How does this affect the length contraction paradox?

[Fantasist]

Hi,

As is well known, Relativity claims that a rod of a given proper length will appear length contracted when measured by a moving observer. Now consider a rod of length L and observer initially at rest relatively to it, with one end of the rod at the observer's coordinate x1(0) and the other at x2(0) (with x2(0)-x1(0) = L ). If the observer is then accelerated with the constant acceleration 'a' towards the rod (along its axis), the coordinates of the rod's ends should then change according to

x1(t) = x1(0) - 1/2*a*t^2
x2(t) = x2(0) - 1/2*a*t^2

But this means the difference

x2(t)-x1(t) = x2(0) -x1(0) = L ,

so the length of the rod as measured by the accelerating (i.e. moving) observer is unchanged and still amounts to the proper length L.

How can this be reconciled with the length contraction claim of Relativity, according to which the measured length should get progressively shorter in this case?

 

[Dale]

The coordinates for a uniformly accelerating observer are known as Rindler coordinates. They are different from the equation you posted.

To see why the equation you posted is wrong, consider $t>c/a$

However, the usual length contraction formula is only for inertial coordinates, and an accelerating observer has non inertial coordinates.

 

[dauto]

1st) Relativity doesn't claim that the rod appears length contracted. It claims that it IS length contracted.

2nd) The equations you posted describe a situation where both ends of the rod start accelerating simultaneously. But according to relativity, simultaneity is relative and an observer that is moving with respect to you will see the front end of the rod start accelerating before the back end. That gives the front end a head start which means the rod is slowly stretched. That stretching exactly cancels the Lorentz-Fitzgerald contraction for the first observer leading to a rod of constant length L.

 

[WannabeNewton]

The problem is that what you wrote down is not the trajectory of a uniformly accelerating observer in SR. It can be shown that the uniformly accelerating observer's worldline as expressed in a global inertial frame with associated coordinates $(t,x)$ (assuming that the accelerating observer is initially comoving with the global inertial frame) is given by $t = a^{-1}\sinh a\tau$ and $x = a^{-1}\cosh a\tau$ where $\tau$ is the proper time along the worldline of the accelerating observer and $a$ his acceleration. In particular, $x^2 - t^2 = a^{-2}$ so $v = \frac{dx}{dt} = t(a^{-2} + t^{2})^{-1/2}$ where $v$ is the 3-velocity of the accelerating observer relative to the global inertial frame.

 

[dauto]

The problem is that what you wrote down is not the trajectory of a uniformly accelerating observer in SR. It can be shown that the uniformly accelerating observer's worldline as expressed in a global inertial frame with associated coordinates $(t,x)$ (assuming that the accelerating observer is initially comoving with the global inertial frame) is given by $t = a^{-1}\sinh a\tau$ and $x = a^{-1}\cosh a\tau$ where $\tau$ is the proper time along the worldline of the accelerating observer and $a$ his acceleration. In particular, $x^2 - t^2 = a^{-2}$ so $v = \frac{dx}{dt} = t(a^{-2} + t^{2})^{-1/2}$ where $v$ is the 3-velocity of the accelerating observer relative to the global inertial frame.

The OP did't claim that the rod is under uniform proper acceleration. (s)he just gave us the equation of motion for both ends of the rod and asked us how to reconcile the motion described by those equations with relativity. The answer is that in a different reference frame the ends of the rod do not necessarily start accelerating simultaneously allowing for the distance between them to change over time in accordance with the space contraction formula as required by relativity.

 

[WannabeNewton]

The OP clearly said the observer is under constant acceleration and listed the equations of motion as a consequence.

 

[dauto]

The OP clearly said the observer is under constant acceleration and listed the equations of motion as a consequence.

The rod is under constant acceleration (as seen from that one observer) even though it is not under constant proper acceleration.

 

[Fantasist]

1st) Relativity doesn't claim that the rod appears length contracted. It claims that it IS length contracted.

2nd) The equations you posted describe a situation where both ends of the rod start accelerating simultaneously. But according to relativity, simultaneity is relative and an observer that is moving with respect to you will see the front end of the rod start accelerating before the back end. That gives the front end a head start which means the rod is slowly stretched. That stretching exactly cancels the Lorentz-Fitzgerald contraction for the first observer leading to a rod of constant length L.

