# B Relativistic Sphere

1. May 2, 2016

### Einstein's Cat

Say there is a theoretical sphere of radius r, at rest, then if it's velocity changes then I assume that the radius is subject to length contraction and thus it's volume would decrease from a stationary observer. Is this assumption true?

2. May 2, 2016

### Staff: Mentor

This might answer it:

http://www.spacetimetravel.org/fussball/fussball.html

The sphere becomes an ellipsoid but due to the finite speed of light and the fact that the light travels different distances to reach your eye you will still see a sphere.

A similar case occurs with a relativistic cube where the cube contracts relative to the stationary observer but the observer actually sees a cube rotated slight toward him/her so that you see the side and the back together because light from the fornt edge gets to your eyes sooner than light from the back edge.

http://www.spacetimetravel.org/tompkins/node1.html

Look up Terrell rotation for more details:

http://www.math.ubc.ca/~cass/courses/m309-01a/cook/terrell1.html

3. May 2, 2016

### pervect

Staff Emeritus
It's easier if you consider a sphere at rest in it's own reference frame, and asks what happens if you ask what it's shape is in some frame moving relative to the sphere's rest frame. That way you don't have to worry about the notion of "rigidity".

The sphere when seen from a moving frame does indeed Lorentz contract in one direction, assuming the shape of an ellipsoid. The volume of the sphere in the moving reference frame is lower, volume is therefore a frame-dependent quantity like length is.

4. May 3, 2016

### Einstein's Cat

What is the relationship between the volume of the ellipsoid and its radius?

5. May 3, 2016

6. May 3, 2016

7. May 3, 2016

### Ibix

The ones in the y and z direction don't change. The one in the x direction scales as $\gamma$.

8. May 3, 2016

### Einstein's Cat

I see because length contraction is in the direction of one dimensional motion. I apologise but what does the quote above mean?

9. May 3, 2016

### Ibix

What you said. Although I should have said $1/\gamma$. At speed v the length is $\sqrt {1-v^2/c^2}$ times its rest length.

10. May 3, 2016

### Einstein's Cat

Sorry to go on but when you say scale do you perhaps mean that "x new"
= 1/Lamda * "x rest"?

11. May 3, 2016

### Ibix

That's a gamma ($\gamma$), not a lambda ($\lambda$), but otherwise yes.

12. May 3, 2016

### Einstein's Cat

Thank you very much and I'll go and revise greek letters!