I understand that we can measure any general number density $n$ as,

$$ n = \frac{N}{V}$$

for total number $N$ and volume $V$. This puts the units of number density as $\text{length}^{-3}$ e.g. $\text{cm}^{-3}$

Now suppose the number density varies with radius as e.g.,

$$ n(r) = N_0 r^{-2}$$

Clearly the units at any particular radius are still $\text{cm}^{-3}$, but then what are the units of $N_0$ and $r$?. It seems to me that for different distributions e.g. $N_0 r^{-3}$ the units would be different. But then how can the units of $n$ always be $\text{cm}^{-3}$

  • 2
    $\begingroup$ $n$ always has dimensions (not units) $L^{-3}$. For $n=N_0 r^{-2}, [N_0]=L^{-1}$ $\endgroup$ – Avantgarde Jul 6 '17 at 13:25
  • 3
    $\begingroup$ You are overthinking this. The units of $N_{0}$ are what is required to make $n(r)$ have units of (per unit volume) at all points, i.e. cm$^{-3}$. $\endgroup$ – Jon Custer Jul 6 '17 at 13:26

In several comments, it was mentioned "just make the units of $N_0$ whatever they need to be, without concern as to why. I thought I might add to that.

Units are a funny thing because in many cases, they serve no deeper purpose than to ensure you did the math correctly. Take Hooke's law as a classic example. Hooke's law is most accurately described in mathematics as $F\propto x$, where F is the force exerted by a spring is proportional to x, which is the amount the spring is extended or compressed. However, this very pure form is inconvenient for doing math. Instead we rewrite that equation as $F=kX$, where $k$ is a measure of the stiffness of the spring. The two expressions convey the same information, but one uses proportionality while one uses equality and a constant multiplier.

When converting proportionality to equality, we have to deal with the fact that proportionality and equality differ in how they deal with units. Proportionality is blind to units. If you do the math, the units always end up canceling out. Equality, however, does require the units to line up. To make them line up we assign a unit to the constant which makes the equation work.

Sometimes we find that these assigned units have a deeper meaning, other times they do not.

So in your case, you write $n(r)=N_0r^2$, but in words you chose to write "...the number density varies with radius," which is a proportionality phrasing $n(r)\propto r^2$. $N_0$ only appears as a side effect of turning that proportionality into an equality, and thus the units of $N_0$ are safely assigned to "whatever they need to be to make the units work." This also points out the answer to your second question: if you used a distribution where $n(r)\not \propto r^2$, then you would have a different equation when you converted it to equality. Thus, you would need to make up a different unit.


Because you need a constant of proportionality with units such that it yields the correct units for $n$. Often this constant is absorbed in the units in which $r$ is measured. For instance, the expression $$ n(r) = N_0 \left( \frac{r}{r_0} \right)^{-2} $$ is equivalent to $$ n(r) = k N_0 r^{-2} $$ with the constant $k \equiv r_0^2$ having dimensions of $L^2$.

Note though that this functional form for the density profile strictly speaking is unphysical as it diverges for $r\rightarrow0$.

  • $\begingroup$ I try as much as possible to reduce everything to dimensionless numbers. In this format it is completely clear (to me atleast) that $n(r)$ has the same dimensions as $N_0$. $\endgroup$ – nluigi Jul 6 '17 at 14:22
  • 1
    $\begingroup$ There's nothing unphysical about this number density. The requirement isn't that $n$ be finite. Just look at the true number density; it's a sum of delta functions, which diverge. The requirement is that $N = \int n \operatorname{d}V$ over any finite volume be finite, and this one satisfies that. $\endgroup$ – Sean E. Lake Jul 6 '17 at 16:35
  • $\begingroup$ I guess it depends on the scale, @SeanE.Lake. On the particle level you might be right, I don't if it makes sense to talk about the density of individual particles at $r=0$. But for a gas cloud 10 pc across, or a dark matter halo, the density profile flattens out for small $r$. $\endgroup$ – pela Jul 7 '17 at 5:39

Number density is always what you stated $\mathrm{m}^{-3}$ and radius is physical length always $\mathrm{m}$, then you just have to solve the units of $N_{0}$. If $n = N_{0}r^{-2}$

$$ N_{0_\mathrm{units}} = \frac{\mathrm{m}^{-3}}{\mathrm{m}^{-2}} $$

You just know the dimension of length and number density. In equations that contain variables of unknown dimension you can just figure out this way what are their units.

  • $\begingroup$ So to confirm: the dimensions of $N_0$ are variable, so as to ensure the dimensions of $n$ are always $m^{-3}$? $\endgroup$ – user1887919 Jul 6 '17 at 14:25
  • $\begingroup$ Yes, you know that radius represents physical distance and it must have dimension of length. Also you know (or should know) that density of something is inversely related to the cube of length. The dimensions on both sides of the equals sign must match and in your example there are three variables with two known dimensions so the final one can be deduced from the two. $\endgroup$ – Communisty Jul 7 '17 at 5:58

So $r$ has units of length, for example $\mathrm{m}$, so $r^{-2}$ has units $\mathrm{m}^{-2}$. to remain dimensionally consistent $N_{0}$ would have to have units $\mathrm{m}^{-1}$. Note that here, we have number density in 3 dimensions, giving a number per unit volume. We might imagine number density on the plane, or on a line. In this circumstance, the dimension of $N_0$ would change. It might be of interest to you to note that $N_0$ in this case is only a dimensionless number when the power law matches the dimension of the space we are considering.


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.