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This is a follow up question to Hydrostatic Equilibrium in a Galaxy Cluster. Is this a reasonable estimate of the hydrostatic equilibrium equation for a spinning mass of gas?

$$\frac{v(r)^2}{r} - \frac{kT(r)}{wm_pr}\left[\frac{d\ln \rho_X(r)}{d\ln r}+ \frac{d \ln T(r)}{d \ln r}\right] = \frac{GM(r)}{r^2}$$ Where $wm_p$ is the mean molecular weight times the mass of a proton, $\rho_X(r)$ is the density of gas at radius $r$, and $T(r)$ and $v(r)$ and $M(r)$ are the temperature, velocity and mass (respectively) at the same radius.

Is there a canonical expression for the accelerations in this kind of body or is the idea of hydrostatic equilibrium invalidated with a rotating body?

Community
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    Planets and stars rotate, so hydrostatic equilibrium is clearly valid for rotating bodies. Out of curiosity, where did you find this equation? – HDE 226868 Oct 14 '15 at 22:49
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    Derived it (added the centripetal acceleration) from the model independent version found in http://arxiv.org/abs/0709.0108. –  Oct 14 '15 at 23:00
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    There is a problem with your equation: The original equation you based it on assumed spherical symmetry. If the gas rotates the symmetry is reduced to cylinder symmetry. So you generalization is certainly not valid, as the centripetal acceleration depends on the distance from the axis $\rho$ not the distance from the center $r$. (But is should be possible to derive a correct generalization, although it will probably depend on two variables: $\rho$ and $z$). – Sebastian Riese Oct 14 '15 at 23:50
  • @SebastianRiese I understand that the velocity will approach zero at the axis and be understated for the equator, but if you're looking at a mass of rotating, hot gas looking for red and blue shifted iron ion spectrum lines, is this a reasonable formula for the maximum velocity you'd expect to find? –  Nov 02 '15 at 20:38
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    BTW All this stuff can be found in any textbook on stellar structure. I first encountered it in Schwartzchild's book---all the data in which is massively dated now, but the basic physics is there and it is relatively inexpensive. – dmckee --- ex-moderator kitten Nov 04 '15 at 15:45
  • @dmckee Thanks for the suggestion. The problem is that I'm investigating a MOND-like model. I've been through all the sources available including Lau et al (2009) where he specifically goes into the dynamics of rotating galaxy clusters. The problem is that all these formulas depend on a Newtonian gravity potential. The formula above is trying to remove the model dependence (you'll notice I've collected the gravity model to one side of the equation where I can substitute other models). –  Nov 04 '15 at 15:48
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    For the purposes of the questions you've been asking it doesn't matter. You'll have to change the gravitational potential eventually which changes the solutions to your boundary value problem, but you are still asking about setting up the boundary value problems and that step simply doesn't change. And there are standard resources that cover that. – dmckee --- ex-moderator kitten Nov 04 '15 at 15:52
  • The force of gravity and the gravitational potential energy are two different things. I don't follow why I'll need the potential energy to solve what is essentially a balance of forces question. Unless you're suggesting that the whole pressure gradient force term needs to be tossed. –  Nov 04 '15 at 16:01
  • "The force of gravity and the gravitational potential energy are two different things." Uh ... they are intimately and unavoidably coupled! And the potential form is usually more mathematically tractable. That's why people use it. – dmckee --- ex-moderator kitten Nov 04 '15 at 18:01

3 Answers3

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The equation you have written cannot be correct, even approximately, except in cases where rotation is unimportant.

If rotation of the gas is important, that rotation occurs around an axis. This breaks any spherical symmetry (and if it doesn't then rotation is unimportant), makes the problem at least two dimensional and an equation featuring a single spatial coordinate won't do.

To make further progress then the relevant expression for hydrostatic equilibrium depends on the rotation law of the gaseous object $\Omega({\bf r})$.

If the cloud rotates as a solid body, such that $\Omega$ is constant, then we can write. $$ \frac{1}{\rho} \nabla P = -\nabla \Phi + \frac{1}{2} \nabla (\Omega^2 r^2\sin^2 \theta)\ ,$$ where $\nabla P$ is the local pressure gradient, $\Phi$ is the gravitational potential, and the second term on the right is the gradient of the centrifugal potential with $r \sin \theta$ representing the distance to the rotation axis at co-latitude $\theta$.

This can be rewritten by assuming a new combined potential $$ \Psi = -\frac{GM_r}{r} - \frac{1}{2}\Omega^2 r^2\sin^2 \theta$$ so that $$ \frac{1}{\rho} \nabla P = -\nabla \Psi $$

In general then, the pressure gradient is no longer in the radial direction, however the isobars are still equipotentials.

An alternative approximation is to consider "shellular rotation", where $\Omega$ is constant along isobars. In this case $$ \frac{1}{\rho} \nabla P = -\nabla \Psi -r^2 \sin^2 \theta \nabla \Omega $$ Here, the surfaces of constant pressure are not equipotentials.

If all you are interested in is establishing whether rotation could alter the pressure gradient, then you just need to compare $GM/r^2$ with $\Omega^2 r$.

