Electric Field seen by an observer in motion

[aliens123]

The correct statements of what the Lorentz transformation does are:

$$(E_G)_\mu \mapsto (E_G)'_\mu$$

$$(E_H)_\mu \mapsto (E_H)'_\mu$$

And, for completeness:

$$F_{\mu \nu} \mapsto F'_{\mu \nu}$$

$$(v_G)^\nu \mapsto (v_G)'^\nu$$

$$(v_H)^\nu \mapsto (v_H)'^\nu$$

In other words: a Lorentz transformation doesn't change which object you are dealing with. It just transforms the components of that object from one frame to another. But going from $E_G$ to $E_H$ changes which object you are dealing with. That is simply a different thing from a Lorentz transformation.

If this is the case, then why isn't the Lorentz transformation given in this link:
https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Related by a "simple" Lorentz transformation? That is, $(E_G)_\mu \mapsto (E_G)'_\mu$
Is just
$$(E_G)_\mu = (E_G)'_\alpha \Lambda_{\mu}^{\ \ \alpha}$$
But this is not what the Lorentz transformation given in the link is. Or, in other words, what should we call
$$F_{\mu 0} \mapsto F'_{\mu 0}$$
This is what I was calling the "Lorentz transformation." But this is equivalent to
$$(E_G)_\mu \mapsto (E_H)'_\mu$$

 

[aliens123]

• The Lorentz transformations for the electric field relates
  the components of $\uE$ according to $u^a$
  with
  the components of $\vE$ according to $v^a$.
• This is different from finding
  the components of $\uE$ according to $v^a$
  and
  the components of $\vE$ according to $u^a$.

This doesn't seem to agree with PeterDonis. He is saying the Lorentz transformation relates
the components of $\uE$ according to $u^a$
with
the components of $\uE$ according to $v^a$.

 

[PeterDonis]

Okay, so the transformation given in this link: https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Does not answer the question: "If I am at rest in frame G, then boost to frame H, what will be the new electric field I measure?"

Let's unpack this question.

If you are at rest in frame G, your 4-velocity is $v_G$, and the electric field you measure is $E_G$.

If you are at rest in frame H, your 4-velocity is $v_H$, and the electric field you measure is $E_H$.

The abstract index notation for these two statements is:

$$(E_G)_a = F_{ab} (v_G)^b$$

$$(E_H)_a = F_{ab} (v_H)^b$$

In component notation in frame G, we have

$$(E_G)_\mu = F_{\mu \nu} (v_G)^\nu$$

$$(E_H)_\mu = F_{\mu \nu} (v_H)^\nu$$

If we want to transform these equations to frame H, we use the Lorentz transformation $\Lambda$. I'll use $\rho$ and $\sigma$ for the indexes in the primed frame, to keep them distinct from the indexes $\mu$ and $\nu$ in the unprimed frame:

$$(E_G)'_\rho = \Lambda^\mu{}_\rho (E_G)_\mu$$

$$(E_H)'_\rho = \Lambda^\mu{}_\rho (E_H)_\mu$$

What you are saying, however, is that we should have

$$(E_G)_\mu = (E_H)'_\rho$$

Which doesn't correspond to anything in the above.

If this is the case, then why isn't the Lorentz transformation given in this link:
https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Related by a "simple" Lorentz transformation? That is, $(E_G)_\mu \mapsto (E_G)'_\mu$
Is just
$$(E_G)_\mu = (E_G)'_\alpha \Lambda_{\mu}^{\ \ \alpha}$$
But this is not what the Lorentz transformation given in the link is

The Lorentz transformation given in the link is the translation into 3-vector notation of

$$F'_{\rho \sigma} = \Lambda^\mu{}_{\rho} \Lambda^\nu{}_\sigma F_{\mu \nu}$$

In other words, it is a description of how you transform the EM field tensor. It is not a description of how you transform either $E_G$ or $E_H$.

 

[PeterDonis]

• The Lorentz transformations for the electric field relates
  the components of $\uE$ (and $\uB$) according to $u^a$
  with
  the components of $\vE$ (and $\vB$) according to $v^a$.

No, it doesn't. The Lorentz transformations for the electric and magnetic fields transform $F_{\mu \nu}$. They don't transform either $\uE$ or $\vE$ (or their magnetic counterparts). The latter objects are not the EM field tensor; they are 4-vectors describing particular observables, which transform as 4-vectors. The EM field tensor transforms as a 2nd-rank antisymmetric tensor, which, when translated into 3-vector notation, gives the Lorentz transformation equations on the page the OP referenced.

 

[aliens123]

What you are saying, however, is that we should have

$$(E_G)_\mu = (E_H)'_\rho$$

I am not saying that these should be equal. The use of the term "Lorentz transformation" perhaps should not have been used. But the formula in the link considers the following "transformation":
$$F_{\mu 0} \mapsto F'_{\mu 0}$$
I should note that the author does not use the term "Lorentz transformation." Nonetheless, that is the transformation he is considering, and he does use the term "transformation."

The transformation
$$F_{\mu 0} \mapsto F'_{\mu 0}$$
is what I (and perhaps others) have been calling the "Lorentz transformation of the Electric field." (Which, is probably not a good idea considering that term already has a meaning).
https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

 

[etotheipi]

No, it doesn't. The Lorentz transformations for the electric and magnetic fields transform $F_{\mu \nu}$. They don't transform either $\uE$ or $\vE$ (or their magnetic counterparts). The latter objects are not the EM field tensor; they are 4-vectors describing particular observables, which transform as 4-vectors. The EM field tensor transforms as a 2nd-rank antisymmetric tensor, which, when translated into 3-vector notation, gives the Lorentz transformation equations on the page the OP referenced.

