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As in the comments, you may omit terms independent of $\varphi$ in the Lagrangian (they typically arise from changing the zero point of potential energy, etc.). You can also integrate by parts / remove total derivatives. Here, we note that the derivative of $\cos(\omega t - \varphi)$ means we can exchange $\dot{\varphi}$ for $\omega$. (NB this does not mean $... 6 You need to understand the definition of$\delta$. First let$q(t)$be given. A variation of$q(t)$is by definition a one-parameter family of curves$\tilde{q}(t;\epsilon)$with the property that $$\tilde{q}(t;0)=q(t)\tag{1}.$$ In that case the first variation$\delta q(t)is defined by the equation $$\delta q(t)=\dfrac{\partial}{\partial \epsilon}\bigg|_{\... 1 We consider a change$$q(t) \rightarrow q'(t) = q(t) + \delta q(t),$$where \delta q(t) is some arbitrary function (maybe subject to some constraints, e.g. we might require that it vanishes at the boundaries of integration). The time derivative of our new function is now$$\frac{d q'}{d t} = \frac{d q}{d t} + \frac{d }{d t}\delta q.$$Thus the change in ... 0 The Lagrangian density is proportional to F_{ab}F^{ab} = g_{ab}g_{cd}F^{ac}F^{bd}. where g is the minkowski metric. Under a rotation, you'd send F^{ab} \rightarrow \Theta^{a}{}_{m}\Theta^{b}{}_{n}F^{mn}, where \Theta is the standard 4D rotation matrix but, from the way this is contracted above, it should be clear that this is mathematically ... 0 You pretty much answered the question on your own. One can understand the very definition of energy (momentum) as "The Noether charge associated with time (space) translations". Consider, for example, the Lagrangian for a free relativistic particle,$$L = \frac{1}{2} U^\mu U_\mu,$$where U^\mu is the four-velocity. If we make a transformation of ... 0 Think of the twin paradox, in which one twin travels out into space in a spaceship at high speed and the returns, while the other twin remains stationary on Earth. In the end, the twin traveling in a spaceship will have aged less because of time dilation. Arguably, the twin who remains on Earth travels along a geodesic, and he will have aged more than his ... 0 There are more variables and hence more coordinate transformations in the Hamiltonian formulation as compared to the Lagrangian formulation. So in order to give a fair/honest comparison, we only analyze coordinate transformations that originate from the Lagrangian formulation in this answer. In the Lagrangian formulation, the Lagrangian L is invariant ... 2 The other answers have explained the issue in a mostly mathematically formal and rigorous way. I want to add to the discussion by instead trying to explain why the Lagrange equations of motion stay the same in an informal yet hopefully intuitive way and then discuss what is different for Hamiltons equations of motion. First let us deal with Lagrange. Let us ... 1 OP is right that if the Lagrangian density remains of 1st order, then the Euler-Lagrange (EL) equations will only be of 2nd order. See also e.g. this & this related Phys.SE posts. However, the Wilsonian effective action$$\begin{align} \exp&\left\{ -\frac{1}{\hbar}W_c[J^H,\phi_L] \right\}\cr ~:=~~~&\int \! {\cal D}\frac{\phi_H}{\sqrt{\hbar}}~\... 2 As it is the case in Classical Mechanics, the conservation of energy in Electromagnetism (which is expressed by means of Poynting's Theorem) is a consequence of time-translation invariance. When working with the Noether Theorem in Field Theory, it is often more interesting to do things covariantly and prove conservation of another quantity, the symmetric ... 1 Your two approaches mean different things, so it's not surprising that they give incompatible results. If you're perturbing the equation of motion, you are keeping the same dynamical system, which will still obey the same equations of motion, and given some solution to that system, asking "what do nearby OTHER solutions look like?" If you're ... 0 In Gaussian normal coordinates (with the surface at constant\lambda$$$ds^2 = \sigma d\lambda^2 + h_{ij}(\lambda,y) dy^i dy^j$$$K_{\mu\nu}$may be written as $$K_{ij} = \frac12 \partial_\lambda h_{ij} \,.$$ From this you should be able to go on (if you still intend to do so, this is a late answer). 1 As a guide to deeper understanding: In classical mechanics we have that theory of motion can be expressed in terms of energy exchange. As we know, newtonian gravity is a conservative force, hence we can define a potential energy. As we know, it is customary to take infinite radial distance (to the source of gravity) as the zero point of potential energy. At ... 