Lorentz Transform Homework: Particle Motion Toward Origin

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SUMMARY

The discussion focuses on calculating the energy E' of a particle moving toward the origin in a system S, using Lorentz transformations. The relevant equations include E^2=(mc^2)^2 + (pc)^2, p=gamma*mv, and E'/c=gamma(E/c-BetaPx). The key steps involve determining the momentum component Px using p_x = gamma m v_x and finding the velocity v of the frame S' such that the particle moves along the y'-axis. The final expression for E' is derived from substituting the calculated values into E'/c=gamma(E/c-BetaPx).

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Homework Statement


Suppose that a particle of mass m and energy E is moving toward the origin of a system S such that its velocity u makes an angle alpha with the y-axis (approaches origin from upper right). Using the Lorentz transformations for energy and momentum, determine the energy E' of the particle measured by an observer in S', which moves relative to S such that the particle moves along the y' -axis.

Please help, I can visualize how the problem is set up, but I don't know how to work it. Thank you.


Homework Equations


I don't know if these are relevant, but:
E^2=(mc^2)^2 + (pc)^2
p=gamma*mv
E=gamma*mc^2
Px'=gamma(Px-(Beta*E/c))
Py'=Py
Pz'=Pz
E'/c=gamma(E/c-BetaPx)


The Attempt at a Solution


I'm not really sure how to apply the equations to the problem. Please help.
 
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There seems to be more information given than necessary... given the mass m and velocity u of the object, we can get its energy... but they've given that as extra... energy is E...

do they say what they want the answer in terms of?

Anyway, this is the equation you need:

E'/c=gamma(E/c-BetaPx)

once you know px... and the velocity of the S' frame, then you can plug into the above...

[tex]\vec{p} = \gamma m\vec{v}[/tex]

so

[tex]p_x = \gamma m v_x[/tex]

so use the above to calculate px for the object... [tex]\gamma = \frac{1}{\sqrt{1-\frac{u^2}{c^2}}}[/tex]

then find the velocity v of the frame S'.

they say that the object should be moving along the y-axis in the S' frame... that means that u'x = 0.

using your velocity transform equations... what velocity should v be in terms of ux... so that u'x = 0

once you have that v... you can get beta = v/c... and [tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] (careful, this gamma is different from the one above used to calculate px)

then you can plug into:

E'/c=gamma(E/c-BetaPx)

and get E'.
 

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