In case of a charged particle which is travelling at a uniform velocity, the electric field due to it at a given point doesn't change instantaneously . The reason for this delay in change of electric field , physically , can be explained as the time the electric field needs to propagate to that point, or, let's say, the change in electric field takes some time to propagate .

I would like to know what exactly does it mean by a propagating electric field, and how exactly does it propagate ? What can we put forth as a reason to explain this propagation at a speed equal to the speed of EM radiations ? (I hope the coming up answers would comprise an important part explaining this physically. ) I guess we can derive the speed of EM radiation from Maxwell's equation ; can we show that this propagation too has the same speed ?

  • $\begingroup$ It is not clear what you ask. When you ask what means "a propagating electric field, and how exactly does it propagate?" you ask how the electrostatic field of the charge moves together with the charge? $\endgroup$
    – Sofia
    Feb 24, 2015 at 21:38

4 Answers 4


Its about Travel Time and Carrier Particles

Particles with electric charge give off a field. That field is made of something; virtual photons, the messenger particle for the EM force. Charged particles give off so many and in such random directions that it basically acts like a continuous thing, a field.

Individual photons must travel whatever distance to reach other particles, so if your source of photons moves, things affected by those photons do not get affected by the new position until those photons come from the new position. Photons which reflect the change in the position "propagate," and we can imagine many of them as a field.

This essentially means that there are two parts to the field; the parts of the field which reflect the original position have not yet updated, and the parts of the field which reflect the new position. The part of the field which reflects the new position travels out (at the speed of light), replacing the field which was caused by the old position.

I find this explanation to be a little hard to digest, so let us try another one.

A Lot of Yelling

If you were to personify this, particles with non-neutral charges are constantly yelling at other particles with non-neutral charges about where they are and how much charge they have. This particle is essentially telling other particles in the universe that "Hey, I'm here and have this charge!" and all the other particles act accordingly. When that particle moves, that particle is now saying "Hey, I'm now over here, and have this charge!" and all the particles must act according to that new information.

There is a time when the particle yells out its new position but the other particles have not heard about it yet. It is because that new information hasn't propagated yet. That message, the "Hey, I'm now over here and have this charge!" takes some time to reach other particles. We mark the boundary between particles that have heard this information and particles which have not. If we watch this boundary, it moves out as that information travels.

Instead of sound waves carrying dialogue, though, these are EM waves/photons. Photons travel at the speed of light, so updates in your field happen at that speed.

  • $\begingroup$ Very nice and amusing explanation, though I would ask a question, because such things were many times discussed here. In emitting the yelling virtual photons the original charged particle doesn't loose energy? Such a continuous emission transports with itself energy, doesn't it? Where from is it taken? Let me say also otherwise: different people in our site claim that the virtual photons "are only on the paper, on Feynman's diagrams". Then, what is the truth? $\endgroup$
    – Sofia
    Feb 24, 2015 at 21:47
  • $\begingroup$ Nonsense. The propagation of electric and magnetic fields with the speed of light is a classical, not a quantum mechanical phenomenon (else we wouldn't have the advanced/retarded solutions in classical electrodynamics!). The field is not made out of virtual photons (it is not even clear that these are actual particles rather than lines in a Feynman diagram), though changes in the field - "waves" - may indeed correspond to actual photons. $\endgroup$
    – ACuriousMind
    Feb 24, 2015 at 21:58
  • $\begingroup$ @ACuriousMind :I guess I would be wrong in saying that the casmir effect is a manifestation of virtual particles.. $\endgroup$ Feb 25, 2015 at 14:59
  • $\begingroup$ @AgniveshSingh: In my view, that is indeed very close to being wrong, although that is a popular misconception, and not in any way your fault. $\endgroup$
    – ACuriousMind
    Feb 25, 2015 at 16:02
  • $\begingroup$ Curious mind should probably open the wikipedia, which clearly says that a force between two particles can be described either as the action of a force field generated by one particle on the other, or in terms of the exchange of virtual force carrier particles between them. Read once more: forces are induced by the virtual (aka messenger) particles. Brain Greene has even made a visualization for you. $\endgroup$ Sep 25, 2016 at 16:57

