Does the idea "Accelerated charges radiate energy as EM waves" come from the Maxwell Equations? I hope it does. Can somebody provide a simple proof of this from the Maxwell equations?


Yes, it does. There is a simple proof of this in Peskin & Schroeder, chapter 6. The book is on Quantum Field Theory, but in my opinion their classical computation is one of the easiest to follow. I will sketch it here to make this post self-contained.

We start with the covariant form of Maxwell's equations:

$$ \partial_\mu F^{\mu\nu} = j^{\nu},$$ where $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$, $A^{\mu}=(\phi,\mathbf{A})$, $j^{\mu}=(\rho,\mathbf{j})$.

For a particle of charge $e$ going through a path in spacetime $y(\tau)$, parametrized by the proper time $\tau$, the current four-vector is

$$ j^\mu(x) = e \int d\tau \frac{dy^\mu(\tau)}{d\tau}\delta^4(x-y(\tau)).$$

The factor $dy^\mu(\tau)/d\tau$ is there to ensure the integral is invariant with respect to changes of variables.

To solve Maxwell's equations, it is easier to work in Lorentz gauge, where $\partial_\mu A^\mu = 0$, and Maxwell's equations become

$$ \partial_\nu \partial^\nu A^\mu = j^\mu.$$

Applying a Fourier transform to this equation we get

$$ \tilde{A}^\mu(k) = -\frac{1}{k^2}\tilde{j}^\mu(k),$$ where we used $k^2=k^\mu k_\mu$.

We will study a very simple trajectory with acceleration. A particle is moving with momentum $p^\mu$, then at $\tau=0$, receives a "kick" that changes its momentum to $p'^\mu$. The trajectory is $y^\mu(\tau)=(p^\mu/m)\tau$ for $\tau<0$, and $y^\mu(\tau)=(p'^\mu/m)\tau$ for $\tau>0$.

To calculate the Fourier transform of $j^\mu$, we need to add epsilons in the exponentials to make the integrals converge:

$$ j^\mu(x) = e \int_0^\infty d\tau \frac{p'}{m}e^{i(k\cdot p'/m+i\epsilon)\tau} + e\int_{-\infty}^0 d\tau \frac{p}{m}e^{i(k\cdot p/m-i\epsilon)\tau},$$

$$ j^\mu(x) = ie\left(\frac{p'^\mu}{k\cdot p'+i\epsilon}-\frac{p^\mu}{k\cdot p -i\epsilon}\right).$$

Using the Fourier-transformed Maxwell's equations, and inverting the transform, we have

$$ A^\mu(x) = \int \frac{d^4k}{(2\pi)^4}e^{-ik\cdot x}\frac{-ie}{k^2}\left(\frac{p'^\mu}{k\cdot p'+i\epsilon}-\frac{p^\mu}{k\cdot p -i\epsilon}\right).$$

This integral is evaluated in the standard way, by turning it into a contour integral in the complex plane, and using Cauchy's theorem, where you must get the contribution from the poles.

Looking at the poles, we see we have poles at $k^0=\pm|\mathbf{k}|$ from the $k^2$ denominator. These poles must be moved to the lower half-plane, since we need to solve Maxwell's equations using retarded boundary conditions to satisfy causality. There are two other poles, at $k\cdot p = 0$ and $k\cdot p'=0$. The first is in the upper half-plane and the second in the lower half, by virtue of the $i\epsilon$ in the denominator.

For $\tau<0$, we close the contour above and only the $k\cdot p=0$ pole contributes. For $\tau>0$, we close the contour below and the other three poles contribute. It is not difficult to prove that the $k\cdot p$ and $k\cdot p'$ poles do not correspond to radiation at all, they are just a Coulomb potential boosted to the frame where the particle has momentum $p$ or $p'$.

The radiation comes from the $k^2=0$ poles, which contribute only for $\tau>0$. That makes a lot of sense, there is radiation only after the kick is applied. So, the field corresponding to radiation is given by

$$A^\mu_{\mathrm{rad}}(x) = \int \frac{d^3k}{(2\pi)^3}\frac{-e}{2|\mathbf{k}|}\left[e^{-ik\cdot x}\left(\frac{p'^\mu}{k\cdot p'}-\frac{p^\mu}{k\cdot p}\right)+\mathrm{c.c.}\right]\bigg|_{k^0=|\mathbf{k}|}.$$

And that is it, really. To finish the calculation you would extract the electric and magnetic fields and compute their energy, this is an interesting calculation and I highly recommend that you check P&S for it, but what is done here should already be sufficient to convince you that radiation happens.

If you do go further you will find out that the energy radiated is actually infinite, but that is because the sudden "kick" is unphysical. Every physical process occurs during a time $T$, and that would provide a cutoff $k_{max} \approx 1/T$ for the integral, making the energy finite. Also, if the kinetic energy of the electron is much larger than its mass, the radiation will be concentrated in the direction of its initial and final speeds $\mathbf{v}$ and $\mathbf{v'}$.

  • $\begingroup$ Is a charged ball hanging from the ceiling radiating? According to GR a frame with gravity is equivalent to an accelerating frame, so the charged ball hanging in gravity should be equivalent to an accelerating ball. $\endgroup$ – md2perpe May 19 '17 at 20:57
  • $\begingroup$ @Felipe Lopes Please can you provide a proof with the form of 4 maxwell equations in the vector calculus format. That's the only one I understand. If you have the time, please. $\endgroup$ – PhyEnthusiast May 21 '17 at 12:44

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