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Relativistic energies

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data
    Two neutrons A and B are approaching each other along a common straight line. Each has a constant velocity v as measured in the laboratory. Show that the total energy of neutron B as observed in the rest frame of neutron A is given by,

    [tex](1+\frac{v^2}{c^2})(1-\frac{v^2}{c^2})^{-1}m_pc^2[/tex]


    2. Relevant equations
    Either:
    [tex]E=\gamma mc^2[/tex]

    or

    [tex]E^2 = m^2c^4 + p^2c^2[/tex]


    3. The attempt at a solution
    I'm completely stuck on this. I think that the best way to get to the required answer is via the first of the two equations that I wrote down. However I can't manipulate the fraction into what is required. I get close I think, but its never quite what is required. I think the closest I get is by doing the following:

    [tex]E = \gamma mc^2 = \frac{mc^2}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}} * \frac{(1-\frac{v^2}{c^2})^2}{(1-\frac{v^2}{c^2})^2} = \frac{mc^2(1-\frac{v^2}{c^2})^2}{(1-\frac{v^2}{c^2})} = mc^2(1-\frac{v^2}{c^2})[/tex]

    From here I'm almost tempted to say that the bracket on the top is the difference of two squares, which would give me 2 brackets with the correct signs, but the v/c part wouldn't have the correct power, and the second bracket wouldn't be raised to the correct power either.

    Another thought I had is that the velocity of B from A's reference frame is 2v (although I'm not 100% convinced about this - it involved some odd hand waving on my part to get to this result!), but i assume that once I have the answer in the correct form I should be able to insert the correct value and the answer will fall out properly.

    If you should decide to help me I would appreciate an early step to help me on the way - I'd quite like the challenge of getting to the result by myself as much as possible. Its just I feel I may have gone wrong somewhere, or that I'm using the incorrect equations.

    Thanks for reading this - I hope you can follow my above working/train of thought!
     
  2. jcsd
  3. Apr 29, 2007 #2
    You are dealing with relativity. Therefore, you should abandon your Galilean ways of doing things. :wink:
     
  4. Apr 29, 2007 #3
    But Galilean is so much easier!!
     
  5. Apr 29, 2007 #4
    Try saying that to your teacher. :devil:
     
  6. Apr 29, 2007 #5
    From that do you mean that the relativistic velocity is [tex]\frac{2v}{1+\frac{v^2}{c^2}}[/tex]?
     
  7. Apr 29, 2007 #6
    Yup!

    10char
     
  8. Apr 29, 2007 #7
    Apart from that was I on the right lines?
     
  9. Apr 29, 2007 #8
    Yes you were.
     
  10. Apr 29, 2007 #9
    So its just a case of algebra - its looking messy so far, I don't seem to able to get things to cancel out just yet.

    Thank you for your help.
     
  11. Apr 29, 2007 #10

    robphy

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    Homework Helper
    Gold Member

    With rapidities and some knowledge of hyperbolic trig identities, the method of solution becomes more obvious and the answer falls out nicely.

    Here's a starting point:
    [tex]v_B=\tanh\theta[/tex]

    [tex]E_B=m\cosh\theta[/tex] in the lab frame. ([tex]E_A=m\cosh(-\theta)[/tex] in the lab frame.)
    [tex]E'_B=m\cosh (\theta +\theta)[/tex] in the A-frame. ([tex]E'_A=m\cosh(-\theta+\theta)=m\cosh(0)[/tex] in the A-frame.)
    Now write [tex]\cosh(2\theta)[/tex] in terms of [tex]\theta[/tex], then in terms of [tex]\tanh\theta[/tex].

    In SR, the velocity is not additive, but the rapidity is additive.
     
    Last edited: Apr 29, 2007
  12. Apr 29, 2007 #11
    Whats a rapidity? I can't seem to find anything on the web - I'm not likely to have used them before under a different name am I?

    Anyway, I seem to be having a few problems with my derivation - could someone take a look for me please?

    [tex]u^2 = (\frac{2v}{1+\frac{v^2}{c^2}})(\frac{2v}{1+\frac{v^2}{c^2}}) = \frac{4v^2}{1+ \frac{2v^2}{c^2} + {v^4}{c^4}}[/tex]

    right?

    Using this I can get to [tex]E = \frac{mc^2}{1-\frac{v^2}{c^2}}[/tex]

    but I can't see where the [tex]1+\frac{v^2}{c^2}[/tex] comes from. Apart from that I think I'm there.

    I realised that when I was doing it before I wasn't getting rid of the square root on the bottom of the fraction by multiplying top and bottom by the bracket squared. Instead this time I multiplied top and bottom by the denominator. Still doesn't explain that other mystery term - its just mystifying!

    Thanks for the info of rapidities - I think its beyond the scope of my course thus far, but I will look into it for my own personal information.
     
    Last edited: Apr 29, 2007
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