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Faraday's law states that "Any change in electric field induces a magnetic field and vice versa". I don't see exactly where these fields are induced, but I assume that these fields are induced at each and every point in space.

Now, let us consider a charge Q, moving with a uniform Velocity V. Obviously this movement of the charge creates a change in the electric field everywhere. As such it should induce a magnetic field. Subsequently, this change in the magnetic field should also bring about an electric field, an induced one, owing to further changes induced in the magnetic field.

Apparently this machinery seems to be similar to an EM radiation. But we know that an EM radiation can only be produced by charges in accelerated movement. Why this chain of events can't be seen as an EM wave propagation?

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  • $\begingroup$ possible duplicate of The propagation of electric field $\endgroup$ – Sofia Feb 24 '15 at 23:59
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    $\begingroup$ @Sofia: how is that a duplicate? $\endgroup$ – Kyle Kanos Feb 25 '15 at 3:08
  • $\begingroup$ @KyleKanos it's the same content: what fields are produced by a moving charge. Please, leave aside the title - the problem is the same. And Timaeus explained the issue by Maxwell's equations, just I insisted that he expresses more clearly the charge density, that in fact should be $q \delta (\vec r - \vec v t)$. I hope to find him tomorrow, or I'll leave some clarifying comments. $\endgroup$ – Sofia Feb 25 '15 at 3:14
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    $\begingroup$ @Sofia: I'm not sure that I would call them duplicates. Your link asks about propagation speeds while this out asks if the time-varying fields make for EM waves. Related, but I don't believe it's a duplicate. $\endgroup$ – Kyle Kanos Feb 25 '15 at 3:18
  • $\begingroup$ @KyleKanos This is the question I quote "In case of a charged particle . . . travelling at uniform velocity, the el. field due to it at a given point doesn't change instantaneously. The reason for this delay . . . can be explained as the time the electric field needs to propagate to that point, . . . . I would like to know what exactly does it mean by a propagating electric field, and how exactly does it propagate ?" Well, what he/she asks in a wrong way, the question is how looks like the generated e.m. field. The speed of a e.m. field is of course $c$. (I continue.) $\endgroup$ – Sofia Feb 25 '15 at 3:41
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I am trying here to answer your comment "I guess, if fields are undergoing a wave-like change in one frame of reference, they must undergo the same kind of change in any other frame of reference."

It's a legitimate remark and I would formulate it even more clearly: in a frame of reference where the charge is at rest it doesn't emit e.m. waves. But it can't be possible that in one frame we have an e.m. wave, and in another one we haven't.

Now, let me show you that although the e.m. field produced by the moving charge is time-dependent, it isn't similar to the e.m. waves which travel with the light velocity. For avoiding solving the equations in my previous answer, I'll use another procedure here. I will consider two frames of coordinate. A frame $O$ where the charge is at rest and in the origin of coordinates, s.t. we have only the Coulomb field,

$$\vec E = \frac {Q}{4\pi\epsilon_0} \frac {\vec r}{r^3}, \ \ \text {and} \ \ \vec B = 0, \tag{i}$$

and a frame $O'$ moving with a velocity $v$ in the direction $-z$. That means, in the frame $O'$ the charge moves with the velocity $v$. Now, we can apply the special relativity transformations for the space and time

$$\begin{cases} t = \gamma \left(t' + \frac {\vec r' \cdot \vec v}{c^2}\right) \\ \vec {r} = r'_{\perp} + \gamma \left(r'_{\|} + \vec v t' \right). \end{cases} \tag{ii}$$

For the electric and magnetic field we have, by taking in consideration that in $O$ there is no magnetic field,

$$\begin{cases} \vec {E'} = \gamma \vec E + (1 - \gamma) \frac {\vec E \cdot \vec v}{v^2} \vec v \\ \vec {B'} = -\gamma \frac {\vec v \times \vec E}{c^2}. \end{cases} \tag{iii}$$

Let's introduce $\text {(i)}$ in $\text {(iii)}$.

$$\begin{cases} \vec {E'} = \gamma \frac {Q}{4\pi\epsilon_0} \left[\frac {\vec r}{r^3} + (\gamma ^{-1} - 1) \frac {r_{\|}}{r^3} \frac {\vec v}{v}\right] \\ \vec {B'} = -\gamma \frac {Q \mu_0}{4\pi} \ \vec v \times \frac {\vec r}{r^3}. \end{cases} \tag{iv}$$

