User:Gaplec/sandbox

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-- = Table of Contents =

Lec 1: Micro (PDF functions/"frictions" N-body) → Macro (Continuity Eq./Momentum Eqs, CE/ME)

Lec 2: Exercise: N-body anisotropic velocity tensor/ellipsoid → measure/balance Gravity (ME).

Lec 3: Exercise: XYZ Dissecting the Melon Galaxy (cont.): → grainy friction, smooth orbits (Invariants)

Lec 4: Invariants in collision-less eq. (cont.) → anisotropic ellipsoid/galaxy shapes (CBE0)

Lec 5/6: Various Derivations of Momentum Eqs.: with/without relaxation (SKIPPED):

Lec 5/6: Connecting the GAP: Viscosity (=Gravitational Friction?), Hydrostatic (=ME?, SKIPPED)

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Conventions < User:Hongshengzhao
In 'Gravitational Physics': we develop a statistical language to model motions of $$N=10^{3-12}$$ macroscopic stars and countless neutrinos etc. Our regime is between solar system dynamics and thermal fluid regimes, with $$ \ln N=O(20)$$.

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Just as Thermal and Statistical lectures use occupation and partition function etc., we in stellar dynamics use Particle Distribution Functions (PDFs) to encapsulate all information of N-body systems in a probabilistic manner, e,g., to compute the average relative velocity between colliding particles.

Literatures ofter various names (phase-space distribution density etc for a point mass $$ m_\bullet = M_\odot$$) and normalisations for the PDF: $$ \int f(t,\mathbf{x},\mathbf{v})dx^3 dv^3 = N, 1, m, m^{-3}, Nm ...$$ by setting something (e.g., particle mass) to unity, $$ m=1 \text{unit} ={M_\text{total}/ N_\text{total} } ={\mathbf{p} /\mathbf{v}} $$.

Resisting these conveniences, we keep the particle mass "m" in all our equations to differentiate its momentum with velocity (except that we continue using traditional phrases "potential" or "energy" for a (test) particle to mean "its potential energy per unit mass").

$$ where $$ dN/N $$ is the chance of locating one of N particles both in a space cell $$d\mathbf{x}^3$$ and in a velocity cell $$d\mathbf{v}^3$$. (If a lecturer slips into another normalisation (say call m f, f/Ntotal, m f/Ntotal, ... as their f), it is worth calling out, although often harmless as most relative "moments" such as relative RMS speed are invariant if shifted to another normalisation, velocity zero point etc.)

The p-subscript $$-_p$$, often omitted for ease of reading, means a "p"article sub-population (say red giants), which provides (small) part of the N-body system's full gravity $$-\nabla \Phi(\mathbf{x}) $$. We generally imply working on a sub-population, unless all gravity is explicitly sourced by this population (aka self-gravity).

Time-dependency is often implied with t omitted, unless in an explicitly time-independent (aka steady-state) system. In hairy equations we shorthand t,x,y,z,vx,vy,vz as t.y.vx.. where each dot is a coordinate obviously implied by the context; or write the (x, y, z) position-dependency as a subscript $$ ._\mathbf{x}$$, or subscript index $$ ._1, ._2, ._3; \text{e.g.} \nabla_3 \equiv \partial_3 \equiv {\partial \over \partial x_3} = \partial_{\mathbf{x}_3} $$.

Overdot means full-time derivative: $$

= Table of Contents =

Lec 1: Micro (PDF functions/"frictions" N-body) → Macro (Continuity Eq./Momentum Eqs, CE/ME)

Lec 2: Exercise: N-body anisotropic velocity tensor/ellipsoid → measure/balance Gravity (ME).

Lec 3: Exercise: XYZ Dissecting the Melon Galaxy (cont.): → grainy friction, smooth orbits (Invariants)

Lec 4: Invariants in collision-less eq. (cont.) → anisotropic ellipsoid/galaxy shapes (CBE0)

Lec 5/6: Various Derivations of Momentum Eqs.: with/without relaxation (SKIPPED):

Lec 5/6: Connecting the GAP: Viscosity (=Gravitational Friction?), Hydrostatic (=ME?, SKIPPED)

Example: molecular isothermal velocity distribution
The coolest brown dwarf star could be modelled as molecules in a constant "room temperature" with equipartition energy $$ {k_B T_0 \over 2} \approx {k_B ~\mathrm{300K} \over 2} = {m_p \mathrm{(1.3km/s)^2} \over 2 }$$; velocity RMS spread of any litre of particles in the cool star is $$ \sigma_r =\mathrm{1.3km/s}$$ everywhere in any direction.

Particles in such cool dwarf have a 3D (iso)thermal (aka Maxwell or Gaussian) PDF, "globally" (in spherical coordinates) as $$ where $$n(r)$$ is the spherical number density of particles, the potential $$ \Phi(r) $$ and kinetic energy per unit mass are collected in the specific energy $$ E(\mathbf{r},\mathbf{v})$$, an implicit function of coordinates.


 * §§§: Is it true: that $$ \langle v_r^2\rangle + \langle v_\varphi^2\rangle + \langle v_\theta^2\rangle = (\sqrt{3} \times \mathrm{1.3km/s} )^2 $$ for every litre of molecules, because particle's PDF implies:

uniform (being globally everywhere with an identical) rms spread of

a 3D isotropic (being the same for all velocity directions)

thermal (Exponential dependence on Kinetic Energy)?

Moments of PDF (particle distribution function)
Consider a small cell of rectangular (aka Cartesian) volume dxdydz, the particle local number density $$ n(x,y,z) $$, mean x-velocity $$ \langle V_x\rangle $$, rms transverse-velocity $$ \langle V_t^2\rangle $$ are so-called "moments" of the DF (distribution function) f, defined as

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Clearly a scalar like energy is independent of rectangular, cylindrical or spherical coordinates; also averaging is a linear operation, one could half both sides.


 * §§§: Is this true, the thermal distributed hydrogens at any point (x,y,z) of a brown dwarf have zero local mean velocity $$ {\mathbf u} \equiv ( \langle v_x\rangle, \langle v_y\rangle , \langle v_z\rangle )= 0  $$?

Decay of velocity anisotropy from microscopic relaxation (friction or collisions)
Particle Distribution Function (phase density) can evolve, from an initial state $$ f_0(x,v)$$ to reach an entropic-favoured state (i.e., a thermal velocity spread $$ f_\infty(x,v)$$), due to either violent mergers of galaxies (called violent relaxation) on macroscopic level, or internal friction such as two-body encounter relaxation.

Boltzmann proved that local collisions always increase the entropy and isotropy in the phase space. For a given energy, velocity isotropy gives more states than anisotropy (meaning particles move in a narrower set of directions).

The PDF in galaxies is not yet thermal, as they are only middle-aged dynamically with only a few dozen collisions.

The PDF evolves, expressed as a full derivative $$\dot{f} \equiv \frac{d f}{dt} \equiv {D f\over Dt}$$, via a diffusion Eq., $$ Here the frictional timescale $$ \tau^\text{frc} \rightarrow \infty $$ in the collisionless limit, the PDF stays essentially unchanged along the trajectory of a parcel of particles.

Let "C" denote a "C"onserved quantity in scattering, e.g., $$C=1, C=m c^2, C=0.5 m v^2, \mathbf{C}=m \mathbf{v}$$. Then any entropy-maximising path (aka relaxation) to thermal PDF is barred from deviating from pre-collision particle number, rest mass energy, KE, and momentum etc!

CBE predicts particles are "isotropically" scattered to diverging trajectories by the randomness of collisions ("internal friction") $$ {(f_t -f_\infty) \over (f_0 -f_\infty) } \rightarrow e^{-t \over \tau_\text{frc}}, 0 \equiv {\dot{f}_{\!\!\infty}\!(E) \over f_{\!\!\infty}(E)} \approx  O({1 \over \tau_\text{frc}}) \approx O({\dot{f}_{E(t),J_1(t),J_2(t),J_3(t),...} \over f})  \approx {\dot{E}(t) \over \sigma_r^2} \approx {\dot{|J|}(t) \over r \sigma_r} \approx {\dot{f} \over f}|_{\tau_\text{frc} \rightarrow \infty} =0, $$ so the end state is a thermal $$ f_\infty $$ isotropic distribution of velocities, with the equipartition of random kinetic energy (temperature) everywhere in 3-directions. Particles tend to shuffle in the PDF, towards the thermal peak at the population's mean velocity $$ \mathbf{u}$$, say, at zero.