Thanks for your reply, but this rather confirms the paradox than resolve it. Does the accelerated observer now measure a progressively shorter length, or does it stay constant? Your points 1) and 2) clearly contradict each other.

 

[Dale]

(s)he just gave us the equation of motion for both ends of the rod and asked us how to reconcile the motion described by those equations with relativity.

Those equations are inconsistent with relativity. You cannot reconcile them, you must correct them.

 

[Fantasist]

Those equations are inconsistent with relativity. You cannot reconcile them, you must correct them.

Can you correct them please?

x1(t)= ?
x2(t)= ?

and in particular

x2(t) - x1(t) = ?

 

[Meir Achuz]

I think you have to start with a course in relativity.

 

[Meir Achuz]

$$L=x_2-x_1=\sqrt{t^2+1/a'^2_2 }-\sqrt{t^2+1/a'^2_1}$$

 

[Dale]

Can you correct them please?

x1(t)= ?
x2(t)= ?

and in particular

x2(t) - x1(t) = ?

Sure:
$x_1(t)=x_1(0) \; sech(t)$
$x_2(t)=x_2(0) \; sech(t)$
$x_2(t)-x_1(t)=(x_2(0)-x_1(0)) \; sech(t)$

The transformations from which this is derived can be found at:
http://en.wikipedia.org/wiki/Rindler_coordinates

Which is what I mentioned in my first reply.

 

[yuiop]

Can you correct them please?

x1(t)= ?
x2(t)= ?

and in particular

x2(t) - x1(t) = ?

Using Rindler coordinates http://en.wikipedia.org/wiki/Rindler_coordinates which assumes constant proper acceleration (a) of the observer,

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(at)^2\right)}$

X2-X1 is the proper length of the rod and this remains constant.

x2(t) - x1(t) is the length of the rod in the accelerating reference frame and this decreases with increasing t as measured in the accelerating reference frame.

See this plot with X2-X1=3 and a=1.

 

[Fantasist]

Using Rindler coordinates http://en.wikipedia.org/wiki/Rindler_coordinates which assumes constant proper acceleration (a) of the observer,

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(at)^2\right)}$

X2-X1 is the proper length of the rod and this remains constant.

x2(t) - x1(t) is the length of the rod in the accelerating reference frame and this decreases with increasing t as measured in the accelerating reference frame.

See this plot with X2-X1=3 and a=1.

Thanks for your replies.

I am not quite sure though how to fit in your 'proper acceleration' here. I defined above the acceleration kinematically with regard to the reference frame in which the rod is at rest. So in this sense the accelerations a1 and a2 of the endpoints of the rod were by definition identical. But according to your formula a1 and a2 would not be identical anymore, and instead you introduce some kinematically undefined 'proper acceleration a'.

Also, if the accelerations of the endpoints of the rod are measured differently by the observer (a1 and a2), shouldn't reversely the observer be measured with different accelerations from the endpoints of the rods as well (i.e. -a1 and -a2)? Yet this would contradict the assumption that the endpoints of the rod are at rest relatively to each other (clearly, two observers at rest relatively to each other must measure the same object with the same acceleration).

 

[Meir Achuz]

Thanks for your replies.
(clearly, two observers at rest relatively to each other must measure the same object with the same acceleration).

It is not clear, because it is wrong.

 

[Dale]

I am not quite sure though how to fit in your 'proper acceleration' here. I defined above the acceleration kinematically with regard to the reference frame in which the rod is at rest.

Since the rod's rest frame is an inertial frame the proper acceleration is the same as the kinematical acceleration when the accelerating observer is momentarily at rest in the inertial frame.

So in this sense the accelerations a1 and a2 of the endpoints of the rod were by definition identical.

That doesn't follow because the frame of the observer is non inertial.

Also, if the accelerations of the endpoints of the rod are measured differently by the observer (a1 and a2), shouldn't reversely the observer be measured with different accelerations from the endpoints of the rods as well (i.e. -a1 and -a2)?

Definitely not. Non inertial frames are not symmetric with inertial frames. In the non inertial frame the endpoints have different coordinate accelerations, in the inertial frame the endpoints have the same coordinate acceleration.

 

[Dale]

$x_2(t)-x_1(t)=(x_2(0)-x_1(0)) \; sech(t)$

I should mention that this assumes that the observer is located at x=1 in the Rindler frame and that we are using units such that c=1 and a=1.

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(at)^2\right)}$

This assumes that the observer is located at x=1 in the Rindler frame and that we are using units where c=1 but a is not necessarily 1.

