Feynman's Lectures, Volume 2, says that the electromagnetic force is invariant in any reference frame, and the magnetic force in one frame becomes the electric field in another.

And Wikipedia says:

That is, the magnetic field is simply the electric field, as seen in a moving coordinate system.

Can we then say that the magnetic field is just a modified "relativistic" version of the electric field?

  • $\begingroup$ Read this and the links within physics.stackexchange.com/q/24010/104696 $\endgroup$
    – Farcher
    Commented Jun 10, 2018 at 15:54
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    $\begingroup$ No. The electromagnetic field is inherently relativistic. There is no "non-reltivistic version" of the electromagnetic field. The field is created by moving charges. Charges moving in space create a magnetic field. Charges moving in time create an electric field. $\endgroup$
    – safesphere
    Commented Jun 10, 2018 at 15:55
  • $\begingroup$ Dear safesphere, can you please tell me more about "Charges moving in time create an electric field. "? An object that is not moving at speed c, so moving slower is always moving in the time dimension. So a charge (moving slower then c in space) is always moving in time. Anything with rest mass (and charge has to have rest mass), is always like that, so a charge is always creating an electric field? $\endgroup$ Commented Jun 10, 2018 at 16:38
  • $\begingroup$ @safesphere Define "Charges moving in time" . Also, moving charges ("in space" obviously, where else) cause both an electric and a magnetic field. $\endgroup$
    – my2cts
    Commented Jun 10, 2018 at 17:04
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    $\begingroup$ @OP "Feynman ... said that the electromagnetic force is invariant in any reference frame" . It said that the electromagnetic force is Lorentz covariant, that is transforms like a four vector. $\endgroup$
    – my2cts
    Commented Jun 10, 2018 at 17:07

1 Answer 1


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In above Figure-02 an inertial system $\:\mathrm S'\:$ is translated with respect to the inertial system $\:\mathrm S\:$ with constant velocity
\begin{equation} \boldsymbol{\upsilon}=\left(\upsilon_{1},\upsilon_{2},\upsilon_{3}\right)=\left(\upsilon \mathrm n_{1},\upsilon \mathrm n_{2},\upsilon \mathrm n_{3}\right)=\upsilon \mathbf n\,, \qquad \upsilon \in \left(-c,c\right) \tag{01} \end{equation} The Lorentz transformation is \begin{align} \mathbf{x}^{\boldsymbol{\prime}} & = \mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{x})\mathbf{n}-\gamma \boldsymbol{\upsilon}t \tag{02a}\\ t^{\boldsymbol{\prime}} & = \gamma\left(t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{x}}{c^{2}}\right) \tag{02b} \end{align} in differential form \begin{align} \mathrm d\mathbf{x}^{\boldsymbol{\prime}} & = \mathrm d\mathbf{x}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathrm d\mathbf{x})\mathbf{n}-\gamma\boldsymbol{\upsilon}\mathrm dt \tag{03a}\\ \mathrm d t^{\boldsymbol{\prime}} & = \gamma\left(\mathrm d t-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathrm d\mathbf{x}}{c^{2}}\right) \tag{03b} \end{align} and in matrix form \begin{equation} \mathbf{X}^{\boldsymbol{\prime}}= \begin{bmatrix} \mathbf{x}^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ c t^{\boldsymbol{\prime}}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} \end{bmatrix} = \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \begin{bmatrix} \mathbf{x}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ c t\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}} \end{bmatrix} =\mathrm L\mathbf{X} \tag{04} \end{equation} where $\:\mathrm L\:$ the real symmetric $\:4\times 4\:$ matrix \begin{equation} \mathrm L \equiv \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \tag{05} \end{equation} and \begin{equation} \mathbf{n}\mathbf{n}^{\boldsymbol{\top}} = \begin{bmatrix} \mathrm n_{1}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\vphantom{\dfrac{}{}} \end{bmatrix} \begin{bmatrix} \mathrm n_{1} & \mathrm n_{2} & \mathrm n_{3} \end{bmatrix} = \begin{bmatrix} \mathrm n_{1}^{2} & \mathrm n_{1}\mathrm n_{2} & \mathrm n_{1}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{2}\mathrm n_{1} & \mathrm n_{2}^{2} & \mathrm n_{2}\mathrm n_{3}\vphantom{\dfrac{}{}}\\ \mathrm n_{3}\mathrm n_{1} & \mathrm n_{3}\mathrm n_{2} & \mathrm n_{3}^{2}\vphantom{\dfrac{}{}} \end{bmatrix} \tag{06} \end{equation}
a matrix representing the vectorial projection on the direction $\:\mathbf{n}$.

The electromagnetic field is an entity and this is more clear if we take a look at its transformation. So, for the Lorentz transformation (02), the vectors $\:\mathbf{E}\:$ and $\:\mathbf{B}\:$ are transformed as follows \begin{align} \mathbf{E}' & =\gamma \mathbf{E}\!-\!\left(\gamma\!-\!1\right)\left(\mathbf{E}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}+\,\gamma\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{B}\right) \tag{07a}\\ \mathbf{B}' & = \gamma \mathbf{B}\!-\!\left(\gamma\!-\!1\right)\left(\mathbf{B}\boldsymbol{\cdot}\mathbf{n}\right)\mathbf{n}\!-\!\dfrac{\gamma}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right) \tag{07b} \end{align} Now, let the Lorentz force 3-vector on a point particle with charge $\:q\:$ moving with velocity $\:\mathbf{u}\:$ with respect to $\:\mathrm S\:$ \begin{equation} \mathbf{f} = q\left(\mathbf{E}+\mathbf{u}\boldsymbol{\times}\mathbf{B}\right) \tag{08} \end{equation} This Lorentz force 3-vector with respect to $\:\mathrm S'\:$ is \begin{equation} \mathbf{f'} = q\left(\mathbf{E'}+\mathbf{u'}\boldsymbol{\times}\mathbf{B'}\right) \tag{09} \end{equation} Note that the value of the charge $\:q\:$ of a point particle is by hypothesis the same in all inertial systems (a scalar invariant), while the velocity 3-vector $\:\mathbf{u}\:$ is transformed as follows \begin{equation} \mathbf{u}^{\boldsymbol{\prime}} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{u})\mathbf{n}-\gamma \boldsymbol{\upsilon}}{\gamma \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot} \mathbf{u}}{c^{2}}\right)} \tag{10} \end{equation} equation proved by dividing equations (03a), (03b) side by side and setting $\:\mathbf{u}\equiv \mathrm d\mathbf{x}/\mathrm d t\:$, $\:\mathbf{u'}\equiv \mathrm d\mathbf{x'}/\mathrm d t'$.

Now, if in (09) we replace $\:\mathbf{E'},\mathbf{B'},\mathbf{u'}\:$ by their expressions (07a),(07b) and (10) respectively, then we end up with the following relation between the force 3-vectors
\begin{equation} \mathbf{f}^{\boldsymbol{\prime}} = \dfrac{\mathbf{f}+(\gamma-1)(\mathbf{n}\boldsymbol{\cdot} \mathbf{f})\mathbf{n}-\gamma \boldsymbol{\upsilon}\left(\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c^{2}}\right)}{\gamma \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot}\mathbf{u}}{c^{2}}\right)} \tag{11} \end{equation} wherein the quantities of the electromagnetic field $\:\color{red}{\bf DISAPPEARED !!!}$

That's why in the early years of Special Relativity transformation (11) was believed to be valid for any force at least of the same type as the Lorentz force (more exactly for any force that doesn't change the rest mass of the particle).
Following the same path by which we construct from (10) the velocity 4-vector $\:\mathbf{U}\:$ \begin{equation} \mathbf{U} =\left(\gamma_{\mathrm u}\mathbf{u}, \gamma_{\mathrm u}c\right) \tag{12} \end{equation} we construct also from (11) the force 4-vector $\:\mathbf{F}\:$ \begin{equation} \mathbf{F} =\left(\gamma_{\mathrm u}\mathbf{f}, \gamma_{\mathrm u}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c}\right) \tag{13} \end{equation} Lorentz transformed \begin{equation} \mathbf{F'} = \mathrm L \mathbf{F} \tag{14} \end{equation} or \begin{equation} \mathbf{F}^{\boldsymbol{\prime}}= \begin{bmatrix} \gamma_{\mathrm u'}\mathbf{f'}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ \gamma_{\mathrm u'}\dfrac{\mathbf{f'}\boldsymbol{\cdot}\mathbf{u'}}{c} \end{bmatrix} = \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \begin{bmatrix} \gamma_{\mathrm u}\mathbf{f}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ \gamma_{\mathrm u}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c} \end{bmatrix} =\mathrm L\mathbf{F} \tag{15} \end{equation}

  • $\begingroup$ Maybe worth mentioning that ${{E}^{2}}-{{B}^{2}}$ is an invariant. If it’s positive, you can find the transformation that makes B from E. If it’s negative, vice versa. $\endgroup$ Commented Jun 12, 2018 at 11:13
  • $\begingroup$ @Bert Barrois I apologize but I don't understand what do you mean with your comment. Please clarify : from one or both Lorentz invariants $\:\left(\left|\!\left|\mathbf{E}\right|\!\right|^{2}-c^{2}\left|\!\left|\mathbf{B}\right|\!\right|^{2}\right)\:$ and $\:\left(\mathbf{E}\boldsymbol{\cdot}\mathbf{B}\right)\:$ we could produce one or both transformation equations (07) ??? $\endgroup$
    – Frobenius
    Commented Jun 12, 2018 at 14:38
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    $\begingroup$ There’s absolutely nothing wrong with your answer. All I’m saying is that the necessary & sufficient conditions for existence of a transformation of an arbitrary combination of E- and B-fields to/from a pure E-field are: $E\cdot B=0$ AND ${{E}^{2}}-{{B}^{2}}>0$. (Vice versa if B is stronger than E.) The OP had seen a statement which seemed to say that the B-field is nothing but a transformed E-field, but that’s not always possible. $\endgroup$ Commented Jun 13, 2018 at 11:32
  • $\begingroup$ love the diagram; are you using Inkscape? $\endgroup$ Commented Jun 21, 2018 at 18:14
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    $\begingroup$ @Larry Harson : Thanks. I use GeoGebra. It's free. $\endgroup$
    – Frobenius
    Commented Jun 21, 2018 at 20:47

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