- #1

latitude

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## Homework Statement

Derive the compton equation.

## Homework Equations

[tex]\lambda[/tex]` - [tex]\lambda[/tex] = h/ mc (1 - cos[tex]\theta[/tex])

E = hf = hc/[tex]\lambda[/tex]

## The Attempt at a Solution

Okay, I'm sorry this is so long, I'll try and make it as concise as it is possible for a whole blather of random crap to be :]

Conservation of momentum components:

h/[tex]\lambda[/tex] = h/[tex]\lambda[/tex]`(cos[tex]\theta[/tex]) + Pe(cos[tex]\psi[/tex])

0 = h/[tex]\lambda[/tex]`(sin[tex]\theta[/tex]) - Pe(sin[tex]\psi[/tex])

After some combining, squaring, and the like (getting rid of [tex]\psi[/tex]):

Pe

^{2}= (h/[tex]\lambda[/tex])

^{2}- (h/[tex]\lambda[/tex]`)

^{2}cos

^{2}[tex]\theta[/tex] + (h/[tex]\lambda[/tex]`)

^{2}sin

^{2}[tex]\theta[/tex] - (h/[tex]\lambda[/tex])(h/[tex]\lambda[/tex]`)cos[tex]\theta[/tex]

E

^{2}= p

^{2}c

^{2}+ E

_{R}

^{2}

So

P

^{2}= (E

^{2}- E

_{R}

^{2})/c

^{2}

So I plug that into my momentum (I'm not going to write the righthand side of the equation while i show what I did w/ that)

(E

^{2}- E

_{R}

^{2})/c

^{2}= ...

((hc/[tex]\lambda[/tex])

^{2}- (mc

^{2})

^{2})/c

^{2}= .

I tried to get rid of the denominator 'c'...

(h/[tex]\lambda[/tex])

^{2}- m

^{2}c

^{2}= ...

(m

^{2}c

^{2}[tex]\lambda[/tex])/h = [tex]\lambda[/tex]/[tex]\lambda[/tex]` - h/[tex]\lambda[/tex]`cos[tex]\theta[/tex]

After some more fiddling I get to this:

[tex]\lambda[/tex]` = h/m

^{2}c

^{2}- (h/[tex]\lambda[/tex])(h/m

^{2}c

^{2})cos[tex]\theta[/tex]

It's kind of close but not really... I can write out all the steps I made if that is necessary, but I'm kind of hoping I made one nice, simple-to-fix error that is glaringly obvious to the more experienced :)

Thank you :)