1
$\begingroup$

In astrophysics and pulsar astronomy, a typical quantity is the dispersion measure or DM, which is the integral of the electron number density along the line of sight,

$$ DM = \int_0^d n \, dl$$

up to some distance $d$.

Now in the literature (e.g. Spitler et al. 2013) the DM is given with units of $\text{pc } \text{cm}^{-3}$. For example, in Spitler a value is given of $DM = 1778\text{ pc } \text{cm}^{-3}$. Does this mean:

  1. The total sum of every density element between the observer and pulsar along the line of sight is $1778$
  2. The total sum of every density element between the observer and pulsar along the line of sight, divided by the total distance in parsecs is $1778$

I think that (1) is the correct way to understand the dispersion measure, but then why do the $\text{pc}$ units matter? If one is just summing along the line of sight then is it not true that $1778 \text{ pc } {cm}^{-3} = 1778 \text{ m } {cm}^{-3} $ ?

$\endgroup$
2
$\begingroup$

There is obviously an ambiguity introduced by the integral - all you can measure is the sum of $n\ dl$ along the path to the pulsar. However, if you have some idea of the distance to the pulsar in parsecs then I suppose it can make sense to express the sum of $n\ dl$ in terms of parsec cm$^{-3}$ because you can immediately translate the dispersion measure into the product of a distance (in parsecs) multiplied by an average electron number density in cm$^{-3}$.

For example, if it turned out that the distance to the pulsar in your question was actually 1778 pc (not an atypical distance for such an object), then the dispersion measure expressed in that way immediately tells you that the average electron number density along that line of sight is 1 cm$^{-3}$.

I think your option #1 is the correct way to think about it, but you are incorrect to think that the units don't matter. A column density of 1778 parsec cm$^{-3} = 5.487\times 10^{21}$ cm$^{-2}$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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