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.
could you show us how it would be without the Lagrange method, only using sum of forces and moments? how it would be if one wall moves with an harmonic function u(t) ?
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').
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.
This is such a great explanation of the Flory-Huggins theory. I am a Chemical Engineering student in my final year and have been doing Polymer Physics research with my professor for the last few years. I remember when I first started in the research group, I looked up this video because we were working on a polymer phase separation problem and I wanted to understand what was happening. I remember not understanding anything in the video. Now three years later, I watched your video again with a much more mature understanding of thermodynamics and stat mech and I understood everything you are saying. All in all I just want to say that I really appreciate the way you explained these difficult concepts.
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.
On the chemical potential, how do we get the value of it? Is it empirically determined? And what would the implication be if the chemical potential is positive or negative?
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 question. The formula as it is used here is correct. If you would instead of an integral use a sum over time intervals Δt, you would need to divide by Δt. Note that a delta-function of a time argument has a dimension of 1/time.
Hi Jos, amazing video and thanks very much for the series you have up here on youtube! At ~17:00 you have that the correlation function for R(t) is q*delta(t1-t2). Should it not be (q / (Delta t)) *delta(t1-t2)?
Excellent explanation, thank you. Wproking towards a different application of the Fokker-Planck equation but even though i can't use all the assumptions, your video made the subject much clearer to me. I'll be going back to this movie
Hello sir, thank you for this video. At 39:00 there is a mistake, in the step where you take the taylor series of the exponential you forget to multiply -1 by m/gamma (but then proceed as if you did, so the result holds)
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?
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.
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.
@JosThijssen, thank you so much for these wonderful videos. If it is not too much trouble, could you identify any books or lecture notes used to develop these videos? I would like to master more of the math behind these models, but it can be tough to parse the notations and argument flow from a different book. If there was a book that corresponds to your lectures, that would be the perfect resource.
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.
I just have a question regarding notations. When we write the dot product like this < ri . rj>, what are the signs < > for ? Are we calculating a temporal mean or a simple arithmetic mean. What is the difference exactly between ri . rj and < ri . rj>?
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.
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.
Just one helpful hint for this lecture - a two particle wave function can be written as the product of individual particle wave functions if they are not entangled. If the particles are distinguishable, they are just the product. If identical, as in the video. Question- Is Pauli's exclusion principle for Fermions, a consequence of this or is it just a consistent representation?
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.
Hello Jos, I am studying a MSc in Physics of Complex Systems (UNED, Spain), where I have a course called Advanced Statistical Mechanics. I was studying the renormalization group techniques through Shang-Keng Ma's "Modern Theory of Critical Phenomena" and was struggling with the concepts. I am super grateful for your videos on these subjects, they have been instructive and inspiring, I hope you are proud of your teaching abilities and of the help you are lending here on RU-vid to students like me. Best regards, Gabriel
Dear Sir, Thank you. However, there is a tiny mistake at 8.40, I think that there shouldn't be the friction constant(drag) \gamma - it is just cancelled. It makes also sense (in my logic) that there shouldn't be \gamma because at the previous video (ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-H9I0PmXwhdo.html) we got that q=2 \gamma k_B T , where q is the "strength" of the random force and \gamma is the friction. Therefore, if t-->0 there is no random force so q=0 and therefore no drag force, \gamma=0. For me, it means that if there is no random force the big particle has no reason to feel any friction because q and \gamma are related in the Langevin equations - random force can be as strong as the friction enables it and friction is as big as the random force push it to be.
Is it not true that the boundaries of the definite integral in 'x' are the same as 'k' (with periodic BCs) running from -infinity to +infinity (defined at around 4:04)? Otherwise (0 to infinity), I think a (1/2)^3 prefactor will appear in the result (at around 6:36).
Dear Somesh, thank you for your kind words. I think the best would be my video on the Ising model, see ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-hX_FrS5XO3I.html Best regards, Jos
I'm studying classical mechanics by myself (becore returning university) and I couldn't find a solution for that problem on internet (to see if my analysis was good), and after watch this video I confirmed my answer! Greetings from Colombia 🇨🇴