Suppose an electron is moving through empty space at speed v.

It produces an electric field because it is a charge. But this field changes as it moves. Changing electric field must give rise to magnetic field.

However, the moving electron also constitutes a transient current through various points in space. This must produce some more magnetic field curling around the electron as it moves.

So, the magnetic field here is due to both changing electric fields and currents. Is this true?

  • 1
    $\begingroup$ A moving point charge cannot build a sustained magnetic field at a point. What an observer sees as electric or magnetic field depends on his frame of reference. $\endgroup$ – UKH Aug 19 '17 at 4:04

The electric $\:\mathbf{E}\:$ and magnetic $\:\mathbf{B}\:$ parts of the electromagnetic field produced by a moving charge $\:q\:$ are :


From Jackson's 'Classical Electrodynamics', 3rd Edition, equations (14.14) and (14.13)

\begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{E}(\mathbf{x},t) & = \frac{q}{4\pi\epsilon_0}\left[\frac{\mathbf{n}-\boldsymbol{\beta}}{\gamma^2(1 - \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R^2} \right]_{\mathrm{ret}} + \frac{q}{4\pi}\sqrt{\frac{\mu_0}{\epsilon_0}}\left[\frac{\mathbf{n}\boldsymbol{\times}\left[(\mathbf{n}-\boldsymbol{\beta})\boldsymbol{\times} \dot{\boldsymbol{\beta}}\right]}{(1 - \boldsymbol{\beta}\boldsymbol{\cdot}\mathbf{n})^3 R}\right]_{\mathrm{ret}} \tag{14.14}\\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf{B}(\mathbf{x},t) & = \left[\mathbf{n}\boldsymbol{\times}\mathbf{E}\right]_{\mathrm{ret}} \tag{14.13} \end{align} where \begin{align} \boldsymbol{\beta} & = \dfrac{\boldsymbol{\upsilon}}{c},\quad \beta=\dfrac{\upsilon}{c}, \quad \gamma= \left(1-\beta^{2}\right)^{-\frac12} \tag{01a}\\ \dot{\boldsymbol{\beta}} & = \dfrac{\dot{\boldsymbol{\upsilon}}}{c}=\dfrac{\mathbf{a}}{c} \tag{01b}\\ \mathbf{n} & = \dfrac{\mathbf{R}}{\Vert\mathbf{R}\Vert}=\dfrac{\mathbf{R}}{R}\equiv\dfrac{\mathbf{r'}}{r'}=\dfrac{\mathbf{r'}}{\Vert\mathbf{r'}\Vert}\equiv \mathbf{e}_{r'} \tag{01c} \end{align}


From Feynman's Lectures, Volume II 'Mainly Electromagnetism and Matter', New Millennium Edition, equations (21.1) \begin{align} \mathbf{E} & = \frac{q}{4 \pi \epsilon_0} \left[ \frac{ \mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'} }{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right] \tag{21.1-Feynman}\\ c\mathbf{B} & = \mathbf{e}_{r'}\boldsymbol{\times}\mathbf{E} \nonumber \end{align}

Without looking at the details it seems from the second equations, those giving $\:\mathbf{B}\:$ from $\:\mathbf{E}\:$, that the former is produced by the latter. But this is not the case : the electromagnetic field is an entity and its separation to the electric and magnetic parts depends upon the observer's inertial frame of reference.

Note that these equations are derived from the so called Liénard-Wiechert potentials which in turn are produced from the Maxwell's equations in empty space
\begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \tag{02a}\\ \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} & = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{02b}\\ \nabla \boldsymbol{\cdot} \mathbf{E} & = \frac{\rho}{\epsilon_{0}} \tag{02c}\\ \nabla \boldsymbol{\cdot}\mathbf{B}& = 0 \tag{02d} \end{align} with electric charge and electric charge current densities \begin{align} \rho(\mathbf{x},t) & = q\cdot \delta\left(\mathbf{x}-\mathbf{x}_{q}\right) \tag{03a}\\ \mathbf{j}(\mathbf{x},t) & = q\cdot \delta\left(\mathbf{x}-\mathbf{x}_{q}\right)\cdot\dfrac{\mathrm d\mathbf{x}_{q}}{\mathrm d t} \tag{03b} \end{align}

