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Also, you can use a sort of change of variables. $\int_a^b \dot{x}^2 dt = \int_a^b v^2 dt = v^2 \int_a^b dt = v^2({t_b}-{t_a})$ where the last part is using the fact v is constant for a line to pull it out of the integral. The result follows.

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The fields satisfy the wave equation. We can therefore write \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + ...

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The equation of motion for a scalar massless relativistic point particle is $$\tag{A} \dot{x}_{\mu}\dot{x}^{\mu}~\approx ~0,$$ where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is $$\tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, ... -3 Particle with zero mass has to have also zero electric charge, otherwise the Lorentz formula for EM force acting on it cannot be used to find its acceleration according to$$ m\mathbf a = q\mathbf{E}_{ext} + q\frac{\mathbf v}{c} \times \mathbf B_{ext}. $$However, particle with zero mass and zero charge has trivial equation of motion$$ 0=0 $$and has zero ... 1 This is an interesting question, and although I don't know a rigorous answer, we can discuss some typical cases. Usually, the inverse exists, but the cases where this inverse does not exist are not necessarily pathological (sound models can have the problem that the inverse does not exist). For standard field theories (say, \phi^4, O(N) models, classical ... 0 For a massless particle we can not have a centre-of-mass frame. Unfortunately I cannot yet add comments. Are you studying Classical Field Theory (CFT) or Quantum Field Theory (QFT)? My guess is CFT since this looks like a line out of a few lectures into a CFT course when you start to make find equations of motion. In that case, for the (massless) photon, ... 2 The quote is taken from just above eq. (1.32) in Ref. 1: [...] If the internal forces are also conservative, then the mutual forces between the ith and jth particles, {\bf F}_{ij} and {\bf F}_{ji}, can be obtained from a potential function V_{ij}. To satisfy the strong law of action and reaction, V_{ij} can be a function only of the distance ... 0 Tong's lecture notes cover this in detail. In particular, your lagrangian is easy to obtain from a complex scalar field that obeys the Klein-Gordon equation in the non-relativistic limit. This lagrangian does lead to a conserved Nöther current of the same form as that of Schrödinger's, but this does not have the interpretation of conserved probability, ... 0 The strong law of action and reaction says that the forces that two bodies exert on each other have the same magnitude, opposite direction and act along the line joining the particles. When you want that last bit to be true and you want to write the force on particle i as -\nabla_i V_{ij}, then the potential has to be a function of the relative distance. ... 2 For a convex function you can do the following: For each point on the graph of the function, draw the line tangent to the function at that point. That point can now be identified by its original x and y=f(x) coordinates, or by specifying the slope of that tangent line and its corresponding y-intercept. Each point maps to one and only one line, and ... 0 This site derives the principle of least action from Newton's laws. http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf 0 Newton's second law, \mathbf{F}_{net}=\dot{\mathbf{p}}, is the definition of force. Lagrangian and action are defined to be  T-V  and \int L\: \mathrm {d} t (and not \mathrm {d} x ) respectively. You don't derive anything from anything here (however we can talk about how  T  and  V  come about). 1 Alternatively, there exists an extended approach to the Legendre transformation between the Lagrangian and Hamiltonian formalism, cf. e.g. Ref. 1. Let us suppress explicit time dependence t from the notation in the following. Consider the extended Lagrangian$$\tag{1} L_E(q,v,p)~:=~ p_i(\dot{q}^i-v^i)+L(q,v)~=~p_i\dot{q}^i-h(q,v,p), \tag{2} ...

