Mass of a composite particle using relativity.

[lukeharvey]

Homework Statement

A photon collides with a stationary particle of rest mass m0 and is absorbed.

Find the mass and velocity of the composite particle

Homework Equations

Conservation of energy:
E + m0 * c^2 = \gamma * mt *c^2 where mt is the total mass

Conservation of momentum:
E/c = \gamma * mt * v

The Attempt at a Solution

I used the attachment as a diagram for the two different frames.

I then solved for the velocity first and found that:

v= E*c/(E+m0*c^2)

I then solved for the mass (mt) and found:

mt = E+m0*c^2/(\gamma*c^2)

However i am not sure if the solution for the total mass is correct because it involves \gamma. If it is correct i was wondering if there was any way to simplify equation, and maybe substitute an expression in for \gamma, so it is not in the equation?

Thanks

 

[gabbagabbahey]

However i am not sure if the solution for the total mass is correct because it involves \gamma. If it is correct i was wondering if there was any way to simplify equation, and maybe substitute an expression in for \gamma, so it is not in the equation?

Thanks

Why not plug your solution for v into the definition of \gamma and simplify?

 

[lukeharvey]

I've tried this but it didnt seem to simplify and when i put it into the equation i found it made it even worse. Any ideas?

 

[gabbagabbahey]

I've tried this but it didnt seem to simplify and when i put it into the equation i found it made it even worse. Any ideas?

It simplifies fairly nicely for me... why don't you show me where you are getting stuck?

 

[lukeharvey]

Im not the best with latex but i can show you the equation i got on here if you can understand it

So i got gamma = 1/(sqrt(1-((((E*c)/(E+m*c^2))^2)/c^2)))

 

[gabbagabbahey]

Im not the best with latex but i can show you the equation i got on wolfram alpha?

I got gamma is equal to the expression shown on wolfram alpha, and it doesn't seem to simplify nicely
wolframalpha.com/input/?i=simplfy+1%2F%28sqrt%281-%28%28%28%28e*c%29%2F%28e%2Bm*c%5E2%29%29%5E2%29%2Fc%5E2%29%29%29

The page won't load properly because of the ascii representations in the address, but I get

\frac{1}{\gamma}= \sqrt{1-\frac{v^2}{c^2}} =\frac{\sqrt{m_0^2c^4+2Em_0c^2}}{E+m_0c^2}

which makes m come out rather nicely as

m=\frac{\sqrt{m_0^2c^4+2Em_0c^2}}{c^2}=m_0\sqrt{1+\frac{2E}{m_0c^2}}

 

[lukeharvey]

But i seem to get:

\sqrt{1-\frac{E^{2}}{E^{2}+M_{0}^{2}c^{2}}}

But now I am stuck thanks for all the help

 

[gabbagabbahey]

But i seem to get:

\sqrt{1-\frac{E^{2}}{E^{2}+M_{0}^{2}c^{2}}}

But now I am stuck thanks for all the help

You are probably expanding v^2 incorrectly, remember

(E+m_0c^2)^2\neq E^2+m_0^2c^4

 

[lukeharvey]

Ah its actually been one of those days, how can i not expand properly? Thanks a lot for the help :)

 

[Morgoth]

why not going to a system where the two initial particles move with equal and antiparallel spatial mommenta?
That way you can easily find the mass of the particle afterwards by asking the square of the 4mommentum to be invariant, because the M1 in that system will be at rest. "CM" system.

Then by a lorentz transformation I think, you will be able to find that system from your initial one, and so the mommentum of M1 as well at the "LAB" system.

Just a general idea, I am too tired to try it out myself right now, to see if it's simpler or not. I generally like more the invariance of 4mommentum during collisions rather than any other method, which would make me get lost with velocities etc...