I am always wondering why only accelerated charges emit electromagnetic waves, while constantly moving ones do not.

I imagine an electron in else empty space moving with speed $v$. The $E$-Field in a given point must change over time due to coulombs law. How can this information propagate when no photons are emitted?

If there are no photons emitted by the moving electron, how fast does the changing $E$-field propagate over time?

Furthermore: Which changes in E (and B respectively) result in electromagnetic waves? In my case the vectors would have shifted in size and in direction.

  • 2
    $\begingroup$ Electric field of moving charge is not given exactly by the Coulomb law. It is similar, but lines of force are compressed in direction of motion. The lines of force move as rigid body, there is no information propagation. $\endgroup$ Sep 20, 2023 at 15:08
  • 1
    $\begingroup$ This is just a problem of wording, whether you count formally zero frequency radiation as “radiation”. Does it matter? What matters is what detectors see, and whether or not a detector is sensitive to such a changing electric field just depends on its design. $\endgroup$
    – knzhou
    Sep 23, 2023 at 21:17
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    $\begingroup$ This is 100% a duplicate of this question: physics.stackexchange.com/questions/65339/… But I'd rather not close it because the most upvoted answer on that original post is a really handwavy and incomplete rendition of the "kinked lines" argument. The second most upvoted answer is completely wrong. Idk was going on with physics stack exchange in 2013-2017. $\endgroup$
    – AXensen
    Sep 28, 2023 at 8:56
  • $\begingroup$ @AXensen, this question is completely different. one is asking how accellerating charges produce EM radiation, while this one is asking why /only/ accellerating charges produce EM waves and why a constantly moving charge doesnt produce photons. (Also pictures are often more helpful than equations) $\endgroup$ Sep 28, 2023 at 12:15
  • $\begingroup$ @StevenSagona I guess you're right. 100% was an exaggeration. But if you really understand why accelerating charges radiate, you should know the answer to this question as well. And I'm not saying there's anything wrong with the kinked lines argument and its associated picture. I'm saying one needs to complete the logical steps - why are kinked electric field lines radiation? why do they mean energy is being carried away from the particle? How are the electric field lines of a moving charge different from the electric field lines of an accelerating charge? etc. $\endgroup$
    – AXensen
    Sep 28, 2023 at 12:38

8 Answers 8


There are two separate issues here:

  1. what kind of electromagnetic fields constitute "waves" or "radiation"?

  2. how does the field of an inertially moving point charge arise, and why is this not a kind of wave?

The first of these questions involves a rather loose use of the word "waves" because you might want to say that any change at all can be regarded as a wave of some kind. Therefore let's look at the second word: radiation. A good way to clarify what we mean by a radiative field is to look at the field energy in some closed region of space. It could be, for example, a spherical shell around some point, or a rectangular box. Also we specify the region of interest such that it changes with time, involving propagation at the speed of light. For example, the spherical shell could propagate outwards with both inner and outer radius moving at $c$. The rectangular box could propagate in a direction orthogonal to one of its faces. If the field energy inside such a region has a contribution which stays constant with time then we have an example of electromagnetic field energy propagating at the speed of light. That is one good way to define electromagnetic radiation.

Note, however, that there remains a subtlety. I said "a contribution". In practice it can happen that the total field energy in the spherical shell example falls, but in such a way that it tends to a constant as the radius tends to infinity. This illustrates that sometimes it is only in the far field that it becomes clear whether there is any radiation, or how much there is.

Now to item 2.

When a charge acts as the source of a field, the equations of electromagnetism make it clear that all influences are local, and influences able to convey information or transport energy have $c$ as maximum speed. So the way the field around a charge is built up is that the charge influences the field right at itself, and then this field influences the field a little further out, and so on, until the whole space receives the influence. All this takes place at the speed of light. In the case where the charge has constant velocity all the field changes that propagate outwards do so in such a way as to make the electric field point to where the charge would arrive later if its velocity did not change! The net result is that whereas we can see it as a field configuration that moves along at the speed and direction of the charge, nevertheless it can also be seen as a sequence of influences propagating outwards as I just described.

