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Relativity - Fast and slow moving objects

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data
    (a) A particle is moving so fast with respect to an observer that the γ factor of its rest frame with respect to the observer’s rest frame is 108. By how much is the velocity of the particle less than that of light?

    (b) A particle is moving very slowly with respect to an observer (β << 1). Find an approximate expression for γ up to order β2 using a binomial expansion.

    (c) If the mass of a particle is m and its velocity is v = βc, then its total energy, which includes its rest energy, mc2 and its kinetic energy, is E = γmc2. Using the same technique as in question (b), find an approximate expression for the difference between its total energy and its rest energy in the case where v << c.


    2. Relevant equations

    γ=1/√(1-β2) [1]
    β=v/c [2]

    3. The attempt at a solution

    I tried to use eq [1] for part (a), but obviously that just leaves me with v=c, which doesn't work as it leaves me with 1/0 = 108

    For (b) I'm not sure how to use binomial expansion when x (or β in this case) is squared inside the brackets.
     
  2. jcsd
  3. Dec 5, 2013 #2

    phinds

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    Unitless numbers are useless in this context. Also, the observers rest frame is not moving at all relative to the observer, so multiplying zero by your 10**8 still gives you zero.
     
  4. Dec 5, 2013 #3

    Doc Al

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    Why do you think it leaves you with v = c?
     
  5. Dec 5, 2013 #4

    Doc Al

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    Gamma may be unitless, but it's certainly not useless.
     
  6. Dec 5, 2013 #5
    Part (a) leaves you with v = c only if you didn't do it right. Try squaring both sides and solving algebraically for β2. Then take the square root of both sides. One of the terms under the square root sign will be very small. You need to figure out how to approximate the effects of this term mathematically.
     
  7. Dec 5, 2013 #6

    phinds

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    Ah ... I misunderstood. I thought he was referring to the Y-axis speed.
     
  8. Dec 5, 2013 #7

    vela

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    It's nothing complicated. Just expand $$\frac{1}{\sqrt{1-y}}$$ and then substitute ##y=\beta^2##.
     
  9. Nov 28, 2015 #8
    So let's come back to this question from 2 years ago ! (I'm now stuck on the same questions)

    For (b) we use the binomial expansion, therefore $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
    Which gives us $$\frac{1}{\sqrt{1-\beta^2}} = (1-\beta^2)^\frac{-1}{2}$$
    And with binomial expansion we obtain $$\gamma=1+\frac{1}{2}\beta^2$$
    Correct?
    However I'm struggling with part (c) of the question.
    We have $$E_{total} = \gamma mc^2 = mc^2+KE = mc^2+\frac{1}{2}mv^2$$
    Where do I go from there?
     
  10. Nov 28, 2015 #9

    nrqed

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    The question asks to calculate ##\gamma m c^2 - mc^2 ## in the non relativistic limit (which gives the usual non relativistic kinetic energy).
     
  11. Nov 28, 2015 #10

    Doc Al

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    Yes.

    Well, what's the rest energy?
     
  12. Nov 28, 2015 #11
    The rest energy is just ##mc^2##
    So it's asking ##\gamma mc^2-mc^2=\frac{1}{2}mv^2##
    Where ##m## can cancel out
    leaving ##\gamma c^2-c^2=\frac{1}{2}v^2##

    I don't understand where the binomial expansion comes in to be honest...
     
  13. Nov 28, 2015 #12

    Doc Al

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    Don't cancel the m.

    The same place it came in in part b: Deriving an approximation for gamma.
     
  14. Nov 28, 2015 #13
    ##\gamma mc^2-mc^2 = \frac{1}{2}mv^2 \rightarrow \gamma mc^2 -mc^2=\frac{1}{2}m\beta^2c^2##
    where ##\gamma = \frac{1}{2}\beta^2+1##
    Therefore ##\frac{1}{2}\beta^2=\gamma-1##

    Which means that $$\frac{1}{2}mv^2=\frac{1}{2}m\beta^2c^2=mc^2(\gamma-1)$$
    $$\gamma mc^2-mc^2 = mc^2(\gamma-1)$$
    Is that correct?
    If so thank you for all the help, the helpers on this forum don't receive enough praise for everything they do, and I'm so grateful !
     
  15. Nov 28, 2015 #14

    Doc Al

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    I think you're overthinking that last question. All you needed to do is get an expression for the difference between the total energy and its rest energy, and realize that it is simply the usual low-speed expression for kinetic energy. You showed that already!
     
  16. Nov 28, 2015 #15

    Doc Al

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    When you wrote:
    I should have said: Done! (That's all you needed to do.)
     
  17. Nov 28, 2015 #16
    Time to close this thread???

    Chet
     
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