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The easiest answer is "because that generates Maxwell's equations". The slightly more difficult answer is that the Lagrangian density has to be gauge invariant and a Lorentz scalar. The objects we have at hand are $F_{\alpha\beta}$, $A_\alpha$ and $J^\alpha$. Now, something like $A^\alpha A_\alpha$ is out, because it is not gauge invariant, $A^\alpha ... 4 Ok, let us start from scratch. A function$g: \mathbb R^n \to \mathbb R$with$f \in C^2(\mathbb R^n)$is said to be convex if its Hessian matrix (i.e. the one with coefficients$\partial^2 f/\partial x_i \partial x_j$) is everywhere (strictly) positive defined. Let$\Omega \subset \mathbb R \times \mathbb R^n$be an open set, and focus on a jointly ... 0 The derivative of a function$f(t)$is the function$\dot{f}(t)$in general different than$f$, and in the general case the two are not even linearly dependent, which is simple to see if you take the Taylor expansion. It is only after you define differential equations with them that they are linked algebraically, and this is what the calculus of variations ... 0 The gist of the response before the edit remains valid. The hamiltonian is defined as $$H(q,p,t) \equiv p \dot{q} - L(q,\dot{q},t),$$ the Legendre trasform of$L$. The Legendre transform takes$p$to$\dot{q}$, because$L$is convex, and this map is defined by$p = \partial L /\partial \dot{q}$. From the latter equation it is obvious that the map is ... 4 I) Lagrangian formalism. Let us suppress position dependence$q^i$and explicit time dependence$t$in the following, and also assume that the Lagrangian$L=L(v)$is a smooth function of the velocities$v^i$, where$i=1, \ldots, n. Define functions $$\tag{1} g_i(v)~:=~\frac{\partial L(v)}{\partial v^i}, \qquad i=1, \ldots, n;$$ $$\tag{2} ... 0 I got some help, I will post it for the few people that ever run into this (please correct me if I abused the notation):$$dW=-dU=-\int_0^{du}Fd(du)=-\int_0^{du}AY\frac{du}{dx}d(du) dU= \int_0^yAY\frac{y}{dx}dy=\frac{1}{2}AY\frac{y^2}{dx}=\frac{1}{2}AY\frac{du^2}{dx}=\frac{1}{2}AY(\frac{du}{dx})^2dx$$1 The force is proportional to the extension:$$ F = kx $$where we subsume all the various constants like Young's modulus and area into the constant k. We know dW = Fdx, so:$$ dW = k x dx $$and integrating this gives:$$ W = \tfrac{1}{2}kx^2 + C $$If we define the work to be zero when the extension is zero the constant C is zero, and we get the ... 0 This can be understood in the general case of Lagrange multipliers in mechanical systems. First, let me stress that in general, the Lagrange multiplier does not have to be a force. In the case of simple action-reaction, it is, but it is e.g. the gradient of pressure which is a force in the case of incompressible fluids. Regarding your questions: a. Why ... 1 The dynamics of a large class of mechanical systems can be described as a geodesic motion in some ambient space. This is the essence of the Kaluza-Klein theory. The basic and most elementary example is the case of a charged particle in 3D coupled to a magnetic field which can be described as a neutral particle geodesically moving in a background metric ... 1 I) OP talks about minimizing curves (rather than higher dimensional objects) so let us concentrate on point mechanics (as opposed to field theory) with Lagrangian L (rather than Lagrangian density {\cal L}). We conventionally call the curve parameter time t, although it doesn't have to correspond to any physical time variable. Let us for simplicity ... 1 You don't need to assume that the path of least action is the path taken. You can show it from Newton's laws. See http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf The path of least action is the path for which F = ma holds at each point. This is the geodesic. This is the shortest path through space-time. You get this path from the Lagrangian. You ... 2 You can derive the equations of motion (equations of geodesics) for a particle in curved spacetime by using the Lagrangian$$L = \frac{1}{2} \sum_{\mu,\nu} g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt},$$so the answer is yes. You could regard the configuration manifold as the manifold, it need not be the physical spacetime. I would like to clarify that ... 2 On one hand, by including the Lautrup-Nakanishi field B^a, we have an off-shell BRST formulation, i.e. we can prove the nilpotency of the BRST transformation without using the (Euler-Lagrange) equations of motion. On the other hand, for some applications, a simpler on-shell BRST formulation (where the Lautrup-Nakanishi field B^a has been integrated ... 0 We use the coordinates x_1 and x_2 for the two blocks. Let \ell be the natural length of the spring. Let us now write down the potential for this system. Note that if x_2 - x_1 > \ell then the spring does not exert any force on either block and hence there is no potential. If x_2 - x_1 < \ell, then the spring is compressed with \Delta x = ... 1 Say the density of the string is \mu and the tension is T. It's clear that the kinetic energy of an infinitesimal piece of string is$$dT = \frac{1}{2} (\mu \, dx) u_t(x)^2The length of the infinitesimal piece of string from (x, u(x)) to (x + dx, u(x + dx)) is \begin{align} d\ell &= \sqrt{dx^2 + (u(x+dx) - u(x))^2} \\ &= \sqrt{dx^2 + ... 0 Comments to the question (v2): It seems that OP assumes that \alpha is independent of x, i.e., OP considers a global quasisymmetry\delta A_{\mu}=\alpha \dot{A}_{\mu}.$The corresponding conserved quantity is energy, cf. Example 1 on the Wikipedia page for Noether's (first) theorem. 2 We may approach the problem via differential forms, or ordinary tensor calculus: Differential Forms: The field strength tensor$F$is a differential form given by the exterior derivative of the 1-form$A$, i.e.$F=\mathrm{d}A$which in components is$\partial_{[\mu}A_{\nu]}$. To add a total derivative to the form$A\$ is equivalent to adding the exterior ...

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