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Relativistic Energy and Lorentz factor

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data

    Two particles of rest mass m0 approach each other with equal and opposite velocity v, in a laboratory frame. What is the total energy of one particle as measured in the rest frame of the other?

    But the question gives a clue which reads "if (v/c)^2 = .5, then E = 3m0c^2."

    3. The attempt at a solution

    My current understanding of this question is that the total energy is simply m0c^2 multiplied by the Lorentz factor. Could you explain to me how the clue is relevant?
  2. jcsd
  3. Oct 11, 2016 #2


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    Maybe the clue is to be used to check your answer for the special case when (v/c)2 = .5
  4. Oct 11, 2016 #3
    When (v/c)^2 = .5 then the equation is just that E = (1.414)(m0c^2). Where did the 3 come from?
  5. Oct 11, 2016 #4


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  6. Oct 11, 2016 #5


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    You are asked to find the energy of one of the particles according to a reference frame moving with the other particle. The statement (v/c)^2 = .5 specifies v for each particle in the lab frame. But you need to work in the frame of one of the particles.
  7. Oct 11, 2016 #6
    It's not a hint per se - it's just providing you with a particular test case to enable you to check if your obtained answer is correct.

    What would be a hint is: what is the velocity of the other particle as measured in the rest frame of one of the particles?
  8. Oct 11, 2016 #7
    Oh I see.
    Could you help me set up the equation in the particle's frame of reference?

    Before I can plug into E = y m0 c^2
    I need to find the m0 when the Lorentz factor is applied first because we are
    observing it from the frame of the particle?
  9. Oct 11, 2016 #8
    ##m_0## is unchanged - it is the rest mass of the particle. What is changed is the Lorentz factor - because the particles move with different velocities in different frames.
  10. Oct 11, 2016 #9
    Does that mean I need to first do velocity transformation
    u = (u' + v) / (1 + u'v/c^2)

    And use the resulting velocity to plug into
    E = y m0 c^2 ?
  11. Oct 11, 2016 #10
    Yes, that is correct - but it would be good if you try not to view it as doing some stuff and then plugging into a formula, because that often hinders with the understanding of the concepts.
  12. Oct 11, 2016 #11
    I see. Thank you!
  13. Oct 11, 2016 #12
    Last edited: Oct 11, 2016
  14. Oct 11, 2016 #13
    Perhaps it would help if you show your final expression for the energy. The transformed velocity ##v'## is a function of ##v##, so your final answer will / should be in terms of ##v##, because well, that is your given initial parameter.
  15. Oct 12, 2016 #14
    u' = [u - (-u)] / [1 - (u)(-u)/c^2] = 2u / [1+(u/c)^2

    Plugging the u' into Lorentz factor for the Energy equation:

    E = y m0 c^2 = m0 c^2 / [1 - ( [2u / [1+(u/c)^2] / c )^2] ^ -1/2

    And with further simplification I get m0 c^2 / (1/3) = 3

    Got it. Thanks!
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