So how often do stars meet and relax?

How long is this friction time scale $$ \tau_\text{frc} $$?

Collision process details skipped, to be discussed at very end
For now, let's come back to the easier problem: conservation of particles and momentum in flux.

Application moments: Continuity Eq. and Momentum Eqs. in XYZ Coordinates
Let's call a conserved quantify $$ C_i$$, such that particle number and C are also conserved for N-body system with internal flow, conserved before and after encounters. E.g., $$ C_0 =1$$ or $$ C_0 =m $$ or $$ C_i =m v_i$$ means number or mass or momentum is conserved in the collisions (apart from the important fact that the centre of mass momentum of colliding particles is changed by the smooth part of gravity during the brief collision).

Consider a cell of mass $$m_\boxdot$$  with the mean flow field or C-flux (as function of time t and position x) are $$

Counting particle C numbers in a cell from inflows
We find that over a time $$ \Delta t$$, the gain of $$ C_i $$ in the cell differs from the rate of C-creation in the cell, so the discrepancy must be equal to influx from total 6 neighbours (three pairs of external sources), $$ where the three square brackets retain the 1st order (Taylor) spatial variation of the flux.

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Collect the momentum and mass terms together
divide the cell volume $$ dx_1 dx_2 dx_3$$ and time $$ \Delta t $$ on both sides, we have the so-called Continuity Eq. and its momentum counterpart. $$

Let $$ u_i \equiv \langle v_i \rangle $$ times the first equation (CE), and be subtracted from the second equation

the simplified Momentum Equation and expanded CE are collected together here, in tensor form valid for all coordinates: $$ where the first bracket term on the rhs of ME and CE is collapsable as a full derivative (Lagrangian form).

Usage of the ME: measure gravity
as with the good old Newtonian Eq. of motion for a box of mass $$ m_\boxdot$$ under gravity and external forces, $$

Unlike for a single free-falling box, three pairs of force are added from differential pressures at the 6 walls.

One goal is to measure the LHS gravity from the RHS "observables", e.g., in spherical coordinates where $$ \nabla \cdot \mathbf{u} = {d (r^2 u_r) \over r^2 dr},   m_{_\boxdot} \rightarrow m_{_\bowtie} \equiv m  [dr (r^2 d\Omega) ] n_\mathbf{r} $$ (Next lecture).

Homework Till Monday: APPLY "XYZ Dissections" of Pressure tensor of a Velocity Ellipsoid
Complete the z-Momentum Eq. RHS expression (end of page) for z-gravity by inserting following pressure gradients (Given, but review their logic) in terms of $$(r,\theta,\varphi),  (\sigma_{rr},\sigma_{\theta\theta},\sigma_{\varphi\varphi}).$$  Finally, remove common factor $$ z/r$$. Have you accidentally re-derived Spherical M.E.? ⇒ ♥ Congratulations ♥.

--- The general Dispersion tensor and cross terms $$\sigma_{ij}$$ (in xyz), are computed (by TEDIOUS spherical projection) from $$ (\sigma_{rr},\sigma_{\theta\theta},\sigma_{\varphi\varphi}) = \langle (V_r^2\rangle -0^2, \langle V_\theta^2\rangle-0^2, \langle V_\varphi^2\rangle -0^2)$$ $$ $$

Lec 3 (Fri): Dissect Melon Exercise: stationary anisotropic velocity ellipsoid → measure/balance Gravity (ME)
With M.E. (of Lec2) Check, if Earth's z-gravity on a box of air is balanced by difference of pressure (random velocity dispersion $$\sigma_{zz}, \sigma_{yz}, \sigma_{xz} $$) acting on 3-pairs of face areas $$ (dxdy, dxdz, dydz) $$:

§§§: Two RHS forces are zero-ed, because atmosphere: (a) is stationary? (b) is rising up? (c) is isothermal? (d) has tilt-less (aka isotropic) pressure tensor: $$ P_{ij}= n_{(x,y,z)} kT \delta_{ij}=0, \text{if ij=xz or yz} ? $$. $$

Excursion: Stress in life (exams, tides, B-pressure) is anisotropic (=directional, tilted)
Stress description needs more than a scalar number (hence a tensor, or a matrix of 2x2, or 3x3, 4x4 with +/- numbers!).

I might apply a $$ P_{z~y} <0 $$, meaning negative-z force, to paint a wall tile $$ \hat{dz} \times \hat{dx} $$, ... $$ P_{y~x} =P_{x~y}=\pm, ~P_{x~z}=P_{z~x}=\pm$$ ...

I might apply a $$ P_{y~z} <0$$, meaning negative-y force, to paint a floor tile $$ \hat{dx} \times \hat{dy} $$, ... $$ P_{y~x} - P_{x~y}=0 $$ (zero torsion, no roller) ...

But I apply a $$ P_{z~z}>0 $$ to paint the xy celling, a $$ P_{y~y} >0 $$ to paint the xz door, a $$ P_{x~x} >0 $$ to paint the yz window.

Also E-M pressure is a 3x3 tensor: A vertical current I creates $$B_\varphi$$-field with a cylindrical radial pressure gradient $$\partial_R [P_{RR}, P_{zz}, P_{\varphi\varphi}] = \partial_R [{\mu_0 I^2 \over 2(2\pi R)^2}, {\mu_0 I^2 \over 2(2\pi R)^2}, {-\mu_0 I^2 \over 2(2\pi R)^2}]$$, confining a parcel of protons to cycle along/against I (aurora?)

Force balance in an $$ \mathbf{u}=0 $$ Stationary Melon (e.g., Spherical Cluster 47 Tucana)
Consider a spherical N-body system with a density $$ \rho(r) \approx m n(r)$$ in equilibrium with spherical symmetry in every sense, except that the radial pressure is smaller than the tangential pressure, i.e.,

No.1: Rule of Thumb in dynamics is force-balancing. For more rigorous statistical balance of force, let's consider first some ...

Here comes an spherical example of momentum balance equation...
Let $$ P_{rr}, P_{\theta\theta}, P_{\varphi\varphi} $$ be the radial, meridional, azimuthal pressure at the midpoint of a cell of mass $$ dM $$; values elsewhere are corrected by an infinitesimal (aka Taylor) expansion term.

For easy of calculation, we express the cell's radial width

Ab Initio Derivation: Count the 3 pairs of r-momentum-flux (radial force) from walls $$ (r \mp 0.5dr),(\theta \mp 0.5d\theta), (\varphi \mp 0.5d\varphi) $$ of $$ dM $$
$$ where we substituted (dr) in favour of (dM).

Radial pressure doesn't balance gravity $$ - {G M dM \over r^2} $$, because of wedge forces
there are two equal forces pressing from the two walls at $$ \theta \pm 0.5 d\theta $$, whose small $$\sin (0.5d\theta)$$ outward projections add up like a wedge of melon or an arch. $$

sum up forces from 6 walls plus gravity
Now we can predict the total acceleration of the cell $$ \dot{u}_r$$via,

where we used the chain-rule for full-time derivative of flow speed.

So we succeeded deriving, from Newton's 2nd law, (aka "Jeans Eqs.") for an equilibrium of a spherical galaxy of potential $$ \Phi(r) $$.

Let's apply our equation to solve a real problem: can we measure the gravity if we know these moments of Particle Distribution Function (PDF)?


 * §§§: Clearly the spherical ME seems more handy for spheres, we have no cross terms, but the physics is the same if coordinates rotated. We know that along the z-axis of a sphere $$\sigma_{zz} \rightarrow \sigma_{rr} \& \sigma_{RR} \rightarrow \sigma_{\theta\theta} $$. Then where in the sphere $$ \sigma_{zz} \rightarrow \sigma_{\theta\theta} \&\sigma_{RR} \rightarrow \sigma_{rr}  $$?
 * §§§: Given the vertical kinetic energy of any system can be expressed as $$ 0.5M\overline{\langle\mathbf{V}_z^2\rangle} = \overline{0.5 \sigma_{zz} + 0.5 u_z^2} M $$, where overline meaning summed over xyz-volume of whole system.  Also given $$ \overline{\langle\mathbf{V}_r^2\rangle  + \langle\mathbf{V}_\theta^2\rangle + \langle\mathbf{V}_\varphi^2\rangle} = k (\overline{\sigma_{yy} + u_{y}^2+\sigma_{xx} + u_{x}^2+ \sigma_{zz} + u_{z}^2}) $$.  What's the value of k?

