# Special relativity and Density?

1. May 31, 2009

### vorcil

A cube has a density of 1900 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 92.0 % of the speed of light in a direction perpendicular to one of its faces?

My attempt,

I honestly have no idea how to do this one, I looked through my book and can't seem to find a way to stitch up a link between relativity and density, all i've ever done is Mechanics of relativity,

Do i somehow work out how much the length of the cube changes? or try to find the length of a cube, from 1900kg/m^3, Density = mass/ volume,

Length seen = Sqrt(1-beta)* length of cube at rest (where beta is velocity^2/ c^2)

length of cube at rest = sqrt( x^2* + x^2 + x^2 )
i'm not sure how to find the length of one side at rest(or i sort of do)

can someone show me how to solve for the rest of this question, it's bloody hard especially for a first year who has only been doing relativity for 4 days ^ ^

2. May 31, 2009

If at rest the cube had a mass M and length l its density would be given by:
density=M/l^3(M/volume).What happens to M and the volume when the cube is moving?

3. May 31, 2009

### vorcil

the volume would decrease because of length contraction,
and mass would stay the same?

4. May 31, 2009

No the mass would increase .Look up the relativistic mass variation equation.

5. May 31, 2009

### vorcil

I'm only doing special relativity, we don't even have the relativistic mass variation equation in our book. the only formula i could find in special relativity that had mass in it was rest energy e=mc^2 and kinetic energy k= (Yp -1)mc^2

i'f what you're saying is that volume decreases and mass increases as it speeds up so that density remains constant, i.e still 1900kg/m^3 i'm not sure it is, is this what you're saying?

6. May 31, 2009

### vorcil

i'm really stuck, can someone help please =]

7. May 31, 2009

### vorcil

i don't think this is what i'm looking for,
there must be some other way around it

8. May 31, 2009

### DaveC426913

Why? What about that formula don't you like?

9. May 31, 2009

### vorcil

1: it's not stated in my book anywhere
2: i found it in a chemistry related page on wikipedia
3: it uses the mass of an electron, where i'm using the density of a cube
4: i have no idea how to use it to solve my problem

10. Jun 1, 2009

Well done vorcil you have done the research and you just need to finish off.A few points:
1.Density = mass /volume so if the mass increase and the volume decreases the density gets bigger.
2.The mass variation equation was derived by Einstein by considering the electron but it is generally true.
3.Write down the equation for the density of the cube when it is at rest
D=M/l^3=1900
4.Using the relativistic equations and the symbols M and l write down the equations for the increased density.
5.You will now have two density equations and with a little bit of mathematical juggling you will be able to find the new density.

11. Jan 24, 2012

### spencek

I know this is 2 and a half years late, but for anyone who happens across this in the future, I hope it helps. I just learned special relativity in my college physics class. I'll give some basic equations for relativistic correction first then show how I used those to create a relativistic density equation.

1: β=$\frac{v}{c}$ ie. if the object is travelling at 75% the speed of light β=.75

2: For the relativistic correction γ=$\frac{1}{\sqrt{1-β^{2}}}$

3: For length contraction l=$\frac{l_{o}}{γ}$ where l$_{o}$ is the rest length

4: For mass increase m=γm$_{o}$ where m$_{o}$ is the rest mass

5: Now for density ρ=$\frac{m}{V}$ So if we want the density in relativistic terms, we can just use the relativistic mass and length. Since we're using a cube, say with side length x, then the volume is x$^{3}$ so the relativistic volume is V=x$^{2}\frac{x_{o}}{γ}$ (you separate the x$_{o}$ and the x$^{2}$ because length contraction is only in the dimension of the velocity vector, the other two dimensions are unaffected) therefore our density equation becomes ρ=$\frac{γm_{o}}{x^{2}\frac{x_{o}}{γ}}$ But we want it in terms of only density since we don't know the total mass or total volume of the cube. To do this we can solve the density equation for mass and for length. So m$_{o}$=ρ$_{o}$V or m$_{o}$=ρ$_{o}$x$^{3}$ and V=$\frac{m_{o}}{ρ_{o}}$ which is x$^{3}$=$\frac{m_{o}}{ρ_{o}}$ or x$_{o}$=$\frac{m_{o}}{ρ_{o}x^{2}}$ Next we plug those into the original relativistic density equation so we end up with ρ=$\frac{γρ_{o}x^{3}}{x^{2}\frac{\frac{m_{o}}{ρ_{o}x^{2}}}{γ}}$ this looks confusing but it turns out nicely. For the denominator of the main fraction the x$^{2}$ multiplies by the m$_{o}$ and then γ on the very bottom, if we multiply by its reciprocal, moves to multiply by the ρ$_{o}$x$^{2}$ therefore the whole denominator becomes $\frac{x^{2}m_{o}}{γρ_{o}x^{2}}$ and you see that the two x$^{2}$'s cancel so the whole equation becomes ρ=$\frac{γρ_{o}x^{3}}{\frac{m_{o}}{γρ_{o}}}$. Now if we get rid of the bottom fraction by multiplying by the reciprocal we get ρ=$\frac{γρ_{o}(γρ_{o}x^{3})}{m_{o}}$. Now, if you look back a few sentences you see that we originally said that m$_{o}$=ρx$^{3}$, so plugging that into the ρ$_{o}$x$^{3}$ in the numerator we get ρ=$\frac{γρ_{o}(γm_{o})}{m_{o}}$. Now the two m$_{o}$'s cancel and the two γ's combine to get the final relativistic density equation of ρ=γ$^{2}$ρ$_{o}$ where ρ$_{o}$ is the rest density. So with your rest density of 1900 $\frac{kg}{m^{3}}$ and the speed being 92% the speed of light you get your β as .92 and plug that in to find γ which comes out to γ=2.551551815 so ρ=(2.551551815)$^{2}$(1900) and your relativistic density is ρ=12369.8 $\frac{kg}{m^{3}}$.

Last edited: Jan 25, 2012