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The question kind of says it all, what I really want to know is are the differences in their forms only due to the co-ordinate transform?

And as such should a suitable spherical system satisfy axisymmetric Jeans equations? (numerically rather than analytically, otherwise it would all get very messy...)

More background: Providing them in sufficient detail to be helpful would be very longwinded (as presumably you'd need to compare all three of the first moment equations in both co-ordinate systems) but for a bit more background people can always look here http://en.wikipedia.org/wiki/Jeans_equations, or in for an in depth discussion look in Galactic Dynamics by James Binney and Scott Tremaine.

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    $\begingroup$ Could you provide both forms of the equations to give future readers some context? $\endgroup$ – Jim Aug 7 '14 at 17:21
  • $\begingroup$ Providing them in sufficient detail to be helpful would be very longwinded (as presumably you'd need to compare all three of the first moment equations in both co-ordinate systems) but for a bit more background people can always look here en.wikipedia.org/wiki/Jeans_equations, or in for an in depth discussion look in Galactic Dynamics by James Binney and Scott Tremaine. $\endgroup$ – Zephyr Aug 7 '14 at 17:34
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The Jeans equations can be a bit tricky. Their simplest form (in cartesian coordinates, with no particular assumptions) is:

$$\frac{\partial\nu}{\partial t}+\frac{\partial(\nu\bar{v_i})}{\partial x_i} = 0$$

$$\nu\frac{\partial\bar{v_j}}{\partial t}+\nu\bar{v_i}\frac{\partial\bar{v_j}}{\partial x_i} = -\nu\frac{\partial\Phi}{\partial x_j}-\frac{\partial(\nu\sigma^2_{ij})}{\partial x_i}$$

Quickly summarizing the notation, $x_i$ and $v_i$ are the components of the position and velocity, respectively and $\sigma_{ij}$ are components of the velocity dispersion tensor. $\nu$ is the spatial density. $\Phi$ is the potential.

The trouble with the Jeans equations is that, assuming you know the potential and density fields, the equations above provide only 4 constraints (notice the second equation is in fact a condensed expression of three independent equations) on 9 unknown functions (3 components of $\mathbf{\bar v}$ and 6 independent components of $\mathbf{\sigma^2}$). Therefore the equations cannot be solved without additional constraints.

Galactic Dynamics derives closed forms (i.e. that can in principle be solved) of the Jeans equations for two special cases. For a spherically symmetric system, the assumptions used are:

  • The system is in a steady state (time independent).
  • The system is spherically symmetric, therefore the distribution function is of the form $f=f(H,\mathbf{L})$

For an axially symmetric system, the assumptions used are:

  • The system is in a steady state (time independent).
  • The system is axisymmetric.
  • The distribution function is of the form $f = f(H,L_z)$.

The assumptions are used in each case to argue that various (different in each case) derivatives vanish, as well as some other terms, so the equations obtained are not equivalent and hold only for systems where the assumptions used to derive them hold good. I'm not going to bother typing out the equations obtained because honestly it's a bit of a mess. If you want the gory details, go find a copy of Galactic Dynamics...

Of course you could also just do coordinate transformations on the original Jeans equations, in which case you would get equations that are equivalent, but this is not what people usually mean by 'Jeans equations for spherical systems' or 'Jeans equations for axisymmetric systems'.

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  • $\begingroup$ my sentiment exactly with typing the equations :) $\endgroup$ – Zephyr Aug 7 '14 at 18:12
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    $\begingroup$ @Zephyr Hang on just saw your comment to the question - you're reading B&T and it didn't answer this question satisfactorily for you? o.O $\endgroup$ – Kyle Oman Aug 7 '14 at 18:13
  • $\begingroup$ you're reasoning above sounds perfect, but leaves the possibility of a slightly more messy and more general form of the equations in both co-ordinate systems based on the same assumptions? Could we assume a less strict symmetry and have equations valid in both? $\endgroup$ – Zephyr Aug 7 '14 at 18:15
  • $\begingroup$ and of course B&T sheds light on all things, but doesn't go into much detail on the stuff in the comment above (also you summarised it very well, which cleared a few points up!) $\endgroup$ – Zephyr Aug 7 '14 at 18:17
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    $\begingroup$ @Zephyr The only symmetry you could assume is axial ($\phi$), and you wouldn't be able to get to a closed system of equations (I think). Probably better off to just use the full equations and coordinate transformations... $\endgroup$ – Kyle Oman Aug 7 '14 at 18:20

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