I Electric Field seen by an observer in motion

[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]

<br /> \def\uE{\stackrel{u}{E^a}}<br /> \def\vE{\stackrel{v}{E^a}}<br /> \def\uB{\stackrel{u}{B^a}}<br /> \def\vB{\stackrel{v}{B^a}}<br />

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{}<br /> \def\hv{\hat v}<br /> \def\hw{\hat w}<br /> \def\hvx{\hat {\bar x}}<br /> \def\hwx{\hat {\tilde x}}<br /> \def\hvy{\hat {\bar y}}<br /> \def\hwy{\hat {\tilde y}}<br /> \def\hvz{\hat {\bar z}}<br /> \def\hwz{\hat {\tilde z}}<br /> \def\vE{\bar E}<br /> \def\vB{\bar B}<br /> \def\vBs{{}^*\bar B}<br /> \def\wE{\tilde E}<br /> \def\wB{\tilde B}<br /> \def\wBs{{}^*\tilde B}<br /> \def\vW{\bar W}<br />

[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).