# Tag Info

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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) - ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 $q$, $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial\dot{q}}\right)=\frac{\partial L}{\partial ... 1 \Phi_c(t) is the directional derivative of L(c(t),\dot{},t) along W(c(t)), so by the chain rule$$ \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}$$There is then an inexplicable notational shift from \frac{\partial\phi^i}{\partial s} to \frac{\partial ... 1 I) Hamiltonian interpretation. Given a Hamiltonian H(z;t) with canonical coordinates$$\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)- ...

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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 ...

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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 $$H=\int d^3x\int\int\frac{d^3p}{(2\pi)^32\omega_p}\frac{d^3p^{\prime}}{(2\pi)^32\omega_p^{\prime}}(A+B+C) ... 1 The expression you have for the stress-energy tensor (the right hand side of the Einstein-equation) for the scalar field is incorrect. We have$$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}} = ...

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 ...

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

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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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: $$\omega = \begin{bmatrix} 0 && -I_{n\times n} \\ I_{n\times n} && 0 \\ ... 1 In the following calculation, I ignore some coefficients. According to J(x)=\int d^4 k_1 e^{ik_1 x} , J(y)=\int d^4 k_2 e^{ik_2 y} and D(x-y)=\int d^4k \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon} We have$$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 ...

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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 ...

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