A question about the equivalence principle.

[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/ \gamma³
The discrepancy due to relative velocity increases over time by, I would guess, by a factor of \gamma³ ,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.}

 

[Austin0]

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.

Hi yuiop you have returned ;-)

i agree with your assumption that it appears that the relative dilation due to coordinate velocity would increase. That was my intuition also but after thought i became less sure.
That was partly my interest in the analysis I described.

But if we are right and the dilation differential would increase up the velocity curve this would seem to present a major problem.

As I said, the two clocks in question could be unconnected.
Independently accelerating clocks that would maintain the proper contracted separation simply as a consequence of the acceleration differential.

It seems clear that the fundamental SR principles must pertain.I.e. The gamma relation between velocity and proper time rates and deltas. And the clock hypothesis.

So any short interval measurements of coordinate velocity and clock comparisons for those intervals must correspond to the fundamental relationship where ever they taken along the course of acceleration.

Simply connecting the two clocks into a single system should not have any effect on this relationship.

So if the results predicted by the Rindler coordinates do not agree with the results predicted by the fundamental principles of SR there would be a real question.

What possible physics would account for this ?

i.e. what would prevent the velocity dilation from occurring?

Which prediction should be considered valid?

Would you agree there would be a question?

It was actually this question which prompted my primitive attempt at calculating the dilation factor I posted earlier.

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

You missed this one. regarding Steve's derivative approach.

So have you calculated the differential due to velocity actually/ or is it just a logical guess? ;-0
thanks

 

[harrylin]

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. [..]

Good one! Yes, there is a very slight difference due to length contraction - and it's always the same for a given distance between clocks and given acceleration. However, if one allows the Earth sensor to free-fall inside or next to the rocket, it can detect no gradient at all, not even in principle.
And I now wonder if one would nevertheless interpret the (incredibly small) difference in accelerometer readings as due to a real field, with what distance to the gravitation source that would correspond, and with what mass - could one really be fooled?

 

[yuiop]

So have you calculated the differential due to velocity actually/ or is it just a logical guess? ;-0

At the time it was a estimate, but I have now done the calculations just to check and can confirm that clock speed differential between front and back increases over time if we use the velocity time dilation approach. For example if we have a constant proper acceleration of 1 at the back of the rocket and 0.1 at the front and use units of c=1, then initially the time dilation ratio (back/front) is 1 at time t=0, 0.71 at t=1 and 0.456 at t=2. This is not what would be observed on board the rocket(s) because the time (t) here is the simultaneous coordinate time in the launch frame and the notion of what is simultaneous is different on board the rocket(s). This is the crux of the matter that I shall come back to.

i agree with your assumption that it appears that the relative dilation due to coordinate velocity would increase. That was my intuition also but after thought i became less sure.

After doing the calculations I am now sure that is what would be observed as coordinate velocity increases relative to any given MCIRF, as measured relative to the simultaneous coordinate time in that given MCIRF. As above, it is the notion of simultaneity that is important and this changes with reference frames and you have not been careful to specify one.

That was partly my interest in the analysis I described.

But if we are right and the dilation differential would increase up the velocity curve this would seem to present a major problem.

We are right but it does not present a problem ;)

As I said, the two clocks in question could be unconnected.
Independently accelerating clocks that would maintain the proper contracted separation simply as a consequence of the acceleration differential.

Agree.

It seems clear that the fundamental SR principles must pertain.I.e. The gamma relation between velocity and proper time rates and deltas. And the clock hypothesis.

Agree, these fundamental SR principles do hold.

So if the results predicted by the Rindler coordinates do not agree with the results predicted by the fundamental principles of SR there would be a real question.

What possible physics would account for this ?

i.e. what would prevent the velocity dilation from occurring?

Nothing. Velocity time dilation still occurs, but the effect cancels out. Consider two signals emitted simultaneously from the front and back of a rocket at coordinate time t=0 as measured in a given MCIRF. To give this MCIRF a label I will call it the launch frame (LF), but there is nothing special about this MCIRF and I could do an identical analysis in any other MCIRF. When the signal from the front arrives at the back let us say the velocity in the LF is vb and when the signal from the back arrives at the front, the velocity in the LF is vf. It turns out that with Born rigid acceleration, vb = vf. This means that sqrt(1-(vb)^2) = sqrt(1-(vf)^2) so at the time of the reception of the signals the velocities and velocity based time dilation is identical at back and front and so there is no differential time dialtion between back and front due to velocity time dilation. All the apparent differential time dilation observed on board the rocket is due to classical Doppler shift according to the observer in the inertial LF.

Which prediction should be considered valid?

They are both valid. One is the interpretation according to the simultaneity of an inertial reference frame and the other is the interpretation due to a different notion of simultaneity in an accelerating reference frame.

Would you agree there would be a question?

Not really. We are used to observer dependent quantities such as coordinate velocity, time, distance, time dilation, length contraction, simultaneity etc. being different in different inertial reference frame (by definition) in regular unaccelerated SR.

You missed this one. regarding Steve's derivative approach: "But what I don't understand is why the simultaneity relative to the MCIRF's would be relevant to the accelerating system?"