The fact that the electromagnetic field is an entity is more clear if we take a look at its transformation. So let $\:\mathrm{S}\:$ and $\:\mathrm{S}'\:$ two inertial frames of reference and the frame $\:\mathrm{S}'\:$ moves uniformly with velocity $\:\mathbf{v}=\upsilon\mathbf{n}, \upsilon \in \left(-c,+c\right)\:$ with respect to $\:\mathrm{S}$, see Figure in the bottom. Then from $\:\mathrm{S}\:$ to $\:\mathrm{S}'$ \begin{align} \mathbf{E}' & =\gamma \mathbf{E}\!-\!\left(\gamma\!-\!1\right)\left(\mathbf{E}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}+\,\dfrac{\gamma}{c}\left(\mathbf{v}\boldsymbol{\times}c\mathbf{B}\right) \tag{04a}\\ c \mathbf{B}' & = \gamma c\mathbf{B}\!-\!\left(\gamma\!-\!1\right)\left(c\mathbf{B}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\!-\!\dfrac{\gamma}{c}\left(\mathbf{v}\boldsymbol{\times}\mathbf{E}\right) \tag{04b} \end{align} From equations (04) the six scalar components bind together more strongly in the so-called electromagnetic field tensor \begin{equation} F=F^{\mu\nu} \begin{bmatrix} \hphantom{-}0 & -E_1 & -E_2 & -E_3 \hphantom{-}\vphantom{\dfrac12}\\ \hphantom{-}E_1 & \hphantom{-} 0 & -cB_3 & \hphantom{-} cB_2 \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E_2 & \hphantom{-} cB_3 & \hphantom{-} 0 & -cB_1 \hphantom{-}\vphantom{\dfrac12} \\ \hphantom{-}E_3 & -cB_2 & \hphantom{-} cB_1 & \hphantom{-} 0 \hphantom{-}\vphantom{\dfrac12} \end{bmatrix} \tag{05} \end{equation} transformed under the Lorentz Transformation $\:\Lambda\:$ as follows \begin{equation} F'=\Lambda F \Lambda \tag{06} \end{equation} or by tensors as \begin{equation} F'^{\alpha\beta}=\Lambda^{\alpha}_{\mu} F^{\mu\nu}\Lambda^{\beta}_{\nu} \tag{07} \end{equation}

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The magnetic field is not "due to" the changing electric field as in there is a causal effect based on a fundamental physics law. Nor is the magnetic field "due to" the current.

In the frame-of-reference of the electron (i.e., moving along with the electron at exactly its speed, so the electron seems to be at rest from that viewpoint), there is only the static electric field, no moving charge, no changing electric field, no current.

The magnetic field is how an observer from another frame-of-reference (for example the laboratory in which the electron is seen to be moving at speed v) observes the charge together with its electric field due to Einstein's special relativity.

Refer to this video on Youtube: "How Special Relativity Makes Magnets Work"

  • $\begingroup$ Ok. But my question still holds. Just because magnetic fields are present only in one frame of reference doesn't mean we should not talk about them, ryt!!! The lab frame is also a valid frame and I need to know the form of electric and magnetic fields in the lab frame $\endgroup$ – PhyEnthusiast Aug 28 '17 at 14:15
  • $\begingroup$ What you can do is calculate the electric field in the frame-of-reference of the electron and then perform a Lorentz transformation (see en.wikipedia.org/wiki/Lorentz_transformation and en.wikipedia.org/wiki/…) into the lab frame-of-reference. Notice however that both the electric and the magnetic fields in the lab frame-of-reference will be time-dependent fields (dynamic as opposed to static) as the electron is moving in that frame-of-reference. $\endgroup$ – Beat Nideröst Aug 30 '17 at 14:31

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