Thursday, May 20, 2010

[from James] polar N-body self-gravity

Here's the physics I've worked out for particle-particle gravitational interactions in polar co-ordinates:
We're calculating the total forces acting on a particle located at (R,θ).
Another particle located at (r,φ) will be a distance d from the first particle.

d2 = R2 + r2 + 2*r*R*cos(ψ)

where ψ is angle between the position vectors for the two particles.
The gravitational force between these two particles acts along d; we can split this into the parts acting along θ and R respectively:

F(d)d = F(d)sinβθ + F(d)cosβR

where the bold values are unit vectors and their non-bold counterparts are the corresponding magnitudes (since I don't know how to write those little vector arrow superscripts in html), and β is the angle between R and d,

β = arccos[(d2 + r2 + R2)/(2dR)]

Pretty straightforward so far, right?

This force along d produces acceleration along both R and θ:

F(d)d = d2θ)/dt2 + d2(RR)/dt2

Ok, now working out the derivatives for those accelerations is kinda long-ish, fortunately B&T did it already for the radial acceleration in chapter 3.1. Deriving the angular acceleration by the same method and adding the two gives the total acceleration as

d2θ)/dt2 + d2(RR)/dt2
= [(d2θ/dt2)*(R+1) + (dθ/dt)*(2(dR/dt) - θ*(dθ/dt)]θ
+ [(d2R/dt2) + θ*(d2θ/dt2) - (dθ/dt)2(R+2)]R

Now combining the above with our force equation near the top, separating the R and θ parts into two equations, and re-arranging, we finally get

d2θ/dt2 = [F(d)sinβ - 2(dR/dt)(dθ/dt) - θ(dθ/dt)2] / (R+1)
d2R/dt2 = F(d)cosβ + θ(d2θ/dt2) + (dθ/dt)2(R+2)

I've double-checked my calculations and obtained the same result, but the presence of d2θ/dt2 in the d2R/dt2 equation doesn't really make sense to me. Anyone have any ideas about that?

Of course the above equations only tell us the particle's acceleration due to the gravitational attraction of one other particle in the disk; we still need to add the (similar) forces from all the other particles, plus of course the central mass. Once we do, we'll have a self-gravitating disk.

1 comment:

  1. [from Pawel] Hey, that's probably all true, but... good news, you can skip the selfgravity of the disk altogether. Surely, there is some of that, but very weak compared with the stellar gravity (radial only and equal to -1./r^2 and radiation pressure +beta/r^2)
    where beta = F_rad/F_grav = beta_0 exp(-tau).
    We neglect it here because (i) we don't intend to study those heavy primordial disks that can be subject to gravitational instability (ii) we want to study radiation pressure effect in isolation, in its pure form. We can always try to add other physics later, after we have understood all about the radiative instabilities. Does it make sense?

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