I'm not sure I entirely understand this; I don't see the issue. Given two observers of 4-velocities $u^a$ and $v^a$, the components of the EM tensor relative to their respective bases transform like
$$\tilde{F}^{\tilde{\mu} \tilde{\nu}} = \frac{\partial \tilde{x}^{\tilde{\mu}}}{\partial x^{\mu}} \frac{\partial \tilde{x}^{\tilde{\nu}}}{\partial x^{\nu}} F^{\mu \nu} = {\Lambda^{\tilde{\mu}}}_{\mu}{\Lambda^{\tilde{\nu}}}_{\nu} F^{\mu \nu}$$and you can find the transformation rules of the electric components by identifying $E^i = F^{i0}$ and $\tilde{E}^i = \tilde{F}^{i0}$.

Further, the $E^i$ are precisely the spatial components of the four-vector $\uE$ with respect to the basis of the $u^a$ observer, and the $\tilde{E}^i$ are precisely the spatial components of $\vE$ with respect to the basis of the $v^a$ observer. So what @robphy wrote is correct, it's just another way of looking at the same thing!

 

[PeterDonis]

I am not saying that these should be equal.

Ok.

the formula in the link considers the following "transformation":
$$F_{\mu 0} \mapsto F'_{\mu 0}$$

No, it does not. What you write here doesn't even make sense; $F_{\mu 0}$ is not even a valid geometric object to begin with, so talking about "transforming" it is nonsense.

The link you give talks about transforming the EM field tensor, just as I said. It then derives the 3-vector transformation equations, for the special case of a pure boost, from the 4-vector transformation equation.

You might be confused by the fact that, in the process of deriving the 3-vector equations, the article chooses the direction of the pure boost to be the "1" direction, and then explicitly shows what that implies for the transformed value of $F^{01}$, which is the E field in the "1" direction. That does not mean the article is deriving a transformation equation for $F_{\mu 0}$. It just means the article is leaving the details for the rest of the components of $E$ and $B$, and how they all combine to reach the final result given, for the reader to fill in.

 

[aliens123]

The link you give talks about transforming the EM field tensor, just as I said. It then derives the 3-vector transformation equations, for the special case of a pure boost, from the 4-vector transformation equation.

You might be confused by the fact that, in the process of deriving the 3-vector equations, the article chooses the direction of the pure boost to be the "1" direction, and then explicitly shows what that implies for the transformed value of $F^{01}$, which is the E field in the "1" direction. That does not mean the article is deriving a transformation equation for $F_{\mu 0}$. It just means the article is leaving the details for the rest of the components of $E$ and $B$, and how they all combine to reach the final result given, for the reader to fill in.

I believe that the following transformation (as in the article):

Is obtained by making the identifications
$$E' = F'_{i 0}, E = F_{j 0}, i, j \in \{1, 2, 3\}$$

 

[PeterDonis]

the $E^i$ are precisely the spatial components of the four-vector $\uE$, and the $\tilde{E}^i$ are precisely the spatial components of $\vE$.

Again, let's unpack this.

We have two frames, one in which an observer with 4-velocity $u$ is at rest, the other in which an observer with 4-velocity $v$ is at rest.

We have two "electric field as measured by" 4-vectors, $\uE$ and $\vE$. The equation defining these 4-vectors ensures that they are orthogonal to their respective 4-velocity vectors. (Instead of "orthogonal", @robphy called them "purely spatial"; same thing.)

Because they are purely spatial, we can "interpret" them as 3-vectors in the appropriate frames.

However, we have to be very, very careful in how we describe this "interpretation", because it's extremely easy, as pretty much all of the OP's posts in this thread show, to confuse this "interpretation" with the claim that transforming $\uE$ and $\vE$ from one frame to another is the same thing as "transforming the electric field". Which is wrong.

To transform "the electric field", you need to transform the EM field tensor. Which, in 3-vector notation, gives a pair of equations for $E$ and $B$ in the primed frame in terms of $E$ and $B$ in the unprimed frame. And the OP is looking at those equations and thinking they ought to be the right transformation equations for $\uE$ and $\vE$. Which of course they're not; $\uE$ and $\vE$ are 4-vectors and transform as 4-vectors. But "the electric field" is part of a 2nd-rank antisymmetric tensor, and doesn't transform as a 4-vector, and the 3-vector transformation law for "the electric field" is not a simple vector transformation law.

 

[PeterDonis]

I believe that the following transformation (as in the article):
View attachment 279551

Is obtained by making the identifications
$$E' = F'_{i 0}, E = F_{j 0}, i, j \in \{1, 2, 3\}$$

That, plus transforming the EM field tensor $F$ all as one thing, as a 2nd-rank antisymmetric tensor. Which is not the same as the transformation law for $E$ being "the transformation law for $F_{\mu 0}$".

