/c$$

The ratio of the clock rates of two clocks at r₁ and r₂ respectively is then given by:

$$\frac{T_2}{T_1} = \frac{ atanh(v_2/c)*r_2/c}{atanh(v_1/c)*r_1/c}$$

In any given MCIRF, v is equal for all clocks at rest in that MCIRF, from the point of view of any other inertial reference frame, so we can assume v₁ = v₂ and can conclude:

$$\frac{T_2}{T_1} = \frac{r_2}{r_1} = \frac{r_1+L}{r_1}$$

This is how much more time elapses (from launch) on the front clock relative to the rear clock as seen in any MCIRF and is also a measure of the constant redshift seen by an observer at r₂ of a signal at r₁

===============================

Hopefully I got it right this time :)

 

[yuiop]

Time dilation in Rindler coordinates:
... for an object at rest in the Rindler coordinates v=0 leaving$$\frac{d\tau}{dt} = \frac{g x}{c^2}$$

From the Wikipedia article on Rindler coordinates we get

If all Rindler observers set their clocks to zero at T=0, then when defining a Rindler coordinate system we have a choice of which Rindler observer's proper time will be equal to the coordinate time t in Rindler coordinates, and this observer's proper acceleration defines the value of g above (for other Rindler observers at different distances from the Rindler horizon, the coordinate time will equal some constant multiple of their own proper time).[1] It is a common convention to define the Rindler coordinate system so that the Rindler observer whose proper time matches coordinate time is the one who has proper acceleration g=1, so that g can be eliminated from the equations

From this we can note that g is the same for all observers in Rindler coordinates, so we can rewrite the ratio based on Dalespam's equation:

$$\frac{d\tau_2}{dt_2} * \frac{dt_1}{d\tau_1} = \frac{g x_2}{g x_1}$$

as:

$$\frac{d\tau_2}{d\tau_1} = \frac{x_2}{x_1}$$

which agrees with equation quoted by stevendaryl and myself.

 

[Austin0]

No. There actually is not very much difference between those two, if the length of the rocket is sufficiently small. I was comparing the comoving inertial frame to the launch frame.



I'm not sure that we are disagreeing. I said that long after launching, the velocity approaches c, and so the desynchronization approaches the constant L/c.

I agree , we are not disagreeing. I see now you were saying that the rate of desynchronization increase was diminishing , not that it didn't continue to increase. Yeah?

I'm not sure whether we disagree, or not. In the comoving frame, the front clock always runs at a rate (R+L)/R faster than the rear clock, where L is the length of the rocket, and R is the characteristic length c²/g (where g is the acceleration of the rear). In the launch frame, the rates for both clocks start out the same, but the ratio gradually approaches the same value.

Agreed, that was what I was saying also.

 

[stevendaryl]

I haven't yet grokked the significance of this fact, but I recently discovered (I'm sure it's already well-known, but I didn't know it) an interesting characterization of constant proper acceleration.

First, we pick a launch frame F, and set up an infinite collection of inertial frames, all related by the Lorentz transformation. For every possible value v, there is a corresponding frame F_v with coordinates x_v, t_v defined by:

x_v = (x - vt)/√(1-(v/c)²)
t_v = (t - vx/c²)/√(1-(v/c)²)

where x and t are the coordinates for frame F.

Now, suppose we have a rocket and we want (for whatever reason) to accelerate in such a way that:

We are momentarily at rest in frame F at time t=0.
For every possible value of v, we come to rest in frame F_v at time t_v = 0.

So we're traveling in such a way that the "local" time is always t=0. It's sort of like traveling west so that it takes you 1 hour to pass through a time zone, so you're always entering a time zone at 12:00.

I can't think of any good reason to want to travel this way, but the interesting fact is that if you travel this way, you will be undergoing constant proper acceleration. This gives a more direct route to the equations for proper acceleration (the usual way requires knowing how acceleration transforms, and also knowing how to integrate hyperbolic trigonometric functions).

If we want to come to rest in frame F_v at time t_v = 0, then using the Lorentz transformations, we find:

t_v = (t - vx/c²)/√(1-(v/c)²)

In order for t_v to be zero, we need
t - vx/c² = 0

Now we can write this (multiplying by -1/2):
d/dt (-1/2 t² + 1/2 x²/c²) = 0

which has the solution

x² - (ct)² = R²

where R = some constant of integration. This is hyperbolic motion.

 

[yuiop]

... I can't think of any good reason to want to travel this way, but the interesting fact is that if you travel this way, you will be undergoing constant proper acceleration.

and if all parts of an extended rocket follow this rule then you will be undergoing Born rigid acceleration.

It took me a while to realize that the equation you derived in post 39 ...

