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To see how acceleration-induced pseudo forces enter into Lagrangians consider a partcle that has position $x(t)$ measured from a moving reference point $X(t)$ (the floor of a lift say) with $\ddot X=a$. The particle's position in an inertial frame is therefore $x_{\rm inertial}=x+X$ and so it has action integral $$S=\int \frac 12m \left(\frac{d(x(t)+X(t))... 0 In Leonard Susskind's Special Relativity and Classical Field Theory, he provides a much more intuitive, but informal, way. Imagine that we take a scaler field, which is time dependent. Let this field denote the position of some imaginary particle. Let's call it \phi(t). Now this field describes the position of some imaginary particle of mass 1. Classical ... 0 Your Lagrangian is fine. The good news is that this Lagrangian is independent of the variables \theta and z, and it depends only on their first derivative:$$L = L\Big(\,r,\, \frac{dr}{dt},\, \frac{d\theta}{dt}, \, \frac{dz}{dt}\,\Big) $$Such variables are called cyclic, and therefore this particular Lagrangian is a special case of a Lagrangian with ... 0$$\nabla_{\mu} \left({\frac{\partial G_2}{\partial(\nabla_{\mu} \phi)}}\right)= \nabla_{\mu}\left(\frac{\partial G_2}{\partial X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)}\right)= \nabla_{\mu}\left(G_{2,X}\frac{\partial X}{\partial(\nabla_{\mu}\phi)}\right)= ? $$Since in the bracket is not a tensor, we may change \nabla_\mu with \partial_\mu, so then ... 3 This is indeed true and is what is called the gauge principle. It tells us that if we make a global symmetry local, we need to add a corresponding gauge field such that the total Lagrangian still remains invariant under this local gauge transformation. This is a new dynamical field which has its own equations of motion and can couple to the fermion leading ... 1 So let's first motivate the definitions of :- Hermitian Conjugate of an Operator - VS - Hermitian Conjugate of a Vector Hermitian Conjugate (Adjoint) of an "Operator" :- A Hermitian Conjugate (technically known as Adjoint ) of an operator \hat{\mathrm A} is defined via the rule \langle\phi|\hat{\mathrm A}|\psi\rangle\ =\langle\hat{\mathrm A}^\... 2 There are two possibilities to look at this: Either the exact form of V(B^2) is known and you can simply take the variation with respect to the B^2-dependency in that potential. More general, you can use use the chain rule as follows $$\frac{\delta V(B^2)}{\delta B^2} \frac{\delta B^2}{\delta g^{\mu \nu}}$$ The same ... 3 Yes, your action is of the form $$S=\int\text{d}^{4}x\sqrt{-g}(\mathcal{L}_{\text{EH}}+\mathcal{L}_{\text{M}}),$$ where $$\mathcal{L}_{\text{EH}}=\frac{R}{2k^{2}}$$ is the part whose variation with respect to g_{\mu\nu} gives you the Einstein tensor in the equations of motion, and \... 1 Yes. The field strength tensor is F_{\mu \nu}. The only two fundamental fields in your action are g_{\mu \nu} and A_\mu (unless you use Palatini variation and treat the connection as independent of the metric). So before preforming the variation, you should write the action in terms of these fundamental fields, and vary w.r.t each of them. The only ... 2 Yes. You have to vary g^{ab} everywhere it appears. 1 You need to allow for time reparameterization. In other words, you should include the so-called lapse in your metric $$g_{\mu\nu} dx^\mu dx^\nu = -N(t)^2 dt^2 + a(t)^2 dx^2$$ If you carry N through your calculation of \sqrt{-g} and R and then vary the action with respect to N, you will arrive at the first Friedman equation.... 3 You vary the action with brute force Variation for simplicity. I assume that you can handle the last two terms and the problem is the first one. Varying with respect to \phi you'll obtain:$$\Box \phi + \cfrac{1}{4}R\phi -m^2 \phi =0$$Varying with respect to the metric you have to integrate by parts the first term.$$\delta(\phi ^2 R) = \phi^2 \delta R = \...

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Often we impose regularity conditions on a Lagrangian formulation to simplify the calculations, and/or so that we can work within some mathematical framework, such as e.g., differentiability, that constraints are holonomic, that the rank of the Hessian is maximal, or at least don't jump, other rank conditions, see e.g. this Phys.SE post, locality, etc. ...