Apply the Field Eq. of Gravity to "weigh the mass" of the galaxy
The "homework" (next page) gives a static sphere of uniform density $$ n(r)=n_0 $$ of anisotropic stress, $$ \begin{align} \sigma_{ij=r,\theta,\varphi}(\mathbf{r})= \langle\mathbf{V}_i\mathbf{V}_j\rangle- 0 \times 0 ,\\ = & \begin{bmatrix} \left[1-({r \over r_0})^2\right]\left({\omega r_0\over 2}\right)^2 & 0 & 0 \\ 0 & \left({\omega r_0\over 2}\right)^2 & 0 \\ 0 & 0 & \left({\omega r_0\over 2}\right)^2 \end{bmatrix} \end{align} $$ So given the pressure tensor of a test population, with zero off-diagonal terms because of the symmetric velocity distribution, we can "weigh the gravity and potential" from Momentum Equation, and then "weighs mass and density" via the Gravitational Field Eq. (aka Poisson Eq.)

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Anisotropy and Tracer

 * Gravity in galaxy is balanced in part by an anisotropy force, on top of usual hydrostatic radial pressure gradient. The eq. has been formulated to account for the centrifugal-like force from tangential rotation or random motion with $$ \langle{v_\varphi^2}\rangle +  \langle{v_\theta^2}\rangle \ge 2 \langle{v_r^2}\rangle $$;

§§§: Can $$\langle{v_\varphi^2}\rangle = \sigma_{\varphi\varphi}+ \langle{v_\varphi}\rangle^2$$ come from pressure and rotational KE?

§§§:Momentum Eq. is valid for the decline of Earth atmosphere pressure balancing gravity from the solid core (not atmosphere). Can the M.E. weigh the solid Earth, given measurements of the atmosphere kT and density radial profile?

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Homework Till Wed: Read through this work out example
Here you "measure" a PDF's various moments first, then put moments into (ME) to predict the gravity on the RHS

A example of PDF to "measure its moments" and then "measure gravity"
Let's measure the gravity responsible for this stellar PDF (in spherical coordinates), $$ f(|\mathbf{r}|,V_r,V_\theta,V_\varphi) = {C_0 \over (\omega r_0)^2 (4I_{\mathbf{r},\mathbf{v}})^{1/2} }, 2I \equiv (1-{r^2 \over r^2})(\omega^2 r_0^2 -V_\theta^2 - V_\varphi^2) - V_r^2 \ge 0. $$ Let's use a short-hand unit-vector  $$ {\mathbf u}= (u_r,u_\theta,u_\varphi) $$ being $$ (V_r, V_\theta,V_\varphi)$$ rescaled by their respective ranges such that $$ $$
 * {\mathbf u}|^2 \equiv u_r^2 + u_\theta^2 + u_\varphi^2 \equiv {V_r^2 \over V_e(r)^2} +  {V_\theta^2 + V_\varphi^2  \over V_e(0)^2}  \le 1,  {V_e(r)^2 \over 2} \equiv {\omega^2 (r_0^2-r)^2 \over 2}, I \equiv {V_e(r)^2 \over 2} (1-|{\mathbf u}|^2) \ge 0.

Example of zero streaming motion and crossterms from reflection symmetry
Our DF is of spherical symmetry in position space, and reflection symmetry in velocity. So this PDF puts equal number of stars on both sides of $$ \pm V_\varphi $$. Hence zero cross term $$ \langle\mathbf{(\pm V_r) V_\varphi}\rangle $$ is predicted from the symmetry, because of equal weighting of +/-. $$

Velocity range is anisotropic limited by positivity
The range in three velocities is set because the DF is zero beyond some range. $$

Also note the boundary of integrations. Indeed, the PDF implies a narrow range on radius $$ 0 \le |\mathbf{r}| V_0 \equiv V_e(0) $$.

A worked example on rescaling the velocity ellipsoid
In fact, the positivity carves an ellipsoid in the $$[V_r, V_\theta, V_\varphi]$$ velocity space ("velocity ellipsoid", tangentially anisotropic).

The velocity ellipsoid (in this case) has rotational symmetry around the r-axis or $$ V_r $$ axis. It is partially isotropic, more squashed in the radial direction, because everywhere $$ V_e(r) \le V_0 $$, except at the origin, where the ellipsoid looks fully isotropic. Now we compute the moments of PDF.

Density moment in a spherical cell
$$\begin{align} (r^2 d\Omega) n & = (r^2 d\Omega) \int^{|u| \le 1} \left[ d (V_e u_r) d(V_0 u_\theta) d(V_0 u_\varphi) \right] ~f(|\mathbf{r}|,V_e u_r,V_0 u_\theta,V_0 u_\varphi) = d(No.of.particles),\\ n(r,\theta,\varphi) & = {C_0 V_e(r)V_0^2 \over 2 V_e(r) V_0^2} { \int_0^1 (1-u^2)^{-1/2} (4\pi u^2 du)} = {\pi^2 C_0 \over 2} = n_0 = n(0), du_r du_\theta u_\varphi \rightarrow (4\pi u^2 du) \end{align}$$ Note $$ du_r du_\theta u_\varphi \rightarrow 4\pi u^2 du $$ gives a canceling factor $$ V_e(r)^{-1} V_e(r)$$ to make the density constant inside the edge $$ r_0$$.

§§§ The normalisation constant $$C_0 = 2 \pi^{-2} n(0) $$. What are the correct dimension or SI units of $$ C_0 V_0^{-3} $$?

Velocity 2nd moments: Apply the substitution symmetry
The rms velocity in the azimuthal direction is computed by a weighted mean as follows, E.g., $$\begin{align} \langle\mathbf{V}_\varphi^2\rangle(|\mathbf{r}|) &\equiv {\int f d\mathbf{V}^3 V_\varphi^2 \over \int f d\mathbf{V}^3 } \\ & = {C_0 \over 2n} {\int_{u^2 \le 1} (du_r) (du_\theta) (du_\varphi) (1 - u_r^2-u_\theta^2-u_\varphi^2)^{-1/2} ~(u_\varphi V_0)^2 }  = 0.25V_0^2\\ & = {C_0 \over 2n} {\int_{u^2 \le 1} (du_r) (du_\varphi) (du_\theta) (1 - u_r^2-u_\varphi^2-u_\theta^2)^{-1/2} ~ (u_\theta V_0)^2 }  =\langle\mathbf{V}_\theta^2\rangle(|\mathbf{x}|)  \rangle  \\ & = {C_0 \over 2n} {\int_{u^2 \le 1} (du_\varphi) (du_\theta) (du_r) (1 -u_\varphi^2-u_\theta^2-u_r^2)^{-1/2} ~(u_r V_e)^2  } {V_0^2 \over V_e^2}=  {\langle\mathbf{V}_r^2\rangle(\mathbf{r}) \over 1-r^2/r_0^2} \end{align}$$

Here $$ \langle V_t^2 \rangle = \langle V_\theta^2 + V_\varphi^2\rangle =0.5V_0^2 \ge 2 \langle V_r^2 \rangle, $$, called tangential anisotropy, i.e., larger tangential kinetic energy than that of radial motion. This is often phrased by an anisotropy parameter $$ \beta(r) \equiv 1 - { 0.5\langle {\mathbf V_t}^2(|\mathbf{r}|) \rangle \over \langle{\mathbf V_r}^2\rangle(|\mathbf{r}|) } = 1 -  {\langle {\mathbf V_\theta}^2(|\mathbf{r}|) \rangle \over \langle{\mathbf V_r}^2\rangle(|\mathbf{r}|) } = - {r^2 \over r_0^2 - r^2} \le 0; $$ a positive anisotropy would have meant that radial motion dominated, and a negative anisotropy means that tangential motion dominates (as in this uniform sphere).

Now we have "measured" the Pressure tensor ...
So the dispersion at any position, fully described by a 3x3 matrics (called 3x3 tensor), is $$

Apply M.E. to the stress tensor
ME tells how the pressure gradient of a tracer balances the potential gradient for an equilibrium galaxy.

