First of all, I'm now back to Toronto from Santa Cruz and I have to say ISIMA was excellent. I definitely learned that there's a lot of more complicated problems in astro modeling where it involves lots of hydrodynamics (MHD) and there was one group from UC San Diego working on fusion (plasma physics). So, after meeting a bunch of graduate students from all over the world there's definitely a lot more we could learn in astro modeling.

Here's the site for what's currently happening in the program...

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So while I was at Santa Cruz, I managed to get some work done on my modal analysis with Pawel. Having to solve for the roots of k (wave number) via WKB approximation along with Cardano's method on the third order polynomial; so there are three roots to this polynomial.

This is the first root of k with both real and imaginary parts. Notice the the three lumps on the graph. Those are the Inner Lindblad, corotation, and Outer Lindblad resonances respectively.

This is the second root for k.

This is the third root for k.

By integrating for eta (look back on my previous entry on what eta is...) using the WKB solution exp(ikr) or more specifically exp(i*integral(k_{i} dr)) we get the following three eta's...

This is the first eta using k_1 with both real and imaginary parts. There seems to be a large response between 0.1 to 0.5 and then some response between r=1 to r=1.7-ish.

This is the second eta using k_2. It seems nothing happens after r=0.5.

This is the strangest one. It seems eta acts violently (look at the number on the eta axis) as r grows which I can't seem wrap my head around. Looking back at the roots of k, k3 sort of acts opposite to k1, so there's a sign reversal on the k's which implies that the graph is flipped horizontally but I don't quite understand why there's such large values for eta_3.

[from Josh] Thats great that things went well and that you enjoyed your time. I don't know what the status is on our project but I've only been able to do a bit of work on it during spare time which is limited. Not much to post.

ReplyDelete[from Pawel] so, is m=2 here? it's all quite a bit different for m=1.

ReplyDeletewell, let's look at the asymptotic behavior of D = kappa^2 - (m Omega - omega)^2, since the dispersions relation for non-radiation-affected waves (like those outside at r>>1 in the disk) takes form

D + k^2 c^2 = 0 , unless we include self-gravity, which we don't (for clarity). Then kc = sqrt(-D), and D must be negative for wave propagation

In th limit r>>1 we have D --> -omega^2, so it's negative and the wave is a simple sound wave: kc = +-omega, positive sign corresponds to trailing outgoing waves.

IN the r<<1 limit, except for m=1 (a special case we shouldn't worry too much about, probably), we have D --> -(m^2-1)Omega^2, and

the waves have wavenumber k such that kc = +-sqrt(-D) --> sqrt(m^2-1) Omega.

So I's say a wave that tightens (gets more tightly wrapped) for r-->0,

Now c ~ V_kepl, that's our assumption of a wedg-shaped disk, right?

(z/r ~ c/v_k). then we have c ~ r^(-0.5) and k in the asymptotic limits becomes k ~ +- r^(-1.5 +0.5) for small r, and +- r^{0 +0.5) for large r.

this explains the shapes of red curves in the first figure (k1).

but not in the other figures, which need a bit more analysis.

remember that things plotted in the figs are not really solutions, just visualization of the expected asymptotic forms.

what about the full numerical solutions Antony? are you now trying to start with the three kinds of waves/non-waves on the left and seeing what transpires on the right? can you post some pix?

P

[from Anthony] Oh yes, I seem to forgot to post the full solns. I'm assuming you're referring to the one using leapfrog, right? I'll try to post that tonight. To give you a preview, I integrated from the left and it simply looked like the three 'waves' that I posted earlier with k1, k2, k3 combined.

ReplyDeleteI was also thinking about using laplace and inverse-lapace transform to solve this equation (since you reminded me that this is an integro-differential equation) analytically (and maybe numerically). I managed to do a laplace transform but the inverse is the challenge. I'll try to update that tonight.