I was trying to show that the field transformation equations do hold when considering electric and magnetic fields as 4-vectors. To start off, I obtained the temporal and spatial components of $E^{\alpha}$ and $B^{\alpha}$. The expressions are obtained from the following equations:

$$E^{\alpha}=F^{\alpha\beta}U_{\beta},\: B^{\alpha}=\frac{1}{2c}\epsilon^{\alpha\beta\mu\nu}F_{\beta\mu}U_{\nu}$$

I obtained: \begin{align*} E^{0}&=F^{00}U_{0}+F^{0i}U_{i}=\frac{\gamma(u)}{c}\left(\vec{E}\cdot\vec{u}\right)\\ E^{i}&=F^{i0}U_{0}+F^{ij}U_{j}=\gamma(u)\left[\vec{E}+\left(\vec{u}\times\vec{B}\right)\right]^{i}\\ B^{0}&=\frac{1}{2c}\epsilon^{0\beta\mu\nu}F_{\beta\mu}U_{\nu}=\frac{\gamma(u)}{c}\left(\vec{B}\cdot\vec{u}\right)\\ B^{i}&=\frac{1}{2c}\epsilon^{i\beta\mu 0}F_{\beta\mu}U_{0}+\frac{1}{2c}\epsilon^{i\beta\mu j}F^{\beta\mu}U_{j}=\gamma(u)\left[\vec{B}-\frac{\vec{u}}{c^2}\times\vec{E}\right]^{i} \end{align*}

I interpreted the above components as that of fields observed by a stationary observer. To show that the fields transform correctly I have to show that: $$\vec E' = \gamma \left( \vec E + c\vec \beta \times \vec B\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec E \right )$$ $$\vec B' = \gamma \left( \vec B - \frac{\vec \beta}{c} \times \vec E\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec B \right )$$

i.e. I have to show that I am able to construct the RHS from the components I have found. However, I do not seem to be able to show that using the Lorentz transformation equations under general boost. The Lorentz transformation equations under general boost is given as: $$A^{'0}=\gamma\left(A^0-\vec{\beta}\cdot\vec{A}\right)$$ $$\vec{A}'_{\parallel}=\gamma\left(\vec{A}_{\parallel}-\vec{\beta}A^0\right)$$ $$\vec{A}'_{\perp}=\vec{A}_{\perp}$$

How should I proceed?

  • 1
    $\begingroup$ I think what I wanted to ask was assuming that the first pair of equations are correct, how do I show that the electric and magnetic fields obtained transform correctly under general Lorentz boost. $\endgroup$
    – user161157
    Commented Mar 27, 2021 at 13:45
  • 3
    $\begingroup$ Perhaps I am missing something, but electric and magnetic fields are not four-vectors, so the title makes no sense to me. Also, what is $E^0$ and $B^0$? Also bear in mind that $\epsilon^{\alpha\beta\gamma\eta}$ is not a tensor. It is a relative tensor. To make it into a proper tensor you need to multiply (or divide?) it by \sqrt{g}, where $g$ is the determinant of the metric tensor (does not matter if all your transformations have unity Jacobians, of course) $\endgroup$
    – Cryo
    Commented Mar 28, 2021 at 1:50
  • $\begingroup$ First, can you show the transformation for the special case of relative motion along the x-direction? If so, you should add this to what you "obtained" in your question. $\endgroup$
    – robphy
    Commented Mar 29, 2021 at 12:05
  • 1
    $\begingroup$ @Cryo The expression $E^{\alpha}=F^{\alpha\beta}U_{\beta}$ is a U-dependent 4-vector that is orthogonal to U. (All frames will agree that this is a U-dependent 4-vector that is orthogonal to U.) In U's frame, these are the 3 spatial components of the electric field according to U (since the time-component of this U-dependent 4-vector according to U is zero). $E^0$ is the lab-frame's time-component of this U-dependent 4-vector. (U would obtain a similar $E'^0$ for $E'^{\alpha}=F^{\alpha\beta}V_{\beta}$ where $V^{\beta}$ is the lab-frame's 4-velocity.) $\endgroup$
    – robphy
    Commented Mar 29, 2021 at 12:15
  • $\begingroup$ @robphy, happy to accept that $F^{\alpha\beta}U_\beta$ is a four-vector, provided $U$ is a four-vector. What is $U$? Four-velocity of some kind? In any case, your definition ($F^{\alpha\beta}U\beta$) is not the electric field. Are you trying to define a new kind of electric field? If so, then you can make it a four-vector by definition. If not, then how do you define the electric field? Maxwell's equations are for 3d vectors, anything beyond this extra definitions. $\endgroup$
    – Cryo
    Commented Mar 29, 2021 at 13:59

2 Answers 2


UPDATE: For clarity, here are some references relevant to the OP's question.
There may be differences in conventions among the references.

