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$\Phi_c(t)$ is the directional derivative of $L(c(t),\dot{},t)$ along $W(c(t))$, so by the chain rule $$\Phi_c(t) = \sum_i\frac{\partial L}{\partial \dot{q}^i}W^i = \sum_i\frac{\partial L}{\partial \dot{q}^i}\frac{\partial \phi^i}{\partial s}$$ There is then an inexplicable notational shift from $\frac{\partial\phi^i}{\partial s}$ to $\frac{\partial ... 2 There exist choices of origin for which the angular momentum of the system is not conserved. Consider, for example, an origin at$x_0\,\hat{\mathbf{x}}$, and consider the initial condition in which$m_2$is at the origin standing still, and$m_1$is at$L\, \hat{\mathbf y}$. Mass$m_2$will remain at the origin, and mass$m_1$will oscillate so that ... 2 The fact that momentum is not conserved in this system is an indication that angular momentum is not conserved either. Let's simplify the problem a bit and consider just one mass, m2, which is constrained to move horizontally and is oscillating back and forth through the origin due to the fact that it is connected to a spring at the origin. The mass m2 goes ... 0 I'm posting here an extraction from a paper written by T. Padmanabhan (http://arxiv.org/abs/hep-th/0608120) : Consider a dynamical variable$q(t)$in point mechanics described by a lagrangian$L_q(q,\dot q)$. Varying the action obtained from integrating this Lagrangian in the interval$(t_1,t_2)$and keeping$qfixed at the endpoints, gives the Euler ... 1 Inserting the expansion $$\psi=\int\frac{d^3p}{(2\pi)^32\omega_p}(a_pe^{-ipx}+b_p^\dagger e^{ipx})$$ into the expression for the Hamiltonian $$H=\int d^3x(\dot{\psi}^\dagger\dot{\psi}+\nabla\psi^\dagger\cdot\nabla\psi+m^2\psi^\dagger\psi)$$ we get $$H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C) ... 1 The following is rewritten after @Qmechanic's comment. While his observation was correct, I think the main point below holds on its own. The case @Cham considers is that of a Lagrangian L' = L - \frac{d}{dt}q^ip_i modified by a total derivative for the purpose of implementing a change in the boundary conditions. Originally the p_i-s are assumed to be ... 0 Not that I know of. However if you're fine with considering only small oscillations, then you can replace \sin \theta by \theta and \cos \theta by 1-\frac{\theta^2}{2}. This might make things simpler although the solution you get will be acceptable for small angles. 0 For the complex momentum,$$ T^{\mu\nu}= \partial^{\mu}\phi^{\dagger}(x)\partial^{\nu}\phi(x) + \partial^{\nu}\phi^{\dagger}(x)\partial^{\mu}\phi(x) - g^{\mu}_{\nu}\mathcal{L} Now you can consider two separate cases: T^{0i} which gives the 3-momentum, P^{i} i.e. g^{0}_{i} = (0,0,0) T^{00} which gives the hamiltonian, H = P^{0} i.e. ... 4 The Hamiltonian H(\theta,p_\theta) needs to be formulated in terms of the coordinate \theta and its canonically conjugate momentum p_\theta = \frac{\partial L}{\partial \dot{\theta}} = R^2 \dot\theta. The correct expression for the Hamiltonian is \begin{align} H(\theta,p_\theta) & = p_\theta \dot{\theta}(\theta,p_\theta) - ... 0 It's actually not that complicated for a linear case. Let's derive the 1D wave equation for velocity potential \Phi (v = \partial_x \Phi). As usual, c_0 denotes the speed of sound. The kinetic energy density is obviously \mathcal{T} = \frac{1}{2}\rho_0\mathcal{v}^2 = \frac{1}{2}\rho_0\left(\frac{\partial \Phi}{\partial x}\right)^2 $$Potential ... 1 I) Hamiltonian interpretation. Given a Hamiltonian H(z;t) with canonical coordinates$$\tag{1} (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n), $$the Hamiltonian Lagrangian reads$$\tag{2} L_H(z,\dot{z};t) ~=~p_k \dot{q}^k - H(z;t). $$Then OP's modified Hamiltonian Lagrangian becomes$$\tag{3}\tilde{L}_H(z,\dot{z};t) ~:=~L_H(z,\dot{z};t)- ... 0 Lets define the following action : $$\tag{1} S' = \int_{t_1}^{t_2} L' \, dt,$$ where $$\tag{2} L' = L - \lambda \, \frac{d}{dt} (q^k \, p_k).$$ An arbitrary variation\delta q^kgives the following : \begin{align} \delta S' &= \int_{t_1}^{t_2} \Big( \frac{\partial L}{\partial q^i} \; ... 0 Don't think of it as components of KE: rather think about it as that the total KE of the body is the summation of it's angular KE and Linear KE(which happens to be in the radial direction in this case)...Hope that helps.. 