New answers tagged

2

Your reasoning is essentially correct, apart from the last paragraph. To conclude, note that Newton's equation: $$\ddot {\mathbf r}(t) = \mathbf f(\mathbf r (t)),$$ with initial condition $\mathbf x (0)=(x_0,0,0)$, $\dot {\mathbf x} (0)=(0,0,0)$ can be solved by puttin $y(t)=z(t)\equiv 0$, thus reducing to a one dimensional problem: $$\ddot x (t)=f(x(t)),$$ ...


1

An internal symmetry is a transformation acting only on the fields, therefore not transforming spacetime points, and leaving the lagrangian or the physical results invariant. Example of internal symmetries are gauge symmetries. These are local symmetries, which means the transformations are in general spacetime dependent in the sense they are, in general, ...


2

Answer posted by Lubos Motl in the comments; I reproduce most of it here. This answer was posted in order to remove this question from the "unanswered" list. Some (sketches of) answers to your questions, one by one: Physical states have to be invariant under gauge symmetries, so all of them are singlets and there are no nontrivial representations, (and 3....


4

I am answering the question formulated after the "edit" in a newer version of the text because that one seems well-defined. Indeed, a situation with a uniform field $\vec E$ may be said to be "uniform" or translationally invariant in space. Noether's theorem says that this "uniformity" (spatial translational invariance) implies the existence of a conserved ...


1

It is a little different in General Relativity. Let's start with Special Relativity and all the 3 forces of the Standard Model in physics. Then we will talk about gravity and the universe. In The Standard Model spacetime is Minkowski, meaning flat in all 4 dimensions. If it is that way clearly any direction and position is equivalent. That's called ...


0

The final mode corresponds to the reflection symmetry of the system. Using the continuous deformation that you describe you can pinch two of the corners in entirely and invert them, which amounts to a reflection and is inequivalent to any continuous rotation of the system.


2

My hint is too long for a comment, but maybe it is worth writing down. I'll not use symmetry arguments, but I'll try to get Newton's equations out of the non-degeneracy of a certain mathematical object. Let $M$ be manifold and $\omega$ a 2-form on it. If $\omega$ is algebraically closed and non-degenerate, then the dimension of $M$ must be even (let's say ...


0

The main confusion was coming from the $x$ and $y$ axes of the phonon dispersion curves diagram having the same units (cm$^{-1}$), because of the choice of the natural unit system. However, two axes obviously have different meanings. The $x$ axis in the phonon dispersion diagram represents the momenta of phonons, while the $y$ axis represents the energy of ...


3

The corresponding symmetry group is the Lorentz group and yes we can use Noether to derive conserved quantities: Invariance under translations $\rightarrow$ momentum conservation Invariance under rotations $\rightarrow$ spin and angular momentum conservation Invariance under boost $\rightarrow$ some strange, not really useful, conserved quantity


1

The energy levels of the bound-states of a hydrogen atom only depend on the radial quantum number n . This is a special property of a (1/r) type of interaction potential . For a general central potential, V (r ) the quantized energy levels of a bound-state can depend on both n and l values. The property that the energy levels of a hydrogen atom ...


1

The wave function of the hydrogen atom $\psi(r,\theta,\phi)$ is a product of the radial part, the angular part and azimuthal term $$ \psi_{n,\ell,m}(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi). $$ The radial part $R(r)$ obeys Laguerre polynomials or, $$ R_{n\ell}(r) = Ae^{-\rho/2}\rho^\ell L^{2\ell+1}_{n-\ell-1},~ \rho = \frac{2r}{na_0}, $$ and $A$ is a ...


1

The field inside an hollow infinite cylinder is $0$, just like the field inside an hollow sphere. This is because of Gauss' law: the flux of the electric field $\vec E$ through any closed surface $S$ is $$\Phi = \int_S \vec E \cdot d \vec S = \frac Q {\epsilon_0}$$ Where $Q$ is the charge inside the volume enclosed by the surface. Let $R$ be the radius ...


0

Comments to OP's post (v4): OP is trying to prove via Noether's theorem that no explicit time dependence of the Lagrangian leads to energy conservation. OP's transformation seems to be a pure horizontal infinitesimal time translation $$\tag{A} t^{\prime} - t ~=:~\delta t ~=~-\epsilon, \qquad \text{(horizontal variation)}$$ $$\tag{B} q^{\prime i}(t) - q^i(t)...


0

The easier way of doing this is to just consider a generic transformation, G, such that the canonical co-ordinates of the Hamiltonian are shifted as below: $$ \delta p = \frac{\partial G}{\partial q} \delta \lambda$$ and $$ \delta q = - \frac{\partial G}{\partial p} \delta \lambda\,,$$ where $\lambda$ is the transformation parameter determining how much of ...


9

To apply Noether's theorem, which is what you are alluding to here, one needs to look at continuous symmetries of a Lagrangian description of a system's dynamics. The damped oscillation equation you have written, although it is invariant with respect to a time translation as you rightly say, is not a Lagrangian description. If you write the Lagrangian for ...


4

Conservation of energy is related to time translation invariance for systems that can be described by a Lagrangian. Dissipative systems in general are not describable by Lagrangian mechanics (without altering the formalism that is) and so Noether's theorem cannot be applied to check whether or not energy is conserved. EDIT: the dissipative system the OP ...



Top 50 recent answers are included