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Generally speaking, people do not consider the generic class of Lagrangians, since they tend to be interested in physical systems, but yes, there are examples of "bad Lagrangians". A Lagrangian should certainly be an integrable function, as well as $C^1$ (or at least weakly differentiable) in its variables, but more importantly, it should have an ...

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You need to transform the region of integration. It it's bounded by constant-time hyperplanes in the unprimed coordinate system, then it won't in general be bounded by constant-time hyperplanes in the primed coordinate system, but you can still integrate over it. If you restrict your boundaries to constant-time hyperplanes in all coordinate systems then you ...

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You can do it in two ways. The commutation of second derivatives for any vector satisfies: $$\xi_{\mu;\nu\lambda}-\xi_{\mu;\lambda\nu} = R_{\mu\sigma\lambda\nu}\xi^\sigma.$$ If you know it's a Killing vector, then $\xi_{\mu;\nu} = -\xi_{\nu;\mu}$. Contracting $\mu$ and $\lambda$ so as to go from (a variant of) the Riemann tensor to (a variant of) the ...

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This is a good question, and there are probably technicalities that I'm not quite qualified to address. I'll give a non-rigorous answer and try to avoid any absolute statements. Geodesics are defined abstractly on some manifold, but they are a direct generalization of the idea of a straight line. Note that in a flat space a straight line is the "...

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By parameterising a worldline from event $A$ to $B$ by the variable $\sigma$, then, denote the action $$S=\int ds,$$ which should then be written as an integral over the parameter $\sigma$. Next, recall the invariant measure $$ds^2 = g_{\alpha\beta}\,dx^\alpha dx^\beta .$$ Using the rules of total derivatives, we then get $$S = \int \sqrt{-g_{\alpha\beta} \... 1 The corresponding action S=\int_{\sigma_i}^{\sigma_f} \mathrm{d}\sigma~L is the arc length between 2 spacetime events. The principle of stationary action (with Dirichlet boundary conditions) therefore leads to geodesics. See also e.g. this related Phys.SE post. 0 Hartle's Lagrangian is designed so that the corresponding Euler-Lagrange equation coincides with the geodesic equation for the metric g_{\alpha\beta}. That this is so is a standard exercise with Christoffel symbols that appears in most GR books. 1 As you point out: surely the Lagrangian must be capitalizing on the fact that with all interconversion of kinetic energy and potential energy the decrease and increase must match each other. We can restate that as follows: During the entire time the rate of change of kinetic energy is equal to the minus rate of change of potential energy: In mathematical ... 1 The energy function is not always equivalent to the energy of the system. Consider the case when it happens to be: Suppose we have lagrangian given by$$\mathcal{L}= \mathcal{L}_0+\mathcal{L}_1+\mathcal{L}_2 $$where$$\mathcal{L}_0=\mathcal{L}_0(q,t)$$,$$\mathcal{L}_1=\sum_{i}b_i\dot{q}_i$$and$$\mathcal{L}_2=\sum_{i,j}c_{ij}\dot{q}_i\dot{q}_j$$In this ... 1 I've eventually found the answer. So we have :$$ \mathcal{L} = \frac{1}{2} m g_{ij}(\vec{x}(t))\dot{q}^i\dot{q}^j = - U(q,t) $$Computing the different terms in Euler-Lagrange equation for a particle k:$$ \frac{\partial \mathcal{L}}{\partial \dot{q}^k} = \frac{1}{2}mg_{ij}\dot{q}^i\delta^j_k + \frac{1}{2}mg_{ij}\delta^i_k\dot{q}^j = \frac{1}{2} m g_{ik} \...

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OP's heuristic explanation certainly contains the right physical ingredients of both dynamical and kinematic nature, although the important relative minus sign between $T$ and $V$ in the Lagrangian $L=T-V$ could preferably use some further elucidation. For a mathematical proof, see e.g. this related Phys.SE post.

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Correct me if I’m wrong. You try 2 functions $\phi$ and $\phi_2=\phi+\delta\phi$ which both must satisfy boundary condition. Since boundary condition is linear, logically $\delta\phi$ satisfies it too.