It is wrong to assume gravity is balanced only by

The radial pressure gradient
$$ -{d (n \sigma_r^2) \over n dr} = -{d \sigma_r^2 \over dr} - {\sigma_r^2 \over r} {d \log n \over d\log r} = {\omega^2 r \over 2} + 0. $$

The reason is Anisotropy
Gravity is partly balanced by anisotropic pressure $$ \begin{align} {(\sigma_\theta^2 -\sigma_r^2) \over r} &=0.25 \omega^2 r \ge 0 \end{align}$$

Verification of ME
Now we can verify that (in absence of collisions) the Momentum Eq. in r-direction, $$\begin{align} {\partial \Phi \over \partial r} & = 0 + {\bar{V}_\theta^2 +\bar{V}_\varphi^2 -2\bar{V}_r^2 \over r} + \left[-{d (n \sigma_r^2) \over n dr} + {\sigma_\theta^2 + \sigma_\varphi^2 -2\sigma_r^2 \over r} \right]  \\ & = {0 \over r} +  \left[ 0.5 \omega^2 r + {(0.5V_0)^2 + (0.5V_0)^2- 2 \times 0.25  (V_0^2 -\omega^2 r^2) \over r} \right] \\ & = (\omega^2 r), ~\omega \equiv {V_0 \over r_0} \end{align} $$ Hence the underlying potential must be of a harmonic oscillator form, up to $$ r_0$$. Reverse solving for the potential, $$ \Phi(|\mathbf{r}|) - \Phi_0 = \int_{r_0}^r (\omega^2 r) dr = {\omega^2 (x^2+y^2+z^2 - r_0^2) \over 2} \equiv -{V_e^2(r) \over 2}. $$ Interestingly $$ V_e(r) = \omega \sqrt{r_0^2-r^2} =\sqrt{2\Phi(r_0)-2\Phi(r)}$$ takes the meaning of the speed to escape to the edge $$ r_0$$; the speed to "escape from the edge to infinity" is $$\sqrt{2}V_0 = \sqrt{2} \omega r_0$$.

Usage of ME
Many observers have been suspecting a black-hole-like Keplerian gravity within the central 0.1pc. Such models would fit the high random dispersion of Doppler velocity $$ \sigma_r^2 $$ of central stars. They used the simple-minded isotropic and even tangential anisotropic velocity model.

Much ado with anisotropy
As we cautioned, isotropy is theoretically unnecessary assumption. Genzel's group spent much effort to falsify the possibility of a (radial) anisotropy as a masquerade of SMBH's gravity, they actually measured the anisotropy by physical proper motion of stars, a measure of transverse velocity dispersion. They in fact have measured a population's DF (apart from the less certain relative distances between central stars), a Nobel-prize quality evidence for a SuperMassive BH (SMBH).

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Explanation of Terms
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self-gravitating vs tracer: a tracer density/velocity $$ m n(r) $$ follows ME, move under the gravity $$ GM/r^2 \rightarrow \rho(r) $$, produced by (often BH, DM, gas or other) gravitating source's density. Bullets and high-jumpers: tracers of gravity from a self-gravitating Earth core.

$$ PDF_{(E,J)} = \left[{C_0 \over 2 V_0^2}\right] I_{\mathbf{x},\mathbf{v}},  I[\mathbf{x},\mathbf{v}] \equiv  [-V_0^2 - E_{\mathbf{x},\mathbf{v}}  + {J^2_{\mathbf{x},\mathbf{v}} \over 2 r_0^2}]^{-1/2}. $$ More similar our PDF for uniform sphere or to thermal PDF? What's the physics of anisotropy? (next lecture).
 * Is the following a solution for the incompressible CBE0 for our uniform sphere example? steady-state? (an)isotropic?

Excursion: Different galaxies shape like electron clouds in QM (n,l,m)=(3,2,0) (AN)ISOTROPIC states around Copper, Silver, Gold nuclei
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47$$e^-$$'s are distributed only on eigenstates of the time-independent $$\text{Ag}^{47+}$$ potential in a silver atom (47Ag of density $$10g/cm^3$$ inside $$r=1.4\AA$$ ) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s1 in ascending order of the levels and gold atom (79Au of density $$10g/cm^3$$ inside a radius 1.4A). 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d10 6s1

Quantum mechanics says these 47 or 79 electrons are in stead state, can be found only on a discrete set of energy levels (1,2,3,4,5,6,...) and angular momentum levels (s, p, d, ...) with a time-independent number of electrons (2,2,6,2,6,10,1 ...). Integrals says stars are distributed with a time-independent number on levels of these energy-like constants of motion: I_1, I_2, I_3 (whose values can be continuous or discrete).

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Lec 4: Dissecting the Mellon Galaxy (cont.): smooth orbits and grainy friction ⇒ Integral Invariants to build anisotropy
Last lecture, we had much ado with force balancing of tracers.


 * §§§: Is ME similar to how you would weigh Jupiter using its four Moons as tracers?

Oth-Rule: "gravity balancing motion"
A common concept in galaxies is that "gravity balancing motion" for each star in the background of a system of mass $$ m N $$, Can we write this?

$$ \text{Specific K.E. } O(\overline{\langle V^2 \rangle})= O(\sigma^2)= O(Vesc)^2 = O\left({R \sim 10kpc \over t_\text{cross} \sim 100Myr} \right)^2 =O(V_\text{cir}^2) = \text{Virial } O({G Mass \over R}), $$ where we consistently omit factors of unity $$ 4\pi $$ etc for clarity, while being ambiguous of geometry etc. ...

Define a crossing time for the Uniform Melon Galaxy
This model allows us to compute every property of a N-body for illustration.

In a quarter of the period, an orbit makes a one crossing of the sphere's radius, customarily called crossing or dynamical time scale, which relates to various quantities by

where $$ V_0 = \sigma_\text{max} =\sqrt{5} \sigma_\text{rms}$$.

Potential: grainy or smooth
How to correct for the fluctuations on top of analytical gradient of a smooth potential $$\Phi$$? $$ where we integrate a smoothened averaged mass density $$\rho = n m$$ to generates potential (Gravitational Field Eq.).

The number of stars fluctuates for very small cell size $$ s $$ (the super-1-AU "radius of influence", also called simulation softening length for computer particles) of a heavy body when encountering a light body.

The cone or sphere of influence
Everybody has a small sphere of influence. E.g., a tiny body "a" can be trapped by the localised potential of a heavy point mass "B" centred on position $$X_\bullet(t) $$, instead of escaping onto its own loop orbit around the Galaxy. E.g., here incomes a mass "c" crashing in for a dance of three within the mutual influence radius around $$X_\bullet$$, and "the sling-shot effect" ionises "a", and "c" gets trapped in mutual influence sphere with "B" instead. Influencers clearer change orbital distributions of each other, but ...

So how frequent do stars meet for a holiday (briefly off duty around the Galaxy)?
Let $$ \Delta t ~ \dot{N} $$ be the number of bodies enter the direct-influencing cross section $$ \pi s^2$$ of another over time $$\Delta t $$. Then with a standard accounting of bodies within in a pencil volume $$ (\pi s_\bullet^2 ) (\Delta t ~v_\text{rms}) $$ of a medium of locally uniform density $$ n_{_\mathbf X} $$ we get $$ where we must take the average $$ \langle \text{crossection } \text{v}_{rms} \rangle $$ as the "reaction rate" over a range of speed relative to local particles.

What energy balance criteria decides the sphere of influence?
or cross-section to deflect/retain bodies whose Galactic orbits happen to cross this channel by chance? Clearly it depends on relative speed and potential. E.g., weightless photons only subscribe to a BH inside a distance s, where $$ {c^2 \over 2} = { G M_\bullet \over s}$$, a balance of relative kinetic energy to (specific) potential energy.

Consider mutual attraction of massive particles of momentum $$ m\mathbf{v} $$ by a localised potential $$- {G M_\bullet \over s} $$ of a point mass of momentum $$ M_\bullet \mathbf{\dot{X}}_\bullet$$ (say, moving presently in x-direction) when they meet within the cross section $$ \pi s^2_\bullet $$. Then the sphere of influence (aka dynamical radius, Bondi radius) is loosely defined by, $$ where the two-body relative kinetic energy $$ 0.5 v_\text{rms}^2$$ is broken into parallel x-motion part and transverse y-or-z-motion parts.