Well now we are back to the crux of the problem. I have been putting this off because while I feel I understand it intuitively in my head, putting it into words is not so easy :) The first difficulty is how to define a notion of simultaneity for the observers on board the rocket, when there own reference clocks appear to be running at different rates in different locations on board the rocket. If we ask the observers at the front and back of the rocket to send signals simultaneously in there own reference frame, how are they going to arrange that? One way to do this would be to agree a convention that defines "simultaneous" events at the front and back of the rocket as being events that are both simultaneous in a shared MCIRF. If they do this they will observe that the signals are received at the back and front "simultaneously" because the reception events will be simultaneous in a shared MCIRF, (but this shared MCIRF will not be the same shared MCIRF that the signals were emitted simultaneously in). For a practical example, let's say we an inertial rocket ir1 moving at 0.6c and another ir2 moving at 0.8c both relative to the LF. After launch we tell the observers at the front and back to send a signal at the moment they are at rest with ir1, then for a suitable length accelerating rocket, they will both receive signals when they are momentarily at rest with ir2. You might argue that I have cheated here, because I have simply defined, rather than derived a notion of simultaneous for the accelerating rocket observers.

Another approach is to speed up the rear clock (as mentioned in an earlier post) so that it appears to be running at the same rate as the front clock. Now we have an unequivocal method of defining simultaneous in the accelerating rocket reference frame and can synchronise clocks using the usual Einstein clock synchronisation convention. Now when we send signals from the front and back using the on board rocket reference clocks, the signals arrive simultaneously at the back and front respectively according to the rocket clocks and the elapsed time between sending and receiving is equal according to both the front and back accelerating rocket observers. At no point here have we had to refer to the MCIRF's but if we compare results, simultaneous emission and simultaneous reception of signals as measured by the accelerating rocket observers, agrees with simultaneous as measured in the MCIRF's. Now the rocket observers know they had to speed up the rear clock to make it run at the same rate as the front clock so they know the real proper time rate of the rear clock must be slower than the proper time rate of the front clock. If they consider themselves to be stationary in their own reference frame they would ascribe this differential clock rate to a gravitational field and this coincides with the proper acceleration they can feel and and measure. The observers in a given inertial reference frame outside the rocket would ascribe the differential clock rates observed by the rocket observers to classical Doppler shift.

Good one! Yes, there is a very slight difference due to length contraction - and it's always the same for a given distance between clocks and given acceleration. However, if one allows the Earth sensor to free-fall inside or next to the rocket, it can detect no gradient at all, not even in principle.
And I now wonder if one would nevertheless interpret the (incredibly small) difference in accelerometer readings as due to a real field, with what distance to the gravitation source that would correspond, and with what mass - could one really be fooled?

The differences between the Born rigid accelerating rocket and a real gravitational field, due to tidal effects, are significant enough, that we would be only be fooled if limited to measurements in a very restricted local region.

 

[stevendaryl]

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

I'm not sure I understand your question. You're asking why the inertial coordinate system is relevant to the accelerated observer? In some ways, it's just a matter of convention for what the accelerated observer considers two "simultaneous" events. One convention is to use the same notion of simultaneity as a comoving inertial observer.

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.

No, that's not correct. If you wait a short time δt after launch, the rear clock will be traveling at speed v_rear = g_rear δt, and the front clock will be traveling at speed v_front = g_front δt. So just based on that, you would expect a relative rate of the clocks to be
√(1- (g_rear δt)²/c²))/√(1- (g_front δt)²/c²))
but the actual ratio of rates is
g_front /g_rear
those aren't close at all. The first goes to 1 as δt → 0, but the second doesn't.

 

[PeterDonis]

No, that's not correct. If you wait a short time δt after launch, the rear clock will be traveling at speed v_rear = g_rear δt, and the front clock will be traveling at speed v_front = g_front δt. So just based on that, you would expect a relative rate of the clocks to be
√(1- (g_rear δt)²/c²))/√(1- (g_front δt)²/c²))
but the actual ratio of rates is
g_front /g_rear
those aren't close at all. The first goes to 1 as δt → 0, but the second doesn't.

I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, but this doesn't look right; the ratio of clock rates *should* be governed by the first formula, based on the two velocities. The ratio of clock rates is certainly *not* equal to the ratio of accelerations. And the ratio *should* go to 1 as delta t -> 0, because at launch, which is what delta t -> 0 means, the two clocks are at rest relative to each other, so their clock rates are the same.

Some web pages that seem to me to be relevant are Greg Egan's page on the Rindler horizon:

http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

And the Usenet Physics FAQ pages on SR and acceleration (which links to the page on the relativistic rocket equation, another good resource), and on the clock hypothesis:

http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html

 

[stevendaryl]

I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, but this doesn't look right; the ratio of clock rates *should* be governed by the first formula, based on the two velocities. The ratio of clock rates is certainly *not* equal to the ratio of accelerations. And the ratio *should* go to 1 as delta t -> 0, because at launch, which is what delta t -> 0 means, the two clocks are at rest relative to each other, so their clock rates are the same.

But that doesn't give the correct answer, which is that, according to measurements performed by observers aboard the rocket, the front clock always runs faster than the rear clock, the ratio of their rates is always the same: 1 + gL/c².

 

[PeterDonis]

But that doesn't give the correct answer, which is that, according to measurements performed by observers aboard the rocket, the front clock always runs faster than the rear clock, the ratio of their rates is always the same: 1 + gL/c².

It's possible that I'm misunderstanding the scenario you are talking about. When you say that a short time after launch, each clock will be traveling at a velocity g delta t, where g is the acceleration of the clock, that tells me that exactly at time t = 0, i.e., before the time delta t elapses, each clock is at rest in the same "launch frame". If the two clocks are at rest relative to each other at some instant, then their clock rates, at that instant, are the same. See the web page I linked to on the clock postulate. My understanding is that that is the scenario you are talking about.

The formula 1 + gL/c describes something a bit different than what I would call the "ratio of instantaneous clock rates", but I'd rather not get into that until I'm sure I correctly understand the scenario you're talking about.

