@5:00. rho is not defined as a volumetric density (as one expects it should be) but rather as a normalized volumetric density (= density/N). That is why rho(p,q)dV behaves as a PDF. Where N is N-cursive (# of microstates) So, rho(p,q)dV is the probability of finding a microstate into a dV volume centered at coordinates [p.q] of the phase-space. Hence, the more dense a location (the higher the rho), the more probable of finding a microstate at that location. So, we could think of a macrostate as a cloud (of variable density or variable rho(p,q)= density/N) defined in a particular section of the phase-space.
when he write O(p,q) its a function that deoends on p and q of all N particle...so when you have to find average you have to integrate over the p and q of all N particle ..so thats why
The volume of the parallelogram was approximated, wasn't it? Due to small time steps the small angle approximation makes sense, yet I can't help but feel unsatisfied that we approximated a a volume, whose lengths themselves are already approximated...
the infinitesimal volume element (what we are integrating over) is just the product dp'.dq'. It looks like a parallelogram but we are not computing the area of a parallelogram
There can be many reasons for this. The camera person was late or there was a camera problem (battery dead and a power cord has to be run). The speaker could start the class early or without warning the cameraman. Many times there isn't much time between classes, which can make the setup and recording of a class a challenge. Basically, stuff happens. =D
so crazy, is this with out gravity and atmosperic pressures ? im not sure what the six dimensions are ??????????????? time to space barrier deconstruction ? to a vacum ? im no math person but if anyone is good with high presure physics message me.
+John Esposito These are the six dimensions of phase space. We need three numbers to specify the position of a particle and three to specify it's momentum, thus six are needed to specify the mechanical state of the particle.
That was unnecessary confusing. We do not prove, we _require_ rho * dGamma = rho' * dGamma' (makes sense for deterministic systems) and then from Liouville's theorem (i.e. Gamma = Gamma' ) it follows that the total derivative w.r.t. time of rho is zero.
@@kevinpierce4061 why should the number of particles in the phase space volume be conserved though, what if the particles happen to be attractive and the number of particles in a phase space volume can potentially increase
@@azizfall5579 I think that's a nice consideration. You may be conflating phase volume with spatial volume. Sure, attractive potentials increase spatial density. But they don't increase phase space density, since they cause mutually compensating changes in momenta
@@azizfall5579 In phase space (6N mutually orthogonal dimensions) each point represents an instantiation of the whole system (a microstate). So you have cursive-N points or microstates. All of them correspond to a defined Macrostate. So, what is conserved is the density of microstates (each of these points) in phase space. This means that after a dt, the fluctuation of the cloud of points remains constant in its density. Such that the properties of the macrostate (a particular collection of microstate) remain time-invariant.
《粒子的统计力学（共26讲）》
th-cam.com/play/PLUl4u3cNGP60gl3fdUTKRrt5t_GPx2sRg.html
第三章 气体动力学理论
00:00 从微观状态描述宏观系统
21:38 Liouville定理
Thank you MIT for sharing this online. Grateful for these videos. Truly grateful.
@5:00. rho is not defined as a volumetric density (as one expects it should be) but rather as a normalized volumetric density (= density/N). That is why rho(p,q)dV behaves as a PDF.
Where N is N-cursive (# of microstates)
So, rho(p,q)dV is the probability of finding a microstate into a dV volume centered at coordinates [p.q] of the phase-space. Hence, the more dense a location (the higher the rho), the more probable of finding a microstate at that location.
So, we could think of a macrostate as a cloud (of variable density or variable rho(p,q)= density/N) defined in a particular section of the phase-space.
I have question, why do you integrate over 6N coords a nd not general 6 coords ?
Because you have N particles, not just one
when he write O(p,q) its a function that deoends on p and q of all N particle...so when you have to find average you have to integrate over the p and q of all N particle ..so thats why
The volume of the parallelogram was approximated, wasn't it? Due to small time steps the small angle approximation makes sense, yet I can't help but feel unsatisfied that we approximated a a volume, whose lengths themselves are already approximated...
the infinitesimal volume element (what we are integrating over) is just the product dp'.dq'. It looks like a parallelogram but we are not computing the area of a parallelogram
Why the first few minutes lost??
There can be many reasons for this. The camera person was late or there was a camera problem (battery dead and a power cord has to be run). The speaker could start the class early or without warning the cameraman. Many times there isn't much time between classes, which can make the setup and recording of a class a challenge. Basically, stuff happens. =D
51:44 must be one hell of a kid.
the first 9 minutes, please.. thank you
ocw.mit.edu/courses/physics/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/lecture-notes/MIT8_333F13_Lec7.pdf
noiiiicee
so crazy, is this with out gravity and atmosperic pressures ? im not sure what the six dimensions are ??????????????? time to space barrier deconstruction ? to a vacum ? im no math person but if anyone is good with high presure physics message me.
+John Esposito your going crazy calm down
+John Esposito These are the six dimensions of phase space. We need three numbers to specify the position of a particle and three to specify it's momentum, thus six are needed to specify the mechanical state of the particle.
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That was unnecessary confusing. We do not prove, we _require_ rho * dGamma = rho' * dGamma' (makes sense for deterministic systems) and then from Liouville's theorem (i.e. Gamma = Gamma' ) it follows that the total derivative w.r.t. time of rho is zero.
nah. showing that conservative hamiltonian dynamics implies jacobian 1 is conceptually useful. It's not true for dissipative dynamics.
@@kevinpierce4061 why should the number of particles in the phase space volume be conserved though, what if the particles happen to be attractive and the number of particles in a phase space volume can potentially increase
@@azizfall5579 I think that's a nice consideration. You may be conflating phase volume with spatial volume. Sure, attractive potentials increase spatial density. But they don't increase phase space density, since they cause mutually compensating changes in momenta
@@azizfall5579 In phase space (6N mutually orthogonal dimensions) each point represents an instantiation of the whole system (a microstate). So you have cursive-N points or microstates. All of them correspond to a defined Macrostate.
So, what is conserved is the density of microstates (each of these points) in phase space. This means that after a dt, the fluctuation of the cloud of points remains constant in its density. Such that the properties of the macrostate (a particular collection of microstate) remain time-invariant.