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Dear Murat, thank you for your kind words. I wrote this a very long time ago in either F90 or C, don't remember. Meanwhile I wrote a much better python script. If you send me an email (you can find it by searching for me at the TU Delft, I do not want to disclose it at too many places), I can forward it to you.

Thanks for your question. Hmmmmmm. You use Lagrange here in order to facilitate the solution. If you do not want to use this, the solution becomes more tedious, but you will find all the forces enforcing the constraints. So you need to include (i) the spring forces, (ii) a force which keeps the suspension point horizontal (iii) the moment of the forces (gravity, spring forces) w.r.t. either the center of mass or the suspension point of the pendulum. Seems doable but why would you like to do that? If the walls move, you can still use Lagrange, but the potential now involves u(t) and is therefore time-dependent. As in the Lagrange equations you take the full (not partial) derivative with respect to time, the potential yields a contribution there. Would be nice to try and then solve the Lagrange equation on your computer and make an animation.

Thanks for your question. Which minute/second are you writing about? If you mean the discussion after 17:00 minutes, it is due to the property of the Langevin forces, <R(t) R(t')> = q \delta(t-t').

thank you so much professor @@josthijssen6782

We simply want to have an unambiguous number. An additive constant would not matter for analysing processes, whereas a multiplicative constant does. We want to make contact with thermodynamics: E = TS-PV+μN, so there's not too much freedom.

Thanks. Another explanation came to my mind. At thermal equilibrium the entropy is at the maximum value and therefore it stays at Heisenbergs lowest possible value; DpDx=h. @@josthijssen6782

@@josthijssen6782 Thank you for the explanation. I see, a constant wouldn't matter. Especially not in differences between states. It came also to my mind that at thermal equilibrium the entropy is at a maximum. This brings Heisenberg's relation to its minimum possible value Dx.Dp=h.

The Hamiltonian reads -K sum_ij s_i s_j. This does not have to do with a magnetic field (the Bohr magneton is not used, we just have constants K and h), but with the coupling of nearest neighbour spins.

The second term (h sum_i s_i) represents the effect of the magnetic field.

Dear Geoffry, thanks for your question. The concept of a chemical potential is not as clear-cut as many other thermodynamic parameters. However it is measurable: whenever you use a volt meter, you measure the chemical potential difference of electrons! For neutral particles, the measurement principle would be equivalent to that of a volt meter: the difference in chemical potential between two reservoirs is determined as the work needed to transfer a particle from one reservoir to the other. In practice that is not always easy to measure. A typical example of exchanging neutral particles is when you are sitting in an office and the temperature is the same as the outside temperature, but the relative humidity is different. That relative humidity is one-to-one related to the chemical potential of the water molecules, which itself is difficult to measure. Note that I only mention 'difference': as with all energies, the difference in chemical potential is what drives diffusion, there is a freedom in setting their absolute scale (i.e. where the zero chemical potential is).

@@josthijssen6782 Oh i see. The concept is kind of mysterious to me because I can't really tell how the chemical potential is determined for a given substance. Lets say there are two rooms (with the same temperature) separated by a partition. Room A is just filled with hydrogen gas. Room B is filled with the same quantity of hydrogen gas and n moles of (H2O vapor + CH4 gas + Xe gas). If the partition is removed, how would those extra gases diffuse related to their chemical potential? Which one has the higher/lower chemical potential? or are they all the same? can the value be zero? On the case of the solid, you were able to model the quantum mechanical potential well (like in the video). I wonder if we can derive the chemical potential of a substance from its microscopic property (intermolecular forces?) without having to directly measure it.... allowing us to make predictions

@@GeoffryGifari Dear Geoffry, The chemical potentials in your example all depend on the partial densities. For ideal gas mixtures, the chemical potential of species i can be calculated as mu_i = k T ln (P_i lambda^3/k T), where lambda is the thermal wavelength and P_i is the partial pressure: P_i = N_i k T/N, with N is the number of all particles.