 

[stevendaryl]

It's possible that I'm misunderstanding the scenario you are talking about. When you say that a short time after launch, each clock will be traveling at a velocity g delta t, where g is the acceleration of the clock, that tells me that exactly at time t = 0, i.e., before the time delta t elapses, each clock is at rest in the same "launch frame". If the two clocks are at rest relative to each other at some instant, then their clock rates, at that instant, are the same.

To talk about clock "rates", you have to refer to two different points in time. To talk about the relative rates of two clocks, you need 4 events, and you need a notion of simultaneity.

e₁ = the rear clock shows time 0.
Let F = the instantaneous rest frame of the rear clock at this event.

e₂ = some event at the front clock that is simultaneous with e₁ in frame F . Let T₂ = the time showing on the front clock at event e₂.

e₃ = the rear clock shows time δT
Let F' be the instantaneous rest frame of the rear clock at this event.

e₄ = some event at the front clock that is simultaneous with e₃ in frame F'. Let T₄ = the time showing on the front clock at event e₄.

So from the point of view of an observer in the rear of the rocket, the rear clock advanced by δT, and the front clock advanced by: T₄ - T₂

Then the ratio of clock rates, as measured by the rear clock is:

R_front/R_rear
= (T₄ - T₂)/δT

As δT → 0, this ratio does not go to 1, but goes to 1 + gL/c², where L is the length of the rocket, and g is the acceleration of the rear.

 

[PeterDonis]

To talk about clock "rates", you have to refer to two different points in time. To talk about the relative rates of two clocks, you need 4 events, and you need a notion of simultaneity.

I agree with the second sentence given the first. I take the first as your definition of how you are using the term "clock rate"; it is not the only possible definition, but I have no problem using yours for this discussion.

e₁ = the rear clock shows time 0.
Let F = the instantaneous rest frame of the rear clock at this event.

e₂ = some event at the front clock that is simultaneous with e₁ in frame F . Let T₂ = the time showing on the front clock at event e₂.

You've left out a key piece of information here: is the front clock at rest, in frame F, at event e2? (By hypothesis the rear clock is at rest in frame F at event e1.)

e₃ = the rear clock shows time δT
Let F' be the instantaneous rest frame of the rear clock at this event.

e₄ = some event at the front clock that is simultaneous with e₃ in frame F'. Let T₄ = the time showing on the front clock at event e₄.

Same question here. I ask these, again, to make sure I understand the scenario you are talking about. I think the answer to both is "yes", but I would really like confirmation before going into details, since there's no point in my talking about a different scenario than you are talking about.

 

[stevendaryl]

I agree with the second sentence given the first. I take the first as your definition of how you are using the term "clock rate"; it is not the only possible definition, but I have no problem using yours for this discussion.

You've left out a key piece of information here: is the front clock at rest, in frame F, at event e2? (By hypothesis the rear clock is at rest in frame F at event e1.)

That's not really relevant to the question of what is the ratio of clock rates, as measured by an observer in the rear of the rocket. But in the way I've set things up, the answer is "yes".

Same question here. I ask these, again, to make sure I understand the scenario you are talking about. I think the answer to both is "yes", but I would really like confirmation before going into details, since there's no point in my talking about a different scenario than you are talking about.

Whether the front clock is at rest at events e₂ and e₄ doesn't seem relevant to me. But if you work out the details for the constant proper acceleration, it turns out to be true, that in frame F, the front clock is at rest at event e₂, and in frame F', the front clock is at rest at event e₄.

 

[PeterDonis]

That's not really relevant to the question of what is the ratio of clock rates, as measured by an observer in the rear of the rocket. But in the way I've set things up, the answer is "yes".

Whether the front clock is at rest at events e₂ and e₄ doesn't seem relevant to me. But if you work out the details for the constant proper acceleration, it turns out to be true, that in frame F, the front clock is at rest at event e₂, and in frame F', the front clock is at rest at event e₄.

Thanks for the confirmation. The reason it's relevant (other than giving me confirmation that I was correctly understanding your scenario) is the clock postulate, which I referred to before. Consider what you have here: you have two pairs of events, at each of which two objects are mutually at rest. By the clock postulate, then, their "instantaneous clock rates" should be the same, because their velocities, relative to any inertial frame, are the same, therefore their "time dilation factors", relative to any inertial frame, are the same. Yet between their respective events, one object (the front clock) experiences more elapsed time than the other (the rear clock).

In fact it's even weirder than that. Consider the entire worldlines of the two clocks; I'll write their equations in the "launch frame" as follows:

Rear clock: x^2 - t^2 = R^2

Front clock: x^2 - t^2 = (R + L)^2

Now draw any straight line through the origin, in the "launch frame", t = vx, where v >= 0. This is either a horizontal line (for v = 0) or a line sloping up and to the right at less than 45 degrees (for v > 0). Call the event where the line intersects the rear clock's worldline Rv, and the event where the line intersects the front clock's worldline Fv. Then all of the following are true:

(1) For any v, the line t = vx is a "line of simultaneity" in the instantaneous rest frame of the rear clock at Rv *and* of the front clock at Fv. Therefore, Rv and Fv are simultaneous as seen by both the front and the rear clocks.

(2) For any v, relative to the launch frame, the rear clock at Rv and the front clock at Fv are both moving at velocity v. Therefore, the front and rear clocks both see each other as at mutual rest at these events, and they both have the same "time dilation" factor at these events (because of the clock postulate).

(3) For any v, the proper time experienced by the front clock between F0 and Fv is greater, by the ratio (R+L)/R, than the proper time experienced by the rear clock between R0 and Rv.

