How to prove Compton Scattering in terms of frequency?

Question

When attempting to prove the formula for Compton Scattering, I obtained the following equation:

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

where $~k~$ is a constant, $f$ is the frequency of light before the collision and $f'$ is the frequency of light after the collision.

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.

Answer 1

After reading your question for three times, I can't still tell what's really your question, so I will answer the title. Instead of working backward, use conservation of energy and momentum. That's

$$P_i=P_f \to \vec{hf/c}=\vec{p}+\vec{hf'/c}~~~(1)$$

$$E_i=E_f \to h|f|+mc^2=E_e'+h|f'|~~~(2)$$ $$E_e'=\sqrt{|\vec{p}|^2c^2+m^2c^4}$$ From $(1)$ we have:

$$ (\vec{hf/c}-\vec{hf'/c})^2=|\vec{p}|^2 \\h^2|f|^2+h^2|f'|^2-2h^2|f||f'|\cos(\theta)=|\vec{p}|^2c^2~~~(3)$$ So from $(2)$ and $(3)$: $$h|f|+mc^2=\sqrt{h^2|f|^2+h^2|f'|^2-2h^2|f||f'|\cos(\theta)+m^2c^4}+h|f'| \\ (h|f|-h|f'|+mc^2)^2=h^2|f|^2+h^2|f'|^2-2h^2|f||f'|\cos(\theta)+m^2c^4\\ 2h^2|f||f'|+2h(|f|-|f'|)mc^2=2h^2|f||f'|\cos(\theta)\\ |f|-|f'|=\frac{h}{mc^2}|f||f'|(1-\cos(\theta))~~~(4)$$ You can rewrite $(4)$ like this: $$\frac{|f|-|f'|}{|f||f'|}=\frac{h}{mc^2}(1-\cos(\theta)) \\ \frac{\lambda'}{c}-\frac{\lambda}{c}=\frac{h}{mc^2}(1-\cos(\theta)) $$

There is no need for tricks. If you like to obtain this formula directly for $\lambda$ start from these:

$$P_i=P_f \to h/\vec{\lambda}=\vec{p}+h/\vec{\lambda'}~~~(1)$$

$$E_i=E_f \to h|c/\lambda|+mc^2=E_e'+h|c/\lambda'|~~~(2)$$

As you can see Compton scattering formula is derived by two formulas (conservation of energy and momentum), that might be the reason as to why you can't arrive at an obvious answer from

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

$m$ is electron's mass, $\vec{p}$ is its momentum and $E'_e$ is its energy after collision, just in case if you didn't get it.