Tag Info

Hot answers tagged

7

In the absence of magnetic monopoles, Maxwell's equations are $$\text d F = 0,$$ $$\star\text d\star F = J_e,$$ where $J$ is the 4- current 1-form due to electric charges. For cohomological reasons, from the first equation one can asserts that there exists a 1-form $A$ such that $F = \text d A$, and $A$ is the interpreted as the 4-potential $(\phi,\mathbf ...


4

Comments to the question (v2): Noether's (first) Theorem is really not about Lie groups but only about Lie algebras, i.e., one just needs $n$ infinitesimal symmetries to deduce $n$ conservation laws. If one is only interested in getting the $n$ conservation laws one by one (and not so much interested in the fact that the $n$ conservation laws often ...


4

Because that way the Euler-Lagrange equations turn out to be linear and thus the superposition principle holds. Superposition principle means "free theory".


4

Comments to the question (v8) concerning numerical integration: On one hand, to solve a Hamiltonian system numerically, there exist the numerical integration schemes of symplectic integrators (SI), where each (finite) numerical iteration step is a canonical transformation/symplectomorphism, which preserves certain properties, such as, e.g., energy, and ...


3

In general, when working out the Lagrangian, start in coordinates that you know and then rewrite in generalized coordinates. Kinetic energy in this case is proportional to $v^2 = \dot x^2 + \dot y^2 + \dot v_z^2$. In your spherical coordinates $$x = r \sin \alpha \cos \theta;~ y = r \sin \alpha \sin \theta;~ z = r \cos \alpha.$$ Take full time derivatives ...


3

The lagrangian you're dealing with is $\mathcal{L}= \frac12 (\partial_{\mu} \phi)^2 - \frac12 m^2 \phi^2$. When you take the partial with respect to $\partial_{\mu}\phi$, you should be getting $2 * (\frac12 \partial^{\mu}\phi)$. This would make the first term in your expression $\dot{\phi}^2$ instead of $\frac12 \dot{\phi}^2$ and things would work out. If ...


3

Comments to the question (v2): First of all, recall the notion of an (off-shell) quasi-symmetry. It means that the action $S[\phi]$ changes by a boundary integral under the transformation of the fields $\phi$ and spacetime point $x$, cf. e.g. this and this Phys.SE posts. Since the action $S=\int_R \!d^nx~{\cal L}$ is an extensive variable, it is clear ...


3

In NRQM, we represent particles by a localized wave packet $\psi$, called the wave function. We say roughly that the classical particle is located at the "peak" of the wave packet. We say that we cannot know the path because the wave function can be nonzero in multiple places. Instead of being able to deterministically tell where the particle is like in ...


2

Comments to the question (v2): A field $\phi^{\alpha}:[t_i,t_f]\times \mathbb{R}^3\to \mathbb{R}$ is the field-theoretic version of a (generalized) position variable $q^i:[t_i,t_f]\to \mathbb{R}$ in point mechanics. Note that the physical position space $\mathbb{R}^3$ typically plays very different roles in field theory and in point mechanics.$^1$ ...


2

Two points. First, the variation of the GHY term is discussed in detail in this post: Explicit Variation of Gibbons-Hawking-York Boundary Term Second, the GHY term on its own is enough to render the boundary value problem well-defined, but it is not sufficient for a physically interesting variational principle. This requires additional surface terms in the ...


2

Comments to the question (v2): "Find the Lagrangian of the theory" typically means that you are given the (classical) equations of motion (EOMs) of some physical system, and are supposed to find the action functional $S$ so that the EOMs are (parts of) the Euler-Lagrange (EL) equations for $S$. Note that an action principle/Lagrangian formulation does not ...


2

I) Before we get to quantization and path integrals there are problems already at the classical level. The Legendre transformation is not well-defined without knowledge of the CCR. For instance if the CCRs for the complex bosonic scalar $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ is zero, this would mean that OP's Hamiltonian density ${\cal H}$ is a pure ...


2

Suppose $g$ is the generator of a certain symmetry (i.e. the generating function of an infinitesimal canonical transformation) and you are interested to know how the observable $f$ changes after the "action" of $g$. In the Hamiltonian formalism the change is found to be $$\delta f \approx \epsilon\{f,g\}$$ which can be related to the time evolution of an ...


2

When contracting with the metric $\eta_{\mu\nu}$ one has explicitly, $$\eta_{\mu\nu} \frac{\partial \mathcal L}{\partial (\partial_\mu \phi)} \partial^\nu \phi = \frac{\partial \mathcal L}{\partial (\partial_0 \phi)} \partial^0 \phi - \sum_{i=1}^3\frac{\partial \mathcal L}{\partial (\partial_i \phi)} \partial^i \phi$$ For the other term, one has, ...


2

This is a consequence of a theorem from classical mechanics. The magnetic force is non conservative and so it is not suitable for the standard approach to the Lagrange formalism. The form that the Lagrange equations normally take is: $$ {d \over dt} { \partial L \over d \dot q_i} -{ \partial L \over d q_i}=0$$ This only holds for conservative Systems ...


