# Special relativity: What is the final mass of the rocket?

1. Oct 22, 2012

### sabre729

1. The problem statement, all variables and given/known data
Ok, the problem says some rocket of mass M0 has this new technology that allows it to convert photons into fuel. It reaches a speed of .5c. The problem asks us to solve for the final mass of the rocket, and then assume that all the lost mass was converted into photon fuel and calculate the total energy of that.

2. Attempt at a solution
Of course the starting E is just M0c^2 because it's starting from rest.

It's equal to the final energy: E(photons)+E(rocket, final).

The rocket's final mass is going to have some factor of gamma (γ) multiplied in, which in the case of .5c (β=.5) is equal to about 1.15. After factoring in the momentum of the rocket:

M0*c^2=E(initial)=E(final)=E(photon)+((γM0*c^2)^2+(γ*β*M0*c)^2)^(1/2)

First off, is this expression okay? Lorentz transform-wise it makes sense that the rocket's energy and momentum are multiplied by γ and γ*β, since we are looking at the speed in the rocket's frame and then boosting it back to the stationary frame. It also factors in the E(photons), which in this case is equal to n*H*v/c, where v is the frequency and c is the speed of light.

But doesn't this give me too many variables to express the change in mass? Unless the mass is only the part of the equation that doesn't include the photons (hence the "mass defect" goes to the photons).

Does what I did make sense to anyone who isn't me?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 23, 2012

### clamtrox

That looks a little too complicated to me. You should be able to find the result by just doing the standard kinematic calculation with conservation of energy and 3-momentum. You can write the equations in terms of 4 variables: initial mass of the rocket, mass of the fuel converted, energy of the photons and gamma.