# Tag Info

1

Consider a bead sliding on a thin, rigid rod in the $x$-direction. An example of virtual infinitesimal displacement that is inconsistent with the constraints is $$\delta \mathbf r = (0, \delta y, 0), \qquad \delta y \neq 0$$ because this displacement represents the bead moving away from the rod.

1

$$\mathcal{L}_{SM}=\mathcal{L}_{EW(after ~symmetry~breaking)}+\mathcal{L}_{QCD}+\mathcal{L}_{loc}+\mathcal{L}_{gf}$$ $$\mathcal{L}_{EW}=\mathcal{L}_K+\mathcal{L}_N+\mathcal{L}_C+\mathcal{L}_H+\mathcal{L}_{HV}+\mathcal{L}_{WWV}+\mathcal{L}_{WWVV}+\mathcal{L}_Y$$ $$\mathcal{L}_{loc}=\mathcal{L}_{gol}+\mathcal{L}_{int}$$

1

Your proposed path has a VERY LARGE action. As @tparker pointed out, you have to minimize the path subject to the constraint that the average velocity doesn't change. Now, the action is quadratic in velocity. A little fiddling around with the math should convince you that to minimize the integral of $v^2$ subject to the constraint that the average velocity ...

2

You're missing that Dirichlet boundary conditions $$x(t_i)~=~x_i \quad\text{and} \quad x(t_f)~=~x_i$$ are implicitly implied. The stationary action principle is not well-posed without boundary conditions.

