# Tag Info

## New answers tagged boundary-condition

1

Axisymmetry implies that there is no change in anything in the $\theta$ direction, i.e. $$\frac{\partial}{\partial\theta}(\text{anything}) = 0$$ Which would mean \begin{align} \frac{\partial p}{\partial\theta} &= 0 \\ \frac{\partial \vec{V}}{\partial\theta} &= 0 \\ \implies &\frac{\partial v_r}{\partial\theta} = 0 \\ \implies ...

0

The geometric argument is clear: Consider a Lagrangian density ${\cal L}=d_{\mu}F^{\mu}$ that is a total divergence. The action $S[\phi] = \int \! d^dx~{\cal L}$ will then be a boundary integral, due to the divergence theorem. Therefore the corresponding variational/functional derivative, $$\tag{1} \frac{\delta S}{\delta\phi^{\alpha}(x)}$$ which is an ...

0

A term $\int\!d^4x\, \operatorname{Tr}\, F \wedge F$ can be added to a non-Abelian Yang-mills theory (it vanishes trivially for the Abelian case, because of the wedge), and it is a total derivative. This term doesn't influence the equations f motion. However, this is a topological charge that counts something akin to the "winding number" of the gauge field. ...

0

If you have a four divergence inside an integral over all of spacetime (which is what you get when you extremize the action), the result will be a term which will be some product of the field(s) and its/their derivatives, evaluated at the boundary of spacetime. Since we assume that all fields go to zero (sufficiently quickly, so that their derivatives also ...

4

Topological degeneracy is only defined in the thermodynamic limit on a closed manifold. The ground state degeneracy of a finite-sized system or on an open manifold is not "topological", and can not be called topological degeneracy. Considering your examples. (1) The ground state degeneracy is ill-defined with open boundary condition. Because there might be ...

1

In a real-life misshapen blob of metal, strictly speaking the cyclic boundary conditions cannot be applied, since the blob only has a trivial group of spatial symmetries. However, the blob is approximately invariant under lattice translations (with the only mismatch occurring with the extremely small number of atoms at the surface) so it is tempting to ...

1

Yes there is a simple explanation. Think of a metallic sphere that is not connected to anything (i.e. floating), move this conducting sphere close to a positive electrode. Now what happens is, the negative charges are induced on the surface of the sphere closest to the electrode; those negative charges leave positive charges behind. Just like figure (a) ...

Top 50 recent answers are included