1. Nov 23, 2013

### 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?

2. Nov 23, 2013

### Staff: Mentor

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.

Last edited: Nov 23, 2013
3. Nov 23, 2013

### 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.

4. Nov 23, 2013

### 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.

5. Nov 23, 2013

### dauto

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.

6. Nov 23, 2013

### WannabeNewton

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

7. Nov 23, 2013

### dauto

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

8. Nov 23, 2013

### Fantasist

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.

9. Nov 23, 2013

### Staff: Mentor

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

10. Nov 23, 2013

### Fantasist

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

and in particular

x2(t) - x1(t) = ?

11. Nov 23, 2013

### Meir Achuz

12. Nov 23, 2013

### Meir Achuz

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

Last edited: Nov 24, 2013
13. Nov 23, 2013

### Staff: Mentor

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.

Last edited: Nov 23, 2013
14. Nov 23, 2013

### yuiop

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.

Last edited: Nov 23, 2013
15. Nov 24, 2013

### Fantasist

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).

16. Nov 24, 2013

### Meir Achuz

It is not clear, because it is wrong.

17. Nov 24, 2013

### Staff: Mentor

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.

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

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.

18. Nov 24, 2013

### Staff: Mentor

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.

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.

19. Nov 24, 2013

### yuiop

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.

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.

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'.

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.

20. Nov 24, 2013

### Fantasist

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.

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?