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I don't think that is what is normally called "ball lightning", but it could be some other anomalous electrostatic discharge from the storm. The videos I have seen which purport to show ball lighting show a small spot of light that moves very fast between clouds or from the clouds to the...

Thanks for the advice. I just started looking at Ha solar filters so I haven't really decided what to go with. The daystar filters look very nice, but you definitely pay a premium on them... $10000 for 0.3A :))

Currently, I don't, but I have been thinking about getting an Ha setup for my 80mm and a white light filter for the 8" from 1000 oaks optical. I've heard many good things about their products. Here are the links below. http://thousandoaksoptical.com/shop/solar-filters/full-aperture-solarlite/...

Thanks, I've been happy with it so far. The other scope is a celestron EdgeHD 800. I use this one for smaller objects like planets and galaxies.

I have a celestron cgx mount I use for this as well as another scope.

Yep, that's it. Yes. I don't know much about those scopes, but I'm sure they are of good quality. I believe it is a FPL53 doublet...

Nice choice, Skywatcher has been known to put out some good equipment for a reasonable price. I actually just purchased my first refractor a few weeks ago: a Skywatcher esprit 80mm. But of course, I haven't had a clear night since I bought it!:frown:

Yes, that was a typo. These should be fixed now. I was originally doing the derivation with dimensionless variables but then later switched to dimensionful variables, but forgot to change this in the article.

You're right that I never proved that using Planck's constant yields the correct discritization of the space space. Rather, I assumed it a priori as a reasonable guess which yields the right answer. It's also true that dividing the volume by Planck's constant doesn't really mean anything...

Yes, I intend to do a few more parts as time permits.

Greg Bernhardt submitted a new PF Insights post Statistical Mechanics Part II: The Ideal Gas Continue reading the Original PF Insights Post.

That's strange, do you have any idea what could be causing this? It looks normal on my computer.

Just to mention, If you are running Windows 10, it comes with a built in Linux subsystem that you can turn on very easily. I've been using it for over a year to compile Python and C++ code with no problems. https://docs.microsoft.com/en-us/windows/wsl/install-win10

The chemical potential of an ideal gas is a function of ##N##, ##V##, and ##T## or simply ##P## and ##T##. $$\mu=kT\ln\left(\frac{\lambda^{3}N}{V}\right)=kT\ln\left(\frac{\lambda^{3}P}{kT}\right)$$ where ##\lambda=h/\sqrt{2\pi mkT}##

This is generally justified a priori by stating that there is no directional preference in the momentum, so the points are uniformly distributed on the sphere. You may be right. I think I have heard of people trying to prove the a priori arguments given and they always fail miserably.

It looks like when using the correct Boltzmann counting, without the factor ##N##, the number of particles in the ##i##th state is a function of the temperature only. This may be reasonable in the thermodynamic limit, but I'm not sure. Correct me if I'm wrong, but it looks like you are holding...

It looks like you are starting to derive the Grand Canonical Ensemble because you have introduced the chemical potential ##\mu##. In that case, ##N## as you have defined it, does not exist because in this ensemble the particle number is not fixed. What you have is not exactly the grand canonical...

What I meant to say was each particle has the same average energy in the microcanonical ensemble. This definition is reasonable. As I mentioned before, a microstate is a set of ##N## positions in a ##6N## dimensional phase space. In the microcanonical ensemble, these points are restricted to...

It's not just the total energy is constant, but that each particle has a constant average energy. This is sufficient because equilibrium statistical mechanics is a time independent construction of the particle behavior. The ansatz of the microcanonical ensemble is that all the particles lie on...

It may be helpful to look back at posts 59 and 60. There, a simple example was given showing how to count the states of two identical particles. The Bose-Einstein counting is the exact counting, but if ##g>>n>>1## then this can be approximated by "correct Boltzmann counting". This is not making...

I think I understand your confusion. The choice of which description of microstate to use depends on the type of statistical ensemble being employed. For example, your definition 2 says to include the labels of which particles occupy which energy levels, but what if all the particles have the...

I think this is how it is justified: $$\frac{(n+g)^{n+g}}{n^{n}g^{g}}=\frac{(n+g)^{n}(n+g)^{g}}{n^{n}g^{g}}\approx\frac{g^{n}(n+g)^{g}}{n^{n}g^{g}}=\frac{g^{n}g^{g}\left(1+n/g\right)^{g}}{n^{n}g^{g}}$$...

Right, but I can't use it to solve for the state of a system until I impose the equilibrium requirement. Even in Zwanzig's book on non-equilibrium systems, I don't think the word entropy even appears in the book; it's not shown in the glossary/appendix. I'm not sure about this. I think if the...

It looks like something went wrong here: After factoring out the ##g^{n}## the expression should read as $$\frac{g^{n}\left(1+n/g\right)^{g}}{n^{n}}$$ then use the approximation $$\left(1+\frac{n}{g}\right)^{g}\approx e^{n}$$ for ##g>>n##.