2 As you noted,$\phi$is cyclic. So the EL equation's LHS is$0$, but one term of it is trivially$0$, and the other is also$0$. And if you delete its$\frac{d}{dm}$, you get something conserved - a conserved momentum, in fact. In this case, that conservation law is$h:=\tfrac12\partial_{\dot\phi }L^2=r^2\sin^2\theta\dot{\phi}$. 0 two-body motion first you transfer the absolute coordinates$~r_1~,r_2~$to the relative coordinate$~r~$and the center of mass coordinate$~R_c~$I use this two equations $$R_c=\frac{m_1\,r_1+m_2\,r_2}{m_1+m_2}\\ r=r_1-r_2$$ the solution for$~r_1~,r_2~$is: $$r_1=\frac{R_c+r\,m_2}{m_1+m_2}\\ r_2=\frac{R_c-r\,m_1}{m_1+m_2}$$ from here you obtain the ... 3 The modern way of looking for things (effective field theory) is that you should include every term in the action consistent with the symmetries and containing the degrees of freedom you are working with. So from that point of view, you would need a reason not to include$\mathcal{G}$. As you mentioned, in four dimensions the Lorentz invariant scalar ... 4$\int \theta(x) {\bf E} \cdot {\bf B}d^4x $occurs in the action for axion electrodynamics. Where have you seen it to higher powers? Perhaps in the quantum corrections due to fermion loops? In higher dimensions there are more invariants than just$|E|^2-|B|^2$and${\bf E} \cdot {\bf B}$. 0 It's not clear to me from your question if you are interested in the gravitational force that two particles exert on each other, or on the gravitational force the two particles experienced if they move in an external field. The way you handle these two cases is different, so below I outline how you can handle both. If you are interested in the gravitational ... 0 The comment above is true. The total differential$\frac{\delta(\partial^2\Phi_a)}{\delta \Phi_b}=\partial^2\delta_{ab}$Euler Lagrange equation can be written as$\frac{\delta \mathcal{L}}{\delta\phi}=0$or in other words$\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}-\frac{\partial \mathcal{L}}{\partial(\phi)}=0$One then need to pay ... 3 The correct starting point is always$^1$$$S~=~\frac{1}{2}\int\! d^4x \left( \mp\partial_{\mu}\phi\partial^{\mu}\phi + \ldots \right) ~=~\frac{1}{2}\int\! d^4x \left( \dot{\phi}^2 + \ldots \right),\tag{1}$$ since the kinetic term should be positive definite. When authors write $$S~=~\frac{1}{2}\int\! d^4x \left( \pm \phi\Box\phi + \ldots\right) ~=~\frac{1}... 6 In general the Legendre transformation^1 from the Lagrangian to the Hamiltonian formulation may be singular, which leads to primary constraints. This is e.g. the case for gauge theories like Yang-Mills (YM) theory with or without matter, which OP mentions. However, in case of a singular Legendre transformation, by performing a so-called Dirac-Bergmann ... 3 It should be \partial_\mu(\phi_1^\dagger e^{-i\theta})\partial^\mu(e^{i\theta}\phi_1), which by the product rule is a sum of four terms which for spacetime-constant \theta simplifies to just one of them, \partial_\mu\phi_1^\dagger\partial^\mu\phi_1. For \theta varying across spacetime, the result is more complicated, motivating gauge covariant ... 1 It's more convenient to have the factor of 1/2 for the real scalar field, while it's more convenient to have the factor of 1 for the complex scalar field. Roughly, the factor of 1/2 eliminates the "symmetry factor" associated with the two real \phi fields being equivalent. The most direct way to see this from elementary QFT is to simply ... 1 start with the position vector to the mass m$$\mathbf R=\left[ \begin {array}{c} x+ \left( r+l_{{0}} \right) \sin \left( \theta \right) \\ \left( r+l_{{0}} \right) \cos \left( \theta \right) \end {array} \right] $$from here you obtain the velocity \mathbf v~ hence the kinetic energy$$T=\frac 12 M\dot x^2+\frac 12 m \mathbf v\cdot \mathbf v$$and the ... 1 The simplest way to keep track of this is to write \ell=\ell_0+\epsilon r and use \epsilon \theta rather than \theta. Expanding your Lagrangian in powers of \epsilon, you can then use \epsilon as a counter to keep track of "smallness". The anharmonic terms are those with 3 or more powers of \epsilon in the Lagrangian, or two or more ... 