Under those conditions, the two equations are clearly equivalent.

 

[yuiop]

I am not quite sure though how to fit in your 'proper acceleration' here. I defined above the acceleration kinematically with regard to the reference frame in which the rod is at rest.

If the accelerating observer is in a rocket, then the proper acceleration is the acceleration measured by an accelerometer attached to the accelerating rocket. This is the acceleration that the accelerating observer feels. Defining acceleration like this makes the equations much simpler. Now let's define the inertial reference frame in which the rod is at rest as S, and the accelerating reference frame of the rocket as S'. If the constant acceleration of the rocket is measured in S as g then the actual acceleration (a) experienced by the observer on the rocket is:

$a = g*(1-v^2/c^2)^{-3/2}$

where v is the instantaneous velocity measured in S. This means the fuel consumed per second by the rocket would have to continually increase, approaching infinite as the speed of the rocket approached the speed of light in S. The rocket would reach the speed of light relative to the rod in a finite time (T =c/g) as measured in S and this is not possible in relativity.

So in this sense the accelerations a1 and a2 of the endpoints of the rod were by definition identical.

In the OP you were clear that it was the observer and not the rod that was being accelerated and you later clarified that you intended the acceleration of the observer to be constant as measured in S. In order for the coordinate accelerations of the endpoints of the rod as measured in S' to be identical, the rod would have to be physically stretched in S or the accelerating observer in S' would have to use rulers that do not have constant length in S'.

Consider an analogous scenario purely in the context of purely Newtonian physics. I define the length of an accelerating brick to be L*t. Does this not prove the Newtonian laws of physics to wrong because the brick is getting longer and that does not happen in Newtonian physics! We could contrive a situation where we physically stretch the brick as it is accelerated so that it's length is L*t and that is basically what you are doing in your scenario in the OP. You have defined a set of non relativistic equations and using them to prove that relativity must be wrong.

But according to your formula a1 and a2 would not be identical anymore, and instead you introduce some kinematically undefined 'proper acceleration a'.

By defining the coordinate acceleration a1 and a2 to be identical in S' you are defining the length of the rod to be constant in S' and then act surprised that S' does not length contract in S'.

... Yet this would contradict the assumption that the endpoints of the rod are at rest relatively to each other

This is your assumption and it contradicts some of your other assumptions/definitions.

(clearly, two observers at rest relatively to each other must measure the same object with the same acceleration).

Are you sure about that? Consider an observer on the surface of the Earth and another observer on top of a tall tower with fixed height. These observers are at rest wrt each other but their clocks are running at different rates relative to each other so they measure the acceleration of a given object differently. They would also measure their own accelerations due to gravity to be different and yet there are at rest wrt each other. Now replace the tower with an accelerating rocket in flat space. The proper accelerations at the nose and tail of the rocket are not equal, but it is possible for the length of the rocket as measured by observers on board the rocket to be constant.

Basically you are using Newtonian equations to prove relativity wrong, when in fact Newtonian equations are provably wrong (by experimental observations) and are only approximations of relativistic equations that better match the reality we observe.

 

[Fantasist]

Since the rod's rest frame is an inertial frame the proper acceleration is the same as the kinematical acceleration when the accelerating observer is momentarily at rest in the inertial frame.

But any reference frame with a constant velocity relative to the rod's rest frame is an inertial frame, and since the accelerated observer always momentarily occupies one of these inertial frames, the proper acceleration should at any moment be identical to the coordinate acceleration.

Definitely not. Non inertial frames are not symmetric with inertial frames. In the non inertial frame the endpoints have different coordinate accelerations, in the inertial frame the endpoints have the same coordinate acceleration.

Assume the following: you have a rocket accelerating away from its base, and on the latter you use a radar speedgun to determine the relative velocity and acceleration of the rocket. Likewise, somebody on the rocket uses a radar speedgun to reversely determine the relative velocity and acceleration of the base. Would the two readouts of the speedguns be the same or not?

 

[Fantasist]

If the accelerating observer is in a rocket, then the proper acceleration is the acceleration measured by an accelerometer attached to the accelerating rocket. This is the acceleration that the accelerating observer feels. Defining acceleration like this makes the equations much simpler.

