How to Calculate Invariant Mass for Negligible Particle Mass?

[tnibbz]

Homework Statement

A 2 GeV electron is incident on a proton with rest mass mpc^2 = .938 GeV at rest. Calculate the invariant mass of the electron proton pair in the center of mass, neglect the mass of the electron.

Relevant Equations

E^2 = p^2*c^2 + m^2*c^4
E = K + E0

So I know that since we are ignoring the mass of the electron, and the proton starts at rest, the proton has no KE and the electron has no rest energy.

So the initial total energy of the system would be

rest energy of proton + KE of electron = 2GeV + .938GeV = 2.938 GeV

and since energy is conserved this would also be the total energy of the final state as well correct?

Now with the equation
E² = p²*c² + m²*c⁴

I can subsitute in 2.938GeV for E, however from here I get confused, because I need both the resulting velocity of the system and the mass.

Or is the mass of the system in this case just the mass of the proton because we are neglecting the mass of electron? But in that case would the invariant mass of the system just be the mass of the proton?

 

[Orodruin]

Homework Statement:: A 2 GeV electron is incident on a proton with rest mass mpc^2 = .938 GeV at rest. Calculate the invariant mass of the electron proton pair in the center of mass, neglect the mass of the electron.
Homework Equations:: E^2 = p^2*c^2 + m^2*c^4
E = K + E0

because I need both the resulting velocity of the system and the mass.

Why would you need the velocity? There is no velocity in that equation.

 

[tnibbz]

well doesn't p=ϒmv?

 

[Orodruin]

That is irrelevant. You already know p.

 

[tnibbz]

Oh because the momentum is conserved? So the final momentum p would be equal to the initial momentum of the system, since the proton was at rest and the electron is essentially massless is the momentum 0? So I would just plug in the E I found earlier and 0 into the equation

E² = p²*c² + m²*c⁴
and solve for m?

 

[Orodruin]

That the electron is massless does not mean momentum is zero. What is the momentum of a massless particle?

 

[tnibbz]

well if it is not zero than it cannot be p= ϒmv. Could we treat it like a photon and use E=pc?

 

[Orodruin]

Could we treat it like a photon and use E=pc?

$E^2 = (pc)^2 + m^2c^4$ is true for all systems. If you just consider a massless particle, you get $E=pc$ so this is true for any massless particle. In general, $E^2 = (pc)^2 + m^2c^4$ is a much more useful relation than $p = m\gamma v$.

 

[tnibbz]

Okay that makes sense, I did not realize that was where the E = pc formula was derived from, for some reason I thought it only held for photons. I believe I can solve the problem from here, thank you for all your help.

 

[PeroK]

Okay that makes sense, I did not realize that was where the E = pc formula was derived from, for some reason I thought it only held for photons. I believe I can solve the problem from here, thank you for all your help.

If $E^2 = p^2c^2 + m^2c^4$, then for a particle whose mass can be neglected, compared to its energy, we have:

$E^2 \approx p^2c^2$

That equation only hold exactly for massless particles. But, it is also a sufficient approximation for particles with small mass.