The influences which I just described are not correctly called photons, because the whole description I just gave was within the terminology of classical electromagnetism.

If you wish to consider photons then you have to have some minimal understanding of what a photon is. One thing that helps a lot is a clear distinction between the real photon and a mathematical abstraction called "virtual photon". A real photon carries energy and momentum obeying the formula $E = pc$ and it can deliver this energy to a detector. A virtual photon is a contribution to a mathematical method to calculate interactions between given charged entities. It has no existence whatsoever outside the calculation that employed it.

If you place an electric field detector somewhere and a charged particle moving at constant velocity flies by, then the detector will register a changing electric field. In order to operate, the detector might require also that it absorbs some energy. If it does absorb energy then that energy comes from the field locally at the detector and consequently there is a change in the field. There will be no immediate effect on the particle (some distance away) but eventually the change will propagate back to the particle and have some influence on it. The photon language does not help very much in getting insight into this sort of situation, but one way to analyse it is to do a Fourier analysis and then you can say that the field is made up of lots of plane waves and each such wave has a large number of real photons, and the detector absorbs some of them.

  • $\begingroup$ "The photon language does not help very much in getting insight into this sort of situation, but one way to analyse it is to do a Fourier analysis and then you can say that the field is made up of lots of plane waves and each such wave has a large number of real photons, and the detector absorbs some of them." This is essentially what I say in my answer, and the bounty was to try to encourage someone to elaborate on this point. $\endgroup$ Sep 28, 2023 at 12:40

When the charge is moving with constant velocity, you can always Lorentz transform into its proper frame - where it is stationary - and from the charge frame of reference you see a constant electric field. A stationary electron or an electron moving with constant velocity are identical situations from physics point of view.

You must have time-varying electric field in order to produce a magnetic field: $$\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\varepsilon_{0}\frac{\partial\vec{E}}{\partial t}$$ That magnetic field will also be time-dependent. So, once you have a time-dependent magnetic field, you will also have an electric field (that will be time-dependent as well): $$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$

You get the idea.

Now you have something that propagates according to: $$\Delta \vec{E}-\frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2}=0$$ or $$\Delta \vec{B}-\frac{1}{c^2}\frac{\partial^2\vec{B}}{\partial t^2}=0$$ which is the electromagnetic wave equation, an equation that describes how electromagnetic wave propagates through space and time. Here $$\Delta=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$$ is the Laplacian or the Laplace operator. And $c$ is the speed of light in vacuum: $$c=\frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$$


Even in the case of a charge moving with constant velocity we have a time-varying electric field (as seen from any other reference frame except the charge proper frame) + a static magnetic field. So coming back to what I said: You must have specific configuration of time-varying electric field in order to produce a specific type of oscillating magnetic field. In order to produce an electromagnetic wave you need an oscillating electric field and an oscillating magnetic field that are perpendicular to each other and to the direction of propagation. That particular configuration is not created when a charge moves with constant velocity.

  • 2
    $\begingroup$ your first argument about reference frames is spot on. But a charge moving with constant velocity does produce an changing $E$ field, which produces a $B$ field, but does not produce radiation. $\endgroup$
    – AXensen
    Sep 28, 2023 at 10:54

I think that the concept of electromagnetic waves is a bit unclear. Every function can be written as the superposition of plane waves.