 

[etotheipi]

To transform "the electric field", you need to transform the EM field tensor. Which, in 3-vector notation, gives a pair of equations for $E$ and $B$ in the primed frame in terms of $E$ and $B$ in the unprimed frame. And the OP is looking at those equations and thinking they ought to be the right transformation equations for $\uE$ and $\vE$. Which of course they're not; $\uE$ and $\vE$ are 4-vectors and transform as 4-vectors. But "the electric field" is part of a 2nd-rank antisymmetric tensor, and doesn't transform as a 4-vector, and the 3-vector transformation law for "the electric field" is not a simple vector transformation law.

Yes, I agree with all of that. The only meaningful components of the $\uE$ and $\vE$ are those with respect to the bases of the $u^a$ and $v^a$ observers respectively; these are simply $F^{i0}$ and $\tilde{F}^{i0}$. In that regard, whilst it's of course mathematically possible to write $\uE$ and $\vE$ in terms of any other bases, the results are not going to tell you anything meaningful physically. It's what makes this "electric-field-4-vector" a little hard to understand at first (or at least, it tripped me up when I first saw it!)

 

[pervect]

In Robert Wald's General Relativity textbook page 64 reads:
__________________________
In prerelativity physics, the electric field $\vec{E}$ and magnetic field $\vec{B}$ each are spatial vectors. In special relativity these fields are combined into a single spacetime tensor field $F_{ab}$ which is antisymmetric in its indices, $F_{ab} = - F_{ba}$. Thus $F_{ab}$ has six independent components. For an observer moving with 4-velocity $v^{a}$, the quantity
$$E_a = F_{ab}v^{b}$$
is interpreted as the electric field measured by that observer, while
$$B_a = -\frac{1}{2} \epsilon_{ab}^{\ \ \ cd}F_{cd}v^b$$
is interpreted as the magnetic field, where $\epsilon_{abcd}$ is the totally antisymmetric tensor of positive orientation with norm $\epsilon_{abcd} \epsilon^{abcd}=-24, \epsilon_{0123}=1.$
__________________________
I am confused by this bit. Of course, in the rest frame we have $v^{b} = (1,0,0,0)$ and so it is not too hard to see that $E_a = F_{a0}$. For an observer not at rest; however, this says that electric field for them (which I'll denote $E'_a$) is given by
$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$
$$\begin{bmatrix} E'_0 \\ E'_1 \\ E'_2 \\ E'_3 \end{bmatrix} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_x & -B_y & B_x & 0 \end{bmatrix} \begin{bmatrix} v_0 \\ v_1 \\ v_2 \\ v_3 \end{bmatrix}$$

Recall that the basis vectors for an observer in motion are not the same as the basis vectors for an observer at rest.

The E-field vector E, like any vector, is expressed as

$\vec{E} = E^0 e_0 + E^1 e_1 + E^2 e_2 + E^3 e_3$

where the $e_i$ are the basis vectors associated with a particular observer.

You have omitted the basis vectors $e_i$ from your formalism, writing only the components. This is common. But this has caused you to become confused, what you want to calculate is the electric field compnents in terms of the basis vectors of the moving observer, while what you've written is the electric field copmonents in terms of the basis vectors of the stationary observer.

But the basis vectors of the stationary observer and the moving observer are different, they're related by the Lorentz transform.

If you transform the basis vectors, you should find that Wald's statements are in agreement with your other statements - I have not done the work myself. Alternatively, you can just transform the components of the Faraday tensor from one set of basis vectors to the others, as other posters have suggested.

$$F'_{cd} = \Lambda^c{}_a \Lambda^d{}_b F_{ab}$$

where $\Lambda$ is the Lorentz transformation matrix.

 

[PeterDonis]

The E-field vector E, like any vector, is expressed as

$\vec{E} = E^0 e_0 + E^1 e_1 + E^2 e_2 + E^3 e_3$

where the $e_i$ are the basis vectors associated with a particular observer.

I'm not sure this way of putting it helps any, because it perpetuates the same confusion about what the term "the electric field" means that I have been trying to clarify in this thread.

Wald's 4-vector $E_a = F_{ab} v^b$ is a perfectly good 4-vector, yes, and in component form, if we include the basis vectors and raise an index, so we have $E^\mu = F^\mu{}_\nu v^\nu$, it looks like the RHS of what you wrote. But the LHS of what you wrote uses the notation $\vec{E}$, which customarily denotes a 3-vector, not a 4-vector.

If the 3-vector $\vec{E}$ is supposed to be "the spatial part" of Wald's 4-vector $E$, you can make it into a 4-vector by noting that "purely spatial" means that it is orthogonal to the basis vector $e_0$, which is also the observer's 4-velocity $v$, which means $E^0 = 0$. But there is still a confusion lurking here: if this 4-vector $E$ is supposed to be the same object as Wald's 4-vector $E$, it does not transform the way "the electric field" is supposed to transform. Wald's 4-vector $E$ transforms like an ordinary 4-vector. But "the electric field" is part of the EM field tensor, and there is no way to write a valid transformation law that just involves the "electric field" part of that tensor. You have to include the magnetic part as well, either by writing two 3-vector transformation laws, or by writing a single 2nd-rank antisymmetric tensor transformation law.

If, OTOH, you intend your $\vec{E}$ to be "the electric field" in the sense of a "piece" of the EM field tensor, then you cannot write it as a 4-vector the way you are doing. It simply is not that kind of object.

 

[Sagittarius A-Star]

Maybe it's easier to transform the electromagnetic field via the four-potential.