That doesn't seem to be a workable definition of acceleration to me. 'Acceleration' is defined as rate of change of velocity, so you need some reference point to define velocity and acceleration. If you had a rocket alone in the universe you would have no concept of velocity and acceleration at all on the basis of an 'accelerometer' readout.

What's more, with that definition, you should not get any length contraction if the observer happens to be accelerated by gravity as the accelerometer wouldn't work here (since unlike externally applied push/pull-forces, gravity directly acts on all parts of a body, so no internal stress forces are set up).

Reversely, an observer sitting on the surface of the Earth should see a stationary rod getting shorter with time (as the 'accelerometer' would indicate an acceleration).

In the OP you were clear that it was the observer and not the rod that was being accelerated and you later clarified that you intended the acceleration of the observer to be constant as measured in S. In order for the coordinate accelerations of the endpoints of the rod as measured in S' to be identical, the rod would have to be physically stretched in S or the accelerating observer in S' would have to use rulers that do not have constant length in S'.

Consider an analogous scenario purely in the context of purely Newtonian physics. I define the length of an accelerating brick to be L*t. Does this not prove the Newtonian laws of physics to wrong because the brick is getting longer and that does not happen in Newtonian physics! We could contrive a situation where we physically stretch the brick as it is accelerated so that it's length is L*t and that is basically what you are doing in your scenario in the OP. You have defined a set of non relativistic equations and using them to prove that relativity must be wrong.

Only that I haven't contrived anything: if you have two objects at rest at the markings x1 and x2 of a ruler, and you accelerate the ruler with a constant acceleration 'a' with regard to said objects, then, by definition, the ruler-coordinates of the latter should change by the same amount over each time interval, i.e. the distance between them should stay constant (unless you assume the ruler to physically expand for some reason). The point is that this conclusion is solely reached on the basis of variables in the ruler's frame, so a Lorentz transformation to the reference frame of the objects (or the rod in the original example) does not come into it at all.

Are you sure about that? Consider an observer on the surface of the Earth and another observer on top of a tall tower with fixed height. These observers are at rest wrt each other but their clocks are running at different rates relative to each other so they measure the acceleration of a given object differently. They would also measure their own accelerations due to gravity to be different and yet there are at rest wrt each other. Now replace the tower with an accelerating rocket in flat space. The proper accelerations at the nose and tail of the rocket are not equal, but it is possible for the length of the rocket as measured by observers on board the rocket to be constant.

I think it is not appropriate to bring in General Relativity here.

 

[WannabeNewton]

That doesn't seem to be a workable definition of acceleration to me. 'Acceleration' is defined as rate of change of velocity, so you need some reference point to define velocity and acceleration. If you had a rocket alone in the universe you would have no concept of velocity and acceleration at all on the basis of an 'accelerometer' readout.

Utterly incorrect on every account. Have you even learned SR yet? You have tons of misconceptions about the foundations of SR that simple thread posts cannot fix. You need to pickup a proper (no pun intended) textbook on the subject and start coming to terms with the foundations of SR.

 

[Dale]

But any reference frame with a constant velocity relative to the rod's rest frame is an inertial frame

Yes.

and since the accelerated observer always momentarily occupies one of these inertial frames

Yes. This is called the "momentarily co-moving inertial frame" or MCIF. It is a fairly important concept. As you mention, for any particle at any point along its worldline it is possible to construct a MCIF.

, the proper acceleration should at any moment be identical to the coordinate acceleration.

Yes, the proper acceleration is identical to the coordinate acceleration in the MCIF. The proper acceleration is not identical to the coordinate acceleration in any non-comoving inertial frame, specifically, the proper acceleration is not identical to the coordinate acceleration in the rod's frame except for the moment where the rod's frame is also the MCIF of the uniformly accelerating observer.

Assume the following: you have a rocket accelerating away from its base, and on the latter you use a radar speedgun to determine the relative velocity and acceleration of the rocket. Likewise, somebody on the rocket uses a radar speedgun to reversely determine the relative velocity and acceleration of the base. Would the two readouts of the speedguns be the same or not?

I don't know. I would have to work out the math to figure that out. In any case, I find that threads where the OP throws out multiple scenarios tend to get confusing very quickly. It is usually best to just stick with one scenario until you understand it.

 

[Dale]

That doesn't seem to be a workable definition of acceleration to me. 'Acceleration' is defined as rate of change of velocity, so you need some reference point to define velocity and acceleration. If you had a rocket alone in the universe you would have no concept of velocity and acceleration at all on the basis of an 'accelerometer' readout.