So we have two clocks, which remain a constant distance apart at mutual rest, and have the same "time dilation factor" at any pair of corresponding events, and yet experience different proper times between corresponding events. I realize all this is true; I just point it out to explain why one has to be careful using the term "clock rate" without more explanation. Many people will think "time dilation factor", as in 1/sqrt(1 - v^2), when they see "clock rate" (after all, that's what I initially thought when I saw it), but that's not what you mean by the term. In particular, I think this may be one source of confusion in some of Austin0's posts.

 

[stevendaryl]

Thanks for the confirmation. The reason it's relevant (other than giving me confirmation that I was correctly understanding your scenario) is the clock postulate, which I referred to before. Consider what you have here: you have two pairs of events, at each of which two objects are mutually at rest. By the clock postulate, then, their "instantaneous clock rates" should be the same, because their velocities, relative to any inertial frame, are the same, therefore their "time dilation factors", relative to any inertial frame, are the same. Yet between their respective events, one object (the front clock) experiences more elapsed time than the other (the rear clock).

For problems involving acceleration, time dilation is not the only consideration. Let's consider a variant of the old twin paradox. We have two twins that are the same age, but live on different planets, light-years apart. One twin on his 20th birthday hops into a rocket and accelerates rapidly to nearly the speed of light and gets to his other twin in just one year, according to the clock in his rocket. But the other twin has aged 20 years during the trip. How did that happen? From the point of view of the traveling twin, the trip only lasted a year. How could the other twin age 20 years?

Well, what you have to take into account is that simultaneity is relative. Let e₁ be the traveling twin celebrated his 20th birthday. Let e₂ be the event at which the distant twin celebrated his 20th birthday. Let F be the "launch" frame of the traveling twin, and let F' be the "traveling" frame. Events e₁ and e₂ are simultaneous in frame F, but not in frame F'; in frame F', e₂ took place long, long before e₁. So in jumping from frame F to frame F', the traveling twin changed his notion of what time "now" is on the distant planet, and so he changed his notion of how old the distant twin is.

So for an accelerated observer, the time on a distant clock changes both because that clock advances, and also because the observer's notion of what is "now" for the distant clock changes.

 

[PeterDonis]

So for an accelerated observer, the time on a distant clock changes both because that clock advances, and also because the observer's notion of what is "now" for the distant clock changes.

Yes, the "simultaneity lines" t = vx that I described illustrate this: they change slope as v increases, meaning that the direction between the rear clock's "now" and the front clock's "now" changes. That means a given segment of the rear clock's worldline includes a set of simultaneity lines that sweep over a larger segment of the front clock's worldline. It's the hyperbolic equivalent of concentric circles, where the same angle sweeps out a longer arc on the circle with a larger radius.

 

[yuiop]

I haven't been following this thread closely, and I can see a lot of formulas have been written down that I have not looked at in detail, ...

Just to summarise, the various formulas written down by Dalespam, stevendaryl and myself for the relative red shift of clock rates on board the accelerating rocket, as measured by observers on board the accelerating rocket are all equivalent, i.e.:

\frac{\Delta\tau_2}{\Delta\tau_1} = (1+g_1L/c^2) = \frac{R_2}{R_1} = \frac{g_1}{g_2}

To the accelerated observers on board the rocket, the clocks remain at constant distance from each other, so from their point of view, none of the observed red shift is due to classical Doppler shift caused by relative velocities.

 

[Austin0]

This is not what would be observed on board the rocket(s) because the time (t) here is the simultaneous coordinate time in the launch frame and the notion of what is simultaneous is different on board the rocket(s). This is the crux of the matter that I shall come back to.
After doing the calculations I am now sure that is what would be observed as coordinate velocity increases relative to any given MCIRF, as measured relative to the simultaneous coordinate time in that given MCIRF. As above, it is the notion of simultaneity that is important and this changes with reference frames and you have not been careful to specify one.

Only one. The launch frame.

Nothing. Velocity time dilation still occurs, but the effect cancels out. Consider two signals emitted simultaneously from the front and back of a rocket at coordinate time t=0 as measured in a given MCIRF. To give this MCIRF a label I will call it the launch frame (LF), but there is nothing special about this MCIRF and I could do an identical analysis in any other MCIRF. When the signal from the front arrives at the back let us say the velocity in the LF is vb and when the signal from the back arrives at the front, the velocity in the LF is vf. It turns out that with Born rigid acceleration, vb = vf.

i am not sure if this actually tracks. If the signals are simultaneous in the emission MCIRF it appears unlikely the reception ,in a frame which would be many frames and spatial distance removed, would be simultaneous. I will have to work on this.

This means that sqrt(1-(vb)^2) = sqrt(1-(vf)^2) so at the time of the reception of the signals the velocities and velocity based time dilation is identical at back and front and so there is no differential time dialtion between back and front due to velocity time dilation. All the apparent differential time dilation observed on board the rocket is due to classical Doppler shift according to the observer in the inertial LF.

even if your assumption is correct and the reception is simultaneous wrt the MCIRF how does this imply no velocity dilation?? by definition there is no motion relative to the MCIRF's
so any calculation based on the MCIRF couldn't reveal relative velocity between front and back.

