# Tag Info

0

The gravitational force acting on the particle as function of radius $r$ is [Gauss's law] $$F(r) = -\frac{GmM(r)}{r^2}$$ where $M(r) = \int_0^r4\pi \rho(r')r'^2dr'$ is the mass contained within a radius $r$. Note that for $r>R$ we have $\rho(r) = 0$ and $M(r) = M(R) \equiv M$ the total mass of the whole sphere. The potential energy is the work needed ...

0

Basically recoil, since photons can also carry momentum. The reason why this recoil can also lead to attraction instead of plain repulsion is that the exchanged virtual photon has a definite momentum, so it can not have a definite position at the same time (the position wave of a virtual photon is stretched out to infinity). The momentum can be transfered ...

4

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

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

The brief answer is yes, they do have topological degeneracy. My understanding is that if you only focus on the surface topological order with some symmetry given by the bulk symmetry protected topological (SPT) state, there is nothing wrong with that symmetry enriched topological order (SET) states. That means they also have all the properties as the usual ...

1

I know that the question specifically refers to classical electrodynamics, but I think it is helpful to look at this from a QED perspective. The term $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ is the kinetic term, i.e. from it we obtain the propagator. In a course on QFT, you probably derived the general relation Z[J]=\int\mathcal{D}\Psi\,\exp\left(i\int ...

Top 50 recent answers are included