Maxwell's Equations are usually written as:



$$\vec{\nabla}\times \vec{E}=-\frac{\partial \vec{B}}{\partial t},$$

$$\vec{\nabla}\times \vec{B}=\mu_0\left(\vec{J}+\epsilon_0\frac{\partial \vec{E}}{\partial t}\right).$$

But the last two might be written more clearly as

$$\frac{\partial \vec{B}}{\partial t}=-\vec{\nabla}\times \vec{E}$$ and

$$\frac{\partial \vec{E}}{\partial t}=\frac{1}{\epsilon_0}\left(-\vec{J}+\frac{1}{\mu_0}\vec{\nabla}\times \vec{B}\right).$$

And then they clearly tell you how the fields change. And now we know what it takes for the fields to not change. An irrotational electric field causes the magnetic field to be steady in time. A $\vec{B}$ field whose curl is exactly balanced by current, produces a steady electric field. But this doesn't tell us what the fields are. For instance if a particle was at rest its electrostatic field could be irrotational, and there could be no magnetic field (or current), so all the fields are steady. But there could also be a wave travelling through space that hasn't yet reached the particle (so the particle stays at rest ... for now, until the wave gets to it). The locations of the particles and their motions don't by themselves tell us the fields.

For a particle at rest, there is a natural and very simple solution, an inverse square electric field, unchanging, and zero magnetic field. To another observer moving at a constant velocity, they see that charged particle moving at constant velocity. So the fields they see, are a very natural solution (of the many possible) to the equations for a charged particle moving at constant velocity.

That's the most natural solution, and you can compute it. If you try to use the causal equations I listed above, it turns into a chicken and egg problem, the fields now depend on what they were in the past. Which is reasonable, but where do you stop?

Edit for Sofia

In the frame where the charge is at rest at point $\vec{p}$, the charge density could be $\rho(\vec{r},t)=Q\delta^3(\vec{r}-\vec{p})$ and the current density definitely is $\vec{J}(\vec{r},t)=\vec{0}$. (If the charge is extended, $\rho(\vec{r},t)$ could equal $\frac{Q}{4\pi R^3/3}$ if $\left|\vec{r}-\vec{a}\right|<R$ and zero otherwise, as another example.) There are many possible electric fields, but $\vec{E}(\vec{r},t)=\frac{Q}{4\pi \epsilon_0 \left|\vec{r}-\vec{a}\right|^3}\left(\vec{r}-\vec{a}\right)$ (outside teh charge) is a natural and simple one. And again there are many possible $\vec{B}$ fields, but $\vec{B}(\vec{r},t)=\vec{0}$ is a natural and simple one. There the Lorentz-transformed versions of these $\vec{E}$, $\vec{B}$, $\rho$ and $\vec{J}$ are obviously natural and simple solutions, though there are still many. Just as there are many possible solutions for electric fields when there is no charge present. For example, you could have a plane wave in any direction, with any magnitude, as well as countless other solutions.

Edit for Agnivesh Singh

What I assume is that the fields obey Maxwell. The first two Maxwell (about the divergence) are constraints. At any time, those need to hold. The other two (with the time derivatives) are about how the fields evolve (so very relevant to your question). What's nice about all four, is that if the constraints hold at one time, and the fields evolve by the evolution equations, then the constraints will continue to hold. What's unfortunate is twofold. One, that they are about the total fields, not about the field due to this or the field due to that. Second, they don't include boundary conditions, so there is no unique solution until you specify boundary conditions.

If I didn't assume the fields obey Maxwell, then actually it would violate conservation of energy and momentum. But that's because we assign an energy and momentum density to the fields such their flux through vacuum is conserved and that their flux at charges and currents is exactly the the force and power exerted by the fields on the charges through the Lorentz Force Law. So it's a bit cheating, but if you'd rather take the energy density and momentum density of fields as given, then we need the fields to satisfy Maxwell to conserve energy and momentum.

