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When attempting to prove the formula for Compton Scattering, I obtained the following equation(I have omitted the details on how the equation is obtained as it is not relevant to this post):

$$f = f' + k(f^2 - 2ff'\cos\theta +f'^2)$$

where $~k~$ is a constant (I again omitted the physical meaning of the variable as it is not relevant).

The formula that I would like to obtain is the following: $$\lambda' - \lambda = \frac{k}{c}(2-2\cos\theta)$$ where $\lambda= \frac{c}{f}$.

If working backward, we can also obtain an expression in terms of $f$: $$cf - cf' = \frac{k}{c}ff'(2-2\cos\theta)$$

The answer key derives the formula for first converting $f$ to $\lambda$ and use the trick of $\Delta = \lambda'-\lambda$.

So here are my following questions:

$(1).~~$ Why does using the trick $\Delta = \lambda'-\lambda$ work to prove the formula? Looking back on the problem, is there something I would have been able to see that suggests that this is the way to go?

$(2).~~$ If I don't do the conversion(from $f$ to $\lambda $), I couldn't seem to get the formula that I wish to prove although these two quantity should be equivalent... why?

Below are my attempt to prove the formula in terms of $f$:

with the starting equation, divide everything by $f^2$:

$$\frac{1}{f} - \frac{f'}{f^2} = k(1-\frac{2f'}{f}\cos\theta + \frac{f'^2}{f^2})$$.

And now we replace $f'$ with $f+ \Delta$:

$$\frac{1}{f} - \frac{\Delta + f}{f^2} = k(1-\frac{2(f+\Delta)}{f}\cos\theta + \frac{(f+\Delta)^2}{f^2})$$

If we clear things up, we get:

$$-\frac{\Delta}{f^2} = k(2-\frac{2(f+\Delta)}{f}\cos\theta + \frac{2\Delta}{f} + \frac{\Delta^2}{f^2})$$

Here I am stuck - I couldn't seem to simplify this expression any further.

Yiyang Zhi
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  • Note that if $\lambda= \frac{c}{f}$ (you don't what $c$ is, but I assume it's a constant), then $\lambda' = \frac{-cf'}{f^2}$, so $\lambda' - \lambda = \frac{-c}{f}\left(\frac{f'}{f} + 1\right)$. I believe you have changed the $+1$ part to $-1$. I haven't yet checked the rest of your argument, but this may affect things. – John Omielan Aug 09 '19 at 02:58
  • Sorry, I couldn't see which part of the proof you are referring to? Is it this line? $\lambda' - \lambda = \frac{k}{c}(2-2\cos\theta)$ – Yiyang Zhi Aug 09 '19 at 03:07
  • No, it's in your next line of $cf - cf' = \frac{k}{c}ff'(2-2\cos\theta)$. Actually, using the correct expression for $\lambda' - \lambda$ and multiplying the line in your comment by $f^2$ should give $-cf - cf' = \frac{k}{c}f^2(2-2\cos\theta)$, so your $f'$ on the right also should be $f$ instead. – John Omielan Aug 09 '19 at 03:15
  • Okay, maybe there is some misunderstanding - $\lambda' = \frac{c}{f'}$., where $c$ is a constant. Sorry if I haven't made this clear. – Yiyang Zhi Aug 09 '19 at 03:26
  • Thanks for clarifying. I assumed the $'$ part in $\lambda'$ meant a derivative. Does $f'$ mean the derivative of $f$? If yes, then using $\lambda'$ can be quite confusing. I suggest using a different variable, like $\lambda_1$, and explicitly stating it means $\frac{c}{f'}$. – John Omielan Aug 09 '19 at 03:29
  • Okay, I completely see your point now. $f'$ is not derivative but it's like $f1$ and $f2$. Thank you for pointing this out. – Yiyang Zhi Aug 09 '19 at 03:38

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