# Tag Info

7

The following passage has been extracted from Bohr's Nobel lecture: While in contradiction to the classical electromagnetic theory no radiation takes place from the atom in the stationary states themselves, a process of transition between stationary states can be accompanied by the emission of electromagnetic radiation, which will have the same ...

11

You are missing nothing. The Bohr model of the atom is false, and nowadays we replace the idea of the semi-classical "orbit" of Bohr with the fully quantum mechanical notion of orbitals or electron clouds, which give a probability distribution for the position of the electron around the nucleus, but do emphatically not imply that the electron is moving in ...

1

That is a unusual case but it can be achieved in a particle accelerator. when they give electrons enough kinetic energy to overcome the electric force they have on each other. keep in mind that electrons, small as they are, do have size. so for them to "touch" you will need A LOT of energy, but not infinite.

7

There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy $W$ of a uniform spherical charge distribution of radius $a$ and total charge $Q$ $$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$ Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ...

11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.

5

$$E^2 = E_x^2 + E_y^2 + E_z^2$$ therefore $$E_x^2 - \frac{1}{2}E^2 = \frac{1}{2}(E_x^2 - E_y^2 - E_z^2).$$

1

It's because magnetic field has zero divergence combined with the symmetry of the problem. At each point on our Ampere's loop we define three orthogonal unit vectors: $\hat{t}$ which is tangential to the loop, $\hat{r}$ which points radially outward from the center of the loop, and $\hat{z}$ which is parallel to the wire. Using these direction, we write the ...

1

Only from that equation, you cannot deduce that the solution is spherically symmetric. Take, for example $E_\phi=r\sin(\theta)$ and $E_\theta=0$. Usually, you cannot solve this equation, or more precisely, there are many solutions. The point is that, in order to solve the electric field, you need to take into consideration the rest of the Maxwell equations. ...

1

The argument is "boundary conditions are spherically symmetrical". If you have a non zero value for the tangential component, then you have to lose the symmetry. Spherical symmetry means you can rotate in any direction and get the same answer. That is only true if the field is the same regardless of the value of $\theta$ or $\phi$. You should, ...

1

You might find Gauge Fields, Knots, and Gravity by Baez & Muniain useful. It doesn't specifically deal with electromagnetism in curved spacetime, however, the first chapter develops differential geometry with the goal of reformulating Maxwell's equations. It assumes no knowledge of GR. The third and final chapter develops some aspects of GR.

2

The Classical Theory of Fields by Landau and Lifshitz fits the bill reasonably well. It doesn't develop electrodynamics from scratch in the context of a curved spacetime, but it does have a three-page section covering the equations of electrodynamics in a curved spacetime, after spending seven chapters developing electrodynamics in the context of flat ...

0

I suspected that one needed to go back to the definition of the currents and indeed, in doing so one can derive the result. Here's a short version. The electron current is defined as [see equation (6.6) in 1] $$\tag{1}J_\mu(x) = -e\bar{u}_f\gamma_\mu u_i \times\mathrm{exp}[(p_f-p_i)\cdot x]$$ which we write as \tag{2}J_\mu(x) = ...

0

It seems to me that there is a typo in the book. Your starting equation should be the following: $$T_{fi} = -i\int \frac{d\omega \,d^{3}\textbf{q}}{(2\pi)^{4}} \tilde{A}(\textbf{q},\omega)\tilde{B}(-\textbf{q},-\omega) \frac{1}{|\textbf{q}|^{2}},$$ where $\tilde{A}$ denotes the Fourier transform of $A$. Then, using ...

3

Well discrete charges, and in particular point charges, are a consequence of quantum mechanics. If you're considering just the classical theory there are no special conditions on the charge distribution. I'm not sure I'd say charge density was more fundamental than charge, but charge density would be what gives you the divergence of the electric field.

2

Ferromagnetic materials contain magnetic domains within which the electrons spins are aligned to give a net magnetic moment. Bulk magnetisation is done by changing the alignment within the domains so they all align in the same direction and their magnetic fields all reinforce. Anything that puts energy into the crystal lattice can randomise the alignments ...

-2

I think the fundamental problem that many people have with quantum mechanics is, that is seems to be about particles, when, in reality, it is about quanta. A quantum is not the same thing as a particle! A quantum is the exchanged amount of physical quantities between two parts of a physical system. That can be a quantum of energy, a quantum of momentum, a ...

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