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1

Yes, but … it would not be called ADM mass, a proper term would be “quasi-local” rather than “local” mass (energy), there are many prescriptions for defining such mass (energy). Note, while literature often uses “mass” and “energy” interchangeably, sometimes a distinction is necessary: energy is a time component of a vector (energy–momentum), while mass ...


0

The system is described by the Euler-Lagrange equations. So basically the Lagrangian is the description of the system. However the Euler-Lagrange equations can be the same for different Lagrangians. The field will be the solution of the Euler-Lagrange equations. They are differential equations so the boundary conditions can change the solution. Now you can ...


1

Perhaps the simplest argument is the following. Since the Legendre transformation should be $$ L+H ~=~P_1\dot{Q}_1+ P_2\dot{Q}_2 ,\tag{1}$$ then $$\frac{\partial (L+H)}{\partial Q_1}~=~0. \tag{2} $$ Now the higher Lagrange equation reads $$\frac{\partial L}{\partial Q_1} - \frac{d}{dt} \frac{\partial L}{\partial \dot{Q}_1} + \frac{d^2}{dt^2} \frac{\partial ...


1

@SuperCiocia: The context is what you mention. Expressing Hamilton Equations only related to Lagrangian, generalized coordinated/velocity, it is not applied to a specific dynamical system, it's general. Particulary this one $\frac{\partial H}{\partial p_i}=\dot{x_i}$ If my math are correct we have: $$ \begin{align} \frac{\partial H}{\partial p_i} & = \...


1

I guess the exact answer depends on the context where you found this. But mathematically, this could be done with the chain rule: $$ \frac{\partial L}{\partial p_i} = \frac{\partial L}{\partial \dot{x}_i} \frac{\partial \dot{x}_i}{\partial p_i}. $$ Then, as suggested in the comment, you'd have to work out $\dot{x}_i(p_i)$ in order to compute $\partial \dot{...


3

Obviously L&L's conclusion (45.9) is correct as always, but their 2nd proof in the paragraph between eqs. (45.9) & (45.10) seems not entirely convincing, although they do mention all the important ingredients along the way. In contrast, their 1st proof method by direct calculation is a straightforward exercise and quite convincing. In this answer, we ...


1

First of all, I did not have a look at your references, and I did not read them earlier. In your notation $\mathcal{L}_{ADM}$ is Lagrange density, i.e., what you have to play with is $$L_{ADM}=\int_\Sigma \mathcal{L}_{ADM} \, d\mu_h \,,$$ where $d\mu_h$ is the volume form on the hypersurface $\Sigma$ induced by the metric $h$. Keeping this in mind, and ...


2

No, L does not stand for the Lagrangian $\cal L$: it stands for the "energy function" , ("as a rule, a quadratic function of the $\dot{q}$s"), an evident constant of the motion, $$ L(q_i, \dot{q}_i) = \frac{\partial \cal{L}}{\partial\dot{q}_i}\dot{q}_i - \cal{L}, $$ so, basically, $\sim {\cal L} + 2V(q)$ for the quadratic function he is considering, a ...


18

The reason the Hamiltonian is undefined is that converting a Lagrangian to a Hamiltonian requires: Finding $p$ Writing $H=p\dot q-L$ Expressing $H$ in terms of $p$ and $q$, eliminating all dependence on $\dot q$. For the third step to be possible, you need to be able to define a coordinate transformation between the $(q,\dot q)$ coordinate system and the $(...


14

OP's Lagrangian $$L~:=~\frac{\dot{q}}{2}\sin^2(q)~=~\frac{d}{dt}\left(\frac{q}{4}-\frac{1}{8}\sin(2q)\right)\tag{1}$$ is a total time derivative, so the Euler-Lagrange (EL) equation is a triviality $0=0$, cf. e.g. this Phys.SE post, and hence not suitable to determine physical trajectories. In other words, $q$ is a gauge degree of freedom. If $q$ is ...


0

Patrick Hamill, actually, borrowed that problem from Goldstein (see- $3^{rd}$ ed. Chapter 9, derivations 1), and unfortunately, misphrased it. So I'm just quoting the problem from Goldstein, and my interpretation of it, and its solution for your part of that problem with a personal advice. One of the attempts at combining the two sets of Hamilton’s ...


2

First, the scaling you mention amounts to failure to clean up your variables, by defining $$ \tilde p\equiv \frac{p}{m\omega}, $$ likewise for $\tilde P$, and $$ \tilde H \equiv \frac{H}{m\omega^2}= \frac{1}{2} (\tilde p^2+ q^2). $$ It is then apparent that the transformation $$ Q=q \cos \theta -\tilde p \sin\theta\\ \tilde P= \tilde p \cos\theta + q \...


0

From your definition, $$Q^i(\mathbf q,\mathbf p, t) = q^i$$ $$P_i(\mathbf q, \mathbf p, t) =p_i +\frac{g}{c}\frac{\partial \Lambda(\mathbf q,t)}{\partial q^i}$$ This transformation can be immediately inverted to yield $$ q^i(\mathbf Q,\mathbf P, t) = Q^i$$ $$p_i(\mathbf Q,\mathbf P, t) = P_i - \frac{g}{c}\frac{\partial \Lambda(\mathbf{q}(\mathbf Q,\mathbf ...


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The title of the question is not quite correct. I offer a a partial answer and I hope it helps to some extent! I might expand it a bit later. The way the angular momentum is first defined in classical mechanics is $\textbf{r}\times\textbf{p}$ or $\epsilon^{ijk}x_jp_k$. This should really be called orbital angular momentum. However, spin can arise even in ...


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