2 Let us consider the corresponding Hamiltonian theory, so that we have a notion of a commutator that we can use to form a Lie algebra bracket. Moreover, let us consider the classical theory for simplicity. Then the Poisson bracket $$\tag{1} \{\Phi^a({\bf x}),\Pi_b({\bf y})\}_{PB}~=~\delta^a_b~\delta^3({\bf x}-{\bf y}), \qquad \text{etc},$$ plays the role ... 0 If $$\tag{1} \delta\varphi~=~\varepsilon$$ is a global shift symmetry, we can gauge the symmetry, i.e. enhance it to a local symmetry by (i) introducing a gauge fieldA_{\mu}$with gauge symmetry $$\tag{2} \delta A_{\mu} ~=~\partial_{\mu}\varepsilon,$$ and (ii) replace partial derivatives$\partial_{\mu}\varphi$with covariant derivatives $$\tag{3} ... 1 Here we have a holonomic constraint: \theta-\omega t=\theta_0 (or \dot{\theta}=\omega). The question is where we should use it in solving the Lagrange's equations. At first you obtained equations from the Lagrangian for the free particle, solved them, and then used the constraint \dot{\theta}=\omega. But let us remember how to solve the classical ... 1 There are at least two generalizations of Noether's theorem. 1) Assume that the Hamiltonian system with Hamiltonian H(z),\quad z=(p,q) has a one-parameter symmetry group \{g^s_F(z)\} which is generated by a Hamiltonian system with Hamiltonian F. Then F is a first integral for H:\quad \{F,H\}=0, moreover if dF\ne 0 then there are local ... 0 Actually the pendulums swing independently, but one can consider all of them together as a Hamiltonian system. The movie illustrates the Poincare recurrence theorem. It also illustrates quasi-periodic motion on the Liouville torus in a completely integrable Hamiltonian system 1 The expression you have for the stress-energy tensor (the right hand side of the Einstein-equation) for the scalar field is incorrect. We have$$R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R = T^\phi_{\mu\nu}$$where$$T^\phi_{\mu\nu} = -\frac{1}{\sqrt{-g}}\frac{\delta [\sqrt{-g}\epsilon g^{\alpha\beta}\partial_\alpha\phi\partial_\beta\phi]}{\delta g^{\mu\nu}} = ... 3 Within the Newtonian framework of mechanics conservation laws are tricky to develop and are not obvious at first glance. Lagrangian mechanics generalises the concept of conservation laws by exploiting "symmetries". The connection between symmetries and conservation laws is made by Noether's theorem. An object has a symmetry if it is invariant under a ... -5 From the definition of lagrangian mechanics, Noether's theorem shows that conservation of momentum and energy comes from invariance vs time and space. Yes, that's what we can read on websites like this. But note that we define our time using the motion of light, and our space too. Is the reverse true? Are Lagrangian mechanics completely ... 2 Normally, the Lagrangian (in Cartesian coordinates) for an object of mass$m$in a potential$V$would be $$L=T-V\to\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)-V(x,y,z)\tag{1}$$ It then follows that for a coordinates$q$, $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{q}}\right)=\frac{\partial L}{\partial ... 1 The fields of a supersymmetric theory form a representation of the super Poincare algebra. When this representation is restricted to a specific value of the mass operator P^{\mu}P_{\mu} = m^2, the representation is called an on shell representation multiplet. On shell representations are characterized by the equality of the number of bosonic and ... 1 In the following calculation, I ignore some coefficients. According to J(x)=\int d^4 k_1 e^{ik_1 x} , J(y)=\int d^4 k_2 e^{ik_2 y} and D(x-y)=\int d^4k \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} We have$$W(J) = \int d^4x d^4y d^4 k d^4 k_1 d^4 k_1 J(k_1)e^{ik_1 x} J(k_2)e^{ik_2 x} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} W(J)=\int d^4x d^4y d^4 k ... 0 I'm not sure this is quite what you're looking for, as classical statistical mechanics in equilibrium is, by definition, not a properly dynamical system, but min/maxing the function $$S[p_i] = -k_{B}\sum_{i}{\ln{p(x)}} + Z\sum_i{\big(p(x)-1\big)}+\beta\sum_i{p_iE_i-\langle E\rangle}+\mu\sum_i{N_ip_i-\langle N\rangle}+...$$ not only recovers the Boltzmann ... 1 Also, you can write Hamilton's equations of motion in sympletic form: $$\dot\xi_i = \omega_{ij}\frac{\partial H}{\partial\xi_j}$$ Where$\xi_i$are the coordinates in the phase space, that is,$\xi = (\mathbf q, \mathbf p)$. And,$\omega$is the sympletic matrix:$\$ \omega = \begin{bmatrix} 0 && -I_{n\times n} \\ I_{n\times n} && 0 \\ ...