Let us look at any electric field $\mathbf{E}(t,\mathbf{r})$. The complete Fourier transform of $\mathbf{E}$ is equal to $$ \mathbf{\tilde{E}}(w,\mathbf{k}) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathbf{E}(t,\mathbf{r})e^{-i(\mathbf{k}\cdot\mathbf{r} - \omega t)}dtdxdydz $$

by definition. Meanwhile, the inverse Fourier transform states that $$ \mathbf{E}(t,\mathbf{r}) = \dfrac{1}{(2\pi)^{4}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathbf{\tilde{E}}(w,\mathbf{k})e^{+i(\mathbf{k}\cdot\mathbf{r} - \omega t)}dwdk_xdk_ydk_z $$

so you can interpret $\mathbf{E}(t,\mathbf{r})$ as the superposition of (perhaps infinite) plane waves, each with amplitude equal to $\mathbf{\tilde{E}}(w,\mathbf{k})$. You can even prove that, in vacuum, these are transversal waves with perpendicular electric and magnetic fields. Take the divergence-related Maxwell equations (in vacuum) $$ 0=\nabla\cdot\mathbf{E}(t,\mathbf{r}) = \dfrac{1}{(2\pi)^{4}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}i\mathbf{k}\cdot\mathbf{\tilde{E}}(w,\mathbf{k})e^{+i(\mathbf{k}\cdot\mathbf{r} - \omega t)}dwdk_xdk_ydk_z. $$

Since the inverse Fourier transform of $\mathbf{k}\cdot\mathbf{\tilde{E}}$ is equal to $0$ then $$ \mathbf{k}\cdot\mathbf{\tilde{E}} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}0e^{-i(\mathbf{k}\cdot\mathbf{r} - \omega t)}dtdxdydz=0 $$

and the same for the magnetic field, i.e. $\mathbf{k}\cdot\mathbf{\tilde{B}}(\omega,\mathbf{k})=0$ because $\nabla\cdot\mathbf{B}(t,\mathbf{r})=0$ as always.

To see that $\mathbf{E}(\omega,\mathbf{k})$ and $\mathbf{B}(\omega,\mathbf{k})$ are perpendicular, use Faraday's law $$ \begin{array}{rcl} \mathbf{0} & = & \nabla\times\mathbf{E}(t,\mathbf{r}) + \dfrac{\partial \mathbf{B}}{\partial t}(t,\mathbf{r}) \\ & & \\ & = & \dfrac{i}{(2\pi)^4}\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left[\mathbf{k}\times\mathbf{\tilde{E}}(w,\mathbf{k}) - \omega\mathbf{\tilde{B}}(\omega,\mathbf{k})\right]e^{+i(\mathbf{k}\cdot\mathbf{r} - \omega t)}dwdk_xdk_ydk_z \end{array} $$

so $\mathbf{\tilde{B}}(\omega,\mathbf{k})$ is perpendicular to both $\mathbf{k}$ and $\mathbf{E}$. Plus, $|\mathbf{\tilde{E}}(\omega,\mathbf{k})| = \tfrac{\omega}{|\mathbf{k}|}|\mathbf{\tilde{B}}(\omega,\mathbf{k})|$, where $\tfrac{\omega}{|\mathbf{k}|}$ is the velocity of wave propagation.

The last of Maxwell's equations allows us to see that these waves travel at the speed of $c$. $$ \begin{array}{rcl} \mathbf{0} & = & \nabla\times\mathbf{B}(t,\mathbf{r}) - \dfrac{1}{c^2}\dfrac{\partial \mathbf{E}}{\partial t}(t,\mathbf{r}) \\ & & \\ & = & \dfrac{i}{(2\pi)^4}\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left[\mathbf{k}\times\mathbf{\tilde{B}}(w,\mathbf{k}) + \dfrac{\omega}{c^2}\mathbf{\tilde{E}}(\omega,\mathbf{k})\right]e^{+i(\mathbf{k}\cdot\mathbf{r} - \omega t)}dwdk_xdk_ydk_z \end{array} $$

so $\mathbf{\tilde{B}}(w,\mathbf{k})\times\mathbf{k} = \tfrac{\omega}{c^2}\mathbf{\tilde{E}}(\omega,\mathbf{k})$, $|\mathbf{\tilde{B}}(w,\mathbf{k})| = \tfrac{\omega}{|\mathbf{k}|c^2}|\mathbf{\tilde{E}}(w,\mathbf{k})| = \tfrac{\omega^2}{|\mathbf{k}|^2c^2}|\mathbf{\tilde{B}}(w,\mathbf{k})|$ and $\tfrac{\omega}{|\mathbf{k}|}=c$.