1-(v₂/c)² = (1-(v/c)²)(1+2(v/c)²gL/c²)
(ignoring higher powers of L)

√(1-(v₂/c)²) = √(1-(v/c)²)(1+(v/c)²gL/c²)

Letting T₂ be the time on the front clock, and T be the time on the rear clock, we find:

T₂ = ∫√(1-(v/c)²)(1+(v/c)²gL/c²) dt

In the limit v→c (long after launch), this becomes

T₂ = T(1 + gL/c²)

... which I assumed to be an approximation is actually the same as the exact result T₂/T = r₂/r, as follows:

T₂/T = 1 + gL/c²

Using the notation r₂ and r for x₂ and x respectively when t=0 in any given MCIRF and noting that r₂ = r+L and that g is defined as c^2/r, then:

T₂/T = 1 + L/r

T₂/T = (r + L)/r

T₂/T = r₂/rThis is surprising to me because you discarded the higher powers of L presumably introducing a small error and ended up with an exact result. I guess the approximation self canceled out somewhere.

 

[harrylin]

I had a physics test at school recently. One of the questions was based on the equivalence principle, going something like this: Two clocks in a spaceship that is accelerating. One at the bottom and one at the top of the space ship. Now think that the spaceship is so far away from any object in space, that it is not affected by any gravitational force.

It is my understanding that according to the equivalence principle, one can not be able to do any test that suggests that you are no longer in a gravitational field, but in an accelerated system. And according to the theory of relativity, the clock furthest down in the gravitational field will go slower than any clock higher up. Yet it makes no sense to me that the same rules would apply in an accelerated system. The correct answer supposedly is that the clock at the bottom of the ship will slow down. What am I missing? Would the clock at the bottom of the ship really slow down? Making the rules of time dilation apply in an accelerated system, the same way it does in an gravitational field?

Hi Nick, welcome to physicsforums.

Before the discussion here went off into very technical (and very interesting) details, you received two "yes" answers which possibly missed the very point that you were missing. Thus here's my slightly different answer.

No, not really:

- The equivalence principle has that the observable effect will be the same. Einstein calculated (predicted) what the observable effect will be due to gravitation, basing himself on the observable Doppler effect due to acceleration.
Thus the clock at the bottom of the ship will only *appear* to slow down by the gravitational time dilation factor *if* you assume that the ship is not accelerating but at rest in a gravitational field. If instead you assume, as you do, that there is negligible gravitational field, then the clock at the rear will really *not* slow down (at least, by far not by that amount) compared to the one at the front: the effect is then at the start (zero speed) explained as just due to Doppler effect. Next, when picking up speed, there is in addition the effect of time dilation of *both* clocks due to speed, in combination with length contraction which makes that the clocks do not exactly slow down the same; this is where things get tricky. But that is all purely special relativity, without any gravitational time dilation.

- Different from fake gravitational fields as caused by acceleration, real gravitational fields get weaker at greater distance from the source, and that gradient can be used for Earth sensors in a "free-falling" satellite that indicate where the Earth is.

 

[yuiop]

Following on from #64, another perhaps interesting observation is that if we speed up the clock at the rear of a Born rigid accelerating rocket so that the radar length of the rocket as measured at the rear, agrees with the radar length of the rocket as measured at the front, then the clocks at the front and rear will remain permanently synchronised with each other and read the same as each other in any MCIRF.

 

[Austin0]

Let me try to explain what I meant by the two causes for the discrepancy between the two clocks.

Let e₁ be some event taking place at the rear clock. Let T₁ be the time on that clock at that event.

Let e₂ be some event taking place at the front clock that is simultaneous with e₁, according to the launch frame. Let T₂ be the time on that clock at that event.

Let e₃ be some event taking place at the front clock that is simultaneous with e₁, according to the instantaneous inertial reference frame of the rocket. Let T₃ be the time on that clock at that event.

Let δT₁ = T₂ - T₁
Let δT₂ = T₃ - T₂

δT₁ is completely due to length contraction; the front travels less than the rear, and so experiences less time dilation.

δT₂ is an additional effect due to relativity of simultaneity.

In the instantaneous rest frame of the rocket, the discrepancy between the times on the two clocks is the sum of the two δs. What I've been saying is that soon after launch, δT₂ is much bigger than δT₁. Long after launch, δT₁ eventually becomes much bigger than δT₂. (The latter approaches a maximum value of L/c). In the accelerated reference frame of the rocket itself, the only relevant number is the sum of the two, which grows steadily at a constant rate.

I agree with everything here.
The increase from simultaneity starts out at a maximum rate wrt coordinate time in the LF and diminishes at a factor of 1/ $\gamma$³
The discrepancy due to relative velocity increases over time by, I would guess, by a factor of $\gamma$³ ,as coordinate acceleration diminished, coordinate time for increased velocity would increase so the cumulative increase of proper time on the front clock due to the differential would comparably increase.
But what I don't understand is why the simultaneity relative to the MCIRF's would be relevant to the accelerating system?


