- #1
Rubber Ducky
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Homework Statement
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A particle of mass [itex]m[/itex] is moving in the [itex]+x[/itex]-direction with speed [itex]u[/itex] and has momentum [itex]p[/itex] and energy [itex]E[/itex] in the frame [itex]S[/itex].
(a) If [itex]S'[/itex] is moving at speed [itex]v[/itex], find the momentum [itex]p'[/itex] and energy [itex]E'[/itex] in the [itex]S'[/itex] frame.
(b) Note that [itex]E' \neq E[/itex] and [itex]p' \neq p[/itex], but show that [itex](E')^2-(p')^2c^2=E^2-p^2c^2[/itex]
Homework Equations
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[itex]E=p^2c^2+m^2c^4[/itex]
[itex]E'=\gamma mc^2=\frac{mc^2}{\sqrt{1-\frac{(u')^2}{c^2}}}[/itex]
[itex]p'=\gamma mu'=\frac{mu'}{\sqrt{1-\frac{(u')^2}{c^2}}}[/itex]
[itex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/itex]
The Attempt at a Solution
My prof says we need to use (a) to derive (b). It seems like it'd be easier a different way, but whatever. My strategy has been to show that [itex](E')^2-(p')^2c^2=m^2c^4[/itex], which would complete the derivation.
I tried playing around with some algebra and got [itex](E')^2=\frac{m^2c^4(c^2-uv)^2}{(c^2-u^2)(c^2-v^2)}[/itex] and [itex](p')^2=\frac{m^2(u-v)^2}{(1-\frac{uv}{c^2})^2(c^2-u^2)(c^2-v^2)}[/itex]
I got a little overwhelmed with the algebra at this point. I'd like to know if this is right so far, at least, before trying anything else.