Timeline for Is there a Lagrangian formulation of statistical mechanics?
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21 events
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| Nov 26 at 7:23 | history | edited | Qmechanic♦ |
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| Jul 17, 2018 at 10:24 | comment | added | N. Virgo | @juanrga I understand that, more or less, but I am having trouble seeing the connection to my question. Could you be more explicit about that please? | |
| Jul 17, 2018 at 7:22 | comment | added | juanrga | Note that the classical Liouvillian is just the first term in a power series expansion of the quantum Liouvillian in powers of $\hbar$. And the quantum Liouvillian follows from a mixed superposition of quantum pure states. | |
| Jul 17, 2018 at 7:17 | comment | added | juanrga | The generator of time translations in statistical mechanics (either classical or quantum) is the Liouvillian, which is a function of the Hamiltonian, never the Lagrangian. In classical statistical mechanics $\mathcal{L}=\{H, \}$ and the state evolution is $\rho(t) = \exp[\mathcal{L}t] \rho_0$. The Liouvillian is a function of the Hamiltonian because the generator of time translations in quantum mechanics is the Hamiltonian. | |
| Jul 14, 2018 at 7:04 | comment | added | N. Virgo | @juanrga the question is purely about classical physics. | |
| Jul 13, 2018 at 19:18 | comment | added | juanrga | As emphasized by Weinberg in his QFT textbook, the generator of time translations in quantum mechanics is the Hamiltonian, not the Lagrangian. The so-called Lagrangian formulation of quantum mechanics is not a proper Lagrangian formulation, because it is the Hamiltonian which is being really used: "It is the Hamiltonian formalism that is needed to calculate the S-matrix (whether by operator or path-integral methods) but it is not always easy to choose Hamiltonians that yield a Lorentz-invariant S-matrix." | |
| Oct 31, 2015 at 23:48 | answer | added | Oscar Heath-Stephens | timeline score: 2 | |
| May 9, 2013 at 22:28 | answer | added | Dilaton | timeline score: 6 | |
| Mar 23, 2013 at 22:35 | answer | added | gatsu | timeline score: 3 | |
| S Mar 5, 2013 at 10:18 | history | bounty ended | N. Virgo | ||
| S Mar 5, 2013 at 10:18 | history | notice removed | N. Virgo | ||
| Mar 5, 2013 at 4:17 | answer | added | Xiao-Qi Sun | timeline score: 7 | |
| Mar 4, 2013 at 20:56 | comment | added | Vijay Murthy | @Slaviks, Thanks for the comment. One can write an action functional and need not do an RG. The OP asked for a Lagrangian description, not an RG. The MSR action functional can be written for particle systems too. So I dont get your comment. Perhaps I am missing something. | |
| Mar 4, 2013 at 19:23 | comment | added | Slaviks | @VijayMurthy For Wilson RG, all spin models get cast into continuous form which means you get Landau-Ginzburg-type Lagrangian. | |
| S Mar 3, 2013 at 5:05 | history | bounty started | N. Virgo | ||
| S Mar 3, 2013 at 5:05 | history | notice added | N. Virgo | Reward existing answer | |
| Feb 28, 2013 at 12:06 | answer | added | David Bar Moshe | timeline score: 16 | |
| Jan 18, 2013 at 13:03 | history | tweeted | twitter.com/#!/StackPhysics/status/292255889287692290 | ||
| Jan 18, 2013 at 11:27 | comment | added | N. Virgo | @VijayMurthy that looks interesting, and I'll look into it further. From those handwritten notes it looks like they're starting with some stochastic dynamics and then deriving something that looks like a path integral; whereas I'm hoping for something that starts with a classical Lagrangian and then derives a statistical ensemble based on it. But thanks, and I look forward to taking a closer look. | |
| Jan 18, 2013 at 11:16 | comment | added | Vijay Murthy | There is the Response-function / Martin-Siggia-Rose formalism which casts a Langevin description into a path-integral picture. See here for a simpler one-particle description. Not sure if this is what you are looking for. | |
| Jan 18, 2013 at 10:27 | history | asked | N. Virgo | CC BY-SA 3.0 |