Some encounters are more relaxing than others
Compared to the timescale of direct encounters $$1/\dot{N}$$, the time scale $$ \tau^\bullet $$ of relaxation (momentum exchange) is diluted by factors, as the relative velocity $$ v_\text{rms} $$ is split by fractions $$ {m \over m + M_\bullet } $$ and $$ {M_\bullet \over m + M_\bullet} $$ between the BH and the intruders for conservation. If any of the these two "recoil" fraction were zero, then no momentum were exchanged. Hence we define $$ Clearly most relaxing is for "heavy" bodies on "slow" holidays in a "dense" background.

What's the science behind crossection logarithm?
The curious correction factor $$\ln \Lambda \approx \ln {N m + M_\bullet \over m + M_\bullet}$$ (aka, Coulomb factor or crossection logarithm) for an N-body system, is a notoriously vague factor, of order $$ O(\ln N) = O(20) $$.

It accounts for, qualitatively, the additional momentum exchange with the greater number of bodies loafing past a larger area outside $$ \pi s^2 $$, "the meeting area". Because the gain in numbers overweighs the weakness of their distraction, this "loafing" majority dominates the hardcore subscribers $$ \Delta t \dot{N} $$ to the influence sphere in terms of overall impact on the relative momentum.

Dynamical friction on a point mass
Following Newton's law, our BH of momentum $$ M_\bullet \mathbf{V}_\bullet \equiv M_\bullet \dot_\bullet $$ must obey an Equation of Motion $$ as the BH is forced both by the gradient of potential $$ \Phi $$ (1st term) and by impulses averaged over a long friction time (2nd term, say 10 billion Newton in 10 billion years).

both are sourced from a sea of particles of mass density $$ m n_\mathbf{X} $$ of local mean velocity $$ \langle{\mathbf v}\rangle$$.

Note the friction force has a sign $$ \propto {m -M_\bullet \over m + M_\bullet}  \rightarrow \pm 1 $$:  the signed factor is negative (implying losing kinetic energy) for heavy point mass (BH),

but positive (implying heating in the less likely case) when stars were much heavier than our point mass.

This suggests a trend for equipartition, a fast and heavy particle will slow down, falling to centre of an N-body, a slower and light particle will pick up speed, spend time in the suburbs of N-body. Equal mass particles tend to "relax" to the "local mean velocity" of other particles, comoving in the smoothed gravitational gradient.

Clearly no net friction among equal mass particles as there's a balance of losing and gaining of kinetic energy is achieved. But the BH's momentum random walks (aka relaxation, Brownian motion) ... due to fluctuating impulses $$ \pm (m \sigma_x, m\sigma_y, m\sigma_z) $$ (i.e., momentum exchanged without net direction), where the definition of these random velocities $$ \sigma $$ needs concepts of PDF ...

Application: Dynamical friction on a SMBH
The Dynamical friction time depends on $$ G^2 M_\bullet $$ and background density through the relation $$ {1 \over \tau_{M_\bullet,V_\bullet} } = { 4\pi G^2  (M_\bullet) \rho \ln \Lambda \over (V_\bullet^2 + 2 \sigma^2)^{1.5} } $$ where we put $$ |{\mathbf v} - \dot_\bullet |^2 \approx V_\bullet^2 + 2 \sigma^2 $$ in the equation for the encounter rate $$ \dot{N} $$ (see Lec 1).

To illustrate that friction can bring a heavy object spiralling down the galaxy, consider a star cluster or a SMBH doing radial oscillation in the Uniform Sphere galaxy, then its Eq. of Motion becomes (after some algebra) $$ \ddot{\mathbf{X} } = - {4\sigma_0^2 \over r_0^2} \mathbf {X} - {48\sigma_0 \dot{\mathbf{X}} \over r_0 (\dot{\mathbf{X}}_\bullet^2 \sigma_0^{-2} + 2 )^{1.5}} {M_\bullet -m \over N m} \ln {N m + M_\bullet \over m + M_\bullet} $$

We solve this non-linear damped-oscillator eq. using a Mathematica one-line solver, and illustrate the solutions, assuming $$r_0=1\text{kpc}, ~\sigma_0=10\text{km/s}. $$  One can see both oscillations and decay of orbits; the rate depends mostly on the mass-ratio $$ N m :  M_\bullet $$ for a variety of initial position and velocities.

Application: Invariant Integrals for CBE0
Recall our earlier distribution function for the uniform sphere $$ f(|\mathbf{r}|,V_r,V_\theta,V_\varphi) = {dN \over (4\pi r^2 dr) dV_r (dV_\theta dV_\varphi)} = {C_0 \over 2V_0^3} \left[ 2I \right]^{-1/2},   2I \equiv  (1-{r^2 \over r_0^2}) (1 - {V_\theta^2 + V_\varphi^2 \over V_0^2} ) - {V_r^2 \over V_0^2} $$

How do we know if this PDF satisfies the CBE with or CBE0 without collisions?
One way is to brute force doing all the derivatives in CBE0 in XYZ coordinates (or slightly better in spherical coordinates), to check if it needs a collision term to balance the CBE.

The better way is to try to re-express the DF, (here with its suspicious 2I factor),

using known integrals E, J, Ez, Jz etc, E.g., $$ f \rightarrow f(E,J,J_z) $$? Because we know $$ \dot{E}=0, \dot{J}=0, \dot{E}_z =0, \dot{J}_z=0 $$ in Harmonic potential sphere $$ \Phi (r)= {V_0^2 r^2 \over 2 r_0^2} + cst$$ !

The logic from the chain-rule (aka Integrals, Jeans Theorem) $$ \dot{f} = {df(I_1,I_2,I_3,...) \over dt}  = \sum_j \dot{I}_j {\partial f \over \partial I_j}   =  ...\dot{E}  + ...+ \dot{J}  = 0+0 + ... $$ for CBE0.

You may use the fact $$ J^2 \equiv |{\mathbf r} \times {\mathbf v}|^2 =r^2 V_t^2, E \equiv \Phi + \sum_{i=1}^{3} {v_i^2 \over 2} = cst + {V_0^2 \over 2 r_0^2}  r^2  + {V_r^2 +V_t^2 \over 2}, $$

Homework (due Fri): Can you check the substitution of E and J into I or f?
We aim to substitute away all other coordinates $$ (r,V_r,V_\theta,V_\varphi) $$ for a final expression with E and J only.

Indeed ⇒ a solution for the CBE0, written in E, J etc, is $$ f \leftarrow f(E,J) \leftarrow  \left[{I_{\mathbf{x},\mathbf{v}} V_0^3 \over C_0}\right]^{-2} \propto I[\mathbf{x},\mathbf{v}] \equiv  \left(-1 - {E \over V_0^2} \right) + {J^2 \over 2 (r_0 V_0)^2} > 0. $$

Next Lecture: Energy-like Integral invariants keep memory of velocity anisotropy, and non-Gaussianess:

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Lec 5/6: Invariants in collision-less eq. → anisotropic ellipsoid/galaxy shapes (CBE0)
Anisotropy builds the empires of galaxies, not in one day. To understand the rise and fall of anisotropy, let's first try to understand the ...

CBE in Collision, to have or not to have not CBE0
Our life is made a lot simpler if collisions are neglected completely, so we have the CBE (aka Vlasov equation in plasma physics), valid for short scale quasi-equilibrium of stellar systems of long relaxation time.

Let's expand the rhs into full derivative with the chain-rule. $$ \dot{f}(t,\mathbf{x},\mathbf{v}) \rightarrow \frac{\partial f(t,,)}{\partial t} + \sum_{i=1}^3 \frac{\partial f(,\mathbf{x},)}{\partial x_i} \dot{\mathbf{x}}_i + \sum_{i=1}^3  \frac{\partial f(,,\mathbf{v})}{\partial v_i} \dot{\mathbf{v}}_i, $$ we then apply the gravitational Eq. of Motion, $$ \dot{\mathbf{x}}_i \rightarrow \mathbf{v}_i, \dot{\mathbf{v}}_i \rightarrow - {\mathbf{\nabla}}_i \Phi = \ddot{\mathbf{x}}_i, \text{for per particle (not per cell)} $$ then we have the full-fledged Collisional Boltzmann equation: $$

The lhs (replacing Gaussian minus f term) says the tendency for a heavy $$m_\bullet$$ particle to settle to the point of highest phase space density of a Gaussian, say at the mean $$ u_i(\mathbf{x}) \equiv \langle \mathbf{v}_i \rangle $$ on a "relaxation" or "friction" time scale $$ \tau_\text{frc}$$.

