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You should check out Marsden and Hughes's "Mathematical Foundations of Elasticity". It is a book that requires a lot of work (I am going through it myself and it is not easy!), using the very general framework of tensors on manifolds + functional analysis, but it is an invaluable resource. 2 chapters in that book (5. Hamiltonian And Variational ...


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I) OP's question (v2) seems to be essentially a question of mathematical precision vs. the way physicists express themselves concisely when speaking of "infinitesimals" without becoming too technical by introducing epsilons and deltas and what not. See also this related Phys.SE post. Perhaps the easiest and most elementary way to make sense of the ...


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As you state in the comments, $$ \frac{dF}{dt}=\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t} $$ So popping this into the Lagrangian, $$ L'=L+\frac{\partial F}{\partial q}\dot{q}+\frac{\partial F}{\partial t} $$ The Hamiltonian $H=p\dot q-L$ implies $$ H'=p'\dot{q}-L'=p\dot q+something\tag{1} $$ where $something$ is for you to work out. ...


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The Hamiltonian is so useful because it is actually the operator providing translation in time (in autonomous systems). We know that any physical quantity on the phase space in the Hamiltonian formalism is evolved like $$\frac{df}{dt} = \{f,H \}$$ Where $\{\}$ is the Poisson bracket. It is thus natural to say $$\frac{d}{dt} = \{\cdot,H\}$$ and a small ...


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I agree with Qmechanic but just to put a different perspective. While one may write down formulae for the Lagrangian, like $$ L = \frac{p^2}{2m} - U(x) $$ which only differs from the Hamiltonian by the minus sign, and while it's possible to simply put hats above all the operators, unlike the Hamiltonian, the Lagrangian isn't a natural operator in any sense. ...


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Comment to the question (v1): Unlike the Hamiltonian $H$ (which is a constant of motion if there is no explicit time dependence), the Lagrangian $L$, as an observable, is typically not conserved in time. Think e.g. of a harmonic oscillator.


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The Brownian motion $x(t)$ is non-differentiable, so a particular trajectory $x(t)$ can't extremize an action $S$ which would be a functional of $x(t)$ and its derivative, $\dot x(t)$, because the derivative isn't even well-defined and any expression of the type $\int [\dot x(t)]^2 dt$, the usual kinetic term in the action, diverges. (See e.g. middle of page ...


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I can't write comments right now So here is what i wanted to write Coupling constant's value is an indication of how strong or weak the intraction and also for ascribing correct dimensions to lagarangian( as mentioned by @winther) is i.e an interaction lagarangian of two particles with higher value of coupling constant will result in stronger interactions ...


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There is a distinction between points and vectors. Points are positions in space, and vectors are directions. One can easily mix up the two, because in Euklidean space they look rather similar. $\theta$ in this case is a coordinate, i.e. part of the description of a point. The vector associated to that coordinate could be called $\hat{e}_\theta$, and point ...


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John R. Taylor's Classical Mechanics has a couple of chapters on Lagrange and Hamilton that I found very helpful.


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You probably know that while the Cauchy stresses are objective, its stress rate (material derivative) is not. If $Q(t)$ is an orthogonal tensor representing a change of frame, the stress is the new frame is $$ T^* = Q T Q^T $$ However, if you take material derivatives on both sides, you have, $$ \dot{T^*} = \dot{Q} T Q^T + Q \dot{T} Q^T + Q T \dot{Q}^T ...


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$ r^2 \dot{\theta} $ is known as the specific angular momentum. Also, the correct formula for the 2 body Lagrangian is actually: $$ \mathcal{L} = \frac{\mu}{2} (\dot{r}^2+r^2 \dot{\theta}^2) + \frac{GMm}{r} = \frac{Mm}{M+m} \frac{\dot{r}^2+r^2 \dot{\theta}^2}{2} + \frac{GMm}{r} $$ where $ \mu = \frac{Mm}{M+m} $ is the reduced mass.


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To derive the field equations more quickly, consider that all the terms in the Lagrangian are "squares": a tensor contracted with itself on all indices. The variation of a square is $\delta(F_{ab} F^{ab}) = 2 (\delta F_{ab}) F^{ab}$ in analogy with $d/dx\; f^2 = 2ff'$. Using this, we have for the variation of the Lagrangian $$\delta \mathcal L = - (\delta ...


