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 $$E^2 = (mc^2)^2 + (pc)^2$$?