1
$\begingroup$

I have an equation that I have found in several papers which I am currently using for a project, including Waligorski (1986) and this book page 32.

4.2 BUTTS AND KATZ MODEL

Butts and Katz model uses Rutherford's SDCS for production of secondary electrons (Mott, 1929; Bradt and Petters, 1948; ICRU, 1995). Rutherford's secondary electron distribution formula gives the number of secondary electrons per unit of path length having energies in the interval from $w$ to $w + dw$, produced by an incident ion of effective charge2 $Z^*$ moving with speed $\beta c$ $$dn = \frac{2\pi N_e e^4 Z^{*2}}{m_e c^2 \beta^2}\frac{dw}{(w + I)^2}\tag{4.15}$$ where $m_e$ and $e$ are the electron mass and charge, $N_e$ is the density of electrons in the material, $I$ is the mean ionization potential of the material, $c$ is the speed of light and $\beta = v/c$ is the speed of the incident particle with respect to the speed of light. Considering water as the target material $$\frac{2\pi N_e e^4}{m_e c^2} = 8.5\times 10^{-3}\,\mathrm{\frac{keV}{\mu m}}\tag{4.16}$$

I cannot figure out why equation (4.16 )has the units of energy/length. The thing is, if (4.15) has $dn$ with units of $\text{length}^{-1}$ as specified, then (4.16) must have those units in order to be consistent with the rest of equation (4.15). Am I missing something here? Because I don't see how it is possible for (4.16) to have those units.

To me there is $$\frac{\text{charge}^{4}}{\text{energy} \times \text{volume}}$$ where the volume comes from the $N_e$ which is the density of electrons. I see no way to convert this into $\text{energy}/\text{length}$.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for rendering it directly David Z; how did you do it though? Or did you have to type it out by hand? $\endgroup$ Feb 14, 2015 at 17:26

2 Answers 2

1
$\begingroup$

The system of equations belong to the ESU, and presumably gaussian, where charge^2 = force*length^2.

Putting this in, you get $F^2 * L^4 / F * L * L^3$ gives F = E/L.

Hint, the correcting factor for SI is to replace $e^2$ with $e^2/4\pi$.

If you really are going to read a lot of older physics, it is best to be versed in the gaussian units: their units are different and they use only three base units: charge is a derived quantity, variously by ampere's law or coulomb's law: ie

$F = QQ/r^2 $ whence $[Q] = [F^{\frac 12} L]$ or $F/L = 2II/r$ whence $[Q] = [F^{\frac 12} T]$. The relation between the emu charge and the esu charge is a constant which SI renders dimensionless, but in CGS has the dimensions of $[L/T]$, and the value of $c$.

From CGS to SI

There are two separate factors involved in conversions from cgs to SI, $4\pi$ and $c$. The first arises from the way Coulomb's law is defined or derived, in that the flux at unit distance from a charge x C is $4\pi x$ coulombs in unrationalised systems, and x coulomb in rationalised systems. The second is the relation between current defined electrically (ampere) vs magnetically (ampere-turn). This can be found because the constant has the dimensions of velocity in cgs and 1 in SI.

The two operate separately, and dimensional analysis can handle 'c' in all cases.

In the case of the $4\pi$, the rule of substance can be applied. This is handled by a rule of 'ticks and crosses', which really corresponds to a dimensional analysis not part of SI. a tick corresponds to a power of +1, a cross to -1. The matching outcome is the missing constant, which is a ticked $4\pi$ if you are going from SI to cgs, and a crossed $4\pi$ if you are going from cgs to SI.

Densities, flows, gradients of x, moments, have the same form as x.

Mass, any mechanical quantity with mass in it (eg force, energy, pressure, torque), charge, poles, dipoles, and their moments, susceptability, susceptance, conductance, capacitance, are all SUBSTANCES and are thus ticked.

Any kinematic quantity (in L and T only), all fields, fluxes, potentials, permeances, permittivity, are UNMARKED.

Resistances and inductances are CROSSED.

In the equation above, charge (q), mass (m), energy (w), are all ticked, which leaves LHS = 0 tick; RHS = 4 ticks / 1 tick * 1 tick / 2 ticks. This means that the right hand side is 2 ticks / 0 ticks, which means that we have to 'correct' it with a $(4\pi)^2$ under. This cancels out the $2\pi$ on top leaving an $8\pi$ under.

$\endgroup$
2
  • $\begingroup$ This seems like it's the case, I guess I'm too young to have heard of these units. I have a question though; why is the correcting factor $4 \pi$? From what I read in the wiki articles, don't you need to include the $\epsilon_0$ factor somehow as well? $\endgroup$ Feb 14, 2015 at 17:34
  • $\begingroup$ 4pi can be corrected by the rule of substance, which i gave above, the c can be corrected by dimensional analysis, but it helps to reduce L=Q=0, M=T=-1 against some kind of power of c, and do the analysis using that. $\endgroup$ Feb 15, 2015 at 8:40
0
$\begingroup$

Without having access to the book, the best guess I can come up with is that the formula is written in a natural unit system where $c=1$, $\hbar=1$, and $e=1$. In this system, charge is unitless, and energy has the same unit as inverse length. So, using $\equiv$ to mean "congruent to" (i.e. having the same units), $$\frac{\text{charge}^4}{\text{energy}\times\text{volume}} \equiv \frac{1^4}{\frac{1}{\text{length}}\times\text{length}^3} = \frac{1}{\text{length}^2}$$ and $$\frac{\text{energy}^4}{\text{length}} \equiv \frac{\frac{1}{\text{length}}}{\text{length}} = \frac{1}{\text{length}^2}$$ However something like this should be specified explicitly. If the book doesn't make clear somewhere that it is using natural units, that is a significant problem.

The corresponding formula in an SI-like system would probably be $$\frac{2\pi N_e e^4}{m_e c^2}\times(\hbar c)^2 = \frac{2\pi N_e \hbar^2}{m_e} = 160\,\frac{\mathrm{keV}}{\mathrm{\mu m}}$$ given $N_e = 333.679\times 10^{27}\,\mathrm{m}^{-3}$, although the fact that the value is totally different makes me doubt this. If this were what the author meant, again, it should have been stated explicitly, and it's very strange to write it in the notation you've quoted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.