It is a completely standard definition which you should find in any decent-quality textbook on relativity. Even without any coordinate system with which to determine coordinate acceleration you can still determine proper acceleration. In my opinion, proper acceleration is the only physically important acceleration and cooridinate acceleration only tells you about your coordinate system and not about the physics.

The Wikipedia article is decent (with the usual caveats for Wikipedia):
http://en.wikipedia.org/wiki/Proper_acceleration

 

[yuiop]

... I think it is not appropriate to bring in General Relativity here.

Yet you appear to be happy to discuss the effects of gravity in the first paragraph of the same post;

... What's more, with that definition, you should not get any length contraction if the observer happens to be accelerated by gravity as the accelerometer wouldn't work here (since unlike externally applied push/pull-forces, gravity directly acts on all parts of a body, so no internal stress forces are set up).

Reversely, an observer sitting on the surface of the Earth should see a stationary rod getting shorter with time (as the 'accelerometer' would indicate an acceleration).

Then again, perhaps you are right. Introducing gravity probably only confuses the issue here. I will give it one last go and address your above comments.

What's more, with that definition, you should not get any length contraction if the observer happens to be accelerated by gravity as the accelerometer wouldn't work here (since unlike externally applied push/pull-forces, gravity directly acts on all parts of a body, so no internal stress forces are set up).

In this case the observer is free falling. Imagine he is falling past a tall building. He can consider himself stationary and (as you correctly noted) he does not experience any proper acceleration. Each floor passes him at a progressively higher speed. Each floor is length contracted by factor related to the speed the floor passes him. Because he experiences no proper acceleration he is equivalent to an observer at rest with the inertial rod in flat space and the building length contracts in the same way the accelerating rocket does. Accelerometers on the building and on the rocket both indicate proper acceleration and so they are 'equivalent'.

Reversely, an observer sitting on the surface of the Earth should see a stationary rod getting shorter with time (as the 'accelerometer' would indicate an acceleration).

The observer and stationary rod are both experiencing the same proper acceleration so there is no relative motion and no length contraction. They are after all at rest wrt each other. However, consider the case of the rod free falling from a great height past the observer that is looking out of a window of a tower fixed to the Earth. The rod now has relative motion and is length contracted according to this observer. This observer experiences proper acceleration so his observation are equivalent to those of the observer in the rocket accelerating past the inertial rod in flat space. The rod falling past the window is equivalent to the inertial rod in flat space as both rods experience zero proper acceleration. The rod falling past the window is length contracted according to the observer looking out the window as is the inertial rod in flat space according to the observer accelerating past it in the rocket. This is the "equivalence principle" at work.

 

[Fantasist]

Yes, the proper acceleration is identical to the coordinate acceleration in the MCIF.

If that's your definition of 'proper acceleration', what's your definition of the MCIF then?

I don't know. I would have to work out the math to figure that out. In any case, I find that threads where the OP throws out multiple scenarios tend to get confusing very quickly. It is usually best to just stick with one scenario until you understand it.

Well, I would think it is a crucial issue here whether the relative acceleration between two reference frames is a reciprocal quantity (i.e. whether each measure the other with the same coordinate acceleration) or not.

It is a completely standard definition which you should find in any decent-quality textbook on relativity. Even without any coordinate system with which to determine coordinate acceleration you can still determine proper acceleration. In my opinion, proper acceleration is the only physically important acceleration and cooridinate acceleration only tells you about your coordinate system and not about the physics.

The Wikipedia article is decent (with the usual caveats for Wikipedia):
http://en.wikipedia.org/wiki/Proper_acceleration

So you think it is useful if, on the basis of your 'accelerator' reading, you believe your rocket is on its way to the stars, but in reality you may still be sitting firmly on the start ramp?

 

[Fantasist]

Let me first go briefly back to your previous post again:

If the accelerating observer is in a rocket, then the proper acceleration is the acceleration measured by an accelerometer attached to the accelerating rocket. This is the acceleration that the accelerating observer feels. Defining acceleration like this makes the equations much simpler. Now let's define the inertial reference frame in which the rod is at rest as S, and the accelerating reference frame of the rocket as S'. If the constant acceleration of the rocket is measured in S as 'a' then the actual acceleration (A) experienced by the observer on the rocket is:

$A = a*(1-v^2/c^2)^{-3/2}$

where v is the instantaneous velocity measured in S. This means the fuel consumed per second by the rocket would have to continually increase, approaching infinite as the speed of the rocket approached the speed of light in S. The rocket would reach the speed of light relative to the rod in a finite time (T =c/a) as measured in S and this is not possible in relativity.

