2

I'm confused about two conflicting ideas: the first is that monopoles cannot emit EM radiation (see here), and the second is that an accelerating point charge does produce EM radiation (see here). I was under the impression that point charges could not produce EM waves? What is the significance of dipoles in EM radiation if point charges accelerating also produce radiation?

Edit: For more context, this has all really been motivated by some vague statements I've heard about how to have gravitational radiation, you need a quadrupole moment, as opposed to in EM where you just need a dipole moment.

Qmechanic
  • 201,751
Chris
  • 141
  • 1
    The thing with gravitational radiation is that a gravitational dipole moment is just a displacement. No mass distribution can have a dipole moment/first moment in its center-of-mass frame. (That's dimply how the center of mass is defined.) So, in the COM frame, there cannot be a changing gravitational dipole moment, and thus there cannot be radiation from a changing dipole moment. The difference with E&M is that charge comes in two types, so even a distribution in its "center-of-charge" frame can have a electric dipole moment. – Buzz Oct 14 '22 at 04:50

4 Answers4

6

Please read the exact wording in the cited source carefully. The forbidden thing is the monopole radiation, not radiation from a monopole.

"Monopole radiation" is the radiation due to time dependent electric field, which in turn is produced by the monopole term in the multipole expansion of the potential. This is forbidden, as it would require time dependent charge.

On the other hand, one can make a monopole radiate by simply moving it. Think of a moving (positive) charge as a static charge plus a time dependent dipole (the negative charge of the dipole is not moving and is sitting on top of the static positive charge). Then the radiation is coming from the time dependent electric field coming from the dipole term in the multipole expansion.

2012rcampion
  • 635
John
  • 3,481
  • Thanks for the answer. I edited in some information about the motivation/context. I'm a little confused still about what is special about a dipole moment (respectively quadrupole moment) when it comes to EM (respectively gravitational) radiation. Does "dipole radiation" exist in the same sense "monopole radiation" does not? – Chris Oct 14 '22 at 02:24
  • My handwavy understanding is that EM being a vectorial field $A_\mu$ must couple to some other vector to form a scalar term in a Lagrangian $A_\mu j^\mu$. Current is associated with changes in dipole moment ($j_i \propto \partial p_i/\partial t$). Hence the importance of dipole moment for EM. Gravity is a rank-2 tensor field $h_{\mu \nu}$. By analogy with EM you expect gravitational field be coupled to a source which is a rank-2 tensor (to form a scalar). Product of dipole moments would do, but it is not linear in density. In contrast quadrupole moment is, a great source in this sense. – John Oct 14 '22 at 09:31
1

I'm confused about two conflicting ideas: the first is that monopoles cannot emit EM radiation ([see here][1]), and the second is that an accelerating point charge does produce EM radiation ([see here][2]).

I was under the impression that point charges could not produce EM waves?

A stationary point charge certainly won't.

What is the significance of dipoles in EM radiation if point charges accelerating also produce radiation?

The potential due to a moving point charge (for simplicity in the non-relativistic limit, and in some convenient unspecified units) looks like: $$ \phi(\vec r) = \frac{q}{|\vec r - \vec s(t)|}\;, $$ where $\vec s(t)$ is the position of the charge as a function of time.

The multipole expansion of this potential is: $$ \phi(\vec r) = \sum_{\ell=0}^\infty \frac{q r_\lt^\ell}{r_\gt^{(\ell + 1)}}P_\ell(\cos(\theta))\;, $$ where $r_\lt$ is the lesser of $|\vec r|$ and $|\vec s|$, $r_\gt$ is the greater, and $\theta$ is the angle between $\vec r$ and $\vec s$.

Suppose, as an example, that the magnitude of the observation point $|\vec r|$ is always larger than $|\vec s|$. And suppose that $\vec r$ is in the $\hat z$ direction. In this case: $$ \phi = \frac{q}{r} + \frac{qs(t)}{r^2}\cos(\theta_q(t)) + \ldots $$

The first term is a monopole term, which is, as expected, time independent. The next term is a dipole term, and there are higher order terms indicated by the "$\ldots$".

hft
  • 19,536
1

I have an intuitive answer. Accelerating a charge induces a changing magnetic field in response which results in radiation. (There is no way the magnetic field is like a monopole since $\nabla\cdot B = 0$. )

Isaiah Curtis
  • 11
  • 1
    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Oct 14 '22 at 03:30
  • "Accelerating a charge induces a changing magnetic field in response which results in radiation", that is true, this changing magnetic field in turn creates a changing electric field, which creates a changing magnetic field, etc... and this is the way the EM wave propagates radially from the initial radiator. – mins Oct 14 '22 at 15:51
0

It depends on what you mean by a "monopole". If by a monopole you mean a fluctuating amount of charge at a fixed position then that kind of a radiator cannot exist for it violates the most basic conservation law. If by a "monopole" you mean charge(s) moving and accelerating/decelerating, that charge will radiate. If by "monopole" you mean an antenna like this monopole, then that is among the most common antennas. There the charges run around back and forth and radiate because they accelerate.

hyportnex
  • 18,748
  • 2
    A so-called "monopole" antenna is really a dipole, because it always has a "counterpoise" (often ground) serving as the other pole. – John Doty Oct 13 '22 at 23:05