Now neglect interesting encounters
Zero lhs completely, valid for short time scales of million years, we have its simpleton CBE cousin (CBE0), $$ This CBE0, resembling the CE, means our Galaxy's phase space density f is incompressible in its 6D x-v space, hence is conserved along a flow of particles.

Time-independent
Generally for time-independent (called steady-state) systems, we can drop the term $$ {\partial_t f(\mathbf{x},\mathbf{v}}) =0 $$, then a solution to the CBE0 would be $$ f = {\cal F}(I_1, I_2, I_3,...) $$ where $$ {\cal F} $$ are explicit functions of conserved quantities (called integrals) I(x,v), which are functions of phase space coordinates with $$ \dot{I}_j(x,v) =0 $$. We can check the steady-state CBE0 is satisfied by the chain rule $$\dot{f}(t,\mathbf{x},\mathbf{v}) = \sum_{j=1}^\text{a few} \frac{\partial {\cal F}}{\partial I_j} \dot{I}_j(x,v) = \sum_{j=1}^\text{a few} 0  $$.

So we proved theIntegrals: the DF (Distribution Function) $$ f(\mathbf{x},\mathbf{v}) $$ is an implicit function of the position and velocity through a functional dependence on "a few independent constants of motion" along orbits.

Example of Popular integrals of motion
E.g., one of the I(x,v)'s could be defined as a linear shift of the orbital energy, for a star cluster, $$ I_1(x,v) \equiv -1 - E(x,v)/\text{(1km/s)}^2, ~ E \equiv \Phi(x_1,x_2,x_3) + \sum_{i=1}^3 \frac{v_i^2}{2}. $$

$$ I_3(x,v) = (\mathbf{r} \times \mathbf{V}_t)^2 = I_2^2 + I_4,  I_2(..)=(x v_y - y v_x), I_4(..)=(y v_z - z v_y)^2 + (z v_x - x v_z)^2 $$
 * §§§: For a spherical system, are following quantities valid integrals of motion?

Conservations of I2, I3, I4, ... as Origin of lasting Anisotropy and stable disks
Galaxies would be all spherical and boringly isotropic in 3D velocity: if nature puts all stars on an energy-only stratification,

a thermal PDF $$ \propto e^{-E \over \sigma^2}$$ of collisional gas produces a spherical star.

Because collisions are slow in galaxies, galaxies are shaped by Integrals, memories of integrals of motion at the birth of stars.

Our Sun, like many disk stars, remembers the angular momentum of its birth place (some gas clouds), has stayed with that Jz (after our Galaxy's last Integrals -theorem-upsetting violent merger). This Jz prevents the disk orbit from shrinking.

$$DF_a \propto {J_z \over \text{100 kpc} \cdot \text{500km/s}} \times |{-E \over \text{(500km/s)}^2}|, DF_b \propto e^{-E \over \text{(500km/s)}^2}, DF_c \propto \delta( {J^2 \over \text{(100 kpc 500km/s)}^2 } + {E +  \text{(1000km/s)}^2  \over \text{(1000km/s)}^2} ). $$
 * §§§: A spherical galaxy cluster has $$N = 1000 $$ occasionally-colliding galaxies, $$ N=10^{50} $$ collisional-ionised hydrogens, and $$ N=10^{77} $$ radially anisotropic collisionless-neutrinos. DF a, b, c are better descriptions of which components?

Note the third (DFc) distributes all particles on a narrow band of energy and angular momentum, thus producing a radial anisotropy.

Various Derivations of ME/CE: with/without relaxation (ADVANCED MATERIAL, SKIPPED)
There are various derivations of CONTINUITY Eqs. with collisions (FLUIDS or STARS)

E.g., We can apply the CBE after integrating it over the entire velocity space for any $$^p$$opulation (e.g., gas, stars, dark matter) of particles, $$ \frac{\partial \int\! f_p d^3\mathbf{v}}{\partial t} + \!\! \sum_{j=1}^{3} \frac{\partial \int\!\! v_j f_p d^3\mathbf{v}}{\partial x_j} - \!\!\int\!\!\!\!\int \!\!dv_2 dv_3 \cancel{f_p(\infty,v_2,v_3)} \frac{\partial\Phi}{\partial x_1} - ... - ...= \frac{0+0+0}{\tau_\text{frc}}, $$

where the r.h.s. is zero because the relaxations/collisions of particles shuffle particles of different velocity within the same tiny volume in the coordinate space, hence cancel locally. Also the phase density $$ f \rightarrow 0$$ when $$ |v| \rightarrow \infty$$, hence we can drop the last three terms on the l.f.s.. We then obtain the Continuity Eq.: $$ {\partial \over \partial t} \underbrace{n_p(t,\mathbf{x})}{} + \sum_{j=1}^{3} {\partial \over \partial x_j} \underbrace{[n_p u_j(t,\mathbf{x}) ]}{}   \overbrace{=}^{Conservation} 0, $$ $$ u_j(t,\mathbf{x}) \equiv {n_p \langle{v_j}\rangle \over n_p(..)} = {{\int\! \mathbf{v}_j f_p(...) d^3\mathbf{v}} \over \int \! f_p(...) d^3\mathbf{v}}, $$ where $$ n_p(t,\mathbf{x}) $$ and $$ u(t,\mathbf{x}) =(u_1,u_2,u_3) $$ are the number density and the mean flow of particles at the position $$ \mathbf{x} $$. This equation is valid even for gas particles in accretion disks.

In many traditional text, one starts from the " (un-necessary) assumption of collisionless system" setting the LHS $$ \dot{f} =0$$. In fact, in many cases of non-zero $$ {d f\over dt} $$ still won't contribute after averaging over velocity space.

Bottomline ⇒ MASS Continuity Eq. and MOMENTUM Continuity Eqs. ARE TRUE both for collisionless N-BODY and for collisional HYDRODYNAMICS Eqs.

Lec 5/6: Connecting the GAP: Viscosity (=Gravitational Friction?), Hydrostatic (=ME?), (ADVANCED, SKIPABLE)

 * §§§: A galaxy cluster with typical crossing time $$ R/V=(\mathrm{1000 kpc})/(\mathrm{1000 km/s}) =\mathrm{1 Gyr}$$, has 1000 galaxies insides. Is it likely for a few of these galaxies to sink and merge at the cluster centre after a 10Gyrs of relaxation or dynamical friction?

The sphere of influence also resembles the tidal destruction radius and the horizon radius of a BH, which affects momentum transfer and the outer and inner boundary of accretion disk. The viscosity of accretion disk joins smoothly into the angular momentum at the edge of the sphere of influence as well. Gravitational Momentum Eq. reduces to its cousin viscous fluid (aka Navier-Stokes) Eqs. for accretion disk, albeit microscopically very different.

Stellar Dynamics has connections to Accretion Disks
Both borrow mathematical concepts of anisotropic magnetic Pressure (stress) Tensor in physics of magnetohydrodynamics in stellar interior plasma and viscous force in fluid mechanics (aka Navier-Stokes Eqs.), except viscosity is small and less intuitive in stellar dynamics.

We will demonstrate: ME/NS have common continuity Eqs. in mass and momentum in the context of BH-feeding from thin stellar/gas disks.

Excursion: Feeding a Supermassive BH (SMBH) from a dense pc-sized rotating Disk
For a thin stellar disk settled around our SMBH, it's reasonable to assume a steady-state $$ \partial_t =0 $$, rotational symmetry $$ \partial_\varphi =0 $$, no vertical motion nor vertical shear streaming $$\partial_z (u_R, u_z, u_\varphi)=0=u_z $$.

We assume an isotropic pressure $$ \rho\sigma^2$$, and a thin disk means the dispersion is small compared to rotation $$ \sigma \ll u_\varphi = O(100) \text{km/s}$$.

