According to Kolanoski's & Werme's book Particle Detectors. fundamentals and applications, page 63, for electrons, we can write the energy losses for ionization and Bremsstrahlung as follows:

$$ \text{ionization}: \qquad \propto Z \ln E/M $$ $$\text{bremsstrahlung}: \propto Z^{2}E/M^{2}$$

(The first equation we get from the Bethe-Bloch formula, remembering that $\ln M\beta^2\gamma^{2} \overset{\beta = 1}{=}M\gamma^{2} = E^{2}/M $ for electrons.)

Obviously, these are only rough approximations, but okay, let's stick to them.

Now, if we define the critical energy, cf. Eq. (3.92) of the book, to be the energy where bremsstrahlung and ionization are the same (and for $E > E_c$, bremsstrahlung dominates over ionization), then we get:

$$E_{c} \propto \frac{m^{2}}{Z}\ln E/M.$$

However, in Table 3.4 of the book (but also on PDG), the critical energy always has a fixed value. However, as my rough calculation shows, the critical energy depends on the energy of the electron ($E$)? Why then isn't $E_{c}$ as a function of $E$ or sth similar given in the textbook and on PDG?


1 Answer 1


By definition the critical energy is a fixed value, you mixed up the notations.

As you mentioned, the critical energy is the energy value where the ionization term and bremsstrahlung term are equal. So (based on the 1st and 2nd formulas in the question) the following equation holds: $$C_iZ\ln{(E_c)}/M=C_bZ^2E_c/M^2$$ where $C_i$ and $C_b$ are the constant factors int the ionization and bremsstrahlung terms and $E_c$ is the (one and only) critical energy.

To obtain the value of critical energy, you should solve this equation for $E_c$.


The function can be written as: $$E_c=C\ln(E_c)$$ where $C$ includes all the energy independent "constants". This can't be solved with simple ("elementary") functions, leads us to the Lambert's $W$ function. The solution should be: $$E_c=-CW(-1/C)$$

A post on math.stackexchange about $x=c\ln(x)$, and the wiki page about the $W$ function.

  • $\begingroup$ +1. Could you please also show me how to solve this eq. for $E_c$? In that case, I would be more than happy to accept this answer. $\endgroup$
    – user248824
    Commented Mar 1, 2021 at 21:05
  • $\begingroup$ Updated my answer ;) $\endgroup$
    – fanyul
    Commented Mar 1, 2021 at 21:58

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