9
$\begingroup$

Are there likely to be stars for which the average density is greater than the density at the centre? Intuitively, I would say not, as density tends to decrease outwards, but ideally I would appreciate a clear mathematical justification for this.

$\endgroup$
1
10
$\begingroup$

$$\frac{dP}{dr} = - \rho g,$$ is the equation of hydrostatic equilibrium, where $\rho$ and $g$ are the local density and gravity, $P$ is pressure and $r$ is the radial coordinate. This can be rewritten as $$\frac{d\rho}{dr} \frac{dP}{d\rho} = -\rho g.$$

Since $\rho$ and $g$ are positive numbers, the pressure gradient is negative.

For all types of matter $P = f(\rho)$, such that $dP/d\rho$ is also positive (i.e. pressure increases with density).

Thus $d\rho/dr$ is negative and the density increases towards the middle.

EDIT: This argument only obviously works if you ignore the dependence of pressure on other variables - such as the temperature or number of particles per unit mass in your gas. It takes more work to establish that $dP/d\rho>0$ in a more general case. For example you might consider a situation where a sufficiently negative temperature gradient might counterbalance a positive density gradient. i.e. if $P = f(\rho, T)$ then $$ \frac{dP}{d\rho} = \frac{\partial P}{\partial \rho} + \frac{\partial P}{\partial T} \frac{dT}{d\rho}$$
and $dP/d\rho<0$ if $$ \frac{dT}{d\rho} < -\frac{\partial P/\partial \rho}{\partial P/\partial T}$$ Leaving this as a placeholder for the moment.

$\endgroup$
3
  • $\begingroup$ Sorry, but asserting $P = f(\rho)$ for “all types of matter” is just bogus. Case in point, you may have water in your cup at atmospheric pressure with $\rho \approx 1000\:\mathrm{\frac{kg}{m^3}}$. But put that water in an enclosed steam kettle and bring to the boiling point, pressure will increase drastically without $\rho$ increase – on the contrary it decreases. $P = f(\rho,T)$ is closer to being true. So your argument would only hold true if $T$ were constant or increasing with radius, which is not at all true either. $\endgroup$ – leftaroundabout Aug 17 '20 at 10:10
  • 1
    $\begingroup$ The real problem with radially-increasing $\rho$ is that absent sufficient viscosity, it immediately spurs convection which reduces the temperature gradient until density is radially-decreasing. $\endgroup$ – leftaroundabout Aug 17 '20 at 10:15
  • 1
    $\begingroup$ @leftaroundabout It is not bogus to claim that P is a function of density. It is obviously not ONLY a function of density (an ideal gas being a case in point). However, I see your point and will think about it. In real stars convection zones have $P\propto \rho^{5/3}$ and radiative zone have $P \propto \rho^{4/3}$ independent of any consideration of the temperature. $\endgroup$ – ProfRob Aug 17 '20 at 10:15
5
$\begingroup$

Not as a steady state, for sure. If there happen to be a part less dense than its surroundings it would just float to the surface.

This is called buoyancy and the math is basic enough.

Stars don't have solid parts and cannot sustain a state different from the hydrostatic near-equilibrium.

Then again, some transient event like supernova Ia explosion may create inverse density distribution with a void in the center, but I am not sure it is a legitimate star anymore at this point (AFAIR Ia explosions leave no dense remnant).

$\endgroup$
5
  • 1
    $\begingroup$ Apart from “solid parts” or viscosity, magnetohydrodynamic forces can also stabilise a non-hydrostatic density configuration. I think this doesn't happen in stars themselves, but it does happen in space plasma all around. $\endgroup$ – leftaroundabout Aug 17 '20 at 10:30
  • $\begingroup$ @leftaroundabout I think you should write an answer. Your comments are quite valid and you might be able to put something together more comprehensive and succinct. I have paused at present (too busy) before reproducing a whole load of stuff about stability against convection. $\endgroup$ – ProfRob Aug 17 '20 at 10:40
  • $\begingroup$ @leftaroundabout affect yes, stabilise no because these phenomena are dynamic in nature. Sun spots are non-hydrostatic, yes (the whole convection is not hydrostatic by definition, but the spots are even less so). Now that I am thinking about it, there are stars that have convection in their cores so they do have some density variation there. Then again, the convection is extremely efficient mechanism for heat exchange and the goal put in the OP (density less than the average) is unthinkably high. $\endgroup$ – fraxinus Aug 17 '20 at 10:41
  • $\begingroup$ Not to say that it is, but in principle considering different composition of layers one can imagine a lot of combination in which equilibrium exists so that a heavy layer is "floating" above an ignited and active core. No to say it happens, again I want to be clear on this. Pure gymnastics. $\endgroup$ – Alchimista Aug 18 '20 at 11:31
  • $\begingroup$ @Alchimista that's how the convection happens. But the density variation is tiny and limited by the adiabatic gradient (or slightly off, depending on the convection efficiency). Then again, our lovely Sun's average density is ~1.4 and in the core it is 100+. The Sun doesn't have a core convection, but other main-sequence stars that do have one can't be extremely different. You need quite a deal of disturbance in order to create a less than 1.4 bubble in a 100+ density core. That's what the supernova events are for (and they happen in even denser star cores). $\endgroup$ – fraxinus Aug 18 '20 at 13:00

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.