When you say here the observer in the LF do you mean the initial launch frame or the current mCIRF ,,,,earlier you were referring to MCIrfs as LF
In any case Doppler shift is as you say ,apparent dilation, so not really relevant

Well now we are back to the crux of the problem. I have been putting this off because while I feel I understand it intuitively in my head, putting it into words is not so easy :) The first fdifficulty is how to define a notion of simultaneity for the observers on board the rocket, when there own reference clocks appear to be running at different rates in different locations on board the rocket. If we ask the observers at the front and back of the rocket to send signals simultaneously in there own reference frame, how are they going to arrange that? One way to do this would be to agree a convention that defines "simultaneous" events at the front and back of the rocket as being events that are both simultaneous in a shared MCIRF. If they do this they will observe that the signals are received at the back and front "simultaneously" because the reception events will be simultaneous in a shared MCIRF, (but this shared MCIRF will not be the same shared MCIRF that the signals were emitted simultaneously in). For a practical example, let's say we an inertial rocket ir1 moving at 0.6c and another ir2 moving at 0.8c both relative to the LF. After launch we tell the observers at the front and back to send a signal at the moment they are at rest with ir1, then for a suitable length accelerating rocket, they will both receive signals when they are momentarily at rest with ir2. You might argue that I have cheated here, because I have simply defined, rather than derived a notion of simultaneous for the accelerating rocket observers

I think you misunderstood my question. I am aware of the problems implementing simultaneity in this circumstance , i mentioned a few earlier.
the question was why take an approach which had these problems and the inherent ambiguity of the result due to these problems.??

Another approach is to speed up the rear clock (as mentioned in an earlier post) so that it appears to be running at the same rate as the front clock. Now we have an unequivocal method of defining simultaneous in the accelerating rocket reference frame and can synchronise clocks using the usual Einstein clock synchronisation convention. Now when we send signals from the front and back using the on board rocket reference clocks, the signals arrive simultaneously at the back and front respectively according to the rocket clocks and the elapsed time between sending and receiving is equal according to both the front and back accelerating rocket observers.

well I have to disagree here. Simply scaling the clocks does not make it an inertial frame.
It is still an accelerating system.
Even disregarding the acceleration/velocity differential, a synchronization which works for one velocity cannot work for other different velocities. Yeah?? How could it?

If they consider themselves to be stationary in their own reference frame they would ascribe this differential clock rate to a gravitational field and this coincides with the proper acceleration they can feel and and measure. The observers in a given inertial reference frame outside the rocket would ascribe the differential clock rates observed by the rocket observers to classical Doppler shift.

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.

I think you have misunderstood both my approach and point.

regarding measurement of relative clock rates at the front and back from the launch frame. LF accelerating system AF
Simultaneity is not an issue. measurement of clock rates of course requires some interval of time a single event doesn't work.

so in LF at (bx_o,t₀) (fx₀ ,t₀) we get

observations bT'₀ and fT'₀ of AF ..

at some later point at (bx₁, t₁), (fx₁,t₁)

we get observations bT'₁ and fT'₁ of AF

(bx) t₁ - (bx)t₀ = (bx)dt

(fx)t₁ -(fx)t₀ = (fx)dt

bT₁-bT₀ =dbT'

fT₁ - fT₀= dfT

then (fx)dt/\gamma =dfT' and (bx)dt/\gamma=dbT' or not

Would you agree that in this circumstance the simultaneity of the LF clocks at either measurement is not important because there is no direct comparison between the observations between them?
The comparison is between an observation of the back clock with a later observation of the back clock.Etc.

Of course it would be necessary to calculate the proper times for these events using the Rindler coordinates to make these comparisons.

Also the measurement points could be widely separated say 0.7c and 0.8c

More and more i suspect that the velocity dilation would be insignificant and might possibly agree with the Rindler predictions. I.e. would not increase with greater velocities. The acceleration magnitudes you used were totally unrealistic. The back of the rocket quickly passing the front and leaving it in the dust ;-) so I am still unsure.
thanks

 

[yuiop]

i am not sure if this actually tracks. If the signals are simultaneous in the emission MCIRF it appears unlikely the reception ,in a frame which would be many frames and spatial distance removed, would be simultaneous. I will have to work on this.

It does track. See the attached chart showing two rockets with Born rigid acceleration. Each time signals (black diagonal lines) are emitted simultaneously in one MCIRF, they are received simultaneously in a subsequent MCIRF. In the chart green curves represent equal proper time for the rocket clocks, red lines represent lines of equal velocity relative to the LF (or subsequent MCIRFs) and the blue lines represent the worldlines of the rockets in the LF. Note the constant nature of the redshift. Signals sent every 3 ticks from the rear rocket are received every 4 ticks by the leading rocket.

even if your assumption is correct and the reception is simultaneous wrt the MCIRF how does this imply no velocity dilation?? by definition there is no motion relative to the MCIRF's so any calculation based on the MCIRF couldn't reveal relative velocity between front and back.

You have the reasoning reversed. The MCIRFs reveal that there is no relative velocity in the reference frames of the accelerating rockets. You are right that simultaneously at given instant (other than t=0) in the LF, the velocities of the front and rear rockets are different, but this simultaneity does not apply to the accelerating reference frame. We can for example show that if there is no length contraction and the rockets are accelerating identically in the LF, that differential time dilation is still occurring in the rocket reference frame.

When you say here the observer in the LF do you mean the initial launch frame or the current mCIRF

Here I meant LF, as in pick one MCIRF and stick with it, rather than keep switching to subsequent MCIRFs.

,,,,earlier you were referring to MCIrfs as LF

I just meant you could pick any arbitrary MCIRF as a LF. There is nothing particularly special about the LF. The rockets are still accelerating even in the LF.

In any case Doppler shift is as you say ,apparent dilation, so not really relevant

This is the bizarre aspect. The inertial observer attributes the redshift to classical Doppler shift due to velocity differential between emission and reception and yet the time dilation observed by the accelerating observers is real (as in physical) and an observer in the rear rocket really ages slower than an observer in the front rocket.

the question was why take an approach which had these problems and the inherent ambiguity of the result due to these problems.??

