# Homework Help: Relativistic density

1. Nov 28, 2007

### Xaspire88

A cube has a density of 2000 kg/m^3 while at rest in the laboratory. What is the cube's density as measured by an experimenter in the laboratory as the cube moves through the laboratory at 90% of the speed of light in a direction perpendicular to one of its faces?

It's the dimensions of the cube that change not the mass correct? if this is the case would i use the equation
$$L = L^1 \sqrt{1 - v^2/c^2}$$

This would give me how the dimensions of the cube would change. This new length equals .316 m. And so the density would be 2000 kg/.316 m^3? and then in kg/m^3 would be 6324.56 kg/m^3? does this seem logical?

2. Nov 28, 2007

### Staff: Mentor

How did you compute the contraction factor?

If they define density as invariant mass over volume, then your approach (but not your answer) is correct.

3. Nov 28, 2007

### Erwin Kreyszig

In addition to length contraction and time dilation, another consequence of Special relativity is that mass of a moving object appears to increase with speed. This increase is actually proportional to the time dilation factor at very high speeds.

the mass at high speeds is given by

m = m0 / $$\sqrt{1-\frac{v^2}{c^2}}$$

where m0 is your rest mass, so you need to factor this in too i think.

Hope that helps. EK

4. Nov 28, 2007

### Xaspire88

I think i forgot to square my velocity initially. The answer should be 2000 kg/ .436m^3 or
4588.31 kg/ m^3?

5. Nov 28, 2007

### Staff: Mentor

Right. (Assuming the standard usage where mass means invariant mass; if so-called "relativistic" mass is meant, then the answer will be even higher.)

6. Nov 28, 2007

### Xaspire88

More than likely it is the relativistic mass. In which case i would need to solve for the relative mass of the particle while moving at .9c as well as the contraction factor of the cube..

2000/square root(1-0.9^2)= 4588.31 kg
and the calculated relative dimensions of the cube were calculated to be .436m
So the density of the cube would then be 4588.31kg/.436m^3 or 10526.3 kg/ m^3?

7. Nov 28, 2007

### Staff: Mentor

The use of relativistic mass is a bit old-fashioned.
Right. Just tack on another gamma factor.
Yep.

8. Nov 28, 2007

### Xaspire88

You say old-fashioned. Do they now use it in the way relating it to the Energy of the particle. Such as in the case of the $$E^2 = (mc^2)^2 + (pc)^2$$?