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Relativistic momentum and kinetic energy

  1. Sep 19, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the kinetic energy of an electron with a momentum of 40 GeV/c?

    3. The attempt at a solution

    Kinetic energy involves velocity of the particle so my first thought was to write momentum in terms of velocity.

    [tex] p = \frac{mv}{(1-(v/c)^2)^{1/2}} [/tex]

    [tex] p^2 = \frac{(mv)^2}{(1 - (v/c)^2)} [/tex]

    [tex] p^2 - p^2v^2/c^2 = m^2v^2 [/tex]

    [tex] p^2 - v^2(p^2/c^2 - m^2) = 0 [/tex]

    [tex] v = \frac{p}{(p^2/c^2 - m^2)^{1/2}} [/tex]

    From my understanding the statement is saying the electron's momentum is (40/c) GeV. Since 1 GeV = 1.602 x 10^(-10) J, I calculated momentum as (40)(1.602 x 10^(-10))/(c). This cancels the (m/s) unit from energy and now the value has units of momentum.

    After looking up the mass of an electron and plugging in the numbers to find v, I got v = c, which is from my understanding impossible for an electron. The electron would have infinite kinetic energy.

    What's off here? Would appreciate any insight.

    EDIT:

    I just realized I made a sign error. Mass in the final equation should have been + and not -.

    Redid calculations though and v still is equal to c. Could this just be a problem where the actual number v is just ridiculously close to c and my calculator is rounding it up? Or am I just doing something plain wrong?
     
    Last edited: Sep 19, 2014
  2. jcsd
  3. Sep 19, 2014 #2

    Orodruin

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    The mass of the electron is very small in comparison to the given momentum and it should be travelling very close to c. With your sign correction, your formula for the speed is fine. However: You do not need to compute he velocity to compute the kinetic energy. Try instead to answer the following questions: What is the total energy of the electron? How is this related to the kinetic energy?
     
  4. Sep 19, 2014 #3
    Hey Orodruin, you're right, I had completely forgotten about

    [tex] E^2 = p^2c^2 + E_0^2 [/tex]

    I guess since I've only recently learned it, it's not sticking out to me in my mind compared to other more grounded ideas. I guess the electron's speed is just nearly c and my calculator's just showing me c then. Thanks for the help!
     
    Last edited: Sep 19, 2014
  5. Sep 19, 2014 #4

    Orodruin

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    Just to make sure you got it right: What do you get for the kinetic energy in the end?
     
  6. Sep 19, 2014 #5
    I got about 6.408 x 10^-9 J. If the number sounds wrong, I calculated total energy E from

    [tex] E^2 = p^2c^2 + (mc^2)^2 [/tex]

    Then I used the relationship between total energy, rest energy, and kinetic energy to find kinetic energy.

    [tex] K = E - mc^2 [/tex]
     
  7. Sep 19, 2014 #6

    Orodruin

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    So this is what I was really after, the actual number is less important than the conceptual understanding.

    On a side note: I would have given the answer in terms of eV. On a particle physics scale it is a much more convenient unit. Since the electron mass is completely negligible in the problem, I would have approximated ##K \simeq E \simeq pc## right from the start and answered 40 GeV, which should be in the ballpark of the result you obtained.
     
  8. Sep 19, 2014 #7

    Vanadium 50

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    It's a pity you removed template element #2, because one equation that can help you is [itex]p = \beta \gamma m[/itex].
     
  9. Sep 19, 2014 #8

    Orodruin

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    It is however the first equation under his template element #3 and to be honest it is a significant detour which involves gamma factors and velocities unnecessarily.
     
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