# Question about SR and increasing mass

1. May 26, 2009

### LennoxLewis

Let's say you're on a spaceship and you can accelerate to a speed very close to c. As your own time seems to slow down (relative to slow moving observers), your mass also increases.

I know that mass increases in the sense that one requires more energy to give the same acceleration, but does this also mean that your impulse increases ? I assume so. So, if your spaceship would crash into something when you're close to c, it would seem as impulse was violated because (assuming you slow down enough), the mass becomes close to the rest mass again? In that case, would one have to consider a relativistic version of the conservation of impulse?

2. May 26, 2009

### Cyosis

Your own time doesn't seem to slow down. In your own frame you are at rest and therefore you will see the clocks of inertial frames moving with respect to you slow down. In a relativistic collision you will have to use relativistic momentum. Classically you would expect mass to be conserved during a collision, but relativistically this is not true.

Lets do the calculations both classically and relativistically.

A spaceship of mass m going at speed v=4/5c crashes into an object at rest with mass m. (everything is happening in the x-direction)

Classically:
$$mv+0*m=M_*u=(m+m)u \Rightarrow m\frac{4}{5}c=2mu \Rightarrow u=\frac{2}{5}c$$

Relativistically:
Momentum as well as total energy is conserved. Total energy is given by $E=\gamma m c^2$ and momentum is given by $p=\gamma mv$. Combining these into a 4-momentum vector for simplicity yields $p_\mu=((\gamma+1)mc,\gamma m v,0,0)$.

$$p_\mu p^\mu=-(\gamma+1)^2m^2 c^2+\gamma^2 m^2 v^2=-M_*^2 c^2 \Rightarrow M_*=\frac{4}{\sqrt{3}}m$$

And

$$u=\frac{p c^2}{(\gamma+1)mc^2}=\frac{1}{2}c$$

Don't forget that this object with which it collides has rest energy as well. Classically its energy would be zero but in relativity this is not so. The total energy as well as momentum is conserved.

Edit: I don't have a lot of time currently so I didn't explain the calculation very well. If you have any questions about it feel free to ask I will try to answer them at a later time.

Last edited: May 26, 2009
3. May 26, 2009

### LennoxLewis

Thanks a lot for the explanation, Cyosis! I do have a question: in the four-vector, where does the (y+1)mc term come from?

4. May 26, 2009

### Cyosis

The 4-vector is defined as $p_\mu=(E/c,\vec{p})$. With E being the total energy. Before the collision we have the energy of the rocket ship $E_r=\gamma mc^2$ and the energy (rest energy) of the object it collides with $E_0=mc^2$. Therefore the total energy is given by $E_r+E_0=\gamma mc^2+mc^2=(\gamma+1)mc^2$. Plug it into the definition of the 4-vector and one c will drop out.

5. May 27, 2009

### LennoxLewis

It all makes sense now. Thanks again.

6. May 27, 2009

### Staff: Mentor

Note that the $$\gamma+1$$ is not a general term, but specifically for the case where the rocket collides with an object of equal mass. It is correct for this case, but please don't accidentally over-generalize to other cases.

7. May 27, 2009

### Cyosis

Yes DaleSpam raises a good point. I used an equal mass for the the object to simplify the calculation, this however may not have been particular smart for didactic reasons. I emphasize the general 4-momentum is given by $p_\mu=(E/c,\vec{p})$. Now imagine the object it collides to has mass $\mu$ instead. Then E becomes $\gamma m c^2+\mu c^2=(\gamma m+\mu)c^2$. Therefore the 4-momentum becomes $p_\mu=((\gamma m+\mu)c,p_x,p_y,p_z)$. Another thing that is important to realize is that if the object it collides with was originally in motion with a speed other than v then its gamma factor differs from the space ship's gamma factor!

Last edited: May 27, 2009