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Why is the action guranteed to have one unique extrema in classical mechanics?

Even in the simplest classical mechanics problems, there are often multiple paths which make the action stationary. As a simple example, consider a simple harmonic oscillator, $V(x) = \frac{1}{2} m \...
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Are the sources in QFT just particles?

Not really particles. The source terms $J$ are a computational tool. First, they allow you to take the functional Fourier transform of the path phase factor $\exp(i S[\phi] / \hbar)$. Why is this ...
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What's the difference between these two lagrangian?

The Lagrangian (2) can always be rewritten into the form of the Lagrangian (1) [with possibly a different functional dependence]. The other way requires some geometric input, i.e. a connection $\nabla$...
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What's the difference between these two lagrangian?

There might be different habits in different fields of physics (you don't specify which one you're talking about), but for me, coming from quantum field theory, $\partial_\mu$ and $\nabla_\mu$ are ...
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Variation of functional with area and volume term

You are correct that you are going to need an induced metric. Namely: \begin{equation} \int _{\partial M} f\, dS = \int d^{2}y \, \sqrt{|h|} \, f(y^{i}) \end{equation} where $h = \det{(h_{ij})}$, with ...
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Is the action of any interacting 6D CFT known?

When it comes to conformal field theories, it is almost always useful to think about the low-lying scaling dimensions of their local operators. If a CFT has an action, then it has a local operator ...
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Do Legendre transformation form a group?

Think about a system with the so-called $C_2$ rotational axis. That is a system such that a rotation by an angle $\pi$ around that axis corresponds to a configuration indistinguishable from the ...
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2 votes

Calculation of Lagrangian from Hamiltonian $\frac{1}{2}(-i\partial_\phi -A)^2$

We start with the Lagrangian $$ L_M~=~\frac{m}{2}\left(\frac{d\vec{r}}{dt_M}\right)^2 + q \vec{A}\cdot\frac{d\vec{r}}{dt_M}-q\phi_M, $$ for a non-relativistic point particle in Minkowski space ...
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Calculus of variations -- how does it make sense to vary the position and the velocity independently?

Although all the answers seem to cover all the details, I'll just add my treatment for those who are like-minded as me and may find it beneficial. It is useful to break the symbol of partial ...
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Kalb-Ramond current fall-offs at future null infinity

This will not be an answer, as I have some counter-questions regarding your post! However, hopefully it will lead to you answering my counter-questions (presumably in the comments) and me replying ...
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Path Integral of Photon

I guess you want to write the Lagrangian density in the form $\Phi{\cal D}\Phi$ where ${\cal D}$ is some differential operator. In your case the important thing to study is $$F^2=F_{\mu\nu}F^{\mu\nu}=(...
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Does a constant in the action always have unobservable consequences in classical mechanics?

There are three so called adiabatic invariants of single-particle motion under the influence of a quasi-static magnetic field. These are discussed in detail at: https://physics.stackexchange.com/a/...
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Why did Noether use the Lagrangian for her conservation of energy theorem?

The very starting point of Noether's theorem (NT) is an action formulation (and hence a Lagrangian), and quasisymmetries thereof. Noether did not formulate her theorem with only energy conservation ...
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Why did Noether use the Lagrangian for her conservation of energy theorem?

However, I know that the Lagrangian doesn't always equal energy. The Lagrangian never equals the energy, unless there is no potential $U$ so the problem is trivial. In many cases, we can show that ...
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1 vote

Why does the Lagrangian have $O(4)$ symmetry after Wick rotating (previously Lorentz symmetry)?

As long as the Minkowski action is constructed from Lorentz-covariant tensors, then under Wick rotation [where the contravariant and covariant $0$-components of the tensors are Wick-rotated in ...
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Klein-Gordon equation from general relativity?

The Einstein field equations give you the dynamical equations for the metric tensor, which is taken as a second rank tensor field whose evolution describes the evolution of the background. The Klein ...
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2 votes

Klein-Gordon equation from general relativity?

To derive a field equation (equation of motion) one needs an action (or a Lagrangian). In this particular scenario you mention, the action will be $$S = \int d^4x \sqrt{-g} \left(\frac{R}{2} - \frac{1}...
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Is there an infinite amount of conserved currents for a given finite symmetry?

Assuming that we are talking about a single finite 1-parameter global quasisymmetry, it is a flow, which is in 1-to-1 correspondence with a vector field, or equivalently, an infinitesimal 1-parameter ...
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Is there an infinite amount of conserved currents for a given finite symmetry?

The conserved current (under a transformation) is defined to be the variation in the Lagrangian that keeps the equations of motion invariant, under the latter transformation. It's infinitesimal form ...
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1 vote

Is there an infinite amount of conserved currents for a given finite symmetry?