Just for clarification: with my original equations

x1(t) = x1(0) - 1/2*a*t^2
x2(t) = x2(0) - 1/2*a*t^2

I was not implying that the coordinate acceleration 'a' (note that I have changed the notation in your quote correspondingly) is necessarily constant. I merely chose it for convenience. One might as well have a time dependent acceleration a(t), and the equations would then be

x1(t) = x1(0) - $\Delta X(t)$
x2(t) = x2(0) - $\Delta X(t)$

where

$\Delta X(t) = \int_0^t dt' \int_0^{t'} dt'' a(t'')$

But this still results in a constant difference

x2(t)-x1(t) = x2(0)-x1(0) = const.

The observer and stationary rod are both experiencing the same proper acceleration so there is no relative motion and no length contraction. They are after all at rest wrt each other. However, consider the case of the rod free falling from a great height past the observer that is looking out of a window of a tower fixed to the Earth. The rod now has relative motion and is length contracted according to this observer. This observer experiences proper acceleration so his observation are equivalent to those of the observer in the rocket accelerating past the inertial rod in flat space. The rod falling past the window is equivalent to the inertial rod in flat space as both rods experience zero proper acceleration. The rod falling past the window is length contracted according to the observer looking out the window as is the inertial rod in flat space according to the observer accelerating past it in the rocket. This is the "equivalence principle" at work.

So you seem to be saying that the proper acceleration A in your previously given length contraction formula (I changed your 'a' to A to be consistent with my notation)

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At)^2\right)}$

is actually the difference between the proper accelerations of the rod and the observer.

What then if you shoot the rod upwards with a catapult? The rod would also experience zero proper acceleration whilst moving up the tower, so your formula would still predict a shrinking length of the rod, even though the velocity relatively to the observer gets smaller all the time.

 

[Dale]

If that's your definition of 'proper acceleration', what's your definition of the MCIF then?

The MCIF is the inertial frame where the object is momentarily at rest. This is another completely standard definition.

I would think it is a crucial issue here whether the relative acceleration between two reference frames is a reciprocal quantity (i.e. whether each measure the other with the same coordinate acceleration) or not.

Those are two different questions. One is about symmetry of coordinate acceleration. The other is about symmetry of a specific measurement. I would have to calculate the latter, but the former is easy.

Relative acceleration between reference frames is not reciprocal. Consider a rotating observer and an inertial observer at rest relative to the COM of the rotation. The rotating observer assigns a large coordinate acceleration to the inertial observer, but not vice versa.

So you think it is useful if, on the basis of your 'accelerator' reading, you believe your rocket is on its way to the stars, but in reality you may still be sitting firmly on the start ramp?

I don't know why you would think that.

It is pretty apparent that you need to learn some basic SR. I would recommend avoiding non inertial reference frames until you have a good grasp on how to analyze non inertial objects from an inertial frame. In particular, you should try to learn some of the basic terminology and the mathematical framework of four-vectors.

 

[Dale]

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At)^2\right)}$

is actually the difference between the proper accelerations of the rod and the observer.

What then if you shoot the rod upwards with a catapult? The rod would also experience zero proper acceleration whilst moving up the tower, so your formula would still predict a shrinking length of the rod, even though the velocity relatively to the observer gets smaller all the time.

In the equation t=0 is the moment when the rod is momentarily at rest. So in your example t would start out negative, the rod would slow, and is length would increase, until it reached the peak altitude at t=0.

 

[yuiop]

Under those conditions, the two equations are clearly equivalent.

Yes, they are. The duplication was unintentional. There was quite a delay between starting a reply and eventually posting it and your post slipped in, in between.

In the equation t=0 is the moment when the rod is momentarily at rest. So in your example t would start out negative, the rod would slow, and is length would increase, until it reached the peak altitude at t=0.

Scooped me again here too. Only thing I would add is that the moment when the rod is momentarily at rest at the apogee at time t=0 is the MCIF that fantasist was asking about, where the proper length of the rod is momentarily equal to the length measured by the observer with proper acceleration.

 

[Fantasist]

The MCIF is the inertial frame where the object is momentarily at rest. This is another completely standard definition.