(Readers unfamiliar with this spacetime viewpoint may misinterpret the title of the OP's question.)

Robert Geroch's General Relativity 1972 Lecture Notes (p. 53-54):
An electromagnetic field is a (smooth) antisymmetric tensor field, $F_{ab} (= F_{[ab]})$, on space-time $M$. An individual observer resolves this single object – the electromagnetic field – into the separate electric and magnetic fields seen by him. Our first task is to see how this resolution comes about. Let our observer have four-velocity $\xi^a$. Recall that $h_{ab}= g_{ab} + \xi_a\xi_b$ is the projection operator orthogonal to $\xi^a$.
To resolve $F_{ab}$ into "spatial tensors" for our observer, we project the indices of $F_{ab}$ parallel and orthogonal to $\xi^a$. Thus, we obtain four tensors: $F_{mn}\xi^m\xi^n$, $F_{mn}\xi^m h^n{}_a$, $F_{mn} h^m{}_a\xi^n$, and $F_{mn} h^m{}_a h^n{}_b$. Since $F_{ab}$ is antisymmetric, the first vanishes, and the second equals minus the third. The third is called the electric field $$E_a=F_{ab}\xi^b.\qquad(20)$$ (Note that $F_{am}\xi^m =F_{mn}\xi^m h^n{}_a$.) Note that the determination of an electric field from the electromagnetic field involves a choice of observer (i.e., of his four-velocity). The remaining piece of $F_{ab}$ is $F_{mn}h^m{}_a h^n{}_b$, a spatial, antisymmetric tensor.
The spatial vector which gives $F_{mn}h^m{}_a h^n{}_b$ is thus $$B_a =\frac{1}{2} \epsilon_{abcd} \xi^b F^{cd}.\qquad (21)$$ This $B_a$ is called the magnetic field. To summarize, the whole story is described by a single tensor field, the electromagnetic field. An observer resolves this $F_{ab}$ using his four-velocity, into a pair of vectors, $E_a$ and $B_a$, orthogonal to his four-velocity. The $E_a$ and $B_a$ depend on the observer. (In terms of components, the six independent components of $F_{ab}$ become three in $E_a$ and three in $B_a$.)

Robert Wald's General Relativity, (p. 64):
"For an observer moving with 4-velocity $v^a$, the quantity $E_a=F_{ab}v^b$ (4.2.21) is interpreted as the electric field measured by that observer, while $B_a = -\frac{1}{2} \epsilon_{ab}{}^{cd}F_{cd}v^b$ (4.2.22) is interpreted as the magnetic field, where $\epsilon_{abcd}$ is the totally antisymmetric tensor of positive orientation with norm $\epsilon_{abcd}\epsilon^{ abcd}=-24$ (see appendix B) so that in a right-handed orthonormal basis we have $\epsilon_{0123}=1$. "

Sachs & Wu General Relativity for Mathematicians (p. 75):
"Let $F$ be an electromagnetic field on $M$, and $(z, Z)$ be an instantaneous observer;
Since $F$ is antisymmetric, $\tilde FZ \in Z^\bot$. $E = \tilde FZ$ is defined as the electric vector $(z, Z)$ measures for $F$. It is kosher to imagine $(z, Z)$ measuring the electric vector in $Z^\bot$ by essentially Newtonian methods--for example, using a "test charge" (Alonso-Finn [2]). Now by exterior algebra there is a unique vector $B\in Z^\bot$ such that $4! \Omega(X, Y, B, Z) = F(X, Y), \ \forall X,Y \in Z^\bot$,
where $\Omega$ is the metric volume element (cf. the discussion of Hodge duality in Bishop-Goldberg). $B$ is defined as the magnetic vector $(z, Z)$ measures for $F$."

DeFelice Relativity on Curved Manifolds Eq 9.9.4 (p.299):
Let us now define (Misner et al., 1973):
$E^j = F^j{}_r u^r$; (9.9.4)
from the skew-symmetry of $F^i{}_j$ it follows that:
$E^j u_j=0 \qquad h^i{}_j E^j = E^i$ . (9.9.5)

See also (implicitly) Misner,Thorne,& Wheeler Box 3.1(p.72) and Ex 3.6 (p.78).

These are spacetime-formulations (as opposed to 3-vector formulations in a spatial slice) of the "observer-dependent electric and magnetic parts" of the electromagnetic field.