3

You have to minimize the integral subject the the constraint that the initial and final positions $x(t_i)$ and $x(t_f)$ are held fixed. In particular, $\Delta x = \int_{t_i}^{t_f} v(t)\, dt$ is held fixed. If the particle slowed down than sped up as you suggested, the action would be less, but it wouldn't have a high enough average speed to cover the full $... 3 This is a partial answer which I will hopefully come back to and expand. The property of being its own Legendre transform is unique to the pure quadratic kinetic energy$T(v)=\frac12 mv^2$. As a simple example, consider$T(v)=\frac14Av^4$. Here the Legendre momentum is $$p=\frac{\partial L}{\partial v}=\frac{\partial T}{\partial v}=Av^3,$$ so the velocity ... 3 Recall that the path integral formulation comes in (at least) two versions: Lagrangian & Hamiltonian. It is often argued that the Hamiltonian version is more fundamental, cf. e.g. this Phys.SE post. Thus we should compare the Polyakov (P) Hamiltonian Lagrangian density $${\cal L}_{P,H}~=~P^{\alpha} \cdot \partial_{\alpha}X +\frac{\gamma_{\alpha\... 7 The path integral involving the Nambu-Goto square root in the exponent is a very complex animal. Especially in the Minkowski signature, there is no totally universal method to define or calculate the path integrals with such general exponents. So if you want to make sense out of such path integrals at all, you need to manipulate it in ways that are ... 1 It isn't "obtained". It's a definition of the function g and this definition is useful because it leads to the nice symmetric relationships on the rest of the page 3. They exchange the two Legendre-dual variables. So the definition of g wasn't really "derived" in any straightforward way. It was a clever guess that Legendre made at some point of his life.... 0 The invariant interval is the object to be extremized. The infinitesimal interval$$ ds^2~=~c^2dt^2~-~dx^2~-~dy^2~-~dz^2 $$is also$$ ds^2~=~\left[c^2~-~\left(\frac{dx}{dt}\right)^2~-~\left(\frac{dy}{dt}\right)^2~-~\left(\frac{dx}{dt}\right)^2\right]dt^2~=~(c^2~-~v^2)dt^2 $$We may then take the square root of this and multiply by the invariant mass$$ mc^... 3 For generic initial conditions, the answer is Yes, due to Lagrange equations $$\frac{dp_i}{dt}~\approx~ \frac{\partial L}{\partial q^i}, \qquad p_i~:=~\frac{\partial L}{\partial \dot{q}^i}.$$ [Here the$\approxsymbol means equality modulo eom.] 2 There is no mistake. The Hamiltonian satisfies the relation $$\frac{\text{d}H}{\text{d}t} = \frac{\partial H}{\partial t}.$$ This follows immediately from Hamilton's equations: \frac{\text{d}H}{\text{d}t} = \dot{H} = \frac{\partial H}{\partial q}\dot{q} + \frac{\partial H}{\partial p}\dot{p} + \frac{\partial H}{\partial t} = \frac{\partial H}{\partial q}... 0 Feynman is using definite small quantities (inches) in place of infinitessimals \delta x etc. Probably he wanted to avoid non-essential mathematical formality, in line with his casual, hand-waving persona. The Principle of Virtual Work requires the structure to undergo infinitessimal displacements (hence "virtual"). He could instead have used units of ... 1 In general field theories Lagrangian are not derived: they are instead assigned (postulated) and proven against the equations of motion. Every Lagrangian that gives rise to the correct equations of motion is in principle a good Lagrangian for the system. One can prove that for mechanical systems described by conservative forces \textbf{F} = - \textrm{grad}\... 2 No, you get a separate Euler-Lagrange equation for each individual degree of freedom, i.e. a system of simultaneous equations. So in your example, \begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right)-\frac{\partial L}{\partial \theta} &= 0, \,\mathrm{and} \\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right)-\frac{... 3 For a (sufficiently nice) expression f(X), where X ranges over a vector space), the directional derivative Ydf(X) with respect to Y (in the same vector space) is the coefficient of \epsilon in an expansion f(X+\epsilon Y)-f(X) where \epsilon is a formal variable with \epsilon^2=0. The functional derivative \delta f(x)/\delta(X) is the ... 4 You already got your answer, all right, several times over, but I will emphasize the central puzzle of your question which you only got indirect answers for, connected to the peculiar special structure of SO(4). Any self-respecting text introducing the standard model more or less has it. I'll skip all superfluous issues like lagrangian terms, the U(1)s, etc.... 1 Answer of this question is quite subtle. First let us consider the most general Higgs potential which is renormalizable and invariant under SU(2)_{L}\otimes U(1)_{Y} gauge transformations, which has the form $$V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2}$$ Where \phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+... 0 After a bit of discussion I believe there is actually a SU(2)\times SU(2) symmetry in a sense. So in principle there is a U(2) symmetry if \phi=(\phi_1,\phi_2)^T, \phi^\dagger=(\phi_1^*,\phi_2^*) and the lagrangian\mathscr{L}=\partial_\mu \phi^\dagger\partial^\mu \phi-m\phi^\dagger\phi-\lambda(\phi^\dagger\phi)^2,$$simply sent \phi\to U\phi, ... -1 If the field is a simple complex scalar field, than the symmetry is just U(1). For a higher symmetry, \phi need to be higher dimensional too, for instance you can add a vector index \phi_i with i=1,2 for simplicity, which means that you add an additional complex field. If these two fields interact, you can have two cases now: Each field has a U(1) ... 2 This is just supplementing Qmechanic's answer. I think the notations here need to be addressed. OP might be confusing Lagrangian (normal L) with Lagrangian density (\mathcal{L}). Formally, we have three fundamental relations:$$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3xS = \displaystyle\int dt \space L = \... 3 Yes, OP is right. In the field-theoretic case, the partial derivatives in OP's first formula (1) should be replaced with functional derivatives $$\delta S~=~\int_{t_1}^{t_2}\!\mathrm{d}t\left(\frac{\delta L}{\delta q}~\delta q+\left. \frac{\delta L}{\delta v}\right|_{v=\dot{q}}~\delta \dot{q}\right),\tag{1'}$$ where the Lagrangian $$L[q(\cdot,t),v(\... 6 There is also the routhian formalism of mechanics which is described as being a hybrid of lagrangian and hamiltonian mechanics. The routhian is defined as$$R = \sum_{i=1}^n p_i\dot{q}_i - L\$ You can learn more about it by clicking this link for wikipedia's description of it. Reading more in regards to the routhian because I was bored, I realized it is ...

3

It's worth pointing out that the Hamiltonian and Lagrangian formalisms are independent, even though they're usually taught as if the former were a filtering of the latter (here enter Legendre transforms). Both formalisms are as independent as the notions of tangent and cotangent bundles in differential geometry: independent, but intrinsically connected. ...

Top 50 recent answers are included