@@josthijssen6782 I see. I pick those examples because water is polar, methane nonpolar, and xenon is the closest to ideal gas to see whether or not intermolecular forces have an effect. I was thinking in cases where both solid and vapor phases are involved (like in the video) that might cause a difference. It makes sense that for ideal gas it depends on partial pressure. Thank you for your insight.

@@GeoffryGifari Yes, I made it easy considering ideal gases. If you take interactions into account then expansions, mean field theory, fluid density theory, computer simulations.... will be necessary.

Thanks for your remark. The exponential is not expanded as t is large; it is neglected and only the term -1 survives in the parenthesis. Am I right or did I overlook something?

@@josthijssen6782 I mean as t goes to 0. 38:30 aprox.

I see, you're right. Thanks.

Thanks for your question. The interaction between the t^k depends on the Hamiltonian of the s_i. It is not always easy to find it, but for the Ising model as in this case, it is assumed that the 'renormalised' interaction (between the t^k) only acts between nearest neighbours and then you can work it out, e.g. numerically. That is not the point of this movie. To see how such a procedure is carried out for the Ising model on a triangular lattice, see e.g. my video on real space renormalisation.

@@josthijssen6782 Thanks a lot! It’s really helpful!

Thanks for your question; I find it hard to answer as the steps are straightforward and explained in the video. Is it the step leading to m^2 ~ qJ-1 or the conclusion from that? At Tc qJ=1, so qJ is expected to vary like A(Tc-T), where A is an arbitrary constant. I hope this helps.

Dear Krishna, Thank you for your kind words. I have a set of lecture notes with links to videos on youtube. You can download them from surfdrive.surf.nl/files/index.php/s/xEJQuCxD6BOIgdg

@@josthijssen6782 Oh this is wonderful. yes, I will certainly read these notes to understand the derivations. I really appreciate what you have done in these videos. I found reading other textbooks extremely dense and difficult to parse, and I am a mathematically mature statistician. So many books seem to privilege the rigor before the explanation. So listening to your lectures I understand the substance which makes the mathematics much easier to follow. Thanks again.

Hi, thank you for your questions. Expectation values are taken in the canonical ensemble, where each configuration r_1, ... r_N is weighted by a Boltzmann factor exp(-beta H). You sum (integrate) r_i . r_j times exp(-beta H) over all r_k and divide by the partition function. I hope this helps.

From the expression for the partition function you can immediately see that the deriv wrt h yields sum_i s_i. Beta=1/kT is incorporated into h. If you would miss an extra factor kT, that would not change the exponent.

@@josthijssen6782 Ah okay, I didn't realize that beta was included inside of h, thank you 🙏

I am afraid he is!

@@josthijssen6782 your book is really awesome. Helped me a lot. Thanks.

@@adwaitnaravane5285 Thanks for your kind words.

Thank you for your question. The only ingredients going into the calculations are the first and second moments of the distribution, the first moment being zero. If you have an alternative distribution with first moment zero, you obtain the same form. For a non-vanishing first moment, you can probably do a similar calculation, where the first moment would appear as an extra contribution to the drift force.

Not sure whether I get your point. The Pauli exclusion principle is a direct consequence of the antisymmetry of the wavefunction under particle swaps. The anti-symmetry property follows from the spin-statistics theorem and is in this analysis the fundamental property.

Dear Maria, Thanks for your remark. You're right, the gamma should go out, i.e. <x^2> = kB T/m t^2.

Thanks for pointing this out. Yes, indeed the integrations over the components of x should run over the entire real axis.

Thanks for your kind words. The notes can be found at surfdrive.surf.nl/files/index.php/s/oZH84S1nHUWyTEW

@@josthijssen6782 Thank you very much!

Dear Somesh, thank you for your kind words. I think the best would be my video on the Ising model, see ruclips.net/video/hX_FrS5XO3I/видео.html Best regards, Jos

너 때문에 먼 타국까지 왔다

Exactly!!!

The factor 2pi is wrong. Thanks for pointing this out. The factor which you keep is q, so it doesn't matter for the sequel