I would think that given the proper acceleration values for front and back for the initial length , that simply calculating velocities from coordinate acceleration would give relative gamma between the front and back directly.

For v_f,v_b
$\gamma$_b= 1/√ 1-v _b²/c²

and $\gamma$ _f = 1/√ 1-v _f<sup>2[/SU/c2 for any coordinate t in the LF ,,,so $\gamma$_f/$\gamma$_b should be it , no?

with no necessity of considering length contraction directly or including it in calculations. Likewise for simultaneity as you are only interested in the proper rates and delta t's of the two clocks which can be regarded as independent entities in this circumstance.
SO is this idea too out of the park??
Whats the hidden snag? Of course I have no idea of how to do the math to make this work ;-)</sup>

 

[Nick20]

Hi Nick, welcome to physicsforums.

Before the discussion here went off into very technical (and very interesting) details, you received two "yes" answers which possibly missed the very point that you were missing. Thus here's my slightly different answer.

No, not really:

- The equivalence principle has that the observable effect will be the same. Einstein calculated (predicted) what the observable effect will be due to gravitation, basing himself on the observable Doppler effect due to acceleration.
Thus the clock at the bottom of the ship will only *appear* to slow down by the gravitational time dilation factor *if* you assume that the ship is not accelerating but at rest in a gravitational field. If instead you assume, as you do, that there is negligible gravitational field, then the clock at the rear will really *not* slow down (at least, by far not by that amount) compared to the one at the front: the effect is then at the start (zero speed) explained as just due to Doppler effect. Next, when picking up speed, there is in addition the effect of time dilation of *both* clocks due to speed, in combination with length contraction which makes that the clocks do not exactly slow down the same; this is where things get tricky. But that is all purely special relativity, without any gravitational time dilation.

- Different from fake gravitational fields as caused by acceleration, real gravitational fields get weaker at greater distance from the source, and that gradient can be used for Earth sensors in a "free-falling" satellite that indicate where the Earth is.

Thank you. I really find physics quite interresting, and I have been doing some thinking of my own. I don't have much of an education though. I'll be 20 years old in october, and I've had 2 years of basic physics. Now this particular task puzzled me and really got me to wonder what was behind it all, because it made no sense to me that the "bottom" clock would experience time dilation as if it was in a gravitational field. So my answer to that task was that there will be no time dilation (concidering that the spaceship is not yet traveling in a relativistic speed.) But appearently that was not correct. I do not believe that they thought too proper about it when they wrote that one.

Anyway I want to thank you all. I have gotten many interresting answers to my question. Reaching over 65 answers is a good first impression to a new forum

Nick

 

[harrylin]

[...] it made no sense to me that the "bottom" clock would experience time dilation as if it was in a gravitational field. So my answer to that task was that there will be no time dilation (concidering that the spaceship is not yet traveling in a relativistic speed.) [..]

As I elaborated, that's roughly correct.

Anyway I want to thank you all. I have gotten many interresting answers to my question. Reaching over 65 answers is a good first impression to a new forum
Nick

Hehe you can surely say that!

Cheers,
Harald

 

[yuiop]

- Different from fake gravitational fields as caused by acceleration, real gravitational fields get weaker at greater distance from the source, and that gradient can be used for Earth sensors in a "free-falling" satellite that indicate where the Earth is.

The apparent field inside an accelerating rocket also gets weaker at greater distance from the apparent source. Accelerometers attached to the nose of the rocket indicate less proper acceleration than accelerometers attached to the tail of the rocket.

But what I don't understand is why the simultaneity relative to the MCIRF's would be relevant to the accelerating system?

I would think that given the proper acceleration values for front and back for the initial length , that simply calculating velocities from coordinate acceleration would give relative gamma between the front and back directly.

For v_f,v_b
$\gamma$_b= 1/√ 1-v _b²/c²

and $\gamma$ _f = 1/√ 1-v _f<sup>2[/SU/c2 for any coordinate t in the LF ,,,so $\gamma$_f/$\gamma$_b should be it , no?

with no necessity of considering length contraction directly or including it in calculations. Likewise for simultaneity as you are only interested in the proper rates and delta t's of the two clocks which can be regarded as independent entities in this circumstance.
SO is this idea too out of the park??
Whats the hidden snag? Of course I have no idea of how to do the math to make this work ;-)</sup>

^{The snag is that the by calculating the relative time dilation by using the relative velocities at the front and back of the rocket one comes to the conclusion that the relative rates of clocks on board the rocket as measured by observers on board the rocket increases over time which is simply not true. The relative rates of clocks on board the rocket as measured by observers on board the rocket is constant over time.}