Particle Physics: Collision

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1. Oct 17, 2014

Matt atkinson

1. The problem statement, all variables and given/known data
Electrons of energy 5.00 GeV are incident upon target protons and a search is made for neutral particles X0 produced in the reaction

e− + p → e− + p + X0

What is the maximum mass of X0 which can be produced? Use any reasonable approximations, but justify your arguments.

[Mass of electron is 0.511 MeV/c2; that of proton is 938.3 MeV/c2.]
[hint: Consider C.O.M frame]
2. Relevant equations
$$W^2=(∑E_i)^2-(∑p_i)^2$$
Fact that the effective mass $$W^2$$ is invarient.
also $$E=(∑m_i)^2+(∑T_i)^2$$

3. The attempt at a solution
As the question asks for the maximum mass of X0 so I assumed all particles at rest after the collision occured.
So before the collision in the COM frame:
$$W=(m_e+m_p)+500GeV$$
As I assumed particles at rest after the collision then:
$$W^2=E^2-0$$
but from here I just get that $$m_{X^o}=500GeV$$ which Seems too easy... and wrong.
Could anyone just point me in the right direction?

2. Oct 17, 2014

Orodruin

Staff Emeritus
This relation is not correct. It is only correct if you remove the squares on the right hand side, otherwise it even has the wrong dimensions.

Where did this relation come from? I suggest you use the definition of W from your relevant equations. What is the momentum of an electron of energy 5 GeV? (It is also unclear from the question whether the 5 GeV are the total or kinetic energy of the electron.)

3. Oct 17, 2014

Matt atkinson

About the energy it just says that, I copied the question exactly.
sorry about the energy relation it was a typo.
Should be:
$$E=∑m_i+∑T_i$$ Because I did this from the COM frame in which $$∑p_i=0$$
Then my equation for W should simplify to, where I assumed 500GeV was the kinetic.
$$W=(m_e+m_p)+500GeV$$
As I see it now, there is a side note saying if the question doesnt explicitly say kinetic energy then It should be assumed to be total energy.
so that means:
$$W=m_p+500GeV$$
for
$$∑p_i=0$$
Is this correct?

4. Oct 17, 2014

Orodruin

Staff Emeritus
How do you know the kinetic energy is 500 GeV in the CoM frame? Note that the given energy is not in the CoM frame (and it is 5 GeV rather than 500 GeV).

I find the hint slightly misleading. I would not suggest doing everything in the CoM frame.

5. Oct 17, 2014

Matt atkinson

Okay, thats why i was confused seemed like it could be much easier in the frame where the proton is at rest, and use conservation of energy and momentum to work the mass of the X_o.
Thanks ill have a go at doing it now and post my results.

6. Oct 17, 2014

Orodruin

Staff Emeritus
Note: I am not saying the CoM frame is useless in the problem ... My hint would have been that invariants are nice things.

7. Oct 17, 2014

Matt atkinson

I tried this:
$$p_e^2=E^2-m_e^2$$
$$m_e=0.511MeV/c^2$$
$$p_e=4.999GeV\approx 5GeV$$
so then before
$$\Sigma p=0+5GeV$$
$$\Sigma E=5GeV+m_p$$
and after..
$$\Sigma p=0$$
$$\Sigma E=m_{X^o}+m_e+m_p$$
I just cant seem to see where im going wrong

8. Oct 17, 2014

Orodruin

Staff Emeritus

The electron energy is 5 GeV so the total energy before is 5 GeV + proton mass. Now the electron mass is negligible as you have seen so you will anyway not make a big error adding it in.

For your equations for after collision, you have used the CoM system so you cannot relate your before and after directly. Do you know some quantity that is frame independent?

9. Oct 17, 2014

Matt atkinson

I could use the effective mass W squared?
and work it out for both before and after?

10. Oct 17, 2014

Orodruin

Staff Emeritus
Yes, this would be my suggestion.

11. Oct 17, 2014

Matt atkinson

okay this is what I did (b denotes before, a denotes after):
$$W_b^2=E_b^2-p_b^2=E_a^2-p_a^2$$
$$as \ W_a=W_b$$
$$W=\sqrt{(5GeV+938.3MeV)2-(5Gev)^2}=3204MeV$$
neglecting the mass of the electron as m_e << m_p;
$$W^2=(m_p+m_{X^o})^2=m_{X^o}^2+2m_{X^o}m_p+m_p^2$$
$$m_{X^o}=(-(2*938.3) \pm 6404 )MeV$$
but this doesn't seem right

12. Oct 17, 2014

Orodruin

Staff Emeritus
And what possible roots does this correspond to? Which one is physical?

13. Oct 17, 2014

Staff: Mentor

There is no need for solving a quadratic formula. If W^2 = x^2 for some x and you know the sign of x, then you can simply take the root on both sides.

14. Oct 17, 2014

Matt atkinson

the + root as negative mass is not physical, I see now! thanks :D

15. Oct 17, 2014

Matt atkinson

Oh so i should just have done:
$$W=m_X+m_p$$
by square rooting both sides

16. Oct 17, 2014

Staff: Mentor

Yes.
The other approach is possible but way too complicated (and your result was not right).

17. Oct 17, 2014

Matt atkinson

Thankyou! i got:
$$m_X=2270MeV$$