The following is taken from Cordes [1] on pulsar astronomy.

The dispersive time delay [is] $$t_\mathrm{DM}=\frac{e^2}{2\pi m_e c} \frac {\int_0^Dds \text{ } n_e}{\nu^2} = 4.15\text{ms} \text{ DM } \nu_{\text{GHz}}^{-2} \tag{3}$$ where the dispersion measure [and its] standard units (for $D$ in pc, $n_e$ in $\text{cm}^{-3}$) are $$ \text{DM}=\int_0^D ds \, n_e(s) \quad \quad (\text{pc cm}^{-3})$$

I'm confused about both the value and the units that the constant in equation 3 (4.15 ms) are derived. Since the integral in the numerator ('DM') is expressed in $\text{pc} \text{ cm}^{-3}$ and $\nu^{-2}$ in GHz the temporal unit (ms) should come from the first factor, right?

The first factor is composed of $e$ (electron charge) which has unit Coulombs, $m_e$ (electron rest mass) which is kg and $c$ which is $m/s$. First I naively did the following $$\frac{(1.60\cdot10^{-19})^2}{2\pi\cdot 9.11 \cdot 10^{-31} \cdot 299792458}$$ which does not give the correct answer, I think due to the unit conversion (to seconds, somehow?)

Then I tried a great deal of SI unit conversion of all three terms to get the desired units (ms) but could not find the solution.

I'm sorry if there's some trivialities I missed; I'm new to studying physics.

  1. Cordes, J. M., 2003. Pulsar Observations I. $-$ Propagation Effects, Searching Distance Estimates, Scintillations and VLBI. In S Stanimirovic et al (eds), Single-Dish Radio Astronomy: Techniques and Applications, ASP Conference Proceedings Vol. 278 (San Francisco: Astronomical Society of the Pacific, 2002), pp. 227-250.
  • $\begingroup$ Hello and welcome to PhysicsSE! Can you please provide a transcript of the image? $\endgroup$ Aug 17, 2020 at 11:26
  • 1
    $\begingroup$ Hi there, thanks! $\endgroup$
    – David
    Aug 17, 2020 at 11:34
  • $\begingroup$ Have you tried looking at cgs or Gaussian units (i.e., astronomers often use these instead of SI)? $\endgroup$ Aug 18, 2020 at 12:57

1 Answer 1


The first tell is in the assignment of $\nu_p^2 = \frac{n_e e^2}{\pi m_e}$ as the plasma frequency; a quick comparison to Wikipedia indicates that the paper is in gaussian units (which then implies that using the electron charge in Coulombs, i.e. in SI units, won't work at all).

If you want to keep things simple by working in SI units* (which I will, because I want Mathematica to do all the units for me**), then you need to replace $\nu_p^2 = \frac{n_e e^2}{\pi m_e}$ by its SI version: this is easiest to do for the angular frequency, $\omega_{p,\mathrm{CGS}}^2 = (2\pi\,\nu_{p,\mathrm{CGS}})^2 = \frac{n_e 4\pi e^2}{m_e}$, where replacing $e^2\to \frac{e^2}{4\pi\epsilon_0}$ gives $\omega_{p,\mathrm{SI}}^2 = \frac{n_ee^2}{m_e\epsilon_0}$, and thus $\nu_{p,\mathrm{SI}}^2 = \frac{\omega_{p,\mathrm{SI}}^2}{(2\pi)^2} = \frac{n_ee^2}{4\pi^2m_e\epsilon_0}$.

Now, if you work through the definition of $t_\mathrm{DM}$ in the paper (in SI, for convenience), it is given by $$ t_\mathrm{DM} = \frac{1}{\nu^2} \frac{e^2}{8\pi^2 \epsilon_0 m_e c} \int_0^D n_e(s)\mathrm ds . $$

My approach, as an AMO physicist, would be to define the middle jumble of constants as some single combination (call it $K$) and then say $$ t_\mathrm{DM} = \frac{K \, \mathrm{DM}}{\nu^2}, $$ where $\nu$ has units of frequency, $\mathrm{DM} = \int_0^D n_e(s)\mathrm ds$ has units of inverse area (which you can express as $\rm pc \:cm^{-3}$ if you want), and $$ K = \frac{e^2}{8\pi^2 \epsilon_0 m_e c} = 1.34 \times 10^{-7} \:\rm m^2/s $$ has units of meters-squared-per-second.

However, from what I understand, astronomers are a bit allergic to this type of manipulation, so instead they pre-specify the units of $\mathrm{DM}$ and $\nu$, and they move them off into the numerical constant. Thus, you define $\nu=\nu_\mathrm{GHz} \:\mathrm{GHz}$ and $\mathrm{DM}=\mathrm{DM}_\mathrm{pc/cm^3}\:\mathrm{pc/cm^3}$ and you put them into the above, to give you $$ t_\mathrm{DM} = \frac{K\:\mathrm{pc/cm^3}}{\mathrm{GHz}^2} \frac{\mathrm{DM}_\mathrm{pc/cm^3}}{\nu_\mathrm{GHz}^2}, $$ where the fraction of the right has the numerical values of $\mathrm{DM}$ and $\nu$ in the units given, and the new combination $$ \frac{K\:\mathrm{pc/cm^3}}{\mathrm{GHz}^2} = 4.15\:\mathrm{ms} $$ is precisely the constant, with dimensions of time, that appears in the paper. (Also, to make things worse, Cordes keeps the unit notation in $\nu_\mathrm{GHz}$ but drops it in $\mathrm{DM}$, because... reasons? Honestly, astronomers are weird.)

If you ask me, this way of working is completely bonkers, but hey, it seems to work for astronomers, so use it if you want to (and definitely learn to work it if you'll be reading astronomy literature).

(And, as a final aside, if you really need the numerical constants, note that it's perfectly possible to write $$ t_\mathrm{DM} = 4.15\:\mathrm{ms} \: \frac{\mathrm{DM}}{\mathrm{pc/cm^3}} \frac{\mathrm{GHz}^2}{\nu^2}, $$ without any funky shenanigans: you can put $\mathrm{DM}$ and $\nu$ in any units you need to, and the units simplify completely if you put them in the intended set. Just saying.)

*though honestly, if you want to work in astronomy, you're probably better off ditching the SI and learning to work in gaussian units as early as possible.


  • $\begingroup$ Even though I had figured it out already (sorry, I meant to put my derivations as an answer here today), this is absolutely amazing, thanks. $\endgroup$
    – David
    Aug 20, 2020 at 13:08

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