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Homework Help: Special Relativity help

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data
    What is the energy of a photon whose momentum is the same as that of a proton with a kinetic energy of 10 MeV?

    2. Relevant equations
    K = mc[tex]^{2}[/tex]([tex]\gamma[/tex]-1)
    p = [tex]\gamma[/tex]mv
    E = pc

    3. The attempt at a solution
    I figured I would go at it like this. I used the first equation listed to obtain a value for gamma and v. I put these into the second equation along with the mass of a proton. Then, I would put that value into the third equation. Is that even close?
  2. jcsd
  3. Jan 29, 2010 #2


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    Seems ok if I understood it right. Your equations might help more.
  4. Jan 29, 2010 #3
    As far as I can understand, the procedure you describe is correct. As long as you have done the calculations properly, you should end up with the correct answer.

  5. Jan 29, 2010 #4
    Ok, so I have [tex]\gamma = 1 + 1.18 \times 10^{-19}[/tex] So, um, how am I supposed to square that (so I can find beta)? My calculator won't go that far... :*)
  6. Jan 29, 2010 #5
    use an online calculator
  7. Jan 29, 2010 #6
    or any computer program will do it
  8. Jan 29, 2010 #7
    or you can just square the 1.18 and then double 19
  9. Jan 29, 2010 #8
    They all give me 1
  10. Jan 29, 2010 #9
    the answer is 1.3924e-38
  11. Jan 29, 2010 #10
    so for next time this is what you will do. you should square 1.18. then you should double the exponent. and voila
  12. Jan 29, 2010 #11
    You're misreading the problem. If it was just 1.18e-19, I wouldn't have a problem. I need to square 1+1.18e-19. BTW, I found a calculator that will do it. Google can be tricksy sometimes...
  13. Jan 29, 2010 #12
    if [itex]\gamma=1+1.18\times10^{-19}[/itex] then you should just approximate [itex]\gamma\simeq1[/itex].

    Also, use www.wolframalpha.com It's like having Mathematica available to you with 0 cost
  14. Jan 29, 2010 #13


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    Your value for [itex]\gamma[/itex] is way too small. The proton's kinetic energy is about 1% of the rest energy, so [itex]\gamma[/itex] should be about 1.01.
  15. Jan 30, 2010 #14
    The speed of light = 299,792,458 m/s;
    Proton mass: [tex]1.67262158*10^{-27} kg[/tex];
    [tex]K= 1.602176487*10^{−15} J[/tex].

    So with a simple mind-based calculation you at least can get that the Lorentz factor is around 1. The following shows its exact value up to 9 decimal digits:


  16. Jan 30, 2010 #15
    This is such a complicated way to do this problem. The energy of a proton is its kinetic energy plus its mass, and that squared is p^2 + m^2 (with c=1). Find p. E(photon) = p. Done.
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