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A tip: use the Kronecker Delta in the derivative, and remember that the x coordinates are related to the generalized coordinates q, {q^dot}. The delta will basically kill that 1/2 factor, and the Christoffel symbol arises from the corrdinate transformation (parallel transport)

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There is one rather famous solution, called Fubini instanton by the name of it founder, discovered in 1976. https://link.springer.com/article/10.1007%2FBF02785664 The abstract claims: A new approach to conformal invariant field theories is presented. The physical idea is to introduce a fundamental scale of hadron phenomena by means of a dilatation ...

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Comments to the post (v5): First of all: an integral where the support of the Dirac delta distribution coincides with one of the integration limits is ill-defined. However, in OP's case this can be avoided altogether. Just add the term $i\lambda(\phi^{\prime}(0)-c\phi(0))$ to the action $S$ instead. For consistency, we need 2 boundary conditions (BCs): 1 ...

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The fundamental importance of the sign of the Lagrangian is that the variation of the action is minimal with respect to the real "trajectory" of the system, regardless of the sign of the kinetic term of the Lagrangian. For example, in the Lagrangian of the relativistic free particle, the sign is minus in order to guarantee the minimum of the action,...

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OP has a point: Eq. (9.91) should read $$\pi~=~-\sqrt{-g}g^{0\mu}\nabla_{\mu}\phi~=~-\sqrt{-g}\nabla^0\phi,\tag{9.91'}$$ i.e. the 0-index should be upstairs.

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There is likely not an exhaustive classification for when pairs of classically equivalent theories are equivalent quantum mechanically. The best one can do is probably to present a relatively short list of known examples. The most famous pairs are square root actions vs. non-square root actions, cf. e.g. Nambu-Goto vs. Polyakov action in string theory or the ...

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The Lagrangian $L$ is a function $(q,v,t)\mapsto L(q,v,t)$ where its arguments $(q,v,t)$ are independent. To form the partial derivatives of $L$ wrt. the variables $(q,v,t)$ it is necessary that the arguments $(q,v,t)$ are independent. Above it is implicitly assumed that we are not using the EOMs. If we are using the EOMs, then the variables $(q,v,t)$ are ...

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It depends on the context. In physics the implicit dependence of functions is often let out which makes a lot of expressions easier to read but in cases like this it can get confusing. When $\partial T/\partial x=0$ was written the author was probably considering this: $$H(x,v)=T(v)+U(x)$$ Here $H$ is just a function of two variables. Nothing more. When you ...

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The BRST symmetry encodes the gauge symmetry. Yes. The $x$-dependent/local gauge-parameter $\alpha^a(x)$ in the gauge formulation (which doesn't contain ghosts) is replaced by an $x$-dependent ghost field $c^a(x)$ and an $x$-independent/global Grassmann-odd parameter $\epsilon$ in the BRST formulation. So the BRST symmetry is an $x$-independent/global ...

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If you can choose $F(q,t)$ such that $L'=0$, then $L$ can already be written in the form $$L\bigg(q(t),\dot q(t),t\bigg) = \frac{d}{dt}G\bigg(q(t),t)\bigg)$$ for some $G$. As a result, the action $$S[q] = \int_{t_1}^{t_2} L\bigg( q(t),\dot q(t), t\bigg)dt = G\bigg(q(t_2),t_2\bigg)-G\bigg(q(t_1),t_1\bigg)$$ is independent of the path $q$ (since the endpoints ...

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The action $S$ is a functional of paths $q(t)$, not a function of the variables $(q,\dot{q},t)$, so the equality $F = -S$ makes no sense. By claiming that you can choose $L = -\frac{\mathrm{d}F}{\mathrm{d}t}$, you already have restricted the original Lagrangian to be a total time derivative (of $-F$). This is not the case for Lagrangians that usefully ...

