Basic high school algebra, with physics

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SUMMARY

This discussion focuses on solving a problem involving a head-on elastic collision between a massless photon and a stationary free electron, utilizing principles of conservation of energy and momentum. The equations relevant to the problem include E=pc for massless particles and E=γmc² for particles with mass. The user attempts to derive the final momentum of the photon after the collision, denoted as p, while grappling with the complexities introduced by the Lorentz factor γ. The solution requires substituting momentum expressions and solving algebraically to isolate p.

PREREQUISITES
  • Understanding of basic algebraic manipulation
  • Familiarity with the principles of conservation of energy and momentum
  • Knowledge of special relativity concepts, particularly the Lorentz factor γ
  • Basic understanding of photon and electron interactions in physics
NEXT STEPS
  • Study the derivation of the Lorentz factor γ in special relativity
  • Learn about elastic collisions in quantum mechanics
  • Explore conservation laws in relativistic physics
  • Practice solving problems involving momentum and energy conservation in particle physics
USEFUL FOR

Students studying high school physics, particularly those interested in the principles of relativity and particle interactions, as well as educators looking for examples of elastic collisions involving photons and electrons.

forrealfyziks
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"basic high school" algebra, with physics

Homework Statement


Consider a head-on, elastic collision between a massless photon (momentum po and energy Eo) and a stationary free electron. (a) Assuming that the photon bounces directly back with momentum p (in the direction of -po) and energy E, use conservation of energy and momentum to find p.


Homework Equations


E=\gammamc2
p=\gammamu
massless: E=pc
rest mass: E=mc2
E2=(pc)2+(mc2)2
v/c=pc/E
\gamma=1/\sqrt{1+(v/c)^2}

The Attempt at a Solution


Note:First of all I know that this is relativity, but it boils down to just plain algebra. I can't figure it out and help is hard to find, so if you can help I would really appreciate it.

I assume that p is the momentum of the electron. m=mass of the electron u=velocity of the electron c=speed of light

conserving energy: poc+mc2=pc+\gammamc2
po+mc=p+\gammamc
po=p+\gammamc-mc

conserving momentum: po=p-p=\gammamu-p

Plugging the result I got in conserving energy into the momentum equation:
p-p=p+\gammamc-mc
p=2p+mc(\gamma-1)
 
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The problem with the last line is that gamma has the speed of the electron (an unknown quantity) buried in it. Forget gamma. Write the energy conservation equation as

p_{0}c+mc^2=pc+\sqrt{(p_ec)^2+m^2c^4}

where pe is the final momentum of the electron.

Use the momentum conservation equation to replace pe with what it is equal to, then solve for p.
 

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