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The derivation of ideal gas equation from Hamilton's equations will take the same procedure as what you have seen in Wikipedia. Since you said you haven't understood the way in which the equation is derived I will give you a step by step explanation on it. So we have a system of perfect gas molecules. Of course they are non-interacting. We are going to ...

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A general advice: Before trying to understand Hamiltonian field theory, make sure you understand Lagrangian field theory. Before trying to understand Lagrangian field theory, make sure you understand Lagrangian point mechanics. In Lagrangian point mechanics, the functional derivative of the action is $$\tag{1} \frac{\delta S}{\delta q(t)} ... 2 Consider a map$$S \ni\phi \mapsto F[\phi] \in \mathbb R$$defined on a class S of smooth functions \phi defined on the compact set \Omega \subset \mathbb R^n obtained by taking the closure of an open set with regular boundary \partial \Omega. Thus the map F associates a real number F[\phi] to each function \phi\in S. We say that the ... 0 Let a d-dimensional Hamiltonian system (i.e. 2d-dimensional phase space) be given. Then the existence of d-1 observables c_1,\dots,c_{d-1} that are in involution with each other and with the Hamiltonian means that there are d constants of motion along each orbit - the Hamiltonian itself and the value of c_1,\dots,c_{d-1}. Systems for which such ... 2 Comments to the question (v2): On one hand, the Kuramoto-Sivashinsky (KS) equation is a dissipative differential equation (DE). Each term has an even number of spatial derivatives. It's a non-linear version of the heat equation. Dissipative systems rarely have variational action formulations nor Hamiltonian formulations. On the other hand, in the Korteweg ... 2 Comments to the question (v2): OP is considering the higher-derivative Lagrangian density$$ {\cal L}_1~=~ \frac{1}{2}(\partial\phi)^2 +\frac{g\phi}{2} (\partial^2\phi)^2,\tag{1} $$where g is a coupling constant. We use Minkowski sign convention (+,-,-,-). Quantum mechanically, the model is not unitary and therefore ill-defined, cf. the Ostrogradsky ... 0 I) Here we will assume that OP is taking about a relativistic point particle with zero spin in a Minkowski spacetime with metric \eta_{\mu\nu} of sign convention (−,+,+,+). Also we put c=1 for simplicity. (OP mentioned that the particle has charge but since it is free that is irrelevant.) Note that the relativistic point particle has world-line ... 2 Well, not really. We COULD write hamiltonian as square root - if we know, what is a square root of an operator. Of course we have simple approximation:$$\sqrt{1+x}=1+\frac x2-\frac{x^2}{8}+O(x^3)$$Using this we could write your hamiltonian as:$$\mathcal H=mc^2\sqrt{1+\frac{p^2}{m^2c^2}}=mc^2+\frac{p^2}{2m}+O(p^4).$$The problem is that this form of ... 0 Given a Lagrangian the states of the theory are given by the solutions to the linearised equations of motion, i.e. the quadratic part of the Lagrangian. Assuming that you have a standard kinetic term the solutions are plane waves e^{i k x}. Then an efficient way to find the Feynman rule for a vertex is to replace each field by its plane wave solutions. In ... 1 The Lagrangian is in fact an equation of \vec \Omega, however, in general it will be a quadratic function of \vec \Omega, as the rotational kinetic energy would be given by$$\frac{1}{2}\vec \Omega ^T \mathbf{I}\ \vec \Omega$$This would give you the desired generalized momenta as a function of the general velocity vectors, as the diagonal entries of the ... 0 This is explained in part II of my Phys.SE answer here, which shows that a 2D system always has a Hamiltonian description locally. It turns out, that before the non-canonical transformation (x,y) \to (q,p), from the first pair of eoms (1) alone, the Hamiltonian and non-canonical Poisson bracket can be derived as$$H~=~\gamma \ln x -x +\ln y -y $$and ... 4 Hint: Eq. (6) in its current form (v4) is meaningless since the lhs. depends on x, while the rhs. is integrated over x. The functional derivative$$\tag{A}\frac{\delta F}{\delta u(x^{\prime})}~\stackrel{(B)}{=}~\frac{\delta u(x)}{\delta u(x^{\prime})}~=~\delta(x\!-\!x^{\prime})$$in eq. (6) for the functional$$\tag{B} F[u]~:=~u(x)~=~\int \! ...

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Yes. Not only is it possible to find a Hamiltonian density but it is even possible to find a plain Hamiltonian. I gave the construction in this EPL ArXiv1303.6143: Euro Physics Letters, 103 (2013) 28004 This Letter is short but dense, I can only outline the method here. The reason why you can't write an hamiltonian for the electromagnetic field is that ...

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Take the orbit equation (using $u=\frac1r$): $$u(\theta)=A\left(1+e\cos\theta\right)$$ Where $A$ is some constant that you can work out from the differential equation if desired, and $e$ is the eccentricity of the orbit. Now, what happens when $\theta$ goes through an angle of $2\pi$? Notice that we would not have returned to the original radius if ...

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The answer is no: if you were to write $p_i\propto q_{i+3}$ for $i=1,2,3$, then your quantum Hamiltonian would be $$H\sim \sum_{i=1}^6 q_i^2$$ which looks like the Hamiltonian of a free particle in a six dimensional space, but it is not. It is not because the commutation relations are not $[q_i,q_j]=0$ but  [q_1,q_2]=0\quad [q_1,q_3]=0\quad ...

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