# Two protons colide and i am to calculate newly formed mass

1. Jul 18, 2013

### 71GA

1. The problem statement, all variables and given/known data
Two protons with kinetic energies $W_{k1}=4GeV$ and $W_{k2}=2GeV$ colide and form new particles. What is the mass of newly born particles?

2. Relevant equations
\begin{align}
W_{before}&=W_{after}\\
p_{before}&=p_{after}\\
W^2 &= {W_0}^2 + p^2c^2 \longleftarrow \substack{\text{Lorentz invariant}}
\end{align}

3. The attempt at a solution
First i wrote the energy conservation law:
\begin{align}
W_{before} &= W_{after}\\
W_{k1} + W_{k2} + 2W_{0p} &= \sqrt{{W_{0~after}}^2 +p^2c^2}\longleftarrow\substack{\text{Here the $W_{0~after}$ is a full}\\\text{rest energy after colision}}\\
W_{k1} + W_{k2} + 2W_{0p} &= \sqrt{\underbrace{\left( 2W_{0p} + W_{0m} \right)^2}_{\smash{\substack{\text{after collision we have}\\\text{2 $p^+$ and new mass $m$}}}} +p^2c^2}
\end{align}

Here i am not sure what to do with the $p^2c^2$. Can i just take the mass frame and say $p=0$? Why yes and why not?

Other option i am thinking of is to say that $pc = \sqrt{{W_{k~before}}^2 - 2W_{k~before}W_{0~before}}$. Here i know that $W_{0~before}=2W_{0p}$, but i am not sure what to put in for $W_{k~before}$. Can i do it like $W_{k~before} = W_{k1}+W_{k2}$?

Which of these two options is right/wrong and why? If both are wrong please tell me how to fix them to become ok.

2. Jul 18, 2013

### Simon Bridge

Taking the center of mass frame is useful for simplifying the calculations.
Remember that the kinetic energies will change.

Why do you think there are still two protons around after the collision?

Note - you can also use the form: $W_k=(\gamma -1)mc^2$

3. Jul 19, 2013

### 71GA

How will they change? I guess one will be bigger and other will be smaller than in lab. frame. How do i calculate this? I will also need to calculate the speed of the COM frame.

I have heard that new mass can be formed in the proximity of the massive bodies - protons have mass...

To calculate the speed of COM? Where would this eq. come in handy?

4. Jul 19, 2013

### Simon Bridge

The total energy of a mass moving at speed v is $E=\gamma mc^2$ - it's just the Einstein mass-energy relation. You keep using $E^2=m^2c^4+p^2c^2$ instead. The kinetic energy is the total energy with the rest-mass energy $mc^2$ removed, thus, $W_k=(\gamma-1)mc^2$.

Thus $\gamma^2m^2c^4 = m^2c^4+p^2c*2$
... it can be a useful way to think about your problem in any reference frame.

If you do not have course notes on how to convert to center-of-mass in relativity, it is probably not a good idea to do so.

As it happens - in p-p scattering you get more particle sout than in, so the rest-energy is recovered and two of the particles are protons. This suggests that only the kinetic energy is available make new particles.

The folowing course-notes walks you through the ohysics:
http://galileo.phys.virginia.edu/classes/252/particle_creation.html

5. Jul 21, 2013

Thank you.