Now, I don't know much about quantum electrodynamics, but like @Bohan Xu said, these waves must correspond to virtual photons when the system is not radiating. Free photons correspond to radiation, which I'm going to talk about in a little bit. Virtual photons exist so that charged matter can interact and exchange energy (i.e. so that the electromagnetic field can act as a force). Free photons exist only to transport energy well into arbitrary distances.

So, when do these "pure" electromagnetic waves (free photons) appear? When does the electromagnetic field carry its energy into the great beyond? The answer lies in the law for energy transport, or rather, the energy conservation law.

As always, to get the energy conservation law we must start from the equation of motion. This is the Lorentz force equation.

Let's assume that there are $N$ types of matter, with different masses $m_i$ and charges $e_i$ per particle. The matter densities $\rho_i(t,\mathbf{r})$ and velocity fields $\mathbf{v}_i(t,\mathbf{r})$ go from $i=1$ to $i=N$. Charge densities are equal to $\rho^e_i(t,\mathbf{r})=e_i\rho_i(t,\mathbf{r})$.

The equations of motion for the matter particles are $$ m_i\dfrac{d\mathbf{v}_i(t,\mathbf{r})}{dt} = e_i\mathbf{E}(t,\mathbf{r}) + e_i\mathbf{v}_i(t,\mathbf{r})\times\mathbf{B}(t,\mathbf{r}) $$

where $\tfrac{d}{dt}$ refers to the derivative along the trajectory of the particle. We can multiply by $\cdot\mathbf{v}_i(t,\mathbf{r})$ in order to get $$ m_i\dfrac{d\mathbf{v}_i(t,\mathbf{r})}{dt}\cdot\mathbf{v}_i(t,\mathbf{r}) = \dfrac{d}{dt}\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right] = e_i\mathbf{E}(t,\mathbf{r})\cdot\mathbf{v}_i(t,\mathbf{r}). $$

This means that the electric field parallel to the velocity is responsible for changing the kinetic energy of each particle.

Now, we want to write this as a local conservation law. Let's look into the left part of the equation for a little bit. The change in the kinetic energy of a particle (after going through a trajectory inside a velocity field for a short time $dt$) comes from a Taylor expansion to first order $$ \begin{array}{rcl} d\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right] & = & \dfrac{1}{2}m_iv_i^2(t+dt,\mathbf{r}+\mathbf{v}_i(t,\mathbf{r})dt) - \dfrac{1}{2}m_iv_i^2(t,\mathbf{r}) \\ & & \\ & = & \dfrac{\partial}{\partial t}\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right]dt + \left(\mathbf{v}_i(t,\mathbf{r})\cdot\nabla\right)\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right]dt \end{array} $$

so $$ \dfrac{d}{dt}\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right] = \dfrac{\partial}{\partial t}\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right]+ \left(\mathbf{v}_i(t,\mathbf{r})\cdot\nabla\right)\left[\dfrac{1}{2}m_iv_i^2(t,\mathbf{r})\right] = e_i\mathbf{E}(t,\mathbf{r})\cdot\mathbf{v}_i(t,\mathbf{r}). $$

