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B Energy relativity derivation

  1. Jan 25, 2017 #1
    Hi community,
    There is a step on the wiki page about a derivation.
    IMG_1485326064.992430.jpg
    I don't seem to get the same thing when I try and solve for (u/c)^2 and then plug into Lorentz factor to get the new Lorentz factor so you can get the full energy momentum relation

    My gamma term looks more like 1/ ((mo^2u^2/p^2))^-1/2 when i've subbed for (u/c)^2??

    Can anyone help or any other resources to show easy as poss how to derive the full equation?
    Thanks
    G.
     
  2. jcsd
  3. Jan 25, 2017 #2

    Orodruin

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    If you have a u left in your expression you have not solved for (u/c)^2 ...
     
  4. Jan 25, 2017 #3
    I rearranged the top expression and made (u/c)^2 the subject but then you still have the u^2 that was in the numerator of the momentum expression? Confused.
     
  5. Jan 25, 2017 #4

    Orodruin

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    Yes, so you have not solved for u^2/c^2. You need to actually solve for u^2/c^2.

    Edit: Hint: Solve for u^2. The speed of light is just a constant.
     
  6. Jan 25, 2017 #5
    IMG_1485332435.532491.jpg
    I got this far and became a bit convoluted. Will try for u^2 then.
    Thanks.
    G.
     
  7. Jan 25, 2017 #6

    Orodruin

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    Hint:
    $$
    1 = \frac{m^2 c^2 + p^2}{m^2 c^2 + p^2}
    $$
    ...
     
  8. Jan 25, 2017 #7
    Ooh, that's cheeky. I'll have another play this evening, thanks Orodruin.
     
  9. Jan 25, 2017 #8
    Got it finally, quite satisfying ...
     
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