Derive lorentz transform for energy

AI Thread Summary
The discussion focuses on deriving the relation E' = γ(E + βpc) using Lorentz transformations. It begins with the energy-mass equivalence E = mc² in a stationary frame and transitions to a moving frame, applying the Lorentz transform for momentum. A key point is the distinction between primed and unprimed momentum, which is crucial for the derivation. The solution involves recognizing that E' can be expressed as γm₀c², and simplifying using the Lorentz transformation for velocities. The final result is achieved by expanding γ and substituting the relevant terms.
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Homework Statement


Derive the relation:
E' = \gamma (E + \beta p c)

Homework Equations


p&#039; = \gamma p \\<br /> E^{2} = p^{2} c^{2} + M^{2}c^{4}

The Attempt at a Solution


start off with stationary frame S E=mc^{2}
then in moving frame S' E&#039;^{2} = p&#039;^{2} c^{2} + E^{2}:
lorent transform momentum:
E&#039;^{2} = \gamma^{2} m^{2} v^{2}c^{2} + E^{2}
and that's as far as I get!
 
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##E=\gamma mc^2##

That "p", in the equation you are supposed to prove, is unprimed.
Is that significant?
 
Yes the p is supposed to be unprimed.

I've solved it now. Thanks anyway.
Here it is for anyone who's having the same problem:
You start off (or derive it as I had to, to understand it) with the lorentz transform for velocities in two frames u = \frac{u&#039;+v}{1+frac{uv}{c^{2}}}
Know that E&#039;=\gamma m_{0}c^{2} because in the stationary frame S, only rest mass provides energy.
You expand out gamma with u' given above and recognise that p&#039;=\gamma p
simplify and you get an answer!
 
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