2

A beautifully conceptual derivation of 11d SuGra was given in Riccardo D'Auria, Pietro Fré, Geometric Supergravity in D=11 and its hidden supergroup, Nuclear Physics B201 (1982) 101-140 (nLab) using the excellent supergeometric methods later laid out in their textbook Leonardo Castellani, Riccardo D'Auria, Pietro Fré, Supergravity and Superstrings - A ...


2

Noether theorem tells you that if you can find a (one parameter) group of infinitesimal transformations $\alpha$ and $\beta$ such that: \begin{equation} t'=t+\alpha\epsilon \end{equation} \begin{equation} q'^\mu=q^\mu+\beta^\mu\epsilon \end{equation} and your lagrangian is invariant under this group of transformations, then the quantity \begin{equation} ...


2

Comments to the question (v1): I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$ However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path ...


2

Assuming no quantum gravity, $\eta^{\mu\nu}$ is a constant and can be pulled out of the derivative and what remains looks like a $\delta^k_{\mu}$ or $\delta^k_{\nu}$-type expression (in the sense of a Kronecker $\delta$), pulling the $k$ into the $\partial^\mu$ or $\partial^\nu$ respectively. If you are confused about where the minus sign comes from, I ...


1

The kinetic energy of the particle depends just on the path it is following - if you imagine the cone is suddenly invisible, the particle continues to go around in a circle. That means that there is no reason to add $\sin\theta$ in your expression for the kinetic energy if you used $r$ to mean (as drawn) the distance from the axis of rotation. Note - your ...


1

So, the nice thing about Lagrangian mechanics is generalized coordinates. The crummy thing about it is that you at least in principle need to know about variational calculus to understand why it works and how to get the equations of motion. Finally, Emmy Noether's theorem gives you a really generic sense for conservation laws. The nice thing about ...


1

The story is the following. We start with the simplest Poincare-invariant action that does not depend on the parametrization $$ S=-m\int dl=-m\int\sqrt{-ds^2}$$ here $ds^2$ is the interval. We can rewrite it as $$S=-m\int\sqrt{dX^\mu dX^\nu \eta_{\mu\nu}}$$ here $\eta_{\mu\nu}$ is the Minkowski metric. Now if we suppose that $X^\mu$ depend only on $\tau$ we ...


1

The minimum number of independent generalized coordinates is given by the number of degrees of freedom. Nothing prevents you from using more coordinates (dependent) if you so desire.


1

Think about an action principle as an abstract mapping of trajectories $\mathbf r = (\vec r(t), t_0, t_1)$ to some number $S(\mathbf r)$ which no longer is explicitly time-dependent. Now, sometimes, the action can be computed from another function. The other function just has a bunch of parameters, which we can call $\{\alpha_i\}, ~ \{\beta_i\}, ~ \tau$. ...


1

In this answer we will just make a general conceptional remark about variational/functional derivative (FD), which hopefully implicitly answers OP's specific questions. OP is apparently considering the 'same-spacetime' FD, $$\tag{A}\frac{\delta {\cal L}(x)}{\delta\phi^{\alpha} (x)}~:=~ \frac{\partial{\cal L}(x) }{\partial\phi^{\alpha} (x)} - d_{\mu} ...


1

The functional derivative $\frac{\delta}{\delta \phi}$ acts on functionals, things that map functions to real numbers. That is, they act on actions $S$, not lagrangians $L$. I don't know where you got your original question, but there indeed should be a minus sign! Altogether, I think what you're asking is why: $$ \frac{\delta}{\delta \phi} \int d^4x ...


1

The branes of string theory are all reflected already in the underlying low-energy supergravity theories as "black branes" and particularly as the extremal black branes, the BPS solutions to the sugra equations of motion. A much-cited lecture note on this is Kellogg Stelle, BPS Branes in Supergravity (arXiv:hep-th/9803116) where for instance figure 6 on ...


1

Comments to the question (v1): Concerning the notion of on-shell and off-shell, see also Wikipedia and this Phys.SE post. In the context of an action formulation, on-shell means that the Euler-Lagrange (EL) equations $$\tag{1} \frac{\delta S}{\delta\phi^{\alpha}(x)}~\approx~0$$ are satisfied. It seems that the potential confusing point is the notion of an ...


1

In terms of Feynman diagrams, a "coupling" translates to a vertex factor. The Lagrangian for a free electromagnetic field is $$\mathcal{L}=-\frac{1}{4}F^2$$ as you well know. Now suppose we have an electron field $\psi$ too. We want this electron field to "interact", or couple, with (to) the photon field. The free Dirac Lagrangian is ...


1

I randomly had this typed up in personal notes. Was probably an exercise somewhere. Consider a harmonic oscillator, which is described by the Hamiltonian $$H=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2$$ Doing the Legendre transform, we obtain the action as $$\mathcal{S}=\tfrac{1}{2}m\int_0^t(\dot{q}^2-\omega^2q^2)dt'$$ Now we use the Euler-Lagrange equation to ...



Only top voted, non community-wiki answers of a minimum length are eligible