Special Relativity Question

PsychonautQQ · Oct 13, 2013

1. The problem statement, all variables and given/known data
If an electron and a Positron are said to each have 50GeV of energy (8E-9J) then how do I find their velocity? Do I use E^2=(pc)^2+(mc^2)^2 and solve for velocity in the momentum? Is there an easier way to do this? I did this, I might have done the algebra wrong but I got an answer that was basically zero, lol. Any advice is nice :D

2. Relevant equations

3. The attempt at a solution

UltrafastPED · Oct 13, 2013

You need to find the Lorentz factor, gamma; you can invert this to find the velocity.

Since you have the total energy, E=50GeV note that this is the sum of the rest energy (E0=mc^2) plus the work done on the electron. This is equivalent to E/E0 = gamma - 1; this is the gamma that you need.

PsychonautQQ · Oct 13, 2013

so E = E0gamma - E0...? Energy = (What is this term?) - Rest Energy? How did you derive this? Did you mean to put an addition sign there at all and then E0gamma would represent the kinetic energy or something? If E is the total energy this equation means E0gamma is greater than it, so what could it possibly be?

Also how is this equation related to
E^2 = (pc)^2 + (mc^2)^2 ??
It looks like it's more related to the nonrelativistic approximation of
E = K + mc^2, which would mean it's wrong because I think I'm suppose to be using the actual relativistic equation for this problem.

UltrafastPED · Oct 13, 2013

Explanation: if E0=mc^2 is the rest energy then the total energy is gamma*E0. Then the kinetic energy is:

E_kinetic = gamma*E0 - E0. So (gamma-1) = E_kinetic/E0 which gives provides a value for the Lorentz factor.

You can derive this from E^2 = (pc)^2 + (mc^2)^2.
See http://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies

Is the 50GeV the total energy, or the total energy? It could be either - but for an electron the kinetic energy is the work done on the electron - so if you have placed an electron in a 50 GV electric potential it will have 50 GeV when it reaches the other side; they actually do this in stages for such high energies. Anyway E0 for an electron or positron is about 0.511 MeV, so the difference is insignificant.

So (gamma-1)=50 GeV/0.511 MeV ~= ~100,000. This implies v=.999 c, where you will have to figure out how many nines! 20 keV gives ~c/4 and 30 keV gives ~c/3 which are the speeds I usually work with.

PsychonautQQ · Oct 13, 2013

I thought E_kinetic = Gamma*E0 - E0 was only an approximation of E^2 = (pc)^2+(mc^2)^2 in nonrelativistic situations?? thanks btw :D

UltrafastPED · Oct 13, 2013

No, it is exact.

Ibix · Oct 14, 2013

...however, this expression is on the usual route to deriving the Newtonian formula for kinetic energy. If you Taylor expand the gamma and disregard terms of order v⁴ and higher, out drops mv²/2. That is an approximation, and may be what you're thinking of.

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Special Relativity Question

PsychonautQQ

UltrafastPED

PsychonautQQ

UltrafastPED

PsychonautQQ

UltrafastPED

Ibix

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