A four-current $\mathbf J = \begin{pmatrix} c \rho \\ \vec j \\ \end{pmatrix}$ creates a four-potential

$\mathbf A = \begin{pmatrix} \phi \\ \vec a \\ \end{pmatrix} =\frac {1}{c} \int \, \frac {\mathbf J}{r} dV$

The four-potential $\mathbf A$ can be Lorentz-transformed. In the new frame, the electric and magnetic fields are:

$\vec E' = - \nabla \phi ' - \frac {1}{c} \frac {\partial \vec a '}{\partial t}$
$\vec B' = - \nabla \times \vec a'$.

Source:
https://en.wikipedia.org/wiki/Electromagnetic_four-potential

Helmholtz worked already with the scalar- and vector-potential before SR was created:
https://physics.princeton.edu//~mcdonald/examples/helmholtz.pdf

 

[vanhees71]

If that is true (I have not checked), then it doesn't conflict with Wald.

I don't know, why this should be wrong. It's the standard Lorentz transformation of the electromagnetic field written in terms of $\vec{E}$ and $\vec{B}$. One should be aware that these are not the spatial components of four-vectors but the entries of the antisymmetric Faraday tensor.

See also Sect. 3.2 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Just to make it clear again. In the Ricci calculus tensor components are understood to be defined wrt. a fixed basis (here a Minkowski-orthonormal basis). Now $F^{\mu \nu}$ are the components of the em. field-strength tensor field and $u^{\mu}$ are the components of the four-velocity of an observer. Then one can define the four-vectors
$$E_{\mu} = F_{\mu \nu} u^{\nu}, \quad B_{\mu}=^{\dagger} F_{\mu \nu} u^{\nu} \frac{1}{2} \epsilon_{\rho \sigma \mu \nu} F^{\rho \sigma} u^{\nu}.$$
For an observer at rest $u^{\mu}=1$, and you get the electromagnetic-field components in three-vector notation $\vec{E}$ and $\vec{B}$.

That's why you can say that for a general $u^{\mu}$ the four-vectors $E_{\mu}$ and $B_{\mu}$ are the electric and magnetic fields as observed by the observer moving with this four-velocity, but it's not the $\vec{E}'$ and $\vec{B}'$ components as measured in the restframe of the observer but the four-vector components $E_{\mu}$ and $B_{\mu}$ are the four-vector components in the "computational frame", i.e., to get $\vec{E}'$ and $\vec{B}'$ as measured in the rest frame of the observer, you have to perform the corresponding Lorentz boost to these four-vector components
$$E^{\prime \mu}={\Lambda^{\mu}}_{\nu} E^{\nu}, \quad B^{\prime \mu}={\Lambda^{\mu}}_{\nu} B^{\nu}.$$
Again: It's important to note that the three-vector components $\vec{E}$ and $\vec{B}$ in some fixed frame of reference are not spatial components of a four-vector but the components of the antisymmetric field-strength tensor (in the older literature for this reason also called a "6-vector" $(\vec{E},\vec{B})$, but I think that's even more confusing).

 

[PeterDonis]

Maybe it's easier to transform the electromagnetic field via the four-potential.

The only problem with doing it this way is that the 4-potential is gauge dependent, whereas the E and B fields are not.

 

[robphy]

$\def\uE{\stackrel{u}{E^a}} \def\vE{\stackrel{v}{E^a}} \def\uB{\stackrel{u}{B^a}} \def\vB{\stackrel{v}{B^a}}$

No, it doesn't. The Lorentz transformations for the electric and magnetic fields transform $F_{\mu \nu}$. They don't transform either $\uE$ or $\vE$ (or their magnetic counterparts). The latter objects are not the EM field tensor; they are 4-vectors describing particular observables, which transform as 4-vectors. The EM field tensor transforms as a 2nd-rank antisymmetric tensor, which, when translated into 3-vector notation, gives the Lorentz transformation equations on the page the OP referenced.

By

the components of $\uE$ (and $\uB$) according to $u^a$

I am referring to the components of the field tensor $F_{ab}$ as measured by the observer with $u^a$. It is these six inner products of the form $F_{ab} \hat u^a \hat x^b (=\stackrel{u}{E}_b \hat x^b)$, etc... where the $x^a$, $y^a$, etc are part of the $u^a$-observer's measuring apparatus. This set of components of the field tensor is what is transformed by the Lorentz Transformation.

I am not describing the Lorentz transformation of an observer-dependent 4-vector like $\stackrel{u}{E_a}$ [partially-]decomposed from $F_{ab}$.

 

[PeterDonis]

these six inner products of the form $F_{ab} \hat u^a \hat x^b (=\stackrel{u}{E}_b \hat x^b)$, etc... where the $x^a$, $y^a$, etc are part of the $u^a$-observer's measuring apparatus.

There aren't six (nonzero) inner products of that form. There are only three, the "electric" ones. The "magnetic" inner products are of the form $F_{ab} \hat x^a \hat y^b$ and so forth. In an orthonormal frame in which $u^a$, $x^a$, $y^a$, $z^a$ are the basis vectors, this is obvious, since in that frame the "electric" components are $F_{0i}$ and the "magnetic" components are $F_{ij}$, where $i$, $j$ are the spatial coordinate indexes. (There's also the complication that, for example, $F_{12}$ is the magnetic field in the 3, or $z$, direction; that's what Wald's formula in terms of $\epsilon_{abcd}$ in the OP is capturing.)

