# Relation between electromagnetic field and field tensor

• A
carpinus
hello,
1. according to Robert Wald, General Relativity, equation (4.2.22)
the magnetic field as measured by an observer with 4-velocity ## v^b ## is given by
## B_a = - \frac {1}{2} {ϵ_{ab}}^{cd} F_{cd} v^b ##
where ## {ϵ_{ab}}^{cd}##, the author says, is the totally antisymmetric tensor (for more detail on the latter the reader is referred to the appendix).
I conclude for ## v^b ## = 0 (I am not sure about the sign of ##v^0##):
## B_a = - \frac {1}{2} {ϵ_{a0}}^{cd} F_{cd}
= \frac {1}{2} {ϵ_{0a}}^{cd} F_{cd} ##

2. I assume the common identities
##B_1 =F_{23},B_2 = F_{31}, B_3 = F_{12}##
are true only for cartesian coordinates -
because if they were true for general coordinates then from:
## ∂_1 F_{23} + ∂_2 F_{31} + ∂_3 F_{12} =0 ##
(one of four homogenous Maxwell-equations, true for general coordinates)
it would follow: ## ∂_1 B_1 +∂_2 B_2 +∂_3 B_3 = 0 ## for general coordinates.
The latter, however, I think is true only in cartesian coordinates.

my question:
does equation (4.2.22) allow for a magnetic field B deviating from ##(F_{23},F_{31}, F_{12})## and how? Can B according to (4.2.22) be seen to satisfy div B = 0 (E,B according to (4.2.21)f to satisfy the homogenous equations) ?

thank you for any comment. I just see that an article by robphy covering this sort of question can at present be found at the top of the forum page. So I should study this first. Or is there a somewhat shorter answer to my question?

## Answers and Replies

wnvl2
I think that it is not possible to make ##v^b## equal to 0 as ##v_b v^b## should be ##c^2##.

carpinus
I think that it is not possible to make ##v^b## equal to 0 as ##v_b v^b## should be ##c^2##.
I think you are right
I wanted to refer only to the space like part: ##v^b = (1,\vec v), v= | \vec v | =0 ##

Gold Member
2022 Award
Since ##v_{\mu} v^{\mu}=c^2## for an observer at rest relative to the computational frame you have ##(v^{\mu})=(c,0,0,0)##.

The components of the em. field in the computational frame are
$$F_{0j}=E_j, \quad F_{jk}=-\epsilon_{jkl} B_l.$$
For more details about the manifestly covariant formulation of electrodynamics, see chpt. 4 of

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

Dale and PeroK
carpinus
The components of the em. field in the computational frame are
$$F_{0j}=E_j, \quad F_{jk}=-\epsilon_{jkl} B_l.$$
I am curious: is this true only for cartesian coordinates or for general coordinates?
thank you for the link.

wnvl2
bout the

wnvl2
The electromagnetic field tensor is a tensor. I don't see any reason why it should be expressed in cartesian coordinates. This should be valid for general coordinates.

Gold Member
2022 Award
This is for special relativity in an inertial frame in pseudo-Cartesian coordinates. In

2022 Award
I think that the components of ##F^{\mu\nu}## correspond to components of ##E## and ##B## in a given basis as long as that basis is orthonormal. The spatial basis vectors need to be orthogonal to the timelike one for the separation between ##E## and ##B## to happen, and then any orthonormal basis is related to a local "clocks and rigid rods" frame by a rotation.

If you're thinking of (for example) the ##1/r## terms in the differentials in cylindrical polars, remember that those are dealing with the failure of ##\partial_\phi## to be normalised - a small change ##d\phi## in the ##\phi## coordinate means a change ##rd\phi## in position.

carpinus and vanhees71
Gold Member
2022 Award
Sure, it holds also for local "Lorentzian bases" (tetrads).

Ibix
wnvl2
The electromagnetic field tensor is a tensor. I don't see any reason why it should be expressed in cartesian coordinates. This should be valid for general coordinates.
Apparently everyone disagrees with me. But what happens when you represent the electromagnetic 4 tensor in a non ortho normal reference frame? The components become a mix of what we are used to call electric or magnetic field and the maxwell equations that relate the different components will look different?

2022 Award
Apparently everyone disagrees with me. But what happens when you represent the electromagnetic 4 tensor in a non ortho normal reference frame? The components become a mix of what we are used to call electric or magnetic field and the maxwell equations that relate the different components will look different?
What are you calling Maxwell's equations? If you write them as ##\partial^aF_{ab}=-4\pi j_b## and ##\partial_{[a}F_{bc]}=0## then they're true in any coordinate system. But if you're using strange coordinates the individual terms in the sums (e.g. ##\partial_2F_{13}##) won't necessarily be pure E or B components, and the equations may be linear combinations of what introductory EM would call Maxwell's equations (##\vec{\nabla}\cdot\vec{E}=\rho## etc). But they impose the same constraints.

Last edited:
wnvl2
wnvl2
I referred here to the most simple form of the Maxwell equations such as you assumed already. Your replied confirmed what I was thinking.

carpinus
This is for special relativity in an inertial frame in pseudo-Cartesian coordinates. In
would you like to complete your answer?

ergospherical
The index notation can obscure the idea that the electric and magnetic fields are related to the orthogonal decomposition of the electromagnetic 2-form ##F## with respect to the observer's tangent vector ##u##, viz:\begin{align*}
F = u \wedge E + \star(u \wedge B)
\end{align*}where in this case ##E## and ##B## are expressed as 1-forms. If you act ##F## on a couple of basis vectors then you extract the components
\begin{align*}
F_{ab} &= (u \otimes E - E \otimes u)(e_a, e_b) + \epsilon(u,B, e_a, e_b)
\\
&= u_a E_b - E_a u_b + \epsilon_{cdab} u^c B^d
\end{align*}So far the basis is arbitrary; if the basis is orthogonal (with ##u = e_0##, i.e. ##u_a = - \delta^0_a##) then the previous expression simplifies to
\begin{align*}
F_{ab} &= -\delta^0