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Author gives a clue on the transition: Let us assume that $\delta\vec{A}$ vanishes at infinity and integrate (formula (1)) by parts... This is the usual step in the Lagrangian theory of field (actually, of anything). At first, we have the variation of action written in an awkward form: $$\delta S=\int_{\substack{\text{domain of least}\\\text{action ... 1 I am not sure if this is what you are up to (it is related to what Xiao-Qi Sun said) to but I'll give it a try too ... At the beginning of Chapter V.2 of his QFT Nutshell, Anthony Zee explains how classical statistical mechanics (characterized by the corresponding partition function involving the Hamilton function) in d- dimensional space is related to ... 0 I asked my professor and in a discussion we came up with the following. The process of establishing the effective action for a fluctuation Lagrangian to consist of the functional determinant of the initial differential operator involved, relies on the equality:$$\det(A)=e^{Tr(\log(A))}$$for a matrix A, which is only true for diagonalizable ... 3 The Lagrangian should not only be independent of the direction of \vec{v} but it should also change correctly under a Galilean transformation. For instance, if K and K' are two frames of reference with a relative velocity \vec{V} then the two Lagrangians L and L' should differ only by a total time derivative. If L is a function of fourth power ... 0 Well, you are not specifying the kind of probabilities you are talking about. Therefore, I refer to the tags and assume that we are talking about quantum (not statistical) probabilities. (See the edit at the end) Then, I have to note that your probabilistic definition of independece does not make much sense. Edit Well, according to the comments, I should ... 4 How about path integrals? The probability that a system evolves between state |\phi_1\rangle and |\phi_2\rangle is$$\langle \phi_2|\phi_1 \rangle =\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)$$where the measure \mathcal{D}\phi is suitably defined and the action S(\phi) is the integral of the Lagrangian (over ... 0 Let's look for solutions of the PDEs in a form \eta(x, t)= \eta(z) and \xi(x, t)= \xi(z), where z= x-c_T t. If we substitute these solution to the PDEs, they are reduced to ODEs. But to answer you question, we only need the second equation, which has the form:$$ c_T^2 \eta_{zz} = {\lambda \over \rho_0} [\eta_{zz} ({\tau_0 \over \lambda} + \xi_z + ...

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Your equation is wrong. Intuitively as either $r$ or $\omega$ vanishes you should recover the equation for a simple pendulum which is not true about your equation. In the Lagrangian formalism we equate $\frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i}$ for each generalised coordinates to zero, not to each other. Usually ...

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I) Here is at least a partial answer. Assume the following set-up. Let there be given a classical Lagrangian field theory in $d+1$ spacetime dimensions, with dynamical field variables $\phi^{\alpha}(x,t)$, and with no explicit time dependence. Action $S[\phi]:=\int \! dt~ L[\phi(t,\cdot)]$. Lagrangian functional $L:=T-V$. Energy functional $E=T+V$. ...

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Your answer is indeed what's going on in the lecture, but it doesn't explain what was wrong with your initial argument: you'd expect a model with $l_0>0$ to be a closer representation of reality than one with $l_0=0$, wouldn't you? Actually, your initial reasoning was correct: transverse displacements of springs under zero tension do indeed result in ...

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Interestingly, the quadratic potential can be justified by considering that the springs have natural length zero, i.e. $l_{0}=0$ and not $1$ (this means $l=1$ for $x=0$ just the same, but the spring is not relaxed and the potential energy will have a different dependency). I understand that this is silently assumed behind the explanation in the 1h15min mark. ...

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Consider the first three terms of the Taylor expansion of the potential. The first term is a constant, which will not affect the motion. The next term is the linear term, which will only change the equilibrium point. So the first interesting term is the quadratic term.

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