However, the more important point is that this statement...

I am not describing the Lorentz transformation of an observer-dependent 4-vector like $\stackrel{u}{E_a}$ [partially-]decomposed from $F_{ab}$.

...appears to me to need some clarification, since the inner products you describe are observer-dependent; they are just the (nonzero) components of the observer-dependent 4-vectors you refer to. When you say you are not describing the transformation of an observer-dependent 4-vector, what you mean is that you are including the change of observer in your definition of "Lorentz transformation"; you are changing the vectors $u^a$, $x^a$, $y^a$, $z^a$ in concert with transforming the components of $F_{ab}$ as an antisymmetric 2nd-rank tensor, so that the transformed components match the changed set of inner products with the new basis vectors.

 

[robphy]

There aren't six (nonzero) inner products of that form. There are only three, the "electric" ones. The "magnetic" inner products are of the form $F_{ab} \hat x^a \hat y^b$ and so forth.

Yes, I meant (schematically) the tx,ty,tz,xy,yz,zx (6 pairs) inner products with F (where the role of "t" is the 4-velocity "u").

... you are changing the vectors $u^a$, $x^a$, $y^a$, $z^a$ in concert with transforming the components of $F_{ab}$ as an antisymmetric 2nd-rank tensor, so that the transformed components match the changed set of inner products with the new basis vectors.

I think this is a fair description.

 

[vanhees71]

The electric and magnetic components of the Faraday tensor can be covariantly defined using the four-velocity of the observer, the Faraday tensor and its dual as already given in #1. Of course, using tetrads you get the description of the electromagnetic field as either the invariant Faraday tensor $F$ or equivalently the invariant four-vectors $E$ and $B$. Note that both descriptions correspond to 6 independent field components.

 

[robphy]

Sorry for the long post (there is a lot of $\LaTeX$ here)...

To address @aliens123 's question on the Electric Field seen by an observer in motion, this may help.
It's based on some old notes of mine that I've been meaning to revise.
(It includes other pieces that are not essential to this question.)Consider two observers: Vic (with 4-velocity $v^a$) and Will (with 4-velocity $w^a$).

Some formula to set the scene...
$\def\MACROS{} \def\hv{\hat v} \def\hw{\hat w} \def\hvx{\hat {\bar x}} \def\hwx{\hat {\tilde x}} \def\hvy{\hat {\bar y}} \def\hwy{\hat {\tilde y}} \def\hvz{\hat {\bar z}} \def\hwz{\hat {\tilde z}} \def\vE{\bar E} \def\vB{\bar B} \def\vBs{{}^*\bar B} \def\wE{\tilde E} \def\wB{\tilde B} \def\wBs{{}^*\tilde B} \def\vW{\bar W}$

[We are using the abstract-index notation.]​

Given an antisymmetric tensor $F_{ab}$, a metric $g_{ab}$ with signature $(+,-,-,-)$ [different from Wald's], and a unit timelike vector $\hv^a$ ($g_{ab} \hv^a \hv^b=+1$),​

define​

$E_a=F_{ab} \hv^b$ (note: $E_a \hv^a=0$)​

and​

$B_a={}^*F_{ab} \hv^b=\frac{1}{2} \epsilon_{abcd}F^{cd}\hv^b$ (note: $B_a \hv^a=0$).​

Thus, the vectors $E^a=g^{ab}E_b$ and $B^a=g^{ab}B_b$ are orthogonal to $\hv^a$.​

For convenience, we define two antisymmetric tensors $E_{ab}=2E_{[a}\hv_{b]}$ and $B_{ab}=2B_{[a}\hv_{b]}.$​

Then ${}^* B_{ab}=\epsilon_{abcd} B^c \hv^d$ and ${}^* E_{ab}=\epsilon_{abcd} E^c \hv^d$.​

​

Note:​

$E_{ab}\hv^b=E_a \hv_b \hv^b - E_b \hv_a \hv^b= E_a (\hv_b \hv^b) - 0 =E_a \quad$ since $\hv_b \hv^b =1$​

​

(so, $B_{ab}\hv^b=B_a$, $E_{ab}\hv^b=E_a$, ${}^* B_{ab}\hv^b=0$, ${}^*E_{ab}\hv^b=0$).​

​

Thus, $$F_{ab}=E_{ab}-{}^*B_{ab}$$​

since $E_a=(E_{ab}-{}^*B_{ab})\hv^b$ and $B_a=({}^*E_{ab}-{}^{**}B_{ab})\hv^b=({}^*E_{ab}+B_{ab})\hv^b$​

​

If $F_{ab}$ is a Maxwell field, then $E_a$ and $B_a$ are the electric and magnetic parts of the field according to an observer Vic with four-velocity $\hv^a$ (a future-directed unit timelike vector).​

​

The main question:
What are the electric and magnetic parts according to another observer Will with four-velocity $\hw^a$?
(I'm going to work on the Electric part.)

Our goal is to
compute Will's components of Will's electric field (and magnetic field)
in terms of Vic's components of Vic's electric and magnetic field.
(This is essentially the Lorentz Transformation of the Electric and Magnetic Fields.)​

To distinguish the observer's electric and magnetic parts
and to emphasize the observer-dependence of their decompositions of the Maxwell field tensor,
we refer to

• Vic's parts with a bar as $\vE_a$ , $\vB_a$, $\vE_{ab}$ and $\vB_{ab}$
  and
• Will's parts with a tilde as $\wE_a$ , $\wB_a$, $\wE_{ab}$ and $\wB_{ab}$.
• [To emphasize the observer-dependent decomposition,
  we could have used $'$ and $''$... but not primed and unprimed.]

