Plotting the Bethe-Bloch Formula

Question

We had in our scriptum the following formula for the Bethe-Bloch formula:

$$-\frac{dE}{dx} = K\frac{\rho Z}{A} \frac{z^2}{\beta^2} \left[ \ln\left( \frac{2m\gamma^2\beta^2}{I} - \beta^2 \right) \right], $$

where I set $c = 1$ for convenience, $K$ is a constant, $I$ is the average ionisation potential, and $\rho$ denotes the density of the material.

My question is: I have often seen plots which show on the $x$-axis $\beta\gamma$, but I am not sure how one would plot $-\frac{dE}{dx}$ as a function of $\beta\gamma$, given that in front of the $\ln$, we have the factor $\frac{1}{\beta^2}$ (and not $\frac{1}{\beta^2\gamma^2}$), and also in the $\ln$, we have $-\beta^2$ and not $-\left( \beta\gamma\right)^2$.

I would love hear some opinions on this!

Answer 1

If you know $\beta$ you can compute $\gamma = (1 - \beta^2)^{-1/2}$, so it is just a matter of calculating $\beta \gamma$ for each value of $\beta$ you are interested in. And if you know $\beta\gamma$, you can extract $\beta$ from it, e. g. by plotting $\beta\gamma$ versus $\beta$ (again, $c = 1$):

When you know $\beta$, you can put it back into the Bethe-Bloch Formula.