We can start relating this to the total energy density by the product with $\rho_i(t,\mathbf{r})$. The left-hand side becomes $$ \begin{array}{rcl} \rho_i\dfrac{d}{dt}\left(\dfrac{1}{2}m_iv_i^2\right) & = & \rho_i\dfrac{\partial}{\partial t}\left(\dfrac{1}{2}m_iv_i^2\right)+ \left(\rho_i\mathbf{v}_i\cdot\nabla\right)\left(\dfrac{1}{2}m_iv_i^2\right)\\ & & \\ & = & \dfrac{\partial}{\partial t}\left(\rho_i\dfrac{1}{2}m_iv_i^2\right) - \dfrac{\partial\rho_i}{\partial t}\left(\dfrac{1}{2}m_iv_i^2\right) + \nabla\cdot\left[\rho_i\mathbf{v}_i\left(\dfrac{1}{2}m_iv_i^2\right)\right] - \nabla\cdot\left(\rho_i\mathbf{v}_i\right)\left(\dfrac{1}{2}m_iv_i^2\right) \\ & & \\ & = & \dfrac{\partial}{\partial t}\left(\rho_i\dfrac{1}{2}m_iv_i^2\right) + \nabla\cdot\left[\rho_i\mathbf{v}_i\left(\dfrac{1}{2}m_iv_i^2\right)\right] \end{array} $$

where we used the conservation of matter law $$ \dfrac{\partial\rho_i}{\partial t} +\nabla\cdot\left(\rho_i\mathbf{v}_i\right)=0. $$

So far, we've arrived to $$ \dfrac{\partial}{\partial t}\left(\rho_i\dfrac{1}{2}m_iv_i^2\right) + \nabla\cdot\left[\rho_i\mathbf{v}_i\left(\dfrac{1}{2}m_iv_i^2\right)\right] = \mathbf{E}\cdot\left(e_i\rho_i\mathbf{v}_i\right) $$

which we can sum over $i$ in order to get information about the total energy $$ \dfrac{\partial u_k}{\partial t} + \nabla\cdot\mathbf{S}_k = \mathbf{E}\cdot\mathbf{j}_e $$

where $$ u_k = \sum_{i=1}^N\rho_i\dfrac{1}{2}m_iv_i^2 $$

is the kinetic energy density, $$ \mathbf{S}_k = \sum_{i=1}^N\rho_i\mathbf{v}_i\dfrac{1}{2}m_iv_i^2 $$

is the kinetic energy flux, and $$ \mathbf{j}_e = \sum_{i=1}^Ne_i\rho_i\mathbf{v}_i $$

is the net current density.

We can use Poynting's theorem to re-write the right-hand side as $$ \mathbf{E}\cdot\mathbf{j}_e = -\dfrac{\partial u_e}{\partial t} - \nabla\cdot\mathbf{S}_e $$

with $$ u_e = \dfrac{1}{2}\left(\epsilon_0|\mathbf{E}|^2+\dfrac{1}{\mu_0}|\mathbf{B}|^2\right) $$

and $$ \mathbf{S}_e = \dfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B} $$

so $$ \dfrac{\partial u_k}{\partial t} + \dfrac{\partial u_e}{\partial t} = -\nabla\cdot\mathbf{S}_k - \nabla\cdot\mathbf{S}_e. $$

This means that $u_e$ is a sort of energy stored in the fields, and $\mathbf{S}_e$ is a flux of energy carried by the dynamics of the fields.

If we integrate this equation over all of space, and use the divergence theorem we get $$ \dfrac{d}{dt}Energy = \int_{V_\infty} \left(u_k+u_e\right)dV = -\oint_{S_{\infty}}\left(\mathbf{S}_k+\mathbf{S}_e\right)\cdot\hat{\mathbf{r}}dA. $$

The term $$ \oint_{S_{\infty}}\mathbf{S}_e\cdot\hat{\mathbf{r}}dA $$

is the loss of energy from the system due to it being carried away by the fields. This is the energy of the "free" photons!

