Why is the stopping power in Bethe Bloch-Plot shown against $\beta\gamma$ on the horizontal axis? Respectively what is the meaning of $\beta \gamma$?
Plot taken from here
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$\begingroup$ Without knowing what particular plot you are looking at, this is pretty hard to answer. $\endgroup$– Jon CusterCommented Aug 3, 2017 at 17:12
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2$\begingroup$ Ahhh... $\beta \gamma = p/Mc$ where $M$ is the rest mass of the particle. This means the master curve can be used for muons, pions, protons, etc. $\endgroup$– Jon CusterCommented Aug 3, 2017 at 17:43
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1 Answer
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$\beta\gamma$ appears often in particle physics, and is convenient for plots. As noted, $\beta\gamma = p/Mc$, and at non-relativistic speeds it is simply $\beta$, or at relativistic speeds $\gamma$.
Thus, $\beta\gamma$ characterises whether we are in a relativistic regime. As the incident particle becomes relativistic, $\beta\gamma >1$ known as the relativistic rise, and this can be explained in terms of the transverse field strengthening.
To elaborate, Lorentz contraction compresses the field in the boost direction, and therefore the transverse field is strengthened relative to the longitudinal part.
Furthermore, around $\beta\gamma \approx 3$, there is a region where energy loses are minimised.
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$\begingroup$ Thanks! It sounds like a helpful term to describe some behaviours. When I rearranged it correctly $\beta \gamma$ is the momententum with a relativistic factor. Is that right? So finally it is stopping power against momentum? $\endgroup$– BenCommented Aug 4, 2017 at 8:34