0 A second way to derive the result is through the use of the \delta-symbol, also called "variation". It usually is a bit easier with managing indices, as you do not have to manage Kronecker-deltas (and it also clarifies the relation between symmetries and equations of motion). The basic properties needed are, that \delta is linear$$ \delta(A^\... 4 Everyone knows the difference between a local operator and a nonlocal operator. But the interesting question is what it means for a theory to be nonlocal. This has been given some poor definitions in the past, e.g. by people who say that everything quantum mechanical is nonlocal because of entanglement. A much better definition is that nonlocal field ... 1 It's just a manner of being careful with the indices. Thake the Euler-Lagrange equation in covariant formulation $$\partial_\rho \frac{\delta \mathcal{L}}{\delta \partial_\rho A_\sigma}-\frac{\delta\mathcal{L}}{\delta A_\sigma}=0$$ and let us start from the second bit. Clearly, the only part in the lagrangian which does not depend on the derivative of the ... 0 The Euler–Lagrange equations are more general than the geodesic equation, in fact. They do not apply exclusively to Newtonian Mechanics, but rather to any theory to which you can write a Lagrangian (possibly with some modifications). That includes, as a particular case, the motion of particles on curved spacetime. You'll notice that the Lagrangian $$L = \... 2 The notion of generalized momentum is not the same as the usual notion of momentum. In other words, \frac{\partial L}{\partial \dot{x}} does not equal the mass of the big block times its velocity. In fact, notice that your Lagrangian yields$$\frac{\partial L}{\partial \dot{x}} = M \dot{x} + m(\dot{x} + l \dot{\theta} \cos{\theta}),$$which means your ... 1 OP's question touches upon the very definition of a functional/variational derivative: If we define$$\frac{\delta\phi(x)}{\delta\phi(y)}~=~\delta^d(x-y),\tag{1}$$then we have to live with consequences that the RHS is a density. If the theory has a density \rho [e.g. if there's a metric tensor g, we can construct \rho=\sqrt{|\det g|}], then it is ... 1 If you define T^{\mu\nu} explicitly, by writing$$ \delta S= -\int d^dx \sqrt{g}\, T^{\mu\nu} \delta g_{\mu\nu}, $$then the invariance of S and d^dx \sqrt{g} under coordinate transformations shows that T^{\mu\nu} is a tensor, rather than a tensor density. I think the definition in terms of \delta S/\delta g_{\mu\nu} is ambiguous because some ... 0 If you want to look at individual atoms and molecules, then thermal energy is just particle motion. Thus, if you have represented your particle motion in your Lagrangian, you have represented thermal motion. In the ordinary case it would be seriously impractical to use such. You are talking something in the range of 10^{22} particles in a gram of air. (... 2 It is possible to describe the non-kinetic terms by a matrix$$ \left(\begin{array}{cc} \phi_1 & \phi_2\end{array}\right) \left( \begin{array}{cc} m^2 & -0.5 g \\ -0.5 g & m^2 \end{array} \right) \left(\begin{array}{cc} \phi_1 \\ \phi_2\end{array}\right) = \left( \begin{array}{c} \phi & \varphi\end{array}\right) \left( \begin{array}{cc} m^2 ... 1 As @ConnorBehan noted, the given classical action is relativistic:$$-m\tau=-\frac{mt}{\gamma}=(\beta^2-1)m\gamma t=m\beta\gamma\hat{v}\cdot t\beta\hat{v}-m\gamma t\stackrel{\star}{=}p\cdot x-Et,$$where$\stackrel{\star}{=}$assumes a constant momentum and$x(0)=0$. I'll leave it to you to consider whether IBP justifies it. 3 You can write the action in terms of differential forms as \begin{equation} S = \int \epsilon_{abcd} R^{ab} \wedge e^c \wedge e^d \end{equation} Up to a numerical factor, this is equivalent to the Einstein-Hilbert action \begin{equation} S = \int {\rm d}^4 x \sqrt{-g} R = \int {\rm d}^4 x |\det e| R \end{equation} The fact that the Einstein-Hilbert action is ... 2 There are only two small problems: this approach has nothing to do with gravity, and it is not at all quantum :) Firt, your action doesn't describe gravity; it describes Yang-Mills theory with the group$GL(n) = U(1) \times SL(n)$. Not gravity. There's a formulation of gravity in the gauge theory language but it uses a different action:$$S[e, A] = \int d^4 ... 1 It has been rigorously established that in 3d there are massless interacting$\phi^4$theories. See for example Section 9 (4) of this article by Brydges, Fröhlich and Sokal. Note that the parameters$\lambda$and eventually$m^2$in$S$don't really mean anything. These are bare couplings which can very well become$\pm\infty\$. One usually introduces a UV ...