​

The Electric Field
We will compute the electric part $\wE_a$ of the field tensor according to Will.
Will's 4-velocity can be described by Vic as follows:
$$\hw^a= \gamma \hv^a + \gamma \vW^a,$$
where $\vW^a$ is the spatial-velocity of Will according to Vic ($g_{ab}\hv^a\vW^b=0$), which has square-norm $W^2=-\vW^a\vW_a (=\vec \vW\cdot \vec \vW)$ and $\gamma=\frac{1}{\sqrt{1-W^2}}$.
(I'll use the bar and tilde notation on the 4-vectors, and on other quantities whose observer-dependence decompositions are to be emphasized. I won't use it on $\gamma$.)

So, we have the Electric Field according to Will (expressed in terms of quantities according to Vic)...
$$\begin{eqnarray*}
\wE_a
&=& F_{ab}\hw^b\\
&=& F_{ab}(\gamma \hv^b + \gamma \vW^b)\\
&=& \gamma (\vE_{a} + F_{ab}\vW^b )\\
&=& \gamma (\vE_{a} + (\vE_{ab}-\vBs_{ab})\vW^b )\\
&=& \gamma (\vE_{a} + (\vE_a \hv_b \vW^b - \vE_b \hv_a \vW^b) -
\epsilon_{abcd} \vB^c \hv^d \vW^b )\\
&=& \gamma (\vE_{a} + (\quad 0\quad ) + (-\vE_b \vW^b) \hat v_a -
\epsilon_{abcd} \vB^c \hv^d \vW^b )\\
&=& \gamma (\vE_{a} - (\vE_b \vW^b) \hv_a -
(-1)^3 \hv^d\epsilon_{dabc} \vW^b \vB^c )\\
&=& \gamma (\vE_{a} - (\vE_b \vW^b) \hv_a \ \ + \
\hv^d\epsilon_{dabc} \vW^b \vB^c )\\
&=& -\gamma (\vE_b \vW^b) \hv_a + \gamma (\vE_{a} +
\hv^d\epsilon_{dabc} \vW^b \vB^c )
\end{eqnarray*}$$
where we have isolated the component of Will's electric part $\wE_a$ that is parallel to Vic's 4-velocity; the second term is purely-spatial according to Vic.

Recall: Our goal is to
compute Will's components of Will's electric field (and magnetic field)
in terms of Vic's components of Vic's electric and magnetic field.

A special case: $W^a=W\hvx^a$
Take the special case where Will's spatial-velocity is along Vic's $x$-axis, which we write as $W^a=W\hvx^a$.
Recalling the form of the Lorentz Transformation for relative motion along the $x$-axis,
\begin{eqnarray*}
\hat t'^a &=& \gamma(\hat t^a+V_{rel}\ \hat x^a) \\
\hat x'^a &=& \gamma(V_{rel}\ \hat t^a+ \hat x^a)\\
\hat y'^a &=& \hat y^a\\
\hat z'^a &=& \hat z^a
\end{eqnarray*} we express this transformation with our notation encoding the observer 4-velocities
\begin{eqnarray*}
\hw^a &=& \gamma(\hv^a+W\hvx^a)\\
\hwx^a &=& \gamma(W\hv^a+\hvx^a)\\
\hwy^a &=& \hvy^a\\
\hwz^a &=& \hvz^a.
\end{eqnarray*}

Due to our signature $(+,-,-,-)$ convention,
define $\wE_{(\tilde x)}=-\hwx^a \wE_a = \hwx \cdot \vec \wE$ to be Will's $\tilde x$-component of the electric field $\wE_a$ that Will measures.
Similarly,
define $\vE_{(\bar x)}=-\hvx^a \vE_a = \hvx \cdot \vec \vE$ to be Vic's $\bar x$-component of the electric field $\vE_a$ that Vic measures.We calculate Will's $x$-component of Will's Electric Field (in terms of quantities according to Vic):
\begin{eqnarray*}
\hwx^a \wE_a
&=&
\gamma(W\hv^a+\hvx^a)
\left(
-\gamma (\vE_b \vW^b) \hv_a \ \ + \gamma (\vE_{a} +
\hv^d\epsilon_{dabc} \vW^b \vB^c )
\right)
\\
&=&
\gamma(W\hv^a+\hvx^a)
\left( -\gamma W(\vE_b \hvx^b) \hv_a + \gamma (\vE_{a} +
W\hv^d\epsilon_{dabc} \hvx^b \vB^c ) )\right)
\\
&=&
\gamma^2 \left[ -W^2 (\vE_b \hvx^b) \hv^a\hv_a\ + 0 + 0 +
\hvx^a \vE_a + \hvx^a W\hv^d\epsilon_{dabc} \hvx^b \vB^c \right]\\
\wE_{(\tilde x)}
&\stackrel{*}{=}& \gamma^2 \left[ -W^2 \vE_{(\bar x)} \qquad \quad\ + 0 + 0 +\ \vE_{(\bar x)} \ + \qquad 0 \qquad\right]
\\
&=& \gamma^2 \left[ -W^2 +1 \right] \vE_{(\bar x)}
\\
\wE_{(\tilde x)}
&=& \vE_{(\bar x)}
\end{eqnarray*}
(The * indicates that we have canceled the common factor $(-1)$ on both sides that arose in our definition of the spatial-components (as the dot product of two spacelike vectors) in this $(+,-,-,-)$ convention.)

