Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have thought about this and looked for answers for a long time now, but I do not have any name or label for this problem, which is the reason for the long title of this question.

I have come across this many times, but I have never found any rationale for this, i.e. a clear interpretation or explanation, so here are a few examples of what I mean. I suspect that they have different answers.

I) In e.g. Shapiro and Teukolsy's book Black Holes, White Dwarfs and Neutron Stars, (1983), the Salpeter birthrate function for stars is defined as

$\psi_s\mathrm{d}\left(\frac{M}{M_\odot}\right) = 2\times10^{-12}\left(\frac{M}{M_\odot}\right)^{-2.35}\mathrm{d}\left(\frac{M}{M_\odot}\right)\mathrm{\ stars\ pc}^{-3}\mathrm{\ yr}^{-1}$

Questions: why does it matter that those differentials are included? Why make the statement look messier by multiplying both sides of the equation with something that obviously cancels out? If it is just to show what $\psi_s$ depends on, then it makes mre sense to me to define the function like so:

$\psi_s(M) = 2\times10^{-12}\left(\frac{M}{M_\odot}\right)^{-2.35}\mathrm{\ stars\ pc}^{-3}\mathrm{\ yr}^{-1}$

What is (or may be) the answer?

II) When defining the equation of continuity for brownian motion particles in 1D space, I have a lecture note that reasons in the following way:

Given the setup

         |  n(x,t) |
         |         |    
J(x,t) ---->     ----> J(x + dx, t)
         |         |
         |         |
         x        x+dx

Where $n(x,t)$ is the number density of particles between positions $x$ and $x + \mathrm{d}x$ and $J(x, t)$ is the flux of these particles at given position $x$ and time $t$. The particle number is conserved, and they are assumed identical and non-interacting with each other.

In the notes that I have, the following equation is readily displayed as though it was crystal clear why it was written inthis way:

$n(x, t + \mathrm{d}t)\mathrm{d}x - n(x, t)\mathrm{d}x = - (J(x + \mathrm{d}x, t)\mathrm{d}t - J(x, t)\mathrm{d}t)$

or (multiplying the minus into the parenthesis)

$n(x, t + \mathrm{d}t)\mathrm{d}x - n(x, t)\mathrm{d}x = J(x, t)\mathrm{d}t - J(x + \mathrm{d}x, t)\mathrm{d}t$

Question: I can not myself justify why it makes sense to multiply with the diferentials as they are here. What is the significance, or rather the line of thought that lies behind the way this equation turns out?

I am unsure of my own answer which is only to look at the dimensions of the two functions themselves and then conclude what unit the factor should have on each side in order for the dimensions to make sense.

Just to complete the derivation, rewriting the equation by replacing the functions $n(x, t + \mathrm{d}t)$ and $J(x + \mathrm{d}x, t)$ with their Taylor expansion (to first order) gives

$n(x, t + \mathrm{d}t) = n(x, t) + \frac{\partial n}{\partial t}\mathrm{d}t$,

$J(x + \mathrm{d}x, t) = J(x, t) + \frac{\partial J}{\partial x}\mathrm{d}x$,

which, when inserted into the original equation gives

$\frac{\partial n}{\partial t}\mathrm{d}t\mathrm{d}x = -\frac{\partial J}{\partial x}\mathrm{d}x\mathrm{d}t$

so producing, again, a shared factor ($\mathrm{d}t\mathrm{d}x$) on both sides of the equation which can just be divided away (right?), and we end with the coninuity equation

$\frac{\partial n}{\partial t} = -\frac{\partial J}{\partial x}$

I hope this question makes sense or at least allows you to suggest a label for this kind of problem that I can use to find more information, Or better, help by answering the wuestion here.


share|cite|improve this question
On your question I) - the differentials emphasize the exact form of the (dimensionless) argument. The function is a density thus in order to calculate the number one needs integrate over the right variable, which is the ratio of $M$ to solar mass in this case. – Slaviks Aug 22 '11 at 17:00
up vote 4 down vote accepted

The differential is used to specify that the number is for a "differential range", which is a way to remind you that the notions involved are somewhat fuzzy.

Let me give a purely mathematical example. Suppose I tell you that I am going to pick an arbitrary real number between 0 and 10, with the likely hood of a number being picked being proportional to the number itself. What is the probability that I pick the number 7? The answer is 0, because there are infinitely many numbers between 0 and 10. But what if I ask: what is the probability that the number I picked is between $7$ and $7+\delta$? Then the probability of course depends on how big $\delta$ is, but if $\delta$ is very small, then all numbers between $7$ and $7+\delta$ are roughly equally likely to be chosen, so we can say that the probability, for $\delta$ very small, is roughly linear in $\delta$. And so we can say that the probability of choosing a number between $7$ and $7+\delta$ is equal to $q(7)\cdot\delta$, where $q(x)$ is a function that specifies the "probability density" at the number $x$.

Similarly, the Salpeter function gives a distribution of the number of stars of a given mass related to the mass of the stars. But one should not state it as "the number of stars at a given mass"! Because if one expects, let us say, one star at each possible mass, since there are infinite number of allowed masses, there will then be infinitely (in fact uncountably infinite) many stars overall! What the Salpeter function gives you is that the number of stars between $M$ and $M+\mathrm{d}M$, for $\mathrm{d}M$ very small, the number of stars can be described by some function $\psi(M) \mathrm{d}M$, while the expected number of stars at exactly mass $M$ should be zero.

In other words, the differential serves to remind you that you are dealing with a sort of a density, rather than a straight-up function.

share|cite|improve this answer
"the number of stars between $M$ and $M+\mathrm{d}M$" is ok, but more general and mathematically correct is "the number of stars between $M_1$ and $M_2$, given by $\int\limits_{M_1}^{M_2}\psi(M)\:\mathrm{d}M$. – leftaroundabout Aug 23 '11 at 1:13
Thanks for that answer! But just to be plain, you seem to imply that the Salpeter function is not $\psi_s$ alone, but is rather taken as the product $\psi_s\mathrm{d}M$. This is never clear (to me) in the definition that I refer to, either. And also, arriving at the continuity equation in the manner that I show, does your answer not mean that one should, in principle, <em>not</em> reduce the equation by dividing out the two differentials on each side? To me, this would not matter, if I reminded myself in any accompanying text that refers to the equation what assumptions are required to use it. – Xidus Aug 23 '11 at 23:47
The main difference is as leftaroundabout said, there is a difference between a density and a function. A density is an object that can be integrated in and of itself, but a function is something that can be integrated only if you specify a volume form or a measure. Unfortunately, in much of the (older, and less geometric) literature, the distinction between the terminology is not really drawn. And the word function could equally well refer to a density or a function. – Willie Wong Aug 24 '11 at 0:02
Since the only natural way to use the Salpeter "function" is to integrate it over a mass range, it may be more natural to consider the density as the object. Whether you actually do it that way is a matter of convention, semantics, and personal preference. – Willie Wong Aug 24 '11 at 0:08
Thanks! If you or anyone else have any references to literature that might shed some light on this or similar discrepancies, historic, traditional or depending on school of thought, I welcome them:) Again thanks for your answwer and for the extra clarification (also to you leftaroundabout). – Xidus Aug 24 '11 at 0:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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