So when is it non-zero? When are there such photons being emitted? These answers require the solution of Maxwell equations a long distance away from the charges. They are given by Jefimenko's equations $$ \begin{array}{rcl} \mathbf{E}(t,\mathbf{r}) & = & \dfrac{1}{4\pi\epsilon_0}\displaystyle\int\left[\dfrac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}\rho_e\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right) + \dfrac{\hat{\mathbf{r}}}{|\mathbf{r}|}\dfrac{1}{c}\dfrac{\partial \rho_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right) - \dfrac{1}{|\mathbf{r}|}\dfrac{1}{c^2}\dfrac{\partial \mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right)\right]dV'\\ & & \\ \mathbf{B}(t,\mathbf{r}) & = & -\dfrac{\mu_0}{4\pi}\displaystyle\int\left[\dfrac{\hat{\mathbf{r}}}{|\mathbf{r}|^2}\times\mathbf{j}_e\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right) + \dfrac{\hat{\mathbf{r}}}{|\mathbf{r}|}\times\dfrac{1}{c}\dfrac{\partial \mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right) \right]dV' \end{array} $$

and since $dA = |\mathbf{r}|^2\sin{\theta}d{\theta}d{\phi}$ the only terms from $\mathbf{S}_e$ which survive the integration are those proportional to $\tfrac{1}{|\mathbf{r}|^2}$ instead of $\tfrac{1}{|\mathbf{r}|^3}$ or $\tfrac{1}{|\mathbf{r}|^4}$. This is why we only take $$ \begin{array}{rcl} & - & \dfrac{\mu_0}{(4\pi)^2|\mathbf{r}|^2}\left[\hat{\mathbf{r}}\times\displaystyle\int\dfrac{\partial \rho_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right) dV'\int\hat{\mathbf{r}}\times\dfrac{\partial \mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}''\right)dV'' \right.\\ & & \\ & - & \displaystyle\left. \dfrac{1}{c}\int\dfrac{\partial \mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right)dV'\times\int\hat{\mathbf{r}}\times\dfrac{\partial \mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}''\right)dV''\right] \end{array} $$

from $\dfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$.

The first term is perpendicular to $\hat{\mathbf{r}}$ so it doesn't contribute to the integral. The second term is a triple vector product that can be re-written as $$ \dfrac{\mu_0}{(4\pi)^2|\mathbf{r}|^2c}\left\{\hat{\mathbf{r}}\left[\displaystyle\int\dfrac{\partial\mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right)dV'\right]^2 - \int\dfrac{\partial\mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}''\right)dV''\int\hat{\mathbf{r}}\cdot\dfrac{\partial\mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right)dV'\right\} $$

so $$ \oint_{S_{\infty}}\mathbf{S}_e\cdot\hat{\mathbf{r}}dA = \dfrac{\mu_0}{8\pi c}\displaystyle\int\left\{\left[\displaystyle\int\dfrac{\partial\mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right)dV'\right]^2 - \left[\int\hat{\mathbf{r}}\cdot\dfrac{\partial\mathbf{j}_e}{\partial t}\left(t-\dfrac{|\mathbf{r}|}{c},\mathbf{r}'\right)dV'\right]^2\right\}\sin{\theta}d\theta. $$

What can we conclude from this formula?

  1. To get some radiated electromagnetic energy, we need a current that varies explicitly with time
  2. We also need a $\tfrac{\partial\mathbf{j}_e}{\partial t}$ which is not purely radial (i.e. no monopole antennas)
  3. The radiated energy reaches $|\mathbf{r}|$ after a time which is equal to $\tfrac{|\mathbf{r}|}{c}$

How does this apply to the case of a single charge $e$ which moves with constant velocity? Take the velocity to lie on the $x$ axis. The current density in that case is equal to $$ \mathbf{j}_e(t,\mathbf{r}) = e\mathbf{v}\delta(x-vt)\delta(y)\delta(z) = e\dfrac{\mathbf{v}}{v}\delta(t-\tfrac{x}{v})\delta(y)\delta(z) $$

so $$ \dfrac{\partial \mathbf{j}_e}{\partial t}(t,\mathbf{r}) = e\dfrac{\mathbf{v}}{v}\delta'(t-\tfrac{x}{v})\delta(y)\delta(z) = -\delta(t-\tfrac{x}{v})\dfrac{\partial}{\partial t}\left[e\dfrac{\mathbf{v}}{v}\delta(y)\delta(z)\right] = \mathbf{0} $$

and there is no emitted radiation. I used the appropiate formula for the derivative of the delta function.