Technically "symmetries" are in general not so much about $S$ not changing at all as $\delta S$ being $0$, or in the language of functional derivatives $\frac{\delta S}{\delta\phi}=0$. ...
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1 vote

Steps taken to differentiate action in wave equation

Which then leads to the wave equation. But, how was that result (i) obtained? It is basic multi-variable differential calculus. See, for example, this answer, which explains something quite similar.
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Expression of a Lagrangian in other form

What you need to use are so-called null-Lagrangians, i.e. terms that are a divergence and can be dropped in the Lagrangian because in the action integral they only contribute a surface term, or, to ...
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Charge of antifermions in 1+1d QED, and in 1+1 CED

I just found the answer, I was describing the fermion fields in the classical limit with complex numbers as if they were scalar fields, which in the case of fermions because of they anti commutativity ...
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Equations of motion of $\mathcal{L}= - \frac{1}{4}F^{2}_{\mu \nu} - A_{\mu}J_{\mu}$ in momentum space

@JeanbaptisteRoux is right. Schwartz should have used symbols less ambiguously to prevent your confusion. Denote the functions in Eq. (8.94) as $A_\mu(x),\,J_\mu(x)$, viz.$$\partial^2A_\nu(x)-\...
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Should rotation of a rigid body confined to a sphere couse it to divert from a big circle?

Can we treat the rotation of the body and its motion along the spherical surface as being independent? If so I think all you have to do is take the kinetic energy in spherical polar coordinates and ...
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How fundamental Physics is constructed?

There's been a lot of back-and-forth in the past few centuries in the way physics laws are written. Sometimes laws that were fundamental became mere consequences of deeper ones. Sometimes there were ...
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Solution of the two-body problem

The initial conditions imply $B=0$. This is so, because by your initial conditions. In polar coordinates \begin{equation} \vec{r} = r\vec{e_r} \end{equation} \begin{equation} \dot{\vec{r}} = \dot{r}\...
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Asymmetry of the indices in a Weyl transformation for the Polyakov action

I think the asymmetry between the indices is a convention. We chose $\sigma^{\mu}=(\sigma, \tau)$ as the integration variables. This implies that the determinant of $h_{ab}$ will be the Jacobian. So ...
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Path variations in deriving Euler-Lagrange equation

In my opinion the Euler-Lagrange equation itself provides the answer to your question. The process of deriving the Euler-Lagrange equation is a process of boiling down to the essence. The Euler-...
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Why is the hamiltonian density written in terms of $\phi$ and $\pi$ only?

My question is why the hamiltonian density is defined as the first equation? In classical mechanics of many particle systems: $$ \frac{\partial H}{\partial p_n} = \frac{d x_n}{dt} $$ and $$ \frac{\...
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Path variations in deriving Euler-Lagrange equation

I am not sure if the following helps you but anyways. In ordinary multidimensional calculus if one has a differentiable, scalar function $f: \mathbb{R}^n \to \mathbb{R}$, one can define the ...
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Partial derivatives of canonical momenta in Poisson brackets

The independent coordinates (on the cotangent bundle of the configuration space) for describing the Hamiltonian are $(\theta,\phi,p_{\theta},p_{\phi})$. So, almost by definition, $\frac{\partial (\...
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Partial derivatives of canonical momenta in Poisson brackets

It's just like in thermodynamics, you need to keep track which variables you are using as coordinates for your function. It's always a good idea to formulate things in a coordinate independent ...
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1 vote

Partial derivatives of canonical momenta in Poisson brackets

When evalauating partial derivatives you need to specify what is being kept fixed as well as what is varying. The Poisson bracket is $$ \{F(p,q),G(p,q)\}=\sum_i \left(\frac{\partial F(q,p)}{\partial ...
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5 votes

Conceptual question Einstein-Hilbert action and QFT in curved spacetime

I think that your question is about backreaction. One can perfectly have an action like $$S = \int d^4x\sqrt{-g}\left(R-2\Lambda -\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi - V(\phi)\right)$$ ...
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3 votes

Conceptual question Einstein-Hilbert action and QFT in curved spacetime

1. Relation between the actions The scalar action you wrote is the one for a minimally coupled field. One could use the more general action (I'm working with the $-+++$ convention and the $+++$ MTW ...
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5 votes

Conceptual question Einstein-Hilbert action and QFT in curved spacetime

The Einstein-Hilbert action (1) describes empty spacetime. The action (2) describes matter. So if you want to describe a scalar in general relativity you need to consider $S_{EH} + S_M$. Similarly if ...
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