0
$\begingroup$

In cgs unit system, the radius of gyration of a particle is: $$r=\frac{pc}{ZeB}$$ where $p$ is momentum, $c$ is the speed of light $Ze$ is the charge and $B$ the magnetic field. I'm confused about the units of measure. Let's suppose $B=1G$ (where $G$ is Gauss, the unit for magnetic field in cgs system and $G=10^{-4}T$). And suppose the particle under consideration is a $1GeV$ electron, so that approximately $\frac{pc}{Ze}=10^9V$ (volts). It seems that however the units are in some way wrong, since $V/G$ is not a unit for length. How can I compute the radius correctly?

$\endgroup$

2 Answers 2

2
$\begingroup$

You should look at, for example, the appendix of J.D. Jackson, Classical Electrodynamics where the conversion rules are given. Your problem is that you are mixing in volts which are not a cgs unit, since Gaussian units do not have an independent extra unit of charge.

In your expression, $pc$ has units of ergs, i.e. $\rm g\cdot cm^2/s^2$. For 1 Gev this is about $1.6\times 10^{-3}$ ergs. $e$ has units of $\sqrt{\rm erg\cdot cm}$ and a value of about $4.8\times 10^{-10}$, ($\frac{e^2}{r}$ is an energy and the conversion factor for charge from Coulombs to cgs is $2.99792458\times 10^9$ and related to the numerical value of the speed of light.) Since $B^2$ has units of energy density, $B$ has units of $\rm \sqrt{erg/cm^3}$ So for your example \begin{equation} r = \frac{1.6\times 10^{-3}\rm erg} {Z 4.8 \times 10^{-10}\rm \sqrt{erg\cdot cm}\cdot 1 \sqrt{erg/cm^3}} = \frac{1}{3Z} \times 10^{7}\rm cm \end{equation}

$\endgroup$
4
  • $\begingroup$ The charge unit in cgs units is called esu (also statcoulomb or Franklin) $\endgroup$
    – Thomas
    Jul 1, 2021 at 7:28
  • $\begingroup$ It is incorrect to suggest $pc$ would represent the kinetic energy of the particle. It may be a good approximation for highly relativistic speeds but otherwise you get completely wrong results with this assumption. See my own answer for details. $\endgroup$
    – Thomas
    Jul 2, 2021 at 17:49
  • $\begingroup$ I assumed that 1 GeV meant a momentum of 1 GeV/c, which is a completely standard way of specifying momentum, and was consistent with the questioner's description with pc= 1 GeV. Of course you are correct if 1 GeV is instead the kinetic energy. $\endgroup$
    – user200143
    Jul 2, 2021 at 18:20
  • $\begingroup$ $1.6x10^{-3}$ ergs is an energy not a momentum. Using it the way you did would lead anybody not familiar with this convention to believe that a particle with e.g. an energy of 10 eV would have a Larmor radius 8 orders of magnitude smaller than for 1 GeV. This would be completely wrong. $\endgroup$
    – Thomas
    Jul 2, 2021 at 18:54
0
$\begingroup$

The units are correct in your formula for the Larmor radius. $eB$ has the unit of a force (note that the magnetic field has the same dimension as the electric field in cgs-units). So if you multiply the equation by $eB$ you have on the left side the unit Force x Length = Energy while $pc$ on the right hand side has dimension of energy as well.

However, it is not correct to take $pc$ as the kinetic energy of the particle. This would only be correct for massless particles. In general the relativistic total energy is given by

$$E=\gamma mc^2 = \sqrt{p^2c^2+m^2c^4}$$

and the kinetic energy by

$$K=E- mc^2$$

from which you can see that

$$pc = \sqrt{K^2+2Kmc^2}$$

If you insert $K$ here in cgs-units (erg; 1 eV= $1.6*10^{-12}$ erg) and $e$ and B in cgs units as well (e=$4.8*10^{-10}$ esu; B in Gauss) you will get the correct Larmor radius in $cm$.

$\endgroup$

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.