Yes I'm sorry but I don't think I understand what you are asking. Can you rephrase this? I think the closest I can give to a "consensus definition" is the one given in Kardar's statistical physics book. He says

There are several members here, including myself, who are involved in amateur astronomy. If you have any particular questions, are seeking advice, or just want to show off some of your astro-images, fell free to post a thread!

Those methods will still work, they will produce a solution which eventually converges to the steady state value. If the problem is at steady state then that must mean that you expect a laminar flow. What is the Reynolds number for the system? Can you give more details on the problem you are...

This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_statistics

Entropy is a state variable only for systems at equilibrium. It is difficult to even define entropy for a dynamical system. Saying entropy is only useful for non-equilibrium systems strikes me as naive, since most texts on statistical mechanics and thermodynamics use entropy only when describing...

Pressure is an average macroscopic quantity of an entire system. I think you are envisioning a box with a single particle in it and saying that the walls of the box will experience some average pressure over long time scales as the particle bounces around inside. What @jbriggs444 is telling you...

I agree with your rant that the definition of microstate, or whatever-state, is a bit sloppy. A microstate is generally interpreted to mean a specific configuration of the sub-units of the system. How to deal with those microstates, and what the physical meaning of those states is, depends on...

Strictly speaking this is true. The actual number of states valid for even small numbers of identical particles is $$W=\frac{(N+g-1)!}{N!(g-1)!}$$ where ##g## is the number of states. For the case given above this means $$W=\frac{(2+2-1)!}{2!(2-1)!}=3$$ In the high temperature limit where...

I don't think density based solvers are used to solve the Navier-Stokes equation because of the non-linearitites as you pointed out. For compressible flow, a common explicit method used is the MacCormack method and a common implicit method is the Beam-Warming scheme.

I think your explanation is appropriate for a 'B' level thread, like this one. I was just pointing out that its difficult to justify the behavior of a dynamical system using results from equilibrium statistical mechanics. In your post #40, it looks like you are saying that the convection of the...

I think this is a pretty good answer except that the equipartition theorem only applies at equilibrium. The difficulty in explaining why hot air rises at the microscopic level, is that this system is not at equilibrium. The temperature is not uniform through the gas and not all of the particles...

Remember, the ##1/N!## is not strictly a product of quantum mechanics. It is there because a permutation of the particles on a constant energy manifold in phase space does not alter the microstate. I think the text fumbles this point a bit and might be confusing you. Although all classical...

I don't know, you will probably have to do some googling to find the answer. Since this question is not directly related to the original question though, it is probably more appropriate in a separate thread.

Over 2500km, using the example given, 2.5dB of the signal is lost. The fraction ##f## of power transmitted is then $$f=10^{-2.5/10}=0.56$$ So the fraction of power lost is ##44\%##.

I'm not sure what you mean by saving a percent loss. Do you mean how much light intensity will be lost over a 2500km cable?

I guess you could think of it that way. The more technical description is that fiber optic cables are a type of waveguide (https://en.wikipedia.org/wiki/Waveguide) for visible light. Yes, but they are low for good cables. I think I've seen cables where the loss is around 1dB/km. Of course...

We know that the ##1/N!## is needed for an ensemble of identical and distinguishable (a better word might be classical) particles to make the entropy extensive. The factorial is not needed if the particles are not identical. Such is the case if two different gasses are allowed to mix. The...

No, but if you are looking at a constant energy system (assuming ##N## and ##V## are also fixed) then the microcanonical ensemble is needed. Are you taking about non-classical distributions here? I thought the question was about where the ##N!## came from in the classical case.

It depends on the functional form of the density as a function of distance from the center. If ##\rho(r)## is decays faster than ##1/r^{2}## then the gravity will be stronger underground. If ##\rho(r)## decays slower than ##1/r^{2}## then the gravity will be strongest at the surface. According...

When the position of two particles in phase space are swapped, the total energy of the system is the same. So we should be looking at these particular groups of microstates in the Boltzmann picture (the microcanonical ensemble). For microstates which are not the same energy, the probabilities...

The assumption is that at sufficiently high energies and low densities, the region of phase space excluded by the Pauli exclusion principle is a very small fraction of the total phase space, so we can safely ignore it. Under this assumption the over counting is just a permutation of the...

It is a consequence of the Gibbs entropy formula. For a system with ##\Omega## accessible microstates the entropy is $$S=-k\sum_{i=1}^{\Omega}p_{i}\ln p_{i}$$ The second law states that the equilibrium state is the maximum entropy state. The probability distribution which maximizes ##S## is the...

No, according to Newtons shell theorem (https://en.wikipedia.org/wiki/Shell_theorem) the gravitational field at a point within a uniform spherical mass is only due to the mass at radii below that point. The field produced by the mass at larger radii cancels out.

A microstate is a unique distribution of particles in phase space. Swapping the position and momentum of two identical particles will give the same configuration in phase space and the same microstate. If we didn't get the same microstate, that would imply that some microstates with many...

So what is ##n## in this case?

Reading this, I think I understand the confusion here. This book uses the terms distinguishable and indistinguishable more loosely than I assumed. It appears to assume distinguishable means something like, the particles have a different color or shape, and assumes indistinguishable means they...