# Is length contraction relative or absolute?

1. Aug 29, 2004

### ardenbook

Is length contraction relative or absolute??

This is from the book "Speakable and unspeakable in quantum mechanics" by J.S. Bell.

Three small spaceships, A, B, and , drift freely in a region of space remote from other matter, without rotation and without relative motion, with B and C equidistant from A. On reception of a signal from A the motors of B and C ignited and they accelerate gently. Let ships B and C be identical, and have identical acceleration programmes. Then (as reckoned by an oberver in A) they will have at every moment the same velocity, and so remain displaced one from the other by a fixed distance. Suppose that a fragile thread is tied initially between projections from B to C. If it is just long enough to span the required distance initially, then as the rockets speed up, it will become too short, because of its need to Fitzgerald contract, and must finally break. It must break when, at a sufficiently high velocity, the artificial prevention of the natural contraction imposes intolerable stress. Is it really so ??

2. Aug 29, 2004

### mathman

In the B,C, thread reference frame, there is no length contraction, so nothing breaks. A sees everything getting shorter, but the shrinkages are relativistic, i.e. what A would measure, but not physically real.

3. Aug 29, 2004

### pervect

Staff Emeritus
Yes, the string breaks - this is the famous Bell Spaceship "paradox".

BTW, one will get the same answer to the "paradox" regardless of which paradigm one uses as far as length contraction goes, as long as one does the calculation correctly.

4. Aug 29, 2004

### ardenbook

Could you please elaborate a bit why the string breaks, Thanks....

5. Aug 29, 2004

### Hurkyl

Staff Emeritus
Actually, that's not entirely accurate.

The requirement that the ships B and C "have identical acceleration programmes" only holds in A's reference frame. During the period of acceleration, B will observe C speeding ahead, while C will observe B falling behind. (assuming that C is in front)

Yes, length contraction is relative. It is, by definition the comparison of the length of an object as measured in its rest frame to its length as measured in some other frame. If you don't specify that other frame, then you cannot speak of length contraction.

6. Aug 29, 2004

yes, the contraction is physically real.

7. Aug 29, 2004

### ardenbook

Could you please elaborate a bit why --
During the period of acceleration, B will observe C speeding ahead, while C will observe B falling behind. (assuming that C is in front)

8. Aug 29, 2004

### Hurkyl

Staff Emeritus
Well, the easiest way to see it is simply by doing the math.

If that's not an option, here are a couple ways to see it heuristically.

The distance between B and C is smaller in A's frame than in the frame into which B&C finally settle. So, we know:

Original distance between B & C
= Final distance between B & C in A's frame
< Final distance between B & C in B&C's final frame

Accelerating frames observe peculiar temporal effects: time is running faster for anything towards which you accelerate, and runs slower (or even backwards!) for things away from which you accelerate.

So, B will see time running faster for C, which means B observes C accelerating faster. And vice versa.

9. Aug 29, 2004

### pervect

Staff Emeritus
If one has heard of the "Rindler horizon" of an accelerated observer, it's obvious that the string breaks if it's long enough. For instance, let the string be longer than 1 light year long, and the acceleration be 1g. Then the string crosses the Rindler horizon of the lead ship. You can't dangle a string through an event horizon without it breaking - among other things it requires an infinite proper acceleration to hold the string stationary at the horizon. Otherwise one could drop a string into a black hole, and pull stuff back out.

But, of course, not everyone has heard of the Rindler horizon. Without going into too much detail, I'll discuss it a bit.

Let's let the acceleration of the spaceships be 1 (light-year)/year^2, which is approximately 1 g. Let's let the string be 1 light year long. Let all distances be expressed in terms of light years, and all times be expressed in years. This makes c=1, which is very convenient.

Let's look at this from a stationary frame centered on the second ship. At t=0, the "start" signal arises, and both ships start accelerating. The equation of motion of the first ship can be parametrized as d=cosh(tau), t=sinh(tau) in the stationary frame, where tau is the proper time of the accelerating ship. See for instance:

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
or see MTW's "Gravitation" section on "accelerated motion"

and note that cosh(0)=1, which is the correct value of the initial position of the first spaceship.

Now, let the second spaceship emit a lightbeam towards the first spaceship as soon as it receives the start signal at t=0. When will this lightbeam arrive at the lead ship?

Well, the equation of motion of the light beam is d=t, since it moves at the speed of light and it's emitted at t=0.

So let's solve for cosh(tau)=sinh(tau)

Multiplying both sides by two, and substituting in the exponential definitions of sinh and cosh, we get exp(tau)+exp(-tau) = exp(tau)-exp(-tau), which happens only at tau=infinity!

Or to put it another way cosh(tau) > sinh(tau) which means d>t.

Therefor, the accelerating spaceship is always ahead of the light beam, though the lightbeam gets very close as tau-> infinity.

The distance from the second ship to the first can be derived by taking the time it takes a light signal to "bounce off" the second ship and arrive back. But - the signal never arrives, so it can't start back. So the distance as computed by the second ship is infinite. We could at this point write down the equations of motion for the second ship if we wished, but I won't bother, since it's pointless.

Since the string obviously can't stretch to an infinite distance, it breaks.

For other approaches, there's also some discussion of the Bell spaceship problem in the sci.physics.faq

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

which may be more straightforwards.

10. Sep 6, 2004

### ardenbook

Here is the paradox. If spaceship B and C hover in a uniform gravitational field using their rocket motor with same power. Then, according to Einstein's equivalence principle, it is the same as ship B and C undergone identical acceleration programmes in free space. But we know for sure that the string will not break in a uniform gravitational field.

11. Sep 6, 2004

### pervect

Staff Emeritus
You'd better think about what you mean by a uniform gravitational field a bit more. The clocks at A and B are going to be running at different rates - if you use the paradigm of a "gravitational field" the clocks will run at different rates because of "gravitational redshift".

Thus an acceleration of 9.8 m/s^2 as seen by an observer at point "A" higher in the "uniform field" will be seen as more by "B", because B's clocks are running slower.

When B gets deep enough in the "uniform" gravitational field, eventually it's clock will stop (from A's viewpoint). This happens when the escape velocity from B to A reaches the speed of light, which occurs when B is 1.03 light years below A in a 1 g field. B will need an infinite proper acceleration to "hold station" at this point.

Of course, if you look back at my other post, you'll see that we are talking about the "Rindler horizon" of an accelerated observer, but are just using different language.

To wrap things up in a neat and simple package to address the "paradox" - when the string does not break, the observer who is lower in the "gravitational field" must have a higher proper acceleration (one measured by his own clocks and rods) than the observer higher in the field. This happens regardless of whether one is in an actual gravitational field, or on an accelerating spaceship.

Also, note that there is a strict limit as to how long the string can be.