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For reference sake, here is the computation of (G) mentioned above in @Qmechanic's answer: The Lagrangian is given by: $$ ...


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Comment to the question (v2): P&S is using the notation of a 'same-spacetime' functional derivative. To illustrate this notation, let us for simplicity stay within first variations, and leave it to the reader to generalize to higher-order variations. I) First of all, functional/variational derivatives should not be confused with partial derivatives. In ...


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It's obvious that the kinetic energy must be compute with respect to the inertial frame. Using Lagrangian method for your problem is very similar to applying this method in order to derive the equations of motion of a gyroscope that can be found anywhere (for example here). The only difference between your case and a gyroscope is that in your case, the ...


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The first thing you should realize is the fact that while $\phi$ has an equation of motion with second time derivatives, it is not the wave function, and therefore there is no problem with QM. The field is just an operator (more or less), not a state. Acting with the fields on the vacuum state you generate the other states which do evolve with an hamiltonian ...


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Hint: If the transformation equations $$\mathbf{r}_i~=~\mathbf{r}_i(q_1,\ldots, q_n ,t) \tag{1.38} $$ do not contain the time explicitly, then the explicit time derivative $\frac{\partial \mathbf{r}_i}{\partial t}=0$ is zero. Note that the position $\mathbf{r}_i$ of the $i$'th point particle also depends implicitly on time $t$ through the generalized ...


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I) Since total divergence terms do not contribute to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post, one could just integrate the Faddeev-Popov $\bar{c}c$ term by part so that there are no more than first derivatives present and the standard form of the EL equations applies. II) Alternatively, in the presence of higher derivatives, the EL ...


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The action you're considering yields Einstein's equations in vacuum, so $R=0$ (this follows immediately from contracting Einstein's equations). Therefore the action vanishes on shell.


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So it turns out that the answer to my question can be found explained in great detail in the following answer to another question: http://physics.stackexchange.com/a/109661/8254 (Taken directly from the linked answer above:) The main message is that in the Euler equation we are considering Langrangian (Material) derivatives of tensor fields $S(t,x)$ in the ...


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It is possible to abstract the notion of a QFT away from the notion of Lagrangians/Hamiltonians, one axiomatic way are the Wightman axioms. As one can see, they reduce the quantum theory to its very heart: A Hilbert space where the states live and a field operator that acts upon it, generating "particles", all of this happening in a Lorentz covariant way. ...


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A large class of QFTs called 2 dimensional conformal field theories, can be defined without an action. The general method for constructing an any dimensional CFT is as follows, you can calculate the action of the symmetry generators of the fields using Ward identities and their commutation relations given by the Virasoro algebra which allows you to ...


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The covariant differential operator $D_\mu$ is $D_\mu = \partial_\mu (Id)- ig A_\mu = \partial_\mu (Id)- ig T^a A_\mu^a$, where $(Id)$ is the identity matrix and the $T^a$ are the generators of a Lie algebra. You have $F_{\mu\nu} = D_\mu A_\nu - D_\nu A_\mu$, which means that $F_{\mu\nu}$ is a covariant quantity. From the expressions above and the ...


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Yes, there are rigorous ways of defining locality in such contexts, but the precise terminology used unfortunately depends on both the context, and who is making the definition. Let me give an example context and definition. Example context/definition. For conceptual simplicity, let $\mathcal F$ denote a set of smooth, rapidly decaying functions ...


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I) Yes, it appears that the sentence [...] the $y$-axis vertically downwards [...] in Ref. 1 p. 81 should have been [...] the $y$-axis vertically upwards [...] II) Let us also mention that Ref. 1 p. 29 eq. (17.9) introduces a function $U$ to be minus the potential energy, however, this $U$ seems unrelated to above. References: C. Lanczos, The ...


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In classical point mechanics with constraints we usually assume that the $i$th position vector $${\bf r}_i~=~{\bf r}_i (q^1, \ldots, q^n;t) $$ of the $i$th point particle is a function of $n$ generalized coordinates $q^1$, $\ldots$, $q^n$, and time $t$. For fixed $t$, the generalized coordinates $q^j$ parametrize the virtual displacements, i.e. the ...



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