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Almost, you're off by a minus sign, but otherwise yes. You're supposed to use (higher-dimensional) integration by parts on the action $$S = - \int_{\Omega} d^4 x\ \big( \partial^\mu \phi^\ast(x) \big) \big( \partial_{\mu} \phi(x) \big) = - \int_{\partial \Omega} d^3 \Sigma\ \phi^\ast(x) n^{\mu} \partial_{\mu} \phi(x) \big) + \int_{\Omega} d^4 x\ \phi^\... 1 The core of OP's question seems to be the following question: What's the difference between (1) the infinitesimal variations to derive EL equations, and (2) the infinitesimal symmetry variations in Noether's theorem? This is explained in this related Phys.SE post. Concerning the Schwinger-Dyson (SD) equations, they are derived using a vertical translation ... 0 First one is only true if Spacetime coordinate is not being transformed but only fields are. This is discussed in great depth at around page number 595 of Goldstein's classical Mechanics. 3 The principle of stationary action always implies the EL equations (1) with partial derivatives, so (1) is a safe bet. By imposing further conditions on the theory, the EL equations (2) with covariant derivatives may hold as well, cf. this related Phys.SE post. 1 Well, if you don't mind explaining before Hamilton-Jacobi's equation then it's not impossible. From the derivation of the Hamilton-Jacobi equation (see for yourself!) I have that $$\mathrm{d}\mathcal{S}=p\ \mathrm{d}q-\mathcal{H}\ \mathrm{d}t$$ Where \mathcal{S} is the action, p the generalized momentum, \mathcal{H} the ... 0 Why not try? for example with a Lagrangian of the specific form "polynomial of second order"$$\mathcal{L}(x^{\mu},X,\partial_{\mu}X):= \sum_{i,j=1}^6\left( A_{ij} X^i X ^j + \sum_{\mu=0}^3 B^{\mu}_{ij} \partial_{\mu}X^i X ^j +\sum_{\mu,\nu=0}^3 C^{\mu\nu}_{ij} \partial_{\mu} X^i \partial_{\nu} X^j \right) \tag{1}\label{1}$$in X=\big(E_x, E_y,E_z,... 3 This will be another entry in my long-running rant series which is (barely) hyperbolically titled "There's no such thing as a total derivative." If you have a well-behaved function of two variables f:\mathbb R\times \mathbb R\rightarrow \mathbb R, then you can define the derivatives with respect to its first and second slots to be$$\partial_1 f :...

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The speed $v\equiv |\vec{\bf v}|\geq 0$ is by definition the magnitude of the velocity $\vec{\bf v}$. Any function of speed $v$, or say $v^4$, can be easily rewritten as a function of $v^2 \equiv |\vec{\bf v}|^2$, or vice-versa. The latter form $f(v^2)$ is preferable when one tries to partial differentiate wrt. a velocity component in order to avoid square ...

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Your definition of the Taylor series in one variable to first-order derivatives is $$f(x) = f(a) + \frac{\partial f}{\partial x}(a)(x-a)$$ Notice we neglect higher-order terms. In two variables this would look like $$f(x_1,x_2) = f(a_1,a_2) + \frac{\partial f}{\partial x_1}(a_1,a_2)(x_1-a_1)+ \frac{\partial f}{\partial x_2}(a_1,a_2)(x_2-a_2)$$ Let us ...

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The potential energy divided by $m\,a^2$ is: $$U={\Omega}^{2} \left( \sin \left( \vartheta \right) \right) ^{2}+2\, \Omega_{{0}}\cos \left( \vartheta \right)$$ the Taylor series for a small $\vartheta$ at $\vartheta_0$ and $\vartheta^n=0~,n=3,4,...~$is: U_T=U(\vartheta_0)+\frac{\partial U}{\partial \vartheta} \bigg|_{\vartheta_0}\left(\vartheta-\...

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Possible conceptional mistakes: Note that the velocity $\dot{x}_{\mu}:= g_{\mu\nu}\dot{x}^{\nu}$ with lower index implicitly depends on the metric. In contrast the velocity $\dot{x}^{\nu}$ with upper index does not depend on the metric. This is important when we vary wrt. the metric. The stress-energy-momentum tensor depends on the sign convention for the ...

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Assuming that the boundary conditions (BCs) of the variational problem are essential/Dirichlet$^1$ BCs, and that the Lagrangian is sufficiently differentiable, then the EL eqs. remain the same for the truncated problem, say in the interval $[t_i,t_m]$. The truncated trajectory $\gamma|_{[t_i,t_m]}$ is still a solution to the EL eqs., so it will still be a ...

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There is no anomaly problem with this system --- except that as written it does not have a continuous $U_A(1)$ symmetry. You need to include a term $i\bar\psi \gamma^5 \psi$ term in addition to the $\bar\psi\psi$ term. With that included it is a simple model that can be be used for illustrating chiral symmetry breaking.

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