  • $\begingroup$ Thanks for the answer, but I'm getting a bit lost in the math here. After the point where you say "This means that the electric field parallel to the velocity is responsible for changing the kinetic energy of each particle." You write how a charges interact with the field, but I can't see the where you write how the E-field changes with the charges. $\endgroup$ Sep 28, 2023 at 16:04
  • $\begingroup$ @StevenSagona That lies within Jefimenko's equations. They are the solution to Maxwell's laws. I concluded that only the $\tfrac{\partial \mathbf{j}_e}{\partial t}$ terms produce radiation $\endgroup$
    – K. Pull
    Sep 28, 2023 at 16:12

This is a good question. I think the answers here aren't dealing with the following issue:

A electron constrained by a device to move at a fixed speed moves towards an "electric field detector," as the electron moves closer the electric-field detector gets exponentially higher in it's value (since it's just measuring the force it recieves from the proximity to the electron). So if there's an increasing infinite amount of force due to the electric field, how on earth does this not consistent of "photons," the supposed mediator of the electric field.

To the question:

When exactly does a change in E or B result in an electromagnetic wave?

One answer is just a technicality:

A change in E or B that results in a wave only occurs in systems where E and B become electromagnetic waves. That is, electromagnetic waves specific solutions to maxwell's equations that have a specific structure, and anythign that is different is not a wave.

The wavelike behavior can be seen in the picture of this answer,

enter image description here,

The biggest difference in your case versus the "wave case" is that the charge is changing in the transverse direction for an E-field, while you are describing a longitudinal direction.

The E-field traverses essentially because the lateral change in E-field strength can be felt over longer distances. And it is because of this oscillation of lateral change that you can get a change in E-field to be seen over longer distances. If you did a "longitudinal" change in E-field, this would just be more challenging to be felt over longer distances. (1/(delta R) is a very small number as R gets bigger!)

But as you have said, in your situation, there is a changing E-field and that does transfer energy, it's just not of the typical kind that is usually useful that people are familiar with photons/E-fields.

Personally, it would not suprise me if the energy of the process you describe could be seen by some type of "photon detector," and would therefore be some kind of "photon" -- but how to do this is not obvious and would be very non-standard. When photons are derived, they are constructed by solving maxwell's equations and getting a superposition of quantum operators. Your change in E-field should be able to be described by maxwells' equations, and I believe there should be a way of representing this as photon operators (but I think this is bit too advanced to get a clear answer here).


"If there are no photons emitted by the moving electron, how fast does the changing e - field propagate over time?" "How can this information propagate when no photons are emitted"

In short, there is some kind of photon even when the election is not moving. They are roughly speaking virtual photons, even though the word virtual photon has more technical meaning in perturbation theory. I think your confusion mostly comes from ambiguity of our language in describing quantum fields theory.

In quantum fields theory, photons simple mean a type of elementary excitation of quantum field, kind of similar to a vibration mode of a piece of string (but a quantum version of it). Those elementary excitation can combine (superposition) in a specific state of quantum field, just like different vibrational mode of a string can sum up to a specific positioning of string (ignoring momentum of the string, and is the quantum version). Even a stationary quantum field is made of those elementary excitation. This is related to people referring static coulomb interaction as "exchanging virtual photons between charged particles".

However, a stationary quantum field from a stationary point charge doesn't do stuff like photoelectric effect. While other quantum field state, like a state that only contain one elementary excitation(ie a photon) would do photoelectric effect (photon behaving like a particle by knocking electrons out of some metal). We call those elementary excitation that do behave like particles as photon. We call those superposition of photons that do have particle behavior as "emitting/containing photons". Other specific superpositions of elementary excitation that doesn't behave like particles (like a quantum field corresponds to a static point charge) are technically still made of elementary excitation which people call photon, but people typically don't use the language of "photon" to describe such overall superposition that doesn't behave like particles. Such field don't have photon in people's everyday definition of the word, but has photon in its technical definition of elementary excitation.

