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I have the following formula:

$$\dfrac{dE}{dx} = \dfrac{4 \pi e^4 z^2}{mV^2}NB,$$

where, for example, in the case of relativistic velocity of a particle, we have

$$B = Z\left[ \ln \dfrac{2mV^2}{I} - \ln(1 - \beta^2) - \beta^2 \right].$$

$E$ is the particle energy;
$ze$ and $V$ are its charge and velocity, respectively;
$M$ and $m$ are the masses of the incident particle and electron, respectively;
$N$ is the number of atoms in the unit volume of a substance;
$Z$ is the charge of nuclei of atoms of a medium substance;
$I$ is the average excitation energy of an atom ($I = 18.5Z \cdot 1.6 \cdot 10^{−19} \ \text{J}$);
$\beta = V/u$;
$u$ is the velocity of light.

$\dfrac{dE}{dx}$ is also known as the “stopping ability” of a substance, and $B$ is called the deceleration coefficient. (But my understanding is that sometimes the quantity $B$ is identified as the “stopping ability” of a substance.)

My understanding is that this formula actually the Bethe formula.

I am told that this version of the Bethe formula shows that, as the particle energy grows, specific losses for ionization decrease at first very rapidly (inversely proportional to energy), but do this more and more slowly as the particle velocity comes closer and closer to the light velocity. This fact is not immediately clear to me from the equation, and I was not provided with any graphs to illustrate the phenomenon. In researching this further, I found this document, which provides the following illustration:

enter image description here enter image description here

As can be seen from the graph, $\beta \gamma$ is plotted on the $x$-axis, so I'm presuming that this is the specific energy? But, if so, then this graph is not describing the aforementioned phenomenon: At first, particle energy decreases very rapidly, and $\beta \gamma$ increases very slowly; and then, at some sudden minimum, particle energy starts to increase very slowly, and $\beta \gamma$ increases very rapidly. So this still leaves me confused.

I would greatly appreciate it if people would please take the time to help me understand this by carefully explaining the equation $\dfrac{dE}{dx}$ and its behaviour as described in the aforementioned description ("as the particle energy grows, specific losses for ionization decrease at first very rapidly ..."), and how this relates to the provided graph.

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  • $\begingroup$ What is plotted on the $x$-axis seems to be the velocity, right? Those A, B, C and D points seem to accurately describe it, I don't exactly understand why you disagree. $\endgroup$ Jul 28, 2020 at 21:26
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    $\begingroup$ @PauloMourão I am not disagreeing with A, B, C, and D. It is this part that I'm interested in: "... as the particle energy grows, specific losses for ionization decrease at first very rapidly (inversely proportional to energy), but do this more and more slowly as the particle velocity comes closer and closer to the light velocity." $\endgroup$ Jul 29, 2020 at 0:45
  • $\begingroup$ Yeah I agree that that does not seem to be what's happening. Maybe there's something we're missing but I don't know what it is $\endgroup$ Jul 29, 2020 at 16:42
  • $\begingroup$ User deleted their answer because they no longer had confidence that it was correct. They linked this interesting document in the comments: pdg.lbl.gov/2009/reviews/… $\endgroup$ Aug 6, 2020 at 9:59

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$\frac{dE}{dx}$ here is the rate of energy loss versus distance traversed by the incident charged particle. The horizontal axis is $\gamma \beta=\frac{v}{\sqrt{c^2-v^2}}$, which makes the graph more convenient since the behavior changes as the particle becomes relativistic. The first part of the curve is saying that slow particles lose energy rapidly. The $1/v^2$ dependence can be derived from the nonrelativistic classical Rutherford cross section. Once the particles become relativistic, the $1/v^2$ dependence is cut off, and the appropriate relativistic calculation gives the logarithmic dependece for large $\beta \gamma$.

J.D. Jackson's Classical Electrodynamics in chapter 13, shows how to derive these equations in section 13.1-13.3 and provides a figure very similar to yours. I suggest reading those 10 or so pages to get more details.

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