tag:blogger.com,1999:blog-2179483401827607038.post5317552016860472312..comments2010-07-30T13:53:13.580-07:00Comments on Astro-HPC Research Team, Summer 2010: [from Anthony] Progress report on modal analysisSummer Research Team 2010http://www.blogger.com/profile/08767435137183387661noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2179483401827607038.post-49561378391950242472010-07-30T13:53:13.580-07:002010-07-30T13:53:13.580-07:00[from Anthony] Oh yes, I seem to forgot to post th...[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.<br /><br />I 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.Summer Research Team 2010https://www.blogger.com/profile/08767435137183387661noreply@blogger.comtag:blogger.com,1999:blog-2179483401827607038.post-60892712833805990802010-07-29T17:11:09.286-07:002010-07-29T17:11:09.286-07:00[from Pawel] so, is m=2 here? it's all quite a...[from Pawel] so, is m=2 here? it's all quite a bit different for m=1.<br /><br />well, 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<br />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 <br />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. <br /><br />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<br />the waves have wavenumber k such that kc = +-sqrt(-D) --> sqrt(m^2-1) Omega. <br />So I's say a wave that tightens (gets more tightly wrapped) for r-->0,<br /><br />Now c ~ V_kepl, that's our assumption of a wedg-shaped disk, right?<br />(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. <br />this explains the shapes of red curves in the first figure (k1).<br /><br />but not in the other figures, which need a bit more analysis. <br /><br />remember that things plotted in the figs are not really solutions, just visualization of the expected asymptotic forms. <br /><br />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?<br /><br />PSummer Research Team 2010https://www.blogger.com/profile/08767435137183387661noreply@blogger.comtag:blogger.com,1999:blog-2179483401827607038.post-68039469036690544312010-07-21T21:42:03.071-07:002010-07-21T21:42:03.071-07:00[from Josh] Thats great that things went well and ...[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.Summer Research Team 2010https://www.blogger.com/profile/08767435137183387661noreply@blogger.com