ProfRob
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  • The thermal profiles of galaxy clusters guarantee that these clouds do not move like a solid. You generally get a peak temperature about 100 to 400 kpc of the radius and then it drops off to zero at the edge. At the edge, the thermal forces will balance the gravitational forces and there's no need for any additional centripetal force. Also, the formula depends on the gravitational potential, $\phi$, which is model dependent (that is, we know this formula doesn't work on something the size of a galaxy without the addition of some virtual mass). –  Nov 02 '15 at 15:02
  • @DonaldRoyAirey The formula I have given is correct for solid body rotation. There is no general equation that you can write down independently of an assumed form for $\Omega(r)$. Of course the formula depends on $\Phi$, as does the version without rotation. – ProfRob Nov 02 '15 at 15:46
  • OK, I'll try my comment in a different way then: what am I missing? Why do you think these clouds rotate like a solid body? I see where this formula works for a star, but I don't see it's application to a gas cloud 2 Mpc in diameter with a range of temperatures from 10 Kev to just above zero. –  Nov 02 '15 at 16:03
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    @DonaldRoyAirey I don't! Postulate an alternative $\Omega({\bf r})$. Let me spell it out. There is no simple formula and the one you have postulated is manifestly incorrect, since it assumes spherical symmetry when there will be an axis of rotation. Gas does not "orbit" like non-interacting bodies in a central potential. – ProfRob Nov 02 '15 at 17:14
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The centripetal force is an aggregate of real forces. $$ F_c = m\frac{v^2}{r} = \sum_i F_i$$ The real forces at work here are the gravitational force inward and the pressure gradient force outward. Using the above equation we have $$-\frac{v^2}{r}\hat{r} = -g(r)\hat{r} - \frac{dP}{dr}\times\frac{1}{\rho(r)}\hat{r}$$Rearranging, removing the unit vector and expanding the terms we get: $$\frac{v^2}{r} = \frac{MG}{r^2} + \frac{kT(r)}{wm_pr}\left[\frac{d\ln \rho_X(r)}{d\ln r}+ \frac{d \ln T(r)}{d \ln r}\right]$$ This, of course, has the disclaimer that spherical symmetry is broken for any significantly large rotation. Also, that the centripetal force component $\frac{v^2}{r}$ is a proxy for all the dynamical forces (rotational, turbulence, etc.) found in a real gas cloud. It's important to note that the centripetal acceleration, $\frac{v^2}{r}$, approaches zero as you get to the cluster edge, so spherical symmetry is restored in the outer annulus. The formula is a 'ballpark' estimate.

  • This is incorrect for (at least) the reason given by Sebastian Rise above. The incorporation of any centrifugal terms will also depend on whether or not co-rotation is established. – ProfRob Nov 02 '15 at 13:53
  • Yes, of course. I could also make the argument that $E = \frac{1}{2}mv^2$ is incorrect, but it is a useful approximation. Is the formula above a useful approximation? –  Nov 02 '15 at 13:58
  • It is ok so long as rotation is unimportant! – ProfRob Nov 02 '15 at 14:26
  • I'm not trying to make the argument that it's unimportant. I'm trying to make the argument that this is a 'ballpark' estimate. We use the Virial Theorem all the time knowing that these bodies are not perfectly spherical but it tells us whether a theory, like MOND, needs some fine tuning or is off by an order of magnitude. Would the above formula work in this context? –  Nov 02 '15 at 16:49
  • A ball park estimate simply calculates whether $\Omega^2 r^3/GM$ approaches unity. QED. Your actual question (which I hesitated to answer because they all end up in an argument like this, where you ask the questions but seem to think you already know the answer) quite specifically asks whether the form of the equation you postulate for hydrostatic eqm. in a rotating gas cloud is correct. It isn't. – ProfRob Nov 02 '15 at 17:25
  • My actual question is "is this a reasonable estimate". I'm not sure why your panties are in a knot because I still haven't got an answer to the question I did ask. None of these clusters are spherically symmetrical, so why do we even bother with Hydrostatic Equilibrium? Because it's close enough to perform reasonable estimates. –  Nov 02 '15 at 18:33
  • You use quote marks to indicate what was said. If you wish to rewrite your question then do so. – ProfRob Nov 02 '15 at 20:26
  • You've been very helpful with these concepts, Rob, thank you. –  Nov 02 '15 at 20:40
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From what I understand and what I have read in text books on the subject, most stars fall under a formula that says the gradient of pressure is equal to the gravity force.

$$\frac{1}{\varrho}\frac{dp}{dr}= -\frac{GM_r}{r^2}$$

This is valid for stars with no fast radial motions. Pulsating stars and stars that have fast radial motions complicate the equation. I can try to dig it up if you would like. I pulled this equation out of 'Physics, Formation and Evolution of Rotating Stars', A. Maeder.

cmontcalm
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  • Thank you. Yes, I agree with the general idea of hydrostatic equilibrium. The problem is that there's a growing body of evidence that these clusters rotate and have turbulence. For example, Perseus, Centaurus and A576 have rotations on the order of 1,000 km s-1. The formula above is trying to make a ballpark estimate of these non-thermal contributions to the mass estimate given that pure HE is an increasingly unreliable assumption. –  Nov 04 '15 at 15:54