2
$\begingroup$

So I know this has been asked before, but I really do not get why the divergence of the gravitational field $\mathbf{g}$ that exists due to a point mass $M$ at the origin is zero. I mean, I get it mathematically, but not the intuition. So take a point mass $M$ at the origin. For any mass $m$ with position vector $\mathbf{r}$ we have $\mathbf{g}(\mathbf{r})=\frac{-GMm}{|\mathbf{r}|^3}\mathbf{r}$. Now if you compute $\nabla\cdot\mathbf{g}(\mathbf{r})$ then this can be easily shown to be zero. My question is what does this mean? If you 'draw' the vector field then you get a lot of arrows pointing at the origin and the magnitude of these vectors decrease the further away from the origin we get. Naively you might think we have a vector field with negative divergence, but this is offset by the magnitude of the gravitational field 'blowing up' as we approach the origin. So I get why there is a problem, but not the intuition of what the statement is saying.

The best I have come up with is this. If we think of the situation where we have a spherically symmetric mass distribution of radius $R$, then the above formula holds provided $|\mathbf{r}|>R$ and we can use the fact that this is a special case of Gauss' law of gravitation, $\nabla\cdot\mathbf{g}=-4\pi G\rho$ where $\rho$ is the mass density at that point where you compute $\nabla\cdot\mathbf{g}$. I have read that if we are 'outside' our spherical mass then $\rho=0$ (so we are in empty space) but why is the mass density zero. I thought when we derive $\mathbf{g}$ above we take a mass $m$ with position vector $\mathbf{r}$, so at that point we have a mass $m$. Therefore, surely the mass density isn't zero. Or do we mean the mass density of the thing that is causing the gravitational field? So in the point mass at the origin case, $\rho=0$ except at the origin (which is infinite as we have a point mass). In the spherically symmetric mass distribution case $\rho=0$ as soon as we go beyond the radius of the mass.

Edit: Upon further thought, could it also be linked with the following intuitive idea. If a mass $m$ is attracted to $M$ then it will pick up speed as it approaches $M$. As it won't suddenly come to a stop then it will continue travelling further on beyond $M$. So if you imagine a sphere encompassing our mass $M$, then for every arrow going into this surface, there is a corresponding arrow going out. So when we are outside this surface, the net divergence is zero? Or am I completely barking up the wrong tree here?

$\endgroup$
  • $\begingroup$ "Or do we mean the mass density of the thing that is causing the gravitational field?" - this, exactly. $\endgroup$ – lisyarus Oct 22 '18 at 10:30
  • $\begingroup$ It simply implies gravitational field does not flow out or vanish in a space without mass. $\endgroup$ – Szeto Oct 22 '18 at 10:30
  • $\begingroup$ @lisyarus so the $\rho$ refers only to the mass $M$ (i.e. the point mass), and not the mass $m$. So does that mean Gauss' law also holds for a spherical mass whose mass is not evenly distributed? So the divergence would vary 'within' the sphere as the mass density varies? $\endgroup$ – Sam Williams Oct 22 '18 at 10:33
  • $\begingroup$ @mathreadler Oh I see! Sorry, I thought you were making some connection with rotation then! $\endgroup$ – Sam Williams Oct 22 '18 at 10:33
  • $\begingroup$ @Szeto I think I am having a problem with this 'space without mass' idea. What do you mean by space without mass? $\endgroup$ – Sam Williams Oct 22 '18 at 10:35
1
$\begingroup$

For me, the most intuitive way to think about the divergence of a field is to think about field lines rather than vector fields. As you say, you can visualise a vector field as a bunch of arrows that are directed radially inwards and getting longer in length as you approach the origin. However, this picture is not all that helpful when considering divergence.

Instead, imagine a spherical mass with radial lines going into it from all directions. Field lines are a very natural thing to consider - I am sure you have seen iron filings around a magnet many times before. It is not hard to convince yourself that this picture is the right one. To see this, notice that the number of field lines that pass through a unit area (with normal parallel to $\hat{r}$) are inversely proportional to the square of the radius. This follows directly from the fact that the surface area of a sphere is $4\pi r^2$ and the field lines being continuous (i.e. the field lines don't suddenly disappear or spawn out of nowhere). An inverse square law is the correct Newtonian description of gravity and here the density of field lines accurately represents the density of the vector field.

If you're comfortable with that then, the statement that the divergence of the gravitational field (in free space) is zero is equivalent to saying that the field lines are continuous. It's as simple as that!

Now for some pictures. Please note that I cannot draw 3D so I have suppressed one dimension. This is actually important because in 2D we would have in inverse law (as opposed to inverse square law) but you get the idea I'm sure:

1) Here we show the gravitational field surrounding a mass. Notice that the field lines are most densely packed close to the origin and drop off as we move further away just like a vector field should. Notice that the field lines are continuous everywhere meaning we have zero divergence. This is the correct description of Newtonian gravity.

Gravitational field with zero divergence

2) Here we can see a field that has negative divergence. Notice that field lines coming in from infinity seem to be disappearing! This should seem very strange indeed. In general, fields with negative divergence seem to... well... vanish. It's a bit spooky really (one interpretation might be that the field is 'flowing out' into higher dimensions but that's a story for another day).

Negative divergence

3) Here we see a field with positive divergence. Notice that the fields lines seem to just appear out of nowhere. I'll let you put your own interpretation on what might cause that but it's not particularly physical for a gravitational field in a vacuum.

enter image description here

Hopefully these pictures give you some intuition for why a gravitational field has zero divergence and what that might represent. Continuous field lines...

$\endgroup$

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.