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1

You're actually dealing with the Potts model, which is a slight generalization of Ising. Not that it really matters, as you won't need any results from Potts. The point of mean field theory is typically to make each site independent of their neighbors, which allows you to evaluate the partition function by only iterating through the possible states of one ...


-1

Point in configuration space represents configuration of the system, i.e. positions of the constituent particles. Point in phase space represents state of the system, i.e. positions and velocities of the constituent particles together. No. Liouville's theorem has no simple analogue in the configuration space. Depends on what is the task at hand and what are ...


1

You should think of the definite integral operation as a function of two arguments: a region over which to integrate (here, $[x_0,x_1]$), and another function $f$ called the integrand (here, $f:\xi \mapsto (E-V(\xi))^{-\frac{1}{2}}$). So first of all, in my definition of $f$ above, we could have used (almost) any other symbol instead of $\xi$ and the ...


1

Here is an outline of the reduction from the Nambu-Goto (NG) action to the light-cone (LC) formulation from a Hamiltonian perspective: The starting point is the Hamiltonian formulation of the NG string, cf. e.g. this Phys.SE post. The Hamiltonian density is of the form "Lagrange multipliers times constraints"$^1$ $$ {\cal H}~=~\lambda^{\alpha} ...


0

The first equation has just explicited the definition of momentum. As a matter of fact, $p=\partial \mathscr{L} / \partial \dot{q}$


3

I) In this alternative answer we resolve the singular Hessian $H_{\mu\nu}$ of the Nambu-Goto string action by introducing two auxiliary variables from the onset, thereby indirectly showing that the Hessian $H_{\mu\nu}$ must have co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. Consider the extended Nambu-Goto ...


1

I'm not so sure, if this is really, what you're looking for, but you can of course solve this easy problem analytically. To do this, it is clever to first analyze the easier Hamiltonian $H_0 = 2g (\vec L \cdot \vec S)$, where the $L_i$ and $S_j$ fulfill independent $SU(2)$-algebrae $$ [L_i, L_j] = i \epsilon_{ijk} L_k\\ [S_i, S_j] = i \epsilon_{ijk} S_k. ...


0

The following might help: $H = \frac{1}{2}(mv^2 + kx^2) + \gamma mkvx$ decays exponentially with time along the solution of the damped system. Check by integrating $H$ with respect to $t$ and using the equations of the system. So the "energy" $H$ decays exponentially instead of remaining constant.


4

I) In this answer we will consider the standard Nambu-Goto string and show that the Hessian has co-rank 2. The target space metric has $(-,+,\ldots,+)$ sign convention, and $c=1=\hbar$. The Nambu-Goto Lagrangian density is $${\cal L}_{NG}~:=~-T_0\sqrt{{\cal L}_{(1)}}, $$ $$ {\cal L}_{(1)}~:=~-\det\left(\partial_{\alpha} X\cdot \partial_{\beta} ...


1

Some of the mathematical aspects of the Liouville operator can be found in the second book by Reed and Simon, in section X.14 (it is not a comprehensive account, but it gives the basic ideas and proofs). In the notes at the end of chapter X, in the part dedicated to section X.14, there is also a quite extensive bibliography that may be useful.


1

Normally we do NOT calculate the phase space density of a system. In the phase space formulation of classical statistical mechanics, the phase space density $\rho(p,q;t)$ has its specified form for different ensembles. Normally for systems at equilibrium the density $\rho$ has no explicit time dependence and thus we work with $\rho(p,q)$. (1) For ...


1

The first thing we can do is to split up $\Gamma$ according to the number of particles in the given states. Let $\gamma_N$ be a state with $N$ particles. The grand canonical partition function is then \begin{align} \mathcal{Z} = & \sum_\Gamma \exp\left(-\beta(\mathcal{H} - \mu N)\right)\\ =& \sum_{N=0}^\infty\exp\left(\beta \mu N ...


1

You are at a point where you'll need $v_1$ and $v_2$. Observe from the original transformation that: $$v_2 = v_1 - v$$ $$\implies V = \frac{(m_1+m_2)v_1 -m_2v}{m_1+m_2}$$ $$v_1 = V + \frac{m_2v}{m_1+m_2}$$ We also get, by a similar procedure, $$v_2 = V - \frac{m_1v}{m_1+m_2}$$ We have expressed $v_1$ and $v_2$ in terms of the new variables, $V$ and $v$. ...


3

The answer is Yes. Define function $g(q):= \frac{1}{f(q)}$ for later convenience. Then the classical Hamiltonian reads $$2h~=~g(q)p^2.$$ One may show that the Weyl-ordered Hamiltonian reads $$2H_W~=~ (g(q)p^2)_W ~=~ \frac{1}{4}P^2 g(Q)+\frac{1}{2} Pg(Q)P+\frac{1}{4} g(Q)P^2$$ $$~=~ Pg(Q)P - \frac{1}{4}\hbar^2g^{\prime\prime}(Q),$$ see e.g. Ref. 1 and this ...


1

This depends on whether the corresponding quadratures have physical meaning in your specific example. This is because if $a=x+ip$, then changing $a\mapsto a'=e^{i\theta}a$ corresponds to the canonical transformation \begin{align} x\mapsto x'= \cos(\theta)\, x -\sin(\theta) \,p, \\ p\mapsto p'=\sin(\theta)\, x +\cos(\theta) \,p. \end{align} This could be ...


0

The momentum is a covector because it is a gradient, and gradients are always covariant. It does what it says on the tin. However, you are right that this is a subtle point and it's not particularly clear at first sight. For a lagrangian of the form $L=T-V$ with $V$ independent of $\dot q$, the canonical momentum is given by $$ p=\frac{\partial L}{\partial ...


-1

I think the normal is always time-like because when you slice your space-time you do it in such a way such that the normal vector to this hyper-surface is time-like. Thus, time components of the original metric are absent in the induced metric. Which reference are you reading from?


2

1) The spacelike hypersurface has three spacelike directions tangent to it. Any vector that is normal to all three spacelike directions in the eneveloping space is necessarily timelike. Equivalently, the spacelike surfaces can be thought to be labeled by a function $\tau$ which gives the "time coordinate"'s value on those surfaces. the normal to the ...


0

At least a partial answer to your question is that commuting hamiltonians help you to solve the physical system described by one of them: in particular, if your system has $N$ degrees of freedom and you have $N$ commuting hamiltonians, there is good hope that you can trivialize the problem and solve it exactly. In classical mechanics, this is known as ...


0

After thinking about Nick P's answer and re-reading the relevant chapter of Sussman's Structure and Interpretation of Classical Mechanics, I came up with the following elaboration of Nick's argument. It's not water-tight, but it convinced me, and perhaps it will help someone else. I will use Sussman's unorthodox but precise notation. The first step (and ...


1

For simplicity consider the 1-d case, with $\psi =\sqrt{n} e^{2i\phi}$, then $$i \psi_t =\frac{i}{2} \frac{\dot{n}}{\sqrt{n}} e^{2i\phi} -\sqrt{n} e^{2i\phi} 2\dot{\phi}.$$ Similarly $$ \frac{\partial H}{\partial \psi^*} = \frac{\partial H}{\partial n}\frac{\partial n}{\partial \psi^*} + \frac{\partial H}{\partial \phi}\frac{\partial \phi}{\partial ...



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