Basically, the definition of photon is confusing in quantum fields theory, because superposition can be confusing. One photon (elementary excitation) by itself behaves like a particle. Superposition of many photons(elementary excitation) might or might not behave like particles, depending how exact that superposition is.


The misunderstanding comes from your false assumption about the empty space. "I imagine an electron in else empty space moving with speed v."

Consider a still electron, it creates a non uniform electric field in the whole Universe. This electric field is given by Coulomb's law.

Now consider the electron moving at constant speed v towards you. By a change of referential, you can say the electron is at rest, so its electric field does not change, but you are moving towards it at speed -v. Obviously, as you travel through different regions, you experience a time varying electric field. If, for instance, you get closer from the source, the electric field increases. Electromagnetism is by essence relativistic, so you are also experiencing a magnetic field, but this is irrelevant here. In fact the same problem appears with gravitation. Imagine a planet moving in space at constant speed, how can the gravitation field change instantaneously in the whole Universe? The simple answer is that it does not change at all. The field permeates the whole Universe, and as the observer moves, it perceives a different field because its distance from the planet changes.


To answer your last question first: "If there are no photons emitted by the moving electron, how fast does the changing E-field propagate over time?"

The changes happen at the speed of light, $c$. If we are one light-second away from the electron then the electric fields at our location point towards where it was 1 second ago. So close to the electron the field lines point to where it is (or close), far from the electron the field lines point to where it used to be.

So the electric field lines are indeed curved slightly, and do change over time. We can ask: "what is the frequency of the EM wave implied by the change?". That is, if we pick a fixed point, how many times per second does the electric field repeat itself.

In this case it never repeats. It changes direction, turning to point towards the moving electron, but is always pointing somewhat behind it due to the information delay. A zero frequency field has zero energy, so their is no energy in this wave. Also notice that in this case the electric field direction is roughly parallel to the direction it moves, so it is a longitudinal thing (not a transverse wave). "Quasi-transverse modes", which have zero frequency, are sometimes important: although I do not think they ever carry any energy or photons.

I do not believe there are any photons present, but this kind of situation can be quite subtle and complicated, for example it may depend on the gauge that is being used. If there are photons their expected frequency (and thus energy) is zero. So their existence is going to be a fairly theoretical/speculative kind of existence rather than something readily measured.

  • $\begingroup$ I don't think this intuition works at all. A linearly increasing signal might not have a well defined "frequency" if you try to break it down in terms of how many times it repeats, but it certainly has a fourier decomposition. This changing e-field signal can certainly be represented as a superposition of cosine waves with different amplitudes and phases. $\endgroup$ Sep 23, 2023 at 20:42

I am tempted to say that EM wave will arise whenever there is no inertial reference frame in which all charges are static (i.e. not moving).

EDIT: In view of the comment by @Andrew, I realize the answer given by me above in not correct. I believe it should read instead:

"EM wave will arise whenever there is no inertial reference frame in which all $fields$ are static "

  • $\begingroup$ This is not correct -- imagine a copper wire with electrons carrying a steady current in the lab frame. Obviously electrons are moving in the lab frame. However, in the frame of the electrons, the protons form a current moving in the opposite direction. This shows there is no frame where all charges are at rest in this example. However, a wire carrying a DC current does not emit electromagnetic radiation. $\endgroup$
    – Andrew
    Sep 20, 2023 at 15:36
  • $\begingroup$ Instead of saying that the charges are static, you should rather say that you have zero current. This follows directly from Maxwell’s equation and avoids examples like the one given by Andrew $\endgroup$
    – LPZ
    Sep 20, 2023 at 16:19
  • $\begingroup$ @Andrew I agree. See the edited answer above. $\endgroup$
    – John
    Sep 20, 2023 at 16:43

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