I am more interested in the mechanism . My question was aimed to know how electric fields propagate ?

The fields evolve according the the equations I provided. If by propagation you mean energy and momentum transport, I think that's a different question (meaning a new question is needed, including the research stage, not an edit to this question). The biggest problem is again that there is no unique solution to Maxwell, even with no charges, there are many possible fields. Throw in a charge and there are again still many possible fields, and the equations don't specify a field as being due to a charge, they are just about the total field due to everything.

There are other difficulties to propagation. If you moved your charge to make a large field nearby and then watched that large deviation propagated, it literally is a source free field (like radiation) and so even if it propagates at $c$, you might object because you wanted to know if fields due to charges propagate at $c$. For that you'd have to trace the disturbance all the way back to the charge. And if your charge is a point charge, the fields themselves blow up there, so now you have infinities to deal with. You can try to make that work, but how convincing is it going to be in the end? And if you have an extended charge, then my example of a simple and natural field (due to the charge) isn't really convincing either since it won't look like a sphere in all frames. If it was a sphere in it's own rest frame it will look pancake like in a moving frame. If it is a sphere in the moving frame, it won't look like a sphere in it's own rest frame. You can try a fluid model, but energy and momentum conservation for a nonpoint charge actually fail unless you have something that holds the extended charge together, charged fluids actually physically spread out.

There is also a completely different way to see causality, which is to use Jefimenko's Equations:

$$ \vec{E}(\vec{P},t)=\iiint \frac{\left(c^2\rho(\vec{r},t_r)+c|\vec{P}-\vec{r}|\dot{\rho}(\vec{r},t_r)\right)(\vec{P}-\vec{r})-|\vec{P}-\vec{r}|^2\dot{\vec{J}}(\vec{r},t_r)}{c^2|\vec{P}-\vec{r}|^34\pi\epsilon_0}d^3\tau$$ $$ \vec{B}(\vec{P},t)=\frac{\mu_0}{4\pi}\iiint \frac{\left(c\vec{J}(\vec{r},t_r)+|\vec{P}-\vec{r}|\dot{\vec{J}}(\vec{r},t_r)\right)\times (\vec{P}-\vec{r})}{c|\vec{P}-\vec{r}|^3}d^3\tau.$$

However if you took those as a given, instead of Maxwell, you not only lose some solutions to Maxwell (like a primordial radiation field) but it also begs the question since the fields are explicitly calculated from the charge and currents in the past (at $t_r$) where $t_r=t-\frac{1}{c}\left|\vec{P}-\vec{r}\right|$.

I'm not expecting anything I wrote to make you completely happy. But maybe learning these things allow you to ask (or find and understand) new and more detailed questions to address your concerns. So hopefully you learned as much as possible based on how you phrased your question.

  • $\begingroup$ you forgot to tell us how the charge $\rho(\vec r, t)$ and how the current $\vec J(\vec r, t)$ look like. $\endgroup$
    – Sofia
    Feb 24, 2015 at 23:51
  • $\begingroup$ @Sofia For the particle at rest, the $\vec{J}=\vec{0}$ everywhere and everywhen and the charge density could be a Dirac Delta, also unchanging in time. What I didn't tell you was how to find out the fields the moving observer sees. If you write the electromagnetic field as a rank two tensor, or as a bivector, then either way you can easily find the components in any inertially moving frame. $\endgroup$
    – Timaeus
    Feb 25, 2015 at 3:35
  • $\begingroup$ the point here is that the charge $\rho$ moves. So is it a charge, or a current? This charged is described by $\rho \delta(\vec r - \vec v t)$. The current is probably $\partial \rho / \partial t = \rho \vec v \cdot \nabla_r \delta$. It's not trivial things, they should be mentioned and discussed. You know how $\delta$ and its derivative look like. $\endgroup$
    – Sofia
    Feb 25, 2015 at 3:55
  • $\begingroup$ you have a point from me because your treatment is the correct one, but please be kind a complete the job. $\endgroup$
    – Sofia
    Feb 25, 2015 at 3:58
  • $\begingroup$ @Timaeus : I am more interested in the mechanism . My question was aimed to know how electric fields propagate ? And how to prove that they propagate at the same speed as that of EM radiations. $\endgroup$ Feb 25, 2015 at 8:48

