# Relativistic Snake

1. Sep 21, 2005

### runevxii

A relativistic snake, of proper length 100cm is traveling across a table at V = 0.6c. To tease the snake, a student holds two ends of a cleaver 100cm apart and plans to bounce them simultaneously on so that the left one lands just behind the snake's tail. The student reasons, "the snake is moving with Beta=0.6 so its length is contracted by the factor gamma=5/4 and its length measure in my frame is 80cm. Therefore, the cleaver in my right hand bounces well ahead of the snake, which is unhurt. The snake reasons "the cleavers are approaching me at B=0.6 so the distance between them is contracted to 80cm, and I shall certainly be cut. Use the Lorentz transformation to resolve this paradox.

2. Sep 22, 2005

### Andrew Mason

You have to use the Lorentz transformation to determine when, in the snake's frame, the two cleavers cut.

$$t' = (t \pm vx/c^2)\gamma$$

If t is the same for each cleaver (ie. in the rest frame) we can see that the t' is different for each of the two events because x differs by 1 m. So the time difference between the cuts in the snakes frame is:

$$\Delta t' = \gamma vx/c^2 = 1.25 * .6 * 1/c = .75/c$$

In that time interval, the cleavers move x' = v\Delta t' = .75 * .6 = .45 m. Add this to the separation observed by the snake (.8 m) to give the distance between cuts (1.25 m) so the cleavers miss the snake at both ends.

AM

[Edit: I was using the reciprocal of gamma instead of gamma. Now corrected]

Last edited: Sep 22, 2005
3. Sep 22, 2005

### Jibobo

I actually have the same problem as well. But when I do out the problem, I don't get the same numbers you do and I don't see what I did wrong.

I call one cleaver A and another B and set A as the origin in both inertial frames so that it's described by the 0 vector in both frames. Then B in the boy's frame is (t, x, y, z) = (0, 1, 0, 0). So if I do the Lorentz Transform with the 4x4 matrix lambda:
(B = beta, y = gamma)
[y -By 0 0]
[-By y 0 0]
[0 0 1 0]
[0 0 0 1]
then I end up with B in the snake's frame = (-By, y, 0, 0). By my calculations, t' = -By/c = -2.5 E -9 sec. I don't understand where the 0.48/c = 1.6 E-9 sec that you got came from.

I also see that I end up with x' = y = 1.25m, but it should be 0.8m I believe. Did I do something wrong and if so, what?

4. Sep 22, 2005

### Andrew Mason

I was using the reciprocal of $\gamma$ so my numbers were not quite right (now corrected). The answer is 1.25 m. That is the distance between cleaver chops in the snake's frame. That distance is greater than the separation between the cleavers observed by the snake because in the snake's frame the cleavers are moving and the chops are NOT synchronous. Fortunately for the snake it is also greater than the snake's length.

AM