That still leaves the 'inertial frame' to be defined.

Relative acceleration between reference frames is not reciprocal. Consider a rotating observer and an inertial observer at rest relative to the COM of the rotation. The rotating observer assigns a large coordinate acceleration to the inertial observer, but not vice versa.

Sorry, I don't see your point. Can you elaborate?

So you think it is useful if, on the basis of your 'accelerator' reading, you believe your rocket is on its way to the stars, but in reality you may still be sitting firmly on the start ramp?

I don't know why you would think that.

Because of what you said before

In my opinion, proper acceleration is the only physically important acceleration and coordinate acceleration only tells you about your coordinate system and not about the physics.

From which one could conclude that you consider it unimportant whether the proper acceleration of the rocket actually gets you somewhere or not.

Evidently, 'proper acceleration' is based on a purely local physical measurement, but in general it is not the best of ideas to completely ignore the global physics around you.

 

[Fantasist]

In the equation t=0 is the moment when the rod is momentarily at rest. So in your example t would start out negative, the rod would slow, and is length would increase, until it reached the peak altitude at t=0.

With the formula as written, that would hardly be an acceptable solution, as it would imply that the observer has to re-synchronize his clock dependent on which rod he measures (or even use different clocks for each rod in the first place). The only thing you could do is to add an initial velocity to the argument A*t i.e.

$x2(t) - x1(t) = (X2-X1)\sqrt{ \left(1- \tanh(At-V0)^2\right)}$


But what about a different scenario? Assume the observer is actually based outside the gravitational field of the Earth (e.g. at one of the Lagrange points) and observes the free-falling rod attracted by the earth. Clearly, both the observer and the rod have zero proper acceleration, so you should not observe any length contraction of the rod, despite the fact that its velocity with regard to the observer increases.

 

[Dale]

That still leaves the 'inertial frame' to be defined.

Go get a textbook. Inertial frames, proper acceleration, MCIF, and all the rest of the standard terminology will be defined there along with detailed explanations and example problems.

From which one could conclude that you consider it unimportant whether the proper acceleration of the rocket actually gets you somewhere or not.

Evidently, 'proper acceleration' is based on a purely local physical measurement, but in general it is not the best of ideas to completely ignore the global physics around you.

I don't know why you would conclude that from what I said.

If you know the global metric then you can use proper acceleration to navigate through it. In fact, cruise missiles do exactly that so that they can approach their target even in the absence of any external information.

If you do not know the global metric then you have no reason to believe that any degree of proper acceleration is leading you any closer or further from a given star.

Furthermore, without the metric, knowledge of the coordinate acceleration doesn't tell you if you are getting further away from or closer to a distant star either. So you cannot say that coordinate acceleration is in any way more parsimonious or informative than proper acceleration.

 

[Dale]

With the formula as written, that would hardly be an acceptable solution, as it would imply that the observer has to re-synchronize his clock dependent on which rod he measures (or even use different clocks for each rod in the first place).

The synchronization was defined from the question in your OP where the rod and observer were at rest at t=0. It seems a little silly to complain about an answer that uses the same convention as your question.

But what about a different scenario? Assume the observer is actually based outside the gravitational field of the Earth (e.g. at one of the Lagrange points) and observes the free-falling rod attracted by the earth. Clearly, both the observer and the rod have zero proper acceleration, so you should not observe any length contraction of the rod, despite the fact that its velocity with regard to the observer increases.

Sure. Rindler coordinates are for flat spacetime only. The spacetime around the Earth is not flat. You would have to use GR, not SR for this scenario.

 

[Fantasist]

Go get a textbook. Inertial frames, proper acceleration, MCIF, and all the rest of the standard terminology will be defined there along with detailed explanations and example problems.

Textbook definitions don't appear to be sufficient to unambiguously answer questions like whether an electron accelerated in an electric field occupies an inertial reference frame or not (assuming that the charge of the electron is distributed homogeneously within its volume (i.e. the electron does not 'feel' the Coulomb force as the latter acts on all parts of it and no internal stress forces are set up)).

If you know the global metric then you can use proper acceleration to navigate through it. In fact, cruise missiles do exactly that so that they can approach their target even in the absence of any external information.

That only works because

1) the external references are programmed into the system in advance
2) it is assumed that the external references don't change
3) any change in proper acceleration is interpreted as a change in coordinate acceleration