Static Charges and the electric field: In classical electromagnetism, first of all thinking about static charges (so we don't have to worry about time dependence or changing fields) the force between two charges is given by coulombs’ law and is proportional to the magnitude of the two charges and inversely proportional to the square of the distance between the charges. This leads to the idea of an electric field which is the force an infinitesimal unit test charge would feel at a point from a source charge. (It’s a debatable point whether the electric field exists or is just an abstract model to help visualise the interactions and do calculations – so questions about ‘how’ it propagates are philosophical).

Charges with constant velocity: Thinking now about moving charges, we find that to describe the force a charge at a constant velocity (relative to a source charge) experiences we need a magnetic field as well the electric field. This can also be derived theoretically from coulombs law and the principles of special relativity. Though in fact, we only need one field – the electromagnetic field – which is invariant under Lorentz transformation and therefore has the same magnitude for all observers and produces the same magnitude of force on a test particle for all observers. The electric field and magnetic field can be thought of as component parts of the electromagnetic field – and people often find this a convenient way to describe the effects of charges - but they are not Lorentz invariant and different observers (moving at relative velocities) will disagree on the magnitude of the electric and magnetic fields.

Accelerating charges: Thinking about accelerating charges, we imagine that as a charge accelerates the electromagnetic field around it moves with the charge but far from the charge the field does not yet know the charge has moved. This leads to a kink in the electromagnetic field as the message about the moving charge travels to far off places. Part of the kink is called Electromagnetic Radiation (the part inversely proportional to the square of the distance from the charge) and for all observers can be broken down into electric and magnetic field components. Following Maxwell’s equations it can be shown that the EM radiation moves at the speed of light in a vacuum.

[Quantisation: The classical electromagnetic field etc. can be quantised leading to quantum electrodynamics onto ideas of photons / virtual photons etc. but I don’t think this helps answer the question. You could still ask how does a photon or virtual photon propagate.]

In summary TL;DR: There is only the electromagnetic field, changes in magnitude of the electromagnetic field travel at the speed of light (in a vacuum). How / why this happens, and if the electromagnetic field exists at all, are philosophical questions.


In the full quantum field theory treatment, photons are the quanta of the electric field. So electric field lines travel at the speed of light because their carriers travel at the speed of light.

In Maxwell's equations, the speed of light and the propagation of the fields are linked, but not in such an obvious way. First, you can solve Maxwell's equations in vacuum to predict that there are electromagnetic disturbances that travel at speed $\frac{1}{\sqrt{\mu_0\epsilon_0}}$=c.

But then there is a minor crisis. When you use Maxwell's equations to solve for an oscillating or accelerating charge, you find that there is no radiation produced. An accelerating charge produces fields, but they die off too fast as you move away from the charge. This contradicts the real world where things like antenna's can produce signals that can be measured far away. The solution of this was to notice that when you look at the electric and magnetic potentials, you find that they are also solutions to the proper equations when you assume the potentials propagate at the $c$. So at least as far as I understand it, Maxwell's equations do predict both the speed of light and speed of propagating fields, but they are not exactly one in the same.

  • $\begingroup$ The concept of the electromagnetic field is not "replaced by virtual photons". The electromagnetic field simply becomes a quantum field like everything else in the theory, and photons are the quanta of the field. Virtual particles are not actual particles at all, but merely lines in a Feynman diagram. Your first paragraph is completely misleading, the other two are on the right track though, I think. $\endgroup$
    – ACuriousMind
    Feb 24, 2015 at 22:04
  • $\begingroup$ @ACuriousMind Hello, I see that there are answers here with opposing comments, and thus making readers to doubt on answers. I would be glad if you provide a complete and correct answer $\endgroup$
    – user326901
    May 10, 2022 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.