For the $y$- and $z$-components of the Electric Field, we have
\begin{eqnarray*}
\hwy^a \wE_a
&=& \hvy^a
\left(
-\gamma (\vE_b \vW^b) \hv_a \ \ + \gamma (\vE_{a} +
\hv^d\epsilon_{dabc} \vW^b \vB^c )
\right)
\\
&=& \gamma \left[ \qquad\quad 0 \qquad\quad\ + \hvy^a \vE_a
+ \hvy^a \hv^d\epsilon_{dabc} \vW^b B^c \right]
\\
\wE_{(\tilde y)}
&\stackrel{*}{=}& \gamma \left[ \phantom{ \qquad\quad 0 \qquad\quad\ +\ } \vE_{(\bar y)} + ( \vW_{(\bar z)} \vB_{(\bar x)} - \vW_{(\bar x)} \vB_{(\bar z)}) \right]\\
\wE_{(\tilde y)}
&=& \gamma (\vE_{(\bar y)}- W \vB_{(\bar z)})
\end{eqnarray*}

\begin{eqnarray*}
\hwz^a \wE_a
&=& \hvz^a
\left(
-\gamma (\vE_b \vW^b) \hv_a \ \ + \gamma (\vE_{a} +
\hv^d\epsilon_{dabc} \vW^b \vB^c )
\right)
\\
&=& \gamma \left[ \qquad\quad 0 \qquad\quad\ + \hvz^a \vE_a
+ \hvz^a \hv^d\epsilon_{dabc} \vW^b B^c \right]
\\
\wE_{(\tilde z)}
&\stackrel{*}{=}& \gamma \left[ \phantom{\qquad\quad 0 \qquad\quad\ +\ \ }
\vE_{(\bar z)} + ( \vW_{(\bar x)} \vB_{(\bar y)} - \vW_{(\bar y)} \vB_{(\bar x)})
\right]\\
\wE_{(\tilde z)}
&=& \gamma (\vE_{(\bar z)}+ W \vB_{(\bar y)})\\
\end{eqnarray*}

The Lorentz transformation of the Electric Field in 3-vector form
In the following, my goal is motivation---not an exhaustive proof.

Let's collect the results
\begin{eqnarray*}
\wE_{(\tilde x)}
&=& \vE_{(\bar x)}
\\
\wE_{(\tilde y)}
&=& \gamma (\vE_{(\bar y)}- W \vB_{(\bar z)})
\\
\wE_{(\tilde z)}
&=& \gamma (\vE_{(\bar z)}+ W \vB_{(\bar y)})
\end{eqnarray*}

Let's rewrite this in a way to suggest a more general form,
resurrecting some terms that evaluated to zero.
Note what we have to do to the x-component in order to introduce $\gamma \vE^a$.
\begin{eqnarray*}
\wE_{(\tilde x)}
&=& \gamma \vE_{(\bar x)} - ( \gamma -1) \vE_{(\bar x)}
\\
\wE_{(\tilde y)}
&=& \gamma \left(\vE_{(\bar y)} +
( \vW_{(\bar z)} \vB_{(\bar x)} - \vW_{(\bar x)} \vB_{(\bar z)})
\right)
\\
\wE_{(\tilde z)}
&=& \gamma \left(\vE_{(\bar z)} +
( \vW_{(\bar x)} \vB_{(\bar y)} - \vW_{(\bar y)} \vB_{(\bar x)})
\right)
\end{eqnarray*}
...and a little more suggestive rewriting (without proof)...
\begin{eqnarray*}
\wE_{(\tilde x)}
&=& \gamma \vE_{(\bar x)} - ( \gamma -1) \vE_{(\bar x)}
\\
\wE_{(\tilde y)}
&=& \gamma \left(\vE_{(\bar y)} +
( \vec \vW \times \vec \vB )_{(\bar y)}
\right)
\\
\wE_{(\tilde z)}
&=& \gamma \left(\wE_{(\bar z)} +
( \vec \vW \times \vec \vB )_{(\bar z)}
\right)
\end{eqnarray*}

Recall that the relative-velocity was along the $\bar x$-direction: $\vW^a = W\hvx^a$.
(More precisely, Vic's $\hv\hvx$-plane coincides with Will's $\hw\hwx$-plane.)
So, by symmetry, we expect (without proof)...
\begin{eqnarray*}
\wE_{(\tilde x)}
&=& \gamma \left(\wE_{(\bar x)} + ( \vec \vW \times \vec \vB )_{(\bar x)}\right) - ( \gamma -1) \vE_{(\bar x)}
\\
\wE_{(\tilde y)}
&=& \gamma \left(\wE_{(\bar y)} +
( \vec \vW \times \vec \vB )_{(\bar y)}
\right)
\\
\wE_{(\tilde z)}
&=& \gamma \left(\wE_{(\bar z)} +
( \vec \vW \times \vec \vB )_{(\bar z)}
\right)
\end{eqnarray*}
or, more compactly,
\begin{eqnarray*}
\vec \wE=\gamma(\vec {\vE} + (\vec\vW \times \vec\vB) ) -(\gamma-1)\vE_{(\hvx)}\hvx
\end{eqnarray*}

