Recent content by jamesw1

  1. J

    Relativistic elastic proton collision

    I have zero intuition for the second equation (y transformation) but I did something $$ u'_x=\frac{u_x-v_x}{1-\frac{v_xu_x}{c^2}}=\frac{0+\beta c}{1}=\beta c $$ $$ u'_y=\frac{u_y\sqrt{1-\frac{v_x^2}{c^2}}}{1-\frac{v_xu_x}{c^2}}=\frac{\beta c\sqrt{1-\frac{\beta ^2c^2}{c^2}}}{1}=\beta...
  2. J

    Relativistic elastic proton collision

    Is this the correct 4-momentum? Probably not? $$ P_a=\left(\frac{E}{c},0\right)=\left(\frac{2\gamma _1mc^2}{c},0\right) $$ $$ P_b=\left(\frac{E}{c},\gamma _2mv\right)=\left(\frac{mc^2+\gamma _2mc^2}{c},\gamma _2mv\right) $$ $$ \gamma _1=\frac{1}{\sqrt{1-\beta ^2}} $$ $$ \gamma...
  3. J

    Relativistic elastic proton collision

    What is the momentum, then? Could anyone walk through a solution, please?
  4. J

    Relativistic elastic proton collision

    $$ P_a=\left(\frac{E}{c},0\right)=\left(\frac{2\gamma _1mc^2}{c},0\right) $$ $$ P_b=\left(\frac{E}{c},0\right)=\left(\frac{mc^2+\gamma _2mc^2}{c},0\right) $$ Where $$ \gamma _1=\frac{1}{\sqrt{1-\beta ^2}} $$ $$ \gamma _2=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} $$ v is the velocity we are trying to...
  5. J

    Relativistic elastic proton collision

    How should I approach this problem? My first thought was to subtract the velocity of the second proton from the velocity of proton going upwards. However, the velocity vectors are perpendicular to each other, therefore I cannot use the SR velocity addition formula.
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