.. because nature does not give us much choice and there is no natural way to have synchronised clocks and static coordinate system in an accelerating reference frame or in a gravitational field.

well I have to disagree here. Simply scaling the clocks does not make it an inertial frame.
It is still an accelerating system.

Yes, it is still an accelerating system and I never claimed to make it an inertial reference frame. I only claimed it gave us a way to have a sensible coordinate system and a way to synchronise clocks that gives a static reference system with coordinate axes that are not changing over time. Can you suggest another way to synchronise clocks that are running at different rates?

Even disregarding the acceleration/velocity differential, a synchronization which works for one velocity cannot work for other different velocities. Yeah?? How could it?

It does work for other velocities. Have another look at the posted chart. The rear rocket sends signals every 3 ticks and the front rocket receives those signals every 4 ticks, consistently even as the relative rocket velocities constantly change over time. If we speed up the rear clock by a factor of 4/3 then the front observer will see the rear clock ticking at the same rate as his own clock for all time.

Also the measurement points could be widely separated say 0.7c and 0.8c

More and more i suspect that the velocity dilation would be insignificant and might possibly agree with the Rindler predictions. I.e. would not increase with greater velocities. The acceleration magnitudes you used were totally unrealistic. The back of the rocket quickly passing the front and leaving it in the dust ;-) so I am still unsure.
thanks

Again, if you look at the attached chart you see that the front rocket sends a signal when the rocket velocities in the LF are approximately 0.69c and the rear rocket receives that signal when the rocket velocities are approximately 0.81c in the LF and there are no problems with the rear rocket overtaking the front rocket. (Another curious aspect is that a light signal sent from left of the origin can never catch up with the accelerating rockets, even though they never attain light speed. That should bake your noodle! ).

P.S. I think Peter has a pretty good handle on it all in post #81.

 

[harrylin]

[..] This is the bizarre aspect. The inertial observer attributes the redshift to classical Doppler shift due to velocity differential between emission and reception and yet [..] an observer in the rear rocket really ages slower than an observer in the front rocket.

That looks bizarre because it's wrong. Indeed the redshift of accelerating rockets is almost purely (semi)classical Doppler shift, insofar as there is no or negligible difference in acceleration. I see no reason for the claim that in reality it's not so, as I also stressed in post #65:

- 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.

It depends on your choice of inertial reference system if you deem that the clock in the rear ages slower or not; according to the launch frame's POV they age equally.

 

[yuiop]

That looks bizarre because it's wrong. Indeed the redshift of accelerating rockets is almost purely (semi)classical Doppler shift, insofar as there is no or negligible difference in acceleration. I see no reason for the claim that in reality it's not so, as I also stressed in post #65:

Are you talking about the case with no length contraction or the case with length contraction as used in in Rindler coordinates or Born rigid acceleration? In the latter case there is most definitely significant difference in proper acceleration as measured on board the rockets and a significant difference in coordinate acceleration as measured simultaneously in the launch frame

It depends on your choice of inertial reference system if you deem that the clock in the rear ages slower or not; according to the launch frame's POV they age equally.

They only age equally from the launch frame's POV if there is no length contraction. Even then, the rear rocket observers will see blue shift of clocks at the front of the rocket and any experiment they carry out will convince them that the rear clocks are tangibly running slower than the front clocks. In this thread we are mainly discussing the equivalence between measurements in an artificially accelerated system and a gravitational system, so we are more interested in what the accelerated observers measure. Also, the case where there is length contraction such that the accelerated observers consider themselves to be at constant distance from each other, is more relevant to a typical gravitational field such as that of the Earth. When we stand on top of a tower and look towards the base, we generally consider the height of the tower to remain constant.

P.S. Some reservations about the none length contracting case have occurred to me. I will try to analyse that in detail later.

 

[harrylin]

Are you talking about the case with no length contraction or the case with length contraction as used in in Rindler coordinates or Born rigid acceleration? [..]

I thought that you were talking about the case with no length contraction as seen from the launch pad frame. Else a little correction is needed, as I also hinted at in my post #65 which I also cited again. I thought (and still think) that you were not talking about that small effect of length contraction when you made your claim about "really slower aging", as that redshift is very small compared to "the redshift" that you discussed. If I misunderstood you, please clarify.

[...] any experiment they carry out will convince them that the rear clocks are tangibly running slower than the front clocks.[..]

Only if, as I pointed out, he is fooling himself into thinking that he his not accelerating. However, that would not be reasonable for someone in a rocket with firing rocket engines - as you also seemed to realize in your answer in post #73. For some reason that escapes me, you replaced "fooled"(=not real) by "real" (=true) between that post and post #86. Someone's instrument reading is not necessarily identical to "what really happens", nor does a smart rocket pilot accept everything at face value.

 

[stevendaryl]

I thought that you were talking about the case with no length contraction as seen from the launch pad frame. Else a little correction is needed, as I also hinted at in my post #65 which I also cited again. I thought (and still think) that you were not talking about that small effect of length contraction when you made your claim about "really slower aging", as that redshift is very small compared to "the redshift" that you discussed. If I misunderstood you, please clarify.

I'm not sure exactly what you are claiming. The differential aging of someone in the front and rear of a rocket is real. We can make it operational as follows:

1. Take a pair of clocks to the rear of the rocket.
2. Set them to the same time, t=0.
3. Move one of the clocks to the front of the rocket.
4. Wait a year.
5. Move it back to the rear.
6. Compare the two clocks.

The prediction is that if we allow length contraction (Born rigid acceleration) then the moving clock will be ahead of the clock that was always in the rear by a factor of 1+gL/c². So it's not simply some kind of illusion.