The final thing is to replace $\vec r$ by the expression in $\text {(ii)}$,

$$\begin{cases} \vec {E'} = \gamma \frac {Q}{4\pi\epsilon_0} \frac {\vec {r'_{\perp}} + \left[\gamma + (1 - \gamma)\frac {\vec v}{v}\right]\left(\vec {r'_{\|}} + \vec v t' \right)}{\left[ r'^2_{\perp} + \gamma^2 \left(r'_{\|} + v t' \right)^2\right]^{3/2}} \\ \vec {B'} = -\gamma \frac {Q \mu_0}{4\pi} \ \frac {\vec v \times \vec {r'_{\perp}}}{\left[r'^2_{\perp} + \gamma^2 \left(r'_{\|} + v t' \right)^2\right]^{3/2}}. \end{cases} \tag{v}$$

As one can see, these are not travelling waves of type ~ $e^{i(kr - \omega t)}$ with $\omega = ck$. The formulas $\text {(v)}$ represent just a Coulomb field, that in the frame $O'$ appears as dragged after the charge $Q$, however, when expressing the position dependence of the field in terms of the coordinates of the frame $O'$ according to the special relativity, dependence of time enters into play. And, as we know, the moving charge produces also a magnetic field. But we have no e.m. waves.

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  • $\begingroup$ Could you tell me what do you mean by "dragging a field " ? $\endgroup$ – Agnivesh Singh Mar 2 '15 at 18:16
  • $\begingroup$ :I guess you have posted your views on another question of mine under this question .Could you transfer it to there ? $\endgroup$ – Agnivesh Singh Mar 2 '15 at 18:24
  • $\begingroup$ @AgniveshSingh "dragging the field"? Thank you for asking. I introduced more suitable words. Indeed the Coulomb field moves together with the charge, but since we have to express its dependence of coordinates in terms of the coordinates in $O'$ times dependence appears as requires the relativity. $\endgroup$ – Sofia Mar 2 '15 at 18:45
  • $\begingroup$ @AgniveshSingh what is the question where to you think that I should move my answer? Give me the site. $\endgroup$ – Sofia Mar 2 '15 at 18:47
  • $\begingroup$ physics.stackexchange.com/questions/167480/… $\endgroup$ – Agnivesh Singh Mar 2 '15 at 18:49
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I might elaborate upon Sofia's excellent answer in two respects.

Not all changing EM fields are wave like. Maxwell's equations have lots of solutions involving changing E and M fields. These can be broken down into two components, the "near field" and the "far field". The difference is that the far field components have a classical wave solution, and hence fall off according to a square law rule. The "near field" is changing E and M components which do not propagate. A charge moving at a constant speed produces only near field solutions - yes there are E and M fields, but they don't propagate and hence do not take energy away from the system.

Secondly, as others have alluded to, magnetism is not considerd a "real" force in Special Relativity. This was spelled out by Einstein in his original paper on SR, hence its title "On the electrodynamics of moving bodies". Consider a reference frame where the charged particle is at rest. Clearly it can't emit radiation. But as the choice of reference frame is arbitrary, a particle moving at constant speed cannot emit radiation. Einstein argued that you only see a magnetic field around a charged particle if it is moving relative to the observer. But all observers should see the same thing. He concluded (and demonstrated) that the magnetic field is a "virtual force" (like centrifugal force) that derives from your choice of what you consider "at rest", an artefact of your coordinate system. This is why the magnetic field exists in some frames of reference but not in others - in SR, the magnetic field doesn't really exist at all, it is a consequence of relativistic effects applied to electric fields. The net result is that all observers see the same thing; observers at rest interpret the repulsion of fixed charges as due to electric fields, observers in motion interpret this as a combination of electric and magnetic fields, and the numbers always end up being the same (as Einstein showed). Its just as easy to assume magnetism doesn't exist at all, and it is simply a virtual force. This may help explain why different observers see different strengths of magnetic fields around moving particles - the magnetic field is a computational byproduct of measuring the electric field in different reference frames.

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