Some suggestions:

  • For simplicity, start with simple case where $\vec \beta=\beta \hat x$
  • Unwrap $$\vec E' = \gamma \left( \vec E + c\vec \beta \times \vec B\right) - \frac{\gamma^2}{\gamma +1} \vec \beta \left( \vec\beta \cdot \vec E \right )$$ by writing this in components:
    "$\vec E'$-parallel to the relative-spatial-velocity" (say, along the x-axis) and
    "$\vec E'$-perpendicular to the relative-spatial-velocity" (in the yz-plane).
    By doing this,
    you will see why the weird term involving $\gamma^2\beta^2/(\gamma+1)$ appears [which can also be written as $(\gamma-1)$].
  • Your goal should be to write
    "the lab-(xyz)-components-of-the-lab's Electric and Magnetic Fields"
    in terms of
    "the traveler's-(xyz)-components-of-the-travelers's Electric and Magnetic Fields"
  • After seeing the situation in this simple case for $\vec \beta=\beta \hat x$,
    try to generalize the result... then re-package in the 3-velocity-form.
  • Rather than use "transformation laws" in component-form (which may lead to one to messing up with accounting), I found it useful to express the transformation in terms of one 4-velocity (a unit vector) in terms of another 4-velocity... then take dot-products as needed.... only interpreting the result in terms of 3-vectors at the end.

[I worked in terms of $F_{ab}$ and 4-vectors as much as possible... in the spirit of Minkowskian Spacetime geometry.

It would be good to specify your signature conventions.
I expressed "components" by using the dot-products of two 4-vectors: the 4-vector I want to study, with the basis unit-4-vectors associated with the observer.]


If $\:\mathbf{f}\:$ is a pure force 3-vector applied on a particle of velocity 3-vector $\:\mathbf{u}\:$, then the 4-dimensional vector \begin{equation} \mathbf{F}\boldsymbol{=}\gamma_{\mathrm u}\left(\mathbf{f}\:,\:\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c} \right) \tag{01}\label{01} \end{equation} is a Lorentz 4-vector. We use the term $''$pure force$''$ for a force which doesn't change the rest mass of the particle.

The Lorentz force per unit charge 3-vector (a pure force) \begin{equation} \mathbf{f}_{_{\mathrm L}}\boldsymbol{=}\mathbf{E}\boldsymbol{+}\mathbf{u}\boldsymbol{\times}\mathbf{B} \tag{02}\label{02} \end{equation} is represented by the Lorentz 4-vector \begin{equation} \mathbf{F}_{_{\mathrm L}}\boldsymbol{=}\gamma_{\mathrm u}\left(\mathbf{f}_{_{\mathrm L}}\:,\:\dfrac{\mathbf{f}_{_{\mathrm L}}\boldsymbol{\cdot}\mathbf{u}}{c} \right)\boldsymbol{=}\gamma_{\mathrm u}\left(\mathbf{E}\boldsymbol{+}\mathbf{u}\boldsymbol{\times}\mathbf{B}\:,\:\dfrac{\mathbf{E}\boldsymbol{\cdot}\mathbf{u}}{c}\right) \tag{03}\label{03} \end{equation} Under the duality transformation of the electromagnetic field produced by the replacements \begin{equation} \begin{matrix} \hphantom{c}\mathbf{E}&\boldsymbol{-\!-\!\!\!\longrightarrow}&\boldsymbol{-}c\mathbf{\tilde{B}}\\ c\mathbf{B}&\boldsymbol{-\!-\!\!\!\longrightarrow}&\hphantom{\boldsymbol{-}c}\mathbf{\tilde{E}} \end{matrix} \tag{04}\label{04} \end{equation} the $''$Lorentz force$''$ of the dual field \begin{equation} \mathbf{\tilde{F}}_{_{\mathrm L}}\boldsymbol{=}\gamma_{\mathrm u}\left(\boldsymbol{-}c\mathbf{\tilde{B}}\boldsymbol{+}\mathbf{u}\boldsymbol{\times}\dfrac{\mathbf{\tilde{E}}}{c}\:,\:\boldsymbol{-}\mathbf{\tilde{B}}\boldsymbol{\cdot}\mathbf{u}\right) \tag{05}\label{05} \end{equation} is of course a Lorentz 4-vector.

I find the terms $''$electric field measured by that observer$''$ and $''$magnetic field$''$ for $\mathbf{F}_{_{\mathrm L}},\mathbf{\tilde{F}}_{_{\mathrm L}}$ inappropriate.


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