 

[yuiop]

I thought that you were talking about the case with no length contraction as seen from the launch pad frame. Else a little correction is needed, as I also hinted at in my post #65 which I also cited again. I thought (and still think) that you were not talking about that small effect of length contraction when you made your claim about "really slower aging", as that redshift is very small compared to "the redshift" that you discussed. If I misunderstood you, please clarify.

Generally when I talk about accelerating rockets in this thread I am talking about the the length contraction version and when I am talking about the more unusual and perhaps less useful none length contracting version I usually make it clear that I am talking about that version. In my last post I mentioned that I intend to analyse the none length contraction version more closely as that might be interesting.

Only if, as I pointed out, he is fooling himself into thinking that he his not accelerating. However, that would not be reasonable for someone in a rocket with firing rocket engines - as you also seemed to realize in your answer in post #73. For some reason that escapes me, you replaced "fooled"(=not real) by "real" (=true) between that post and post #86. Someone's instrument reading is not necessarily identical to "what really happens", nor does a smart rocket pilot accept everything at face value.

When Einstein introduced the equivalence idea he described comparing measurements in a closed accelerating box so that the observers inside would be unaware of whether they were stationary in a gravitational field or accelerating artificially in flat space. Without the luxury of being able to look out the window he would no be aware of his rocket engines fireing away. In both cases he would measure proper acceleration and redshift of signals from below him and in a small enough enclosure whereby tidal effects are negligable, he would be "fooled", in the sense that he would be uncertain as to whether he was being artificially accelerated in flat space or stationary in a gravity field. In both the artificially accelerated case and when stationary in a gravity field, clocks lower down really and unambiguously run slower than clocks higher up. No one is being fooled about whether the clocks run at different rates or not.

 

[Austin0]

Generally when I talk about accelerating rockets in this thread I am talking about the the length contraction version and when I am talking about the more unusual and perhaps less useful none length contracting version I usually make it clear that I am talking about that version.
When Einstein introduced the equivalence idea he described comparing measurements in a closed accelerating box so that the observers inside would be unaware of whether they were stationary in a gravitational field or accelerating artificially in flat space. Without the luxury of being able to look out the window he would no be aware of his rocket engines fireing away. In both cases he would measure proper acceleration and redshift of signals from below him and in a small enough enclosure whereby tidal effects are negligable, he would be "fooled", in the sense that he would be uncertain as to whether he was being artificially accelerated in flat space or stationary in a gravity field. In both the artificially accelerated case and when stationary in a gravity field, clocks lower down really and unambiguously run slower than clocks higher up. No one is being fooled about whether the clocks run at different rates or not.

And to what do you attribute the different rates if differential velocity is ruled out??

 

[yuiop]

And to what do you attribute the different rates if differential velocity is ruled out??

I guess the inertial observer would attribute part of the differential rates to differential velocity, but the Rindler observers on board the rockets would not because as far as they are concerned the rockets are stationary with respect to each other and they would have to attribute the differential clocks rates to a real or pseudo force field.

 

[harrylin]

I'm not sure exactly what you are claiming. The differential aging of someone in the front and rear of a rocket is real. We can make it operational as follows:

1. Take a pair of clocks to the rear of the rocket.
2. Set them to the same time, t=0.
3. Move one of the clocks to the front of the rocket.
4. Wait a year.
5. Move it back to the rear.
6. Compare the two clocks.

The prediction is that if we allow length contraction (Born rigid acceleration) then the moving clock will be ahead of the clock that was always in the rear by a factor of 1+gL/c². So it's not simply some kind of illusion.

Your operation is quite different from the one I commented on, and I have not analysed yours. I thought that Yuop was discussing clocks in two rockets, and thus I assumed a similar situation as Bell's spaceships.

[..] he would be uncertain as to whether he was being artificially accelerated in flat space or stationary in a gravity field. In both the artificially accelerated case and when stationary in a gravity field, clocks lower down really and unambiguously run slower than clocks higher up. No one is being fooled about whether the clocks run at different rates or not.

In a case of two rockets such as presented by Bell, according to the launch frame observation the two clocks will age identically, and that observation is as valid as any other one; and note that the Doppler redshift will be nearly the same as in a case with length contraction. If instead we consider a single rocket as viewed from the launch frame then there will be a small effect due to length contraction (I did not calculate it or analyse from all perspectives, but at first sight it gives a slight slowdown of the rear clock according to all observers). You did not reply my question to you if indeed you were talking about the (much bigger?) effect of Doppler redshift.

 

[Austin0]

Your operation is quite different from the one I commented on, and I have not analysed yours (but I did qualitatively analyse a similar one, see next). I thought that Yuop was discussing clocks in two rockets, and thus I assumed a similar situation as Bell's spaceships.

In a case of two rockets such as presented by Bell, according to the launch frame observation the two clocks will age identically, and that observation is as valid as any other one. If instead we consider a single rocket as viewed from the launch frame then there will be a very small effect due to length contraction (I did not calculate it or analyse from all perspectives, but at first sight it gives a slight slowdown of the rear clock according to all observers). You did not reply my question to you if indeed you were talking about the much bigger effect of "gravitational" (Doppler) redshift.

I think we are all talking now about the length contracted case. either as a single rocket or two rockets with the expected contraction effected through differential acceleration.

So there are three questions;

1) How much dilation would be effected purely through length contraction ?[which i think would have to be calculated from the launch frame , not momentarily comoving frames]
]
2) Would this dilation factor increase over time with greater velocities?.

3)how would this figure compare with the expected relative dilation in the accelerating system as calculated using the Rindler coordinates?

1+gL/c². yes i think they are talking about this factor as being actual dilation , not apparent Doppler dilation

 

[harrylin]

I think we are all talking now about the length contracted case. either as a single rocket or two rockets with the expected contraction effected through differential acceleration. [..] 1) How much dilation would be effected purely through length contraction ?[which i think would have to be calculated from the launch frame , not momentarily comoving frames] [..]

