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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$will oscillate so that ... 2 The fact that momentum is not conserved in this system is an indication that angular momentum is not conserved either. Let's simplify the problem a bit and consider just one mass, m2, which is constrained to move horizontally and is oscillating back and forth through the origin due to the fact that it is connected to a spring at the origin. The mass m2 goes ... -1 Hamiltonian can be written as :$ dH=\frac{\partial H}{\partial q_i}dq_i+\frac{\partial H}{\partial p_i}dp_i+\frac{\partial H}{\partial t}dt $. or,$ \frac{dH}{dt}=\frac{\partial H}{\partial q_i}\dot{q_i}+\frac{\partial H}{\partial p_i}\dot{p_i}+\frac{\partial H}{\partial t} $. We lso know that$\frac{\partial H}{\partial p_i}=\dot{q_i}$and$\frac{\partial ...

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 $$H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C) ... 0 For the complex momentum,$$ 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} $$Now you can consider two separate cases: T^{0i} which gives the 3-momentum, P^{i} i.e. g^{0}_{i} = (0,0,0) T^{00} which gives the hamiltonian, H = P^{0} i.e. ... 1 First of all, there are a few problems with your question: J_{ab}^0 = \pi^a \epsilon^{ab} \Phi^b is not a valid expression, since there is a summation on the right hand side of the equation, but a and b are free indices on the left hand side. Your definition of \epsilon is a bit weird, too. What you mean is$$ J_{ab}^0 = \pi^i \epsilon_{ab}^{ij} ...

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 ...

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 ...

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 ...

1

As I see, maybe the problem is energy. So, What is energy? The formal classical definition of energy is: Energy is a dynamical invariant of a system that came from time-translation symmetry. There is also a question here about it. If you want more references about it, let me know. So.. when Bob write, $E = T + V$ in dissipative systems (damped OHS for ...

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