# Finding CM frame velocity in special relativity

1. Dec 31, 2013

### fayled

1. The problem statement, all variables and given/known data
An electron of energy 10 GeV strikes a proton at rest. What is the velocity of the electron-proton CM system? What energy is available to produce new particles?

2. Relevant equations
Conservation of momentum and energy,
E2=p2c2+m2c4

3. The attempt at a solution
I'm really not sure where to start - solving for the velocity in classical mechanics is simple but the equations here would be really messy so I think there must be a simple way I'm missing. If the two particles were identical it would be easy... Then once I've got that velocity, then would the sum of the energies in the CM frame before the collision be the energy available to produce new particles in the lab frame - i.e the extra energy in the lab frame goes toward conserving momentum? Any clues would be appreciated, thanks.

2. Jan 1, 2014

3. Jan 1, 2014

### vela

Staff Emeritus
In the lab frame, say the four-momenta of the electron and proton are given by
\begin{align*}
p_e^\mu &= (E, p) \\
p_p^\mu &= (M, 0)
\end{align*} Let $\beta=v/c$ be the speed of the CM frame. Write down the four-momenta in the CM frame. What do you know about the three-momenta in the CM frame?

4. Jan 1, 2014

### vanhees71

Also have a look at the Mandelstam variables. They are very useful invariants!

5. Jan 1, 2014

### fayled

Ok so I think using this approach gives the four momenta in the CM frame as

Pe'=(E1,p') and Pp'=(E2,-p').

I used the fact that if I take the sums of each of the 4 momenta in each frame, then the square of this is invariant between frames so I can equate them to give

(E+M)2-(pc)2=(E1+E2)2

If I can find E2 then I can get the speed of the proton which will be the same as the speed of the CM frame. I just need another relation between E1 and E2 though.

6. Jan 1, 2014

### vela

Staff Emeritus
Use the Lorentz transformation to find expressions for the momenta in the CM frame.

7. Jan 1, 2014

### vanhees71

The additional relation are the on-shell conditions in the CM system
$$E_1=\sqrt{m_1^2 c^4 + p_{\text{cm}}^2 c^2}, \quad E_2=\sqrt{m_2^2 c^4 + p_{\text{cm}}^2 c^2}.$$
As I already said, using the four-dimensional invariants (the Mandelstam variables), it's a bit simpler, but that was what you did anyway, be it in somewhat implicit form.

To use the rotation-free Lorentz boost directly is also a nice exercise.

8. Jan 2, 2014

### fayled

So I managed to find the two energies:

E1=2116.575821MeV (electron) and
E2=2315.110569MeV (proton)

I checked and these two energies do correspond to equal momenta as expected in the CM frame.

The velocities from these are

v1=0.9999999709c (electron) and
v2=0.9142439191c (proton)

So as the proton was at rest in the lab frame, I can deduce the speed of the CM frame is simply v2.

However as a further check, I know the speed of the electron in the lab is velab=0.9999999987c. Thus, by using the relativistic velocity addition formula, I would expect that if I add velab to the negative of the CM frame speed (-v2), using the formula, I would obtain the CM speed of the electron...

Vecm=(0.9999999987c-0.9142439191c)/[1+(0.9999999987x-0.9142439191)]=0.999999971c. So clearly either my answer is wrong (which I can't see how it would be given that the momenta come out as equal) or this isn't a valid check (in which case why)... Thanks.

Finally, the energy available to produce new particles in the CM frame is just E1+E2, so how would that relate to the lab frame? I can find the rest mass of the products in the CM frame and these are the same in the lab frame, so surely the energy available for new particle production is just the rest energy in the CM frame?

Last edited: Jan 2, 2014
9. Jan 2, 2014

### vela

Staff Emeritus
Since you apparently used masses with three significant figures, the digits after decimal points are meaningless. You shouldn't be writing them down just because it's what your calculator spit out.

The speeds from the two methods match. Why do you think they don't agree?

10. Jan 2, 2014

### fayled

Yeah I realise that is a bit silly, but not as silly as me failing to see that they agree. That is terrible.

But thanks :)

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