Note that in the derivation of $\wE_{(\tilde x)}$,
the quantity $\vE_{(\bar x)}$ on the right-hand-side is really the component of $\vE^a$ along the $\vW^a$-direction (using the unit -vector $\hat \vW^a$).
Since we can write
$$\vE_{(\bar x)}
=\vE_{(|| \hvx )}
= \hvx \cdot \vec \vE $$
and
$$\vE_{(\bar x)}\hvx
=\vE_{(|| \hvx )} \hvx
= \left( \hvx \cdot \vec \vE\right) \hvx, $$
we can write this more generally for an arbitrarily-directed $\vec\vW$ as
$$\vE_{(|| \vW)} = - \hat \vW^a \vE_a = \hat\vW \cdot \vec \vE $$
so that
$$\vE_{(|| \vW)}\hat \vW = \left(\hat \vW \cdot \vec \vE\right) \hat\vW $$

Thus, we obtain the more general expression
\begin{eqnarray*}
\vec \wE=\gamma(\vec {\vE} + (\vec\vW \times \vec\vB) ) -(\gamma-1)\left(\hat \vW \cdot \vec \vE\right) \hat\vW
\end{eqnarray*} which agrees with the last set of equations for the Electric Field in OP's question ( Electric Field seen by an observer in motion ).

The magnetic field is handled in a similar way... likely easier using (or invoking) $B_a={}^*F_{ab} \hv^b$.

 

[PeterDonis]

The electric and magnetic components of the Faraday tensor can be covariantly defined using the four-velocity of the observer, the Faraday tensor and its dual as already given in #1. Of course, using tetrads you get the description of the electromagnetic field as either the invariant Faraday tensor $F$ or equivalently the invariant four-vectors $E$ and $B$. Note that both descriptions correspond to 6 independent field components.

You don't have two invariant 4-vectors without any constraints; that would be eight independent components, not six. The 4-vectors you get by forming the contractions described in the OP do not describe "the electric field" and "the magnetic field" in general. They only describe those fields as observed by the particular observer whose 4-velocity we put into the contractions. Those two 4-vectors are constrained, by the contractions, to both be orthogonal to the 4-velocity; that is what reduces the number of independent components to 6 instead of 8. But transforming those 4-vectors, as 4-vectors, into some other frame does not tell you anything useful about what an observer at rest in that other frame would observe. To know that, you need to transform $F$ itself, as an antisymmetric 2nd-rank tensor; or, equivalently, you need to form different contractions, of $F$ with the 4-velocity of the new observer. (@robphy is basically taking the latter approach.)

 

[vanhees71]

Of course $E_{\mu} u^{\mu}=B_{\mu} u^{\mu}=0$, and that makes the constraints. In both equivalent descriptions of the electromagnetic field of course you have 6 (real) field-degrees of freedom. Since the constraints are manifestly covariant they are valid in any frame and $E_{\mu}$ and $B_{\mu}$ of course transform as four-vectors. There's a one-to-one mapping between $E_{\mu}$ and $B_{\mu}$ and $F_{\mu \nu}$:
$$F_{\mu \nu}=E_{\mu} u_{\nu} - E_{\nu} u_{\mu} - \epsilon_{\mu \nu \rho \sigma} B^{\rho} u^{\sigma} \; \Leftrightarrow \; E_{\mu} =F_{\mu \nu} u^{\nu}, \quad B_{\mu} = ^{\dagger} F_{\mu \nu} u^{\nu} = \frac{1}{2} \epsilon_{\mu \nu \rho \sigma} F^{\rho \sigma} u^{\nu}.$$

 

[PeterDonis]

Since the constraints are manifestly covariant they are valid in any frame and $E_{\mu}$ and $B_{\mu}$ of course transform as four-vectors.

Yes, but they're different 4-vectors for different observers (different 4-velocities $u^\mu$). So just transforming the 4-vectors $E_\mu$ and $B_\mu$ doesn't do what most people are thinking of when they talk about "transforming the electric and magnetic fields". Most people are thinking of transforming between observers, i.e., they want to go from the electric and magnetic fields observed by one observer, with 4-velocity $u^\mu$, to the electric and magnetic fields observed by a different observer, with 4-velocity $v^\mu$. Transforming the 4-vectors $E_\mu$ and $B_\mu$ does not do that; those 4-vectors, even when transformed, are still contractions with $u^\mu$, not $v^\mu$. To transform between observers, you need to form different contractions; you need to go back and start again from $F_{\mu \nu}$ and contract it with $v^\mu$ instead of $u^\mu$.

 

[vanhees71]

Sure, that reflects the physical fact that the decomposition of the electromagnetic field into electric and magnetic field components depends on the chosen frame of reference. The discussed formalism just makes the usual decomposition into electric and magnetic components wrt. a given reference frame manifestly covariant.

An application of this formalism of current interest in my research community (relativistic heavy-ion collisions) is the manifestly covariant formulation of (viscous) special and general relativistic magnetohydrodynamics to describe the medium created in relativistic heavy-ion collisions as well as in the astrophysical context like the simulation of the gravitational and em-wave signal of neutron-star mergers (two topics that are surprisingly closely related).