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Special Relativity Question

  • #1
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Homework Statement


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


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
UltrafastPED
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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.
 
  • #3
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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.
 
  • #4
UltrafastPED
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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.
 
  • #5
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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
 
  • #6
UltrafastPED
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No, it is exact.
 
  • #7
Ibix
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...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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