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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 ...


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 ...


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....


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\...


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

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 $\approx$ symbol means equality modulo eom.]


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(\...


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

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. ...


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 ...


2

That the Hamiltonian is zero is completely correct. The system is time-reparametrization invariant - changing $\tau$ to $\xi(\tau)$ transforms $$ n(q(\tau))\mapsto n(q(\xi)),\quad \dot{q}\mapsto \frac{\mathrm{d}\xi}{\mathrm{d}\tau}q', \quad \mathrm{d}\tau\mapsto \frac{\mathrm{d}\tau}{\mathrm{d}\xi}\mathrm{d}\xi$$ and the action is invariant under this ...


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{...


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: $$L = \displaystyle\int \mathcal{L}(\phi(x,t),\dot \phi(x,t),x,t) \mathrm d^3x$$ $$S = \displaystyle\int dt \space L = \...


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}...


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.


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 ...


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

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....


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}\...


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 \begin{equation} V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2} \end{equation} Where \begin{equation} \phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+...



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