The problem statement:

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? There are as many as possible new particles.

Relevant equations here are: \begin{align} E_{before}&=E_{after}\\ p_{before}&=p_{after}\\ E^2 &= {E_0}^2 + p^2c^2 \longleftarrow \substack{\text{Lorentz invariant}} \end{align}

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

At this point I am not sure what to do with the last part $p^2c^2$. The only thing I came up with was to set:

\begin{align} pc &= \sqrt{{E_{k~before}}^2 - 2E_{k~before}E_{0~before}}\\ pc &= \sqrt{\left(E_{k1} + E_{k2}\right)^2 - 2\left(E_{k1}+E_{k2}\right)2E_{0p}}\\ \end{align}

If I continued with this calculation I got $W_{0m}=1.87GeV$ while in the solutions it is said to be $W_{0m}=5.634GeV$. Where did I go wrong?

  • 1
    $\begingroup$ Do the protons collide heads on? Anyway, in this case I'm pretty sure moving to COM frame will simplify things dramatically! $\endgroup$ – Ali Jul 18 '13 at 19:15
  • $\begingroup$ Also, how many particles are made? are they identical? $\endgroup$ – Ali Jul 18 '13 at 19:16
  • $\begingroup$ I don't know how many they are as it is not specificaly said. I only need to calculate the rest energy of all of them. In these types of problems i have never used COM before. If i write the Lorentz invariant for COM i can assume $p = 0$ right? so i get: $ W_{k1} + W_{k2} + 2W_{0p} = \sqrt{\left( 2W_{0p} + W_{0m} \right)^2 + 0}$ Is this the simplification you were talking about? Is this correct? $\endgroup$ – 71GA Jul 18 '13 at 19:19
  • $\begingroup$ It really depends on what happens. The question is still ambiguous to me, do the protons collide heads on? Do they stick together after collision, i.e. all the particles move with the same speed(honestly this is what I expect to be asked, otherwise there are some free parameters which have to be fixed)? $\endgroup$ – Ali Jul 18 '13 at 19:28
  • 1
    $\begingroup$ Also, it's good practice if you describe your notation when it's necessary(e.g. for $W$'s). $\endgroup$ – Ali Jul 18 '13 at 20:15

As I said in the comments it really depends on what happens. Here I will assume the protons are colliding heads-on and after collision, all the particles will move with the same speed(so the square of their total momentum will be minimum).

$$E_1=\sqrt{m_p^2c^4+p_1^2 c^2} \\ E_2=\sqrt{m_p^2c^4+p_2^2 c^2} \\ \Rightarrow E=E_1+E_2$$

After collision, as I stated the momentum will be $p_1-p_2$(heads-on collision), so we will have:

$$E^2- \left(p_1 -p_2\right)^2c^2=m_{after}^2c^2=(2m_p+m)^2c^4 \\ \Rightarrow mc^2= \sqrt{E^2-(p_1c-p_2c)^2}-2m_p c^2$$

which I calculated to be $4.1 \text{GeV}$. I think this shows the spirit of solutions to these type of questions, although its details depends on what is exactly stated in the problem.


Presumably what you want to do is think in the center-of-momentum frame. In this frame, there is no net momentum, just energy. So presumably you want to convert to a system of particles at rest, two of which are protons and the rest of which are unspecified. How much energy is left over to go into the unspecified other stuff?

Although I think it's most convenient to think about this in the center-of-momentum frame, you can also do the calculation in another frame.

I don't really understand your notation, and it's kind of odd to talk about "kinetic energy" for a relativistic particle, but presumably it means the total energy $E = mc^2 + W$ where $W$ seems to be your notation for "kinetic energy." In that case, what are the momenta of the two particles? What is the total momentum in the frame you're working in? The total energy? The invariant mass?

  • $\begingroup$ I fixed my notation and used $E$ for energy $E_k$ for kinetic energy and $E_0$ for rest energy. I hope this helps. $\endgroup$ – 71GA Jul 18 '13 at 20:31
  • $\begingroup$ I think your mistake is with your calculation of $pc$. Try answering the questions I asked, starting with: what is the momentum of each particle at the beginning? $\endgroup$ – Matt Reece Jul 18 '13 at 21:35
  • 1
    $\begingroup$ I calculated both momentums to be $p_1=4.84Gev/c$ and $p_2=2.78GeV/c$. I calcalated the sum to be $p=p_1-p_2=2.06Gev/c$ this will be the same before/after collision if I am not wrong. So this is the start :) $\endgroup$ – 71GA Jul 19 '13 at 8:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.