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Homework Help: Particle energy and momentum in different reference frames

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle is observed in a certain reference frame (i.e. observer A) has a total energy of 5GeV and a momentum of 3GeV/c.
    (a) What is the energy of the frame in which its momentum is equal to 4GeV/c.

    2. Relevant equations

    [tex]E'={\gamma}E-{\gamma}{\beta}(cp)[/tex]
    [tex]cp'=-{\beta}{\gamma}E+{\gamma}(cp)[/tex]

    3. The attempt at a solution

    I plugged in cp'=4GeV, cp=3GeV, and E=5GeV, and solved the second equation for v. Then I plugged in v to the applicable variables in the first equation, and I got 4GeV, but the answer is 5.62GeV. And a different part of the question asks for the relative velocity of the two frames, and for that I get .6 c, as opposed to .19 c.
     
  2. jcsd
  3. Sep 28, 2010 #2

    vela

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    You might find the invariant m2 = E2-p2 useful.

    Also, the Lorentz transformations can be written in terms of the hyperbolic trig functions:

    [tex]\begin{align*}
    E' &= E\cosh \theta + p\sinh \theta \\
    p' &= E\sinh \theta + p\cosh \theta
    \end{align*}
    [/tex]

    The velocity is then given by [itex]\beta = \tanh \theta[/itex].
     
  4. Sep 28, 2010 #3
    Is the invariant some other form of the equation

    [tex]E^{2}=(cp^{2})+(mc^{2})^{2}[/tex]?
     
  5. Sep 28, 2010 #4

    vela

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    Yeah, it's the same equation with c=1.
     
  6. Sep 28, 2010 #5
    okay, i managed to get the velocity correct, but i still cant manage to get E' to come out correct
     
  7. Sep 28, 2010 #6
    when the question says "total energy", does it mean E^2=(cp)^2+(mc^2)^2 or does it mean E=ymc^2?
     
  8. Sep 28, 2010 #7

    vela

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    Either way should give you the same answer. Show us your calculations so we can spot your error.
     
  9. Sep 28, 2010 #8

    fzero

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    Both E's are the same since

    [tex] \sqrt{ (pc)^2 + (mc^2)^2} = \gamma m c^2. [/tex]

    I was actually able to get E' to work out, but found [tex]\beta = 0.92 c[/tex] which disagrees with the relative velocity you quoted.
     
  10. Sep 28, 2010 #9

    vela

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    I got β=0.186, so you should be able to get it to work out.
     
  11. Sep 28, 2010 #10
    i managed to get that for beta as well. but when i plug it into the transformation equations to find E', it doesnt come out to be correct
     
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