I was still editing my answer when you answered, as just after answering I got the impression that although I wasn't commenting on the calculations, someone (perhaps me) may have made an error related to the numbers. But if so, I haven't yet figured out where...

In any case, answers to your questions will also clarify that point for me! (now I have now no time to look at it myself).

 

[stevendaryl]

Your operation is quite different from the one I commented on, and I have not analysed yours. I thought that Yuop was discussing clocks in two rockets, and thus I assumed a similar situation as Bell's spaceships.

Well, the difference between the two rocket case and the one-rocket case is length contraction. In the two-rocket case (with identical accelerations), the clocks will always show the same time in the "launch" frame, but the front clock will run ahead of the rear clock in the instantaneous comoving frame of the rear clock.

In a case of two rockets such as presented by Bell, according to the launch frame observation the two clocks will age identically, and that observation is as valid as any other one;

Well, sort of. I thought you were saying that the differential aging was a kind of illusion, which I interpreted as saying that they were really the same age. The relative age of distant twins (or clocks--I forget which we're talking about) is a frame-dependent quantity, but I wouldn't call that an illusion.

and note that the Doppler redshift will be nearly the same as in a case with length contraction. If instead we consider a single rocket as viewed from the launch frame then there will be a small effect due to length contraction

It's not a small effect, when you consider the case of the rocket accelerating for long periods of time. As I have pointed out in a different post, the time difference between the times on the front and rear clocks can be broken down into two contributions:

Let e₁ be the event at which the rear clock shows time T₁. Let e₂ be the event at the front clock that is simultaneous with e₁, according to the "launch" frame. Let T₂ be the time on the front clock at event e₂. Let e₃ be the event at the front clock that is simultaneous with e₁ in the comoving frame of the rocket. Let T₃ be the time on the front clock at event e₃.

Let δT₁ = T₂ - T₁.
Let δT₂ = T₃ - T₂.

δT₁ is purely due to length contraction; it's equal to 0 if there is no length contraction (the two-rocket case).

δT₂ is an additional contribution due to relativity of simultaneity; what's simultaneous in the launch frame is not simultaneous in the comoving frame.

δT₁ starts off zero and gradually gets bigger and bigger, growing without bound, if the two clocks continue accelerating.

δT₂ starts off nonzero, and approaches a maximum value.

The total discrepancy between the two clocks, as viewed by the comoving frame of the rocket, is the sum of the two δT = δT₁ + δT₂. That sum grows at a constant rate of gL/c₂; that is, δT/T₁ = gL/c² at all times.

The two effects, length contraction and relativity of simultaneity, are both important in explaining the discrepancy between the two clocks. Relativity of simultaneity is the dominant effect soon after launch, and length contraction is the dominant effect long after launch.

 

[stevendaryl]

I think we are all talking now about the length contracted case. either as a single rocket or two rockets with the expected contraction effected through differential acceleration.

So there are three questions;

1) How much dilation would be effected purely through length contraction ?[which i think would have to be calculated from the launch frame , not momentarily comoving frames]
]
2) Would this dilation factor increase over time with greater velocities?.

I think those questions have already been answered. The effect due to length contraction starts off zero, and increases without bound. Long after launch, the ratio of the time on the front clock to the time on the rear clock approaches the value 1+gL/c², as measured in the launch frame.

3)how would this figure compare with the expected relative dilation in the accelerating system as calculated using the Rindler coordinates?

1+gL/c². yes i think they are talking about this factor as being actual dilation , not apparent Doppler dilation

Well, "actual" versus "apparent" is a fuzzy distinction. The time difference is real, in the operational sense that I gave: If you synchronize two clocks in the rear, take one clock to the front and let it sit for a year, and then bring it back to the rear, the clock that was in the front will show more elapsed time. And the difference will be exactly what the Doppler shift showed.

 

[harrylin]

[..] I thought you were saying that the differential aging was a kind of illusion, which I interpreted as saying that they were really the same age. The relative age of distant twins (or clocks--I forget which we're talking about) is a frame-dependent quantity, but I wouldn't call that an illusion.

Instead I was saying that the pseudo gravitational field is a kind of illusion, and I simply tried to clarify in post #87 that as the inertial observer attributes the redshift at low relative velocity to classical Doppler shift, an observer in the rear rocket cannot be said to really age slower by this red shift factor than an observer in the front rocket - and thus there is nothing "bizarre" going on here.

The two effects, length contraction and relativity of simultaneity, are both important in explaining the discrepancy between the two clocks. Relativity of simultaneity is the dominant effect soon after launch, and length contraction is the dominant effect long after launch.

Thanks for the analysis with which I agree (only what you call "relativity of simultaneity", I call Doppler shift). It's interesting to see that at very high speeds the effect is mainly attributed to length contraction, indeed I had not realized that.

 

[yuiop]

Instead I was saying that the pseudo gravitational field is a kind of illusion, and I simply tried to clarify in post #87 that as the inertial observer attributes the redshift at low relative velocity to classical Doppler shift, an observer in the rear rocket cannot be said to really age slower by this red shift factor than an observer in the front rocket - and thus there is nothing "bizarre" going on here.

If we have twins at the front of the rocket (initially the same age) and one free falls to the rear of the rocket and some time later the other free falls to the rear of the rocket, the twin that spent the most time at the rear of the rocket will have physically aged less than the twin that spent the most time at the front of the rocket. When we compare twins side by side and observe differential ageing, that is as real as it gets, as far as time dilation is concerned.