Recent content by jamesw1

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...

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...

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

$$ P_a=\left(\frac{E}{c},0\right)=\left(\frac{2\gamma _1mc^2}{c},0\right) $$ $$ P_b=\left(\frac{E}{c},0\right)=\left(\frac{mc^2+\gamma _2mc^2}{c},0\right) $$ 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...

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.