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I am currently studying the Jeans equation for a system in spherical equilibrium. Since the distribution function $f(x,v)$ can be written in depency of energy and angular momentum $f(E, L)$ it seems to follow that most of the velocity moments disappear:

$$ \tag{1} \left<\upsilon_r\right> = \left<\upsilon_\vartheta\right> = \left<\upsilon_\varphi\right> = 0 $$

and

$$ \tag{2} \left<\upsilon_r \upsilon_\vartheta\right> = \left<\upsilon_r \upsilon_\varphi\right> = \left<\upsilon_\vartheta \upsilon_\varphi\right> = 0. $$

Where $f(x,v)$ denotes the number density of particles in phase space and $\left<Q\right> = \frac{m}{\rho(x)} \int Q\,f(x,v) \text{d}^3v$,$\;\;$ $\rho(x) = m \int f(x,v)\text{d}^3 v$.

I don't find it intuitive why these velocity moments are zero. What ist the mathematical way to get these results?

Gradgar
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1 Answers1

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Jeans equation is just an analog to the Euler equations.

In the limit you describe, the results of your Equation 1 imply you are working in the center of momentum frame, i.e., the bulk flow rest frame. The results of equation 2 state the pressure tensor can be diagonalized, i.e., there is no viscosity. Viscosity arises from off-diagonal terms in the pressure tensor.

I have not heard the term "spherical equilibrium" before but I assume you are implying spherial symmetry? Regardless, in the bulk flow rest frame there is no first velocity moment, i.e., it is zero. The lack of off-diagonal terms implies there are no stress/strains in the flow, e.g., azimuthal flow will not be transported across a radial plane.

honeste_vivere
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