# Different acceleration in an accelerating rocket?

1. May 15, 2010

### Ascenxion

I read from a physics website that says that an accelerating object experiences different acceleration throughout its structure according to special relativity. I never thought of that!

"In special relativity, one finds that when a spacecraft maintains its structural integrity, so that the distance from tail to nose is constant for passengers in the spacecraft, it experiences different rates of acceleration throughout its structure. The nose of the spacecraft accelerates at a lower rate than the tail of the spacecraft; the amount of acceleration depends solely on position along the direction of acceleration. The acceleration increases as one moves back. Because time dilation is determined by the rate of acceleration, a passenger in the spacecraft finds that a clock placed in the nose of the spacecraft moves faster than a clock placed in the tail."

http://www.astrophysicsspectator.com/topics/specialrelativity/Accelerated.html

If that is the case, is it true that inside an rocket accelerating in one direction, the ceiling and and the floor have different rate of acceleration (less acceleration for the ceiling and more acceleration for the floor)?

What is the explanation behind this?

2. May 15, 2010

### Fredrik

Staff Emeritus
If the rocket accelerates slowly, every particle will experience approximately constant proper acceleration. Proper acceleration is what an accelerometer would measure, and it's also the coordinate acceleration in a co-moving inertial frame. (Note that the co-moving inertial frame is a different one at each point). When the different parts of the rocket move like this, it's said to be doing Born rigid motion. Note that it's necessary for the rear to have a higher coordinate velocity than the front for the length of the rocket in the launch frame to follow the Lorentz contraction formula. If the rear and the front both start out with the same velocity in the launch frame, e.g. both at rest, then there's no way for the coordinate velocity of the rear to become greater than the coordinate velocity of the front unless the coordinate acceleration also is greater at the rear.

If we're talking about constant proper acceleration, the world lines of different parts of the rocket will look like the hyperbolas in the diagram in this Wikipedia article.

3. May 15, 2010

### Ascenxion

Thanks.

By the way, what is Born Rigidity? I heard it is considered so that the distance between the ceiling and the floor remain constant. Is it considered to keep up with Lorentz Contraction, which may change the distance between the two?

4. May 16, 2010

### Fredrik

Staff Emeritus
It's the type of motion that keeps the object's length (in the original rest frame) in agreement with the Lorentz contraction formula. There's a formal definition that involves Lie derivatives, but I haven't yet made the effort to understand it. This is how I think of it: If an object is doing Born rigid motion, then in every inertial frame that's co-moving with some specific atom, the neighboring atoms have approximately the same velocity at any time. (Note that this means that they have different velocities at any given time in the inertial frame where the whole object started out at rest).

5. May 16, 2010

### DrGreg

Just to clarify, the only reason it's approximate is because of the random, temperature-dependent motion of the atoms. At absolute zero, it would be exactly the same velocity (i.e. zero).

6. May 16, 2010

### Fredrik

Staff Emeritus
That's actually not what I had in mind, but I have to admit that the details aren't perfectly clear to me. I was thinking that in general, even at absolute zero temperature, the approximation is only exact in the limit where the distance between atoms goes to zero. I know that for constant proper acceleration, there's no need to take that limit. We don't even have to talk about adjacent atoms. The statement is true even when we consider the rear and the front of the rocket (which I think of as a 1-dimensional solid rod that's pushed by a small force applied to the rear) instead of two adjacent atoms. But I've been assuming that for a small but otherwise arbitrary proper acceleration, this won't be the case, and we need to take that limit. But as I said, I haven't thought this through to the end.

7. May 16, 2010

### DrGreg

Yes, I was thinking of constant proper acceleration. I can't actually recall seeing the phrase "Born rigid" used in any other context, but I've never done a search, or stumbled across any mathematically rigorous definition. If "Born rigid" can be used for varying accelerations then I can see there might be all sorts of problems of the sort you indicate.

8. May 16, 2010

### Fredrik

Staff Emeritus
There's a rigorous-looking definition on page 36 of this article. I haven't actually studied it, or even verified that it agrees with my intuitive understanding of Born rigidity.

9. May 16, 2010

### Passionflower

Consider a rod A============B with clocks A and B at the left and right ends accelerating in the direction AB away from an observer.

When A and B accelerate the same way, the length of the rod, as inferred from the observer, would stay the same, but instead one would expect a Lorentz contracted length as the rod is in relative motion with the observer. Hence the object must have been stretched or even torn apart. In this case clock A and B would remain synchronized.

But in the case of a Born rigid acceleration the stretching is compensated for by having A accelerate more than B. As a result the rod is no longer stretched and the observer will infer a Lorentz contracted rod. But now the clocks A and B are no longer synchronized since A and B have undergone a different proper acceleration.

It is impossible for this rod to accelerate such that both the length remains and the clocks remain synchronized.

In a rotating case both the length and the clock synchronization cannot be maintained.

Last edited: May 16, 2010
10. May 16, 2010

### DrGreg

Thanks for that. I haven't time to study that in depth, yet, but it sounds plausibly similar to your description. I'm assuming that it is correct to describe as "Born rigid" what the paper says is "rigid". On that assumption, it also seems possible to describe rotation as "Born rigid", which isn't what I'd previously assumed.

11. May 16, 2010

### Passionflower

It is impossible for a rotating rod or disk to move Born rigid.

12. May 16, 2010

### Fredrik

Staff Emeritus
That seems likely, considering that the definition of "rigid" starts with the words "Definition 6 (Born 1909)".

Right after equation (92) on page 42, he says that
This says that the vorticity cannot change along a rigid motion in flat space. It is the precise expression for the remark above that you cannot rigidly set a disk into rotation.​

13. May 16, 2010

### Passionflower

Yes, that is true.

14. May 17, 2010

### yuiop

This is a good example of the equivalence principle. In a tower building on the surface of the Earth, the distance between floors of the tower remains constant, but an accelerometer at the top of the tower measures less acceleration than an accelerometer at the base. A clock at the top of the tower also ticks at a faster rate than a clock at the base, as can be observed by measuring the redshift of a signal sent from the base. This is basically what Pound and Rebka measured in the Havard Tower. http://en.wikipedia.org/wiki/Poundâ€“Rebka_experiment

In the case of the accelerating rocket, an inertial observer sees the back of the rocket catching up with the front of the rocket because the rocket appears to be length contracting in his frame (while the length of the rocket appears constant to the rocket observers undergoing Born rigid acceleration). This means that at any given instant in the initial inertial frame, the back of the rocket has slightly greater coordinate acceleration and is moving slightly faster than the front of the rocket and therefore the clock at the back of the rocket has greater time dilation than the clock at the nose of the rocket, according to the inertial observer.

15. May 17, 2010

### yuiop

It is possible for a ring to have Born rigid acceleration (to a first aproximation and locally) if we do not insist on the radius of the ring remaining constant. As the rotational speed of the ring increases, length contraction of the ring can make the circumference of the ring appear constant to observers on the ring. In the case of a solid disc, there is no natural length contraction in the radial direction and disc becomes physically stressed, making Born rigid acceleration of a solid disc impossible.

Last edited: May 17, 2010
16. May 19, 2010

### Passionflower

That depends on the observer. For instance an inertial observer free falling towards the center of the earth will observe the distance to reduce over time, similar to the case of the accelerating rocket.

Correct, but in addition to a greater coordinate acceleration the proper acceleration is also greater.

Last edited: May 19, 2010
17. May 21, 2010

### Ascenxion

Thank you all for the wonderful explanations.