How did you define the phase angle you are plotting? Note that getting the phase angle of the instantaneous orbit in a frozen potential requires to obtain the map from (x,v) to (J,phi) (action-angle variables), though J is not required. Thus phi is not just some polar angle in (x,v) space.
I did not do such a transformation... the angle in the plots is defined as the arctan(v/x), so in this case it is just some polar angle in phase space. Philosophically I understand that there is an arbitrary scaling in V, so that this angle is not well defined, but on the other hand, if you look at the plot I made of angle vs. x, the density at each x is the projection through all the layers, so the density does somehow depend on this non-meaningful angle. Another way to say this: if all the velocities were scaled, then the windup proceeds at a different rate as a function of time... so a given snapshot would have 1) different angles yes but also 2) a different total number of winds. Therefore I am confused.
what do you think about this: 2010arXiv1010.3008V ?
Whilst this paper is rather confuse, I think in a way they also claim that
mixing (of energies for example) is not relevant or does hardly happen,
similar to our findings. However, I don't think they provide any explanation
for the final profile.
I read this paper, thanks for pointing it out. I had not thought of Scott's objection, but I agree with him. Specifically, I think the energy transfer between particles on various orbits depends a lot on their exact phase relationship as they bounce through the middle (or I guess periapse), and this is not evolving dynamically in their approach, which may make a difference in the final density distribution.
I am not sure I agree with your assessment that they are finding that the energies don't mix. Maybe I am interpreting this wrong, but if you compare the 1st and 2nd panels of figure six, the different colored curves seem to me to be roughly proxies for initial and final energies (for a snapshot at the final time). The first panel shows that while there is no mixing in the outskirts, it's definitely particles that started life in the middle that contribute most to the inner radii. We are finding this too. The fact that in the second panel the ordering of the colored curves (even in the outskirts) is not the same implies that the turnaround radius of particles at the end of the day does not seem to be a monotonic function of the initial radius. Caveat: since I am thinking the initial potential energy as the dominating energy, it's possible that whatever they used for initial velocities messes up this logic.
I think I understand what you mean by mean-field effect. The way I would say it is that the scaling behavior only shows up as long as the system looks cold, i.e. as long as there are well defined streams in phase space.
Well, I think the surprise is that the scaling behavior DOES show up while the system is still cold and there are well defined streams in phase space... and apparently stays that way after the well defined stream disappear.
I think the ISW effect is not a useful analogy for non-cosmologists (!)
Agreed, I won't use it in the paper :)
The situation in three dimensions is that rho/sigma^eps is a power-law over a much larger range than either rho or sigma, if and only if eps=3. Various people claim that this is because eps=3 makes rho/sigma^eps look like a phase space density, but this is a weak argument. It's not at all obvious what value of eps works best in one dimension but if its not eps=1 then we have a counter-example to the arguments about phase space.
Here is a plot of $\rho/\sigma^{eps}$. The curves have been offset by a factor of 100, so their curvature at high r is easier to compare. Eps=1 is indeed not the exponent that maximizes radial range of the power law, but I am not sure that is significant, as I will explain below:
You'll note that eps=3,4,8 all seem to give equally good power law behavior, and there is a reason for this. Jerry pointed out to me that the behavior of particles in the outskirts of the system look to be behaving isothermally, and if that is the case, $\sigma$ will go toward a constant. Here is a plot of $\rho$ and $\sigma$ individually, and I have also plotted $r^{-1/2}$ truncated with something that looks like $e^{-\Phi}$, which indeed seems to fit well (in a chi by eye sense).
The velocity dispersion $\sigma$ seems to go to a constant (if you ignore the last data point), and is furthermore fairly noisy. If you raise a constant (with noise) to a high power, it amplifies the noise. Dividing by this noisy thing makes it harder to detect any rolling off of Q, but if sigma is actually FLAT all powers of sigma are doing the same thing to the curvature of Q... nothing save hiding it in the noise.
Of course $\sigma$ is rolling slowly toward flat, and may be rolling in concert with $\rho$ to keep Q scaling the same way in the rollover region... we might be able to argue from the plot above that $eps=3$ is a legitimately a power law farther out than $eps=1$. But I am not sure.
You say that you have plots of the contribution of each wind to the density and the energy. A related question is this: is the potential gradient near the center dominated by the "small" winds near the center of phase space, or the "large" winds?
This question is confusing, can you clarify? The force from particles outside r is canceling, so only the potential due to particles inside r influence the dynamics of a particle at r, or am I misunderstanding something? Sorry...
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