We assume the relaxation of stars generates a tangential torque,

and such gravitational friction brings lower angular momentum stars in quasi-equilibrium to gather closer towards the SMBH, feeds around 1 solar mass from 1 pc every crossing time of $$ 1\text{pc}/(100\text{km/s}) =10^4 \text{yr}$$ into the SMBH's horizon, so at a steady rate $$ \dot{M}_\bullet = 10^6M_\odot$$ per Hubble time.

To estimate the inflow speed $$ u_R$$, we split the isotropic axisymmetric MEs (in Lec 4) into two eqs.

z-component hydrostatic equilibrium
$$ assuming a vertical Gaussian density profile of surface density $$ \Sigma$$ and of scale height H which flares such that we have a constant disk opening angle $$\beta$$.

R-component of ME
$$ { O(u_R)^2 \over R} + {G M_\bullet \over R^2} = {O(\sigma^2) \over R} + {u_\varphi^2 \over R},\rightarrow u_\varphi  =\sqrt{G M_\bullet \over R} = {\sigma \over \beta} \approx 100 ({R \over 1\text{pc}})^{-0.5} \text{km/s}, $$ i.e., the Keplerian speed near a $$10^6M_\odot$$ SMBH.

Example for Tensor Virial Theorem: rotation flattening
We can compute the Virial in z vs x and y, the latter two sums up to cylindrical radial. Clearly Tensor Virial Theorem predicts a flat disk, with a force ratio $$ {\overline{ z \partial_z \Phi} \over \overline{R \partial_R \Phi} } = {2K_{zz}/M \over \overline{x \partial_x \Phi} + \overline{y \partial_y \Phi} } = {\overline{v_z^2} \over \overline{v_x^2 + v_y^2} } = {\overline{u_z^2 + \sigma^2} \over \overline{v_\varphi^2 + v_R^2} } = {\overline{\sigma^2} \over \overline{(u_\varphi^2 + \sigma^2) + \sigma^2 } } = O(\beta^2) \ll 1 $$


 * Make crude estimate of the opening angle $$ \beta $$ for a hypothetical nuclear disk near SMBH with rotation to dispersion ratio of 10.

We add an Azimuthal-component ME
The azimuthal ME gives the tangential friction torque for moving $$1 M_\odot$$ across a radius $$ \Delta R $$ over a time $$ {M_\odot \over \dot{M}_\bullet} = \Delta t \equiv {\Delta R \over |u_R|} $$ $$ \begin{align} R \overbrace{M_\odot {O(\beta u_\varphi) \over \tau_{*\text{frc}} } }^{F^*_\text{frc}} & = { \Delta R \over \Delta t} {\partial  [ M_\odot~ u_\varphi~ R ] \over \partial R} \rightarrow  (0.5 R |u_R|)  (M_\odot~ u_\varphi), \end{align} $$ where the rhs is the rate of change of angular momentum $$ M_\odot u_\varphi~ R \propto R^{0.5} $$. We may cancel the $$ u_\varphi $$ on both sides to find $$ (R |u_R|) =O(\beta) R^2/t_\text{frc} $$

We add Radial Flow Continuity Eq.
Insert all the derived velocities and friction timescale into the Continuity Eq., we have, $$\begin{align} {\dot{M}_\bullet \over 2\pi R \Sigma } = |u_R| & \rightarrow \beta R \overbrace{O(t^{-1}_{*\text{frc}})}^{O(4\pi G^2 M_\odot) [\rho/\sigma^3] }\left.\right|_{\sigma =\beta ({G M_\bullet \over R})^{0.5} } \\ \rightarrow & O(2\pi R ~2H\rho) ~{M_\odot \over (\beta^2 M_\bullet)^{1.5} } (G R)^{0.5} \\ \rightarrow & \left[{ G \dot{M}_\bullet M_\odot \over M_\bullet \tilde{u}_\varphi}\right]^{0.5} \left. \right|_{\tilde{u}_\varphi = O(\beta^3)({G M_\bullet \over R})^{0.5} } \\ &\approx R_\text{pc}^{0.25} ({4 \over \beta})^{1.5} ({0.1\dot{M}_{\bullet M_\odot/Gyr} \over M_{\bullet M_\odot} })^{0.5} ~\text{cm/s} \end{align},$$ where the 3rd line is from taking the geometric mean of the rhs of 2nd line with the lhs of 1st line to cancel the $$ O(2H \rho) \sim \Sigma $$ surface density term.

This is a neat satisfying result: that fully relaxed isotropic rotating stellar disk with an inflow speed $$ u_R \propto R^{0.25}$$ around a SMBH settles to the famous $$ \rho(R) = \Sigma/O(\beta R) = \dot{M}_\bullet/O(4 \pi \beta R^2 u_R) \approx R_\text{pc}^{-2.25} ({0.1\beta \dot{M}_{\bullet M_\odot/Gyr} M_{\bullet M_\odot} })^{0.5}  M_\odot \text{pc}^{-3}$$ volume density law, well-known in literature on SMBH growth by relaxation for a spherical cluster. Here we generalised the result to a thin and flaring disk with an opening angle $$ 0 < \beta ={H \over R} <{\pi \over 2} $$.

We estimate the loss cone angle is of the order of $$ O({8\text{cm/s} \over 100\text{km/s} }) = O(1'') $$, so stars slowly diffuse to the SMBH by random walk from outside. For million solar mass Bh grown over 10 Gyrs, the disk stellar density of order $$ 1000M_\odot$$ inside 1 pc volume.

Excursion: Connections between star loss cone and gravitational gas accretion physics
Instead of a background of dissipationless particles: if a BH $$ M_\bullet$$ is moving through a dissipational gas particles of thermal pressure $$ m n_\text{gas}(x) v^2$$, then every gas particle of mass m will likely transfer its relative momentum $$ m v_\text{rms} $$ to the BH when coming within a cross-section $$ \pi s^2 $$, with a capture rate (roughly accretion rate) $$  \dot{M}_\bullet = m \dot{N} \approx  \pi \left[ { G (m+M_\bullet) \over v_\text{rms}^2/2} \right]^2 (m n) v_\text{rms},  v_\text{rms} = \sqrt{ \langle v^{-2}\rangle^{-1}+V_\bullet^2}. $$ where we adopt a typical speed by take the inverse rms $$ \langle 1/v^2\rangle$$ for background particles.

These gas streams collide and form an accretion disk around the BH, where gas viscosity drives a slow inflow to grow a small BH to a supermassive BH, which experiences more friction and sinks to galaxy centre as a jet-emitting Active Galactic Nucleus.

Set $$ m \rightarrow 0, ~ \langle v^{-2}\rangle^{-1} =\sigma^2 $$ and $$ V_\bullet=0 $$ for a stationary BH, we obtain the famous (Bondi) spherical accretion rate, assuming equation of state is ISOthermal for dissipational gas but Adiabatic for N-body, $$ \begin{align} \dot{M}_\bullet & \rightarrow \pi \rho_\text{iso.gas} \sigma \left[{(2G M_\bullet) \over \sigma^2}\right]^2 \\ \dot{N}_\bullet & \rightarrow \pi n_\text{nbody} \sigma \left[{(2G M_\bullet) \over 2\sigma^2} \right]^2 \end{align} $$ The difference: because $$ \langle v^{-2}\rangle^{-1} = 2 \sigma^2 $$ for dissipationless particles (i.e., adiabatic gas or N-body particles); a reduction of encounter rate for more rigid dissipationless particles as particles are compressed and heated up down the flow, hence less collimated.

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The centroid $$ \mathbf{v} \approx \mathbf{\omega} \times \mathbf{x} $$ of a rotating population experiences an acceleration $$ \mathbf{A}_i \rightarrow { \mathbf{\omega} \times \mathbf{x} \over t_{\bullet\text{frc}} }, $$ which generates a counter-rotation torque.

in terms of its particle's energy and angular momentum respect to the centre of mass of the background particles, defined as $$ \begin{align} f^\bullet_\text{eff} & \equiv \left[ f^\bullet(t,E,J) + {m \sigma^2_\mathbf{x} \over m_\bullet } {\partial f^\bullet(t,E,J) \over \partial E} - { m \sigma^2_\mathbf{x} \over m_\bullet v^2} {J \partial f^\bullet (t,E,J) \over \partial J} \right], \\ & \approx \left[ 1- {m \sigma^2_\mathbf{x} \over m_\bullet \sigma^2_\bullet} (1+ {\omega J \over v^2})  \right] f^\bullet(t,E,J) , \text{for a rotating Gaussian } f^\bullet \approx f^\bullet_0 ~ e^{-E + \omega J \over \sigma^2_\bullet} \\ & \rightarrow (-{\omega J \over v^2}) f^\bullet(t,E,J), \text{if equipartition everywhere: } \sigma^2_\mathbf{x} = \sigma^2_\bullet. \end{align} $$

\begin{align} f^\bullet_\text{eff} & \equiv \left[ f^\bullet_{t,\mathbf{x},\mathbf{v}}   +  {m \sigma^2_\mathbf{x} \over m_\bullet }  \sum_{j}^3 \left( {3 \over 2} \delta_{ij} -  {\mathbf{v}_i \mathbf{v}_j \over |\mathbf{v}|^2 } \right) \mathbf{\nabla}_{\mathbf{v}_j} f^\bullet_{t,\mathbf{x},\mathbf{v}} \right] \\ &= \left[ f^\bullet(t,E,J) + {m \sigma^2_\mathbf{x} \over m_\bullet } {\partial f^\bullet(t,E,J) \over \partial E} - { m \sigma^2_\mathbf{x} \over m_\bullet v^2} {J \partial f^\bullet (t,E,J) \over \partial J} \right], \end{align}

$$ {Df^\bullet \over Dt}  =\sum_{i}^3 \mathbf{\nabla}_{\mathbf{v}_i} \cdot \left[ \mathbf{A}_{i} f^\bullet_\text{eff}\right] = , \sum_{i}^3 \mathbf{\nabla}_{\mathbf{v}_i} \cdot \left[{ \mathbf{v}_i \over t_{\bullet\text{frc}} } f^\bullet_\text{eff} \right] $$

$$ \sum_{i}^3 \mathbf{\nabla}_{\mathbf{v}_i} \cdot \left[ \mathbf{A}_{i} f^\bullet_{t,\mathbf{x},\mathbf{v}}\right], \mathbf{A}_{i} \equiv { [2\pi G^2 \ln \Lambda] \left[ 2 m_\bullet \mathbf{v}_i \rho_{\mathbf{x}} +  m P_{\!\mathbf{x}} \left( 3 \mathbf{\nabla}_{\mathbf{v}_i} - {\mathbf{v}_i \over |\mathbf{v}|} {\partial \over \partial |\mathbf{v}|} \right) \right] \over \left[ 2\left(\!{P \over \rho}\!\right)_{\!\mathbf{x}} + |\mathbf{v}|^2 \right]^{1.5} }, $$

Here is the microscopic picture. Imagine our test particle "$$m_\bullet$$ initially" in a volume dxdv (say coordinates t,x,v being 0,0,w), the drift term represents the rate it's kicked out of its volume dxdv by any particle $$m$$ of relative speed "w" in the same position dx.

Now imagine our test particle "$$m_\bullet$$ finally" arrives the volume dxdv (at 0,0,w) from an earlier coordinate (0,0,w+k), thanks to a kick by any $$m$$ particle which is located at velocity -k and ends up at velocity "0".

The details here account for the conservations of particle number, particle momentum and energy in an elastic scattering of cross-section $$ \pi s^2 $$. These explain the "-" viscous heat diffusion term from $$m$$ particles, and the "+" friction term for heavy $$m_\bullet$$ particles.

If we limit ourselves only to gravitational scatterings of two low-mass particles of small relative speed "w' and weak impulse "k", and keep only the dominant Taylor expansion in powere small "w" and "k", we would obtain the so-called Fokker-Planck Equation, which is beyond our reach mathematically, but conceptually it's understandable as a competition of above two processes, driving the system towards a thermal PDF.

$$ \begin{align} {df \over dt} & =\sum_{i=1}^3 \partial_{v_i} [{ v_i \over t_\text{fric}^\bullet } f] \\ D&=\sum_{i=1}^{3} \partial_{v_i}\!\! { v_i 4 \pi G^2 (m-M) \int \!\! \int \!\! d^3\mathbf{v} d^3\mathbf{w} \hat{m} \hat{f} \\ \end{align} $$ -->

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Excursion: Rotating ellipsoid
A PDF with a non-trivial flow is the rotating ellipsoid (aka Freeman's bar) of uniform density $$ m n_0$$ up to a height $$ z_\text{max}$$ spinning around the z-axis with

$$ f_\text{bar} = {n_0 \delta(\tilde{v}_2 + {K_1 \over 2 \omega} x_1) \delta(\tilde{v}_1 + {K_2 \over 2 \omega} x_2) \over \pi \sqrt{2\sigma_{33} - \tilde{v}_3^2 }},  K_{1} \equiv (\omega_1^2-\omega^2), K_{2} \equiv (\omega_2^2-\omega^2) , ~n_0={ 3 N_0 \over 4\pi z_\text{max} y_\text{max} x_\text{max} }, $$ where the x and y velocities $$\tilde{v} $$ in the rotating frame conspire to be two integrals, describing a dispersion-less cold flow along concentric ellipses such that their linear Coriolis force $$ -2 \omega \times \tilde{\mathbf{v}}$$ balances the centrifugal force and gravity $$ (-K_1 x_1, -K_2 x_2)$$ for such particles in an anisotropic harmonic potential (up to an overall shift of constant) $$\Phi(x_1,x_2,x_3) ={(\omega_1 x_1)^2 + (\omega_2 x_2)^2 +(\omega_3 x_3)^2 \over 2} $$, or an effective potential $$\Phi_\text{eff} \equiv \Phi - {\omega^2 (x_1^2+x_2^2) \over 2} $$.

The vertical dispersion satisfies a relation $$ E_3 + (E_J-E_3) \gamma^{-1} \equiv {\tilde{v}_3^2  \over 2} - \sigma_{33}(\mathbf{x})  \equiv  {\tilde{v}_3^2 \over 2}  + { (\omega_3 x_3)^2 - (\omega_3 z_\text{max})^2 \over 2} + {(K_1  x_1^2  +  K_2 x_2^2) + ({K_1 x_1 \over 2\omega})^2 + ({K_2 x_2 \over 2\omega})^2 \over 2 \gamma} $$, which is an integral, from superposition of the rotation-frame energy integral $$ E_J$$ and the vertical motion energy integral $$ E_3 = {(\omega_3 x_3)^2 + \tilde{v}_3^2 \over 2} $$. The potential is fixed by the density of the ellipsoid via eq. of gravity. In turn, they fix the pattern speed $$ \omega $$, so that the continuity equation in steady state is satisfied by the cold flow ellipses being parallel to the bar's elliptical boundary with a shape ratio $$ x_\text{max}:y_\text{max}$$. Nevertheless, for any bar density shape many equilibrium solutions exist; each solution is specified by the tuneable weighting coefficient $$ \gamma $$ as long as $$ \sigma_{33}(x_\text{max},0,0) \ge 0, ~\sigma_{33}(0,y_\text{max},0) \ge 0, ~ \sigma_{33}(0,0,z_\text{max})=0. $$

These integrals produce a self-consistent equilibrium for a uniform pattern-rotating ellipsoid with a mass-conserving circulation along the shape of the ellipsoid in the xy cut, and with anisotropic harmonic oscillator potential. All MEs (on the continuity of mass, momentum and energy) are satisfied with heat flux $$q_{ijk}=0$$ and $$ \sigma_{11}=\sigma_{22}=\sigma_{12}=0 \ne \sigma_{33} $$. Both the flow and velocity dispersion are anisotropic; it is a cold divergence-less linear gradient flow in the x and y without dispersion. There is no vertical flow, only a positive vertical pressure.


 * . Can you verify the central dispersion $$ { \int v_3^2 f dv^3 \over \int f dv^3} |_{\mathbf{x}=0} = {n_0 \sigma_{33}(0,0,0) \over n_0} = {(\omega_3 z_\text{max})^2 \over 2} $$ for this model?

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Final fruits for thoughts
Since ME is based on the conservations of particle number and momentum (and energy even during collisions under "local approximation"), is it surprising that a turbulent collisional dissipational accretion disk satisfy a generalised ME (with suitable translation of terminology)?

Thinking on some deep level, there seems to be no boundary of Physics, no dichotomy between Gravitation vs. Accretion Physics.

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