# A Roller Coaster Ride through Relativity/Length Contraction
## Length Contraction

Cast your mind back to the river race. Albert takes longer than usual to cross the river because he has to use part of his velocity aiming upstream. Beatrice, however, takes a lot longer because it is really hard work rowing directly upstream.
Now consider the Michelson-Morley experiment. In fact, suppose you are watching someone else performing the experiment in a fast spaceship travelling at nearly the speed of light. The photon which travels at right angles to the direction of motion is a photon clock and, like Albert, it is slowed down by a factor of
* $\gamma =1 \over {\sqrt {1-v^{2}/c^{2}}}$
The photon which moves parallel to the motion of the ship is like Beatrice and ought to be slowed down even more. But as we know, the Michelson-Morley experiment gives a null result - ie the two photons take exactly the same time for the journey. (Incidentally there can be no disagreement between you and the people on the space ship as to whether the photons arrive at the same time or not. It would be easy to rig up a device to blow up the ship if any difference was detected and it is not possible for you to disagree with your colleagues on the ship as to whether they are blown up or not. At least, not according to the Theory of Relativity. Now Quantum Theory... well that's another story!)
So how do we reconcile these two viewpoints? The answer is that the photon which moves parallel to the direction of motion of the ship does not have to go as far because lengths in that direction are contracted. Here is the formula (for the proof see Appendix B):
* $l={l_{0} \over \gamma }={\sqrt {1-v^{2}/c^{2}}}l_{0}$
We see that, while lengths perpendicular to the direction of motion remain unchanged, lengths parallel to the direction of motion must decrease by a factor of $\gamma$ to preserve the constancy of the velocity of light.
Note that the length contraction factor is just the same as the time dilation factor but you have to divide rather than multiply. For example, if you travel at 80% of the velocity of light past someone at rest, your spaceship will appear to him as if it is only 60% of its proper length. Of course, he will claim that it is your spaceship that is squashed but that's relativity for you!
Is that why the penny went down the crack? you ask.
Absolutely. The penny was going so fast that it became shorter than the crack and so fell down – see the diagram below:
Hey - wait a minute! Suppose I was a flea sitting on the penny! Wouldn't it be the crack that was shrunk and not me? I am prepared to accept that times and lengths might appear to extend and shrink but surely, either the penny goes down the crack or it doesn't! Ha! Ha! Got you now! I knew there was something wrong with this Relativity business!
When you have quite finished crowing I will explain.
You mean to say you have an answer to that one as well?
Uh-uh.
Well, what is it then?
I am afraid you will find this even harder to believe than anything I have told you so far.
Go on - try me
Well. it is like this. From your point of view, as the penny begins to extend over the (contracted) crack, the front of the penny begins to move downwards before the back end leaves the table. Like this:
But the penny can't bend!
Well, it doesn't bend in a physical sense, no more than it contracts in a physical sense. But to the flea sitting on the penny, that's what appears to happen. Correction. To the flea sitting on the penny, that's what really happens. I am afraid we have to accept:
* Bizarre consequence number 3
* Moving objects whose speed changes appear to bend.
Well, I guess you were right after all.
So you do believe it then?
No. I don't believe it. But you were right when you said I wouldn't believe it!
But you accept it?
Yes. I suppose so. But I am getting so confused. Is there anything that two observers can actually agree on?
Yes there is. In fact this is a rather important theorem which I shall call
* Reassuringly normal consequence number 4
* Although two observers in relative motion can each argue that they are stationary and it is the other who is moving, they will both agree on their relative velocity.

## The long straight

After two hair-raising descents, the roller coaster levels out and enters a long straight section of track.
Beside the track is a spare train, identical to ours except that, of course, it looks shorter. As I get out my stop watch to time its passing, I notice my brother sitting in the other train with watch in hand timing us going past.
I suppose he thinks that our train is shorter than his.
Correct. Well done.
So, when he calculates how fast we are going, he will get the wrong answer, won't he? Because he is using the wrong length.
Er – no; he will just use the right wrong length.
The right wrong length! Don't be ridiculous! How can the length be both right and wrong at the same time!
Yes; I am sorry; I shouldn't have said that. What I meant was that he will be correct to use the relativistically contracted length to calculate the speed.
Why? Surely to calculate the correct speed, you should use the correct length?
Absolutely. But don't forget that, to my brother, the length of our train really is shorter. So that is the correct length to use.
But if we do the calculation, won't we get a different answer? because we are obviously going to use the real length of the train.
OK – let's work out the details:
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