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Show that the total relativistic energy of a proton

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data

    The mass of a proton when at rest is m. According to an observer using the
    detector frame, the speed of the anticlockwise moving bunch, A, is such that
    va^2/c^2=24/25
    Show that the total relativistic energy of a proton in bunch A, as observed in
    the detector frame, is exactly 5mc^2, and work out the speed of the proton,
    expressed as a decimal multiple of c, (to 5 significant figures).

    2. Relevant equations
    Right I think its these
    E=mc^2
    Etot=mc^2/√1-v^2/c^2
    Etrans=mc^2/1-v^2/c^2
    and maybe
    E^2tot=p^2c^2+m^2c^4
    p=mv/1-v^2/c^2


    3. The attempt at a solution
    Now I know that to get Etot you need e=mc^2 and Etrans and so maybe combing these equations will give the answer but I think this IS not the right to get 5mc^2.
    So maybe its E^2tot equation I have to use.
    Im just not sure.
     
  2. jcsd
  3. Mar 26, 2012 #2
    well if [itex]\frac{v^{2}}{c^{2}}=\frac{24}{25}[/itex]

    and the mass of the proton is m, and Etot = mc2[itex]\gamma[/itex]

    where [itex]\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

    then why don't you just try plugging the values in?

    I'm not sure what Etrans is supposed to be


    and to solve for the velocity of the proton all you need is [itex]\frac{v^{2}}{c^{2}}=\frac{24}{25}[/itex]
     
  4. Mar 26, 2012 #3
    Yeah getting that wrong about Etrans, just looked at my text books and was reading it wrong . I have an asnwwer for the last bit not sure if rights but this is what I have for that.
    24/25 x 3.00 x10^8 =2.88000 x 10^8 ?

    So are you saying just add the values for a proton = 1.67 x 10^-27
    the speed of light 3.00 x 10^8
    and then add 24=v and 25=c into the gamma part
     
  5. Mar 26, 2012 #4
    if [itex]\frac{v^{2}}{c^{2}}=\frac{24}{25}[/itex]

    then [itex]v=c\sqrt{\frac{24}{25}}[/itex]

    and just leave it as a multiple of c


    for the first part all you need to write is

    [itex]E_{tot}=\frac{mc^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

    and then substitute [itex]\frac{v^{2}}{c^{2}}[/itex] for [itex]\frac{24}{25}[/itex]

    you should be able to do it in your head, no need to put in the mass of the proton or the exact speed of light, since the answer you want is just 5mc2 where m is the mass of the proton and c is the speed of light
     
  6. Mar 26, 2012 #5
    Thank you very much I have now , its taken two days for that to sink in !
     
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