electromagnetism - Are magnetic fields just modified relativistic electric fields?

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?

Answer

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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 $$\begin{equation} \mathbf{F}^{\boldsymbol{\prime}}= \begin{bmatrix} \gamma_{\mathrm u'}\mathbf{f'}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ \gamma_{\mathrm u'}\dfrac{\mathbf{f'}\boldsymbol{\cdot}\mathbf{u'}}{c} \end{bmatrix} = \begin{bmatrix} \mathrm I+(\gamma-1)\mathbf{n}\mathbf{n}^{\boldsymbol{\top}} & -\dfrac{\gamma\boldsymbol{\upsilon}}{c} \vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ -\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c} & \hphantom{-}\gamma \end{bmatrix} \begin{bmatrix} \gamma_{\mathrm u}\mathbf{f}\vphantom{\dfrac{\gamma\boldsymbol{\upsilon}^{\boldsymbol{\top}}}{c}}\\ \gamma_{\mathrm u}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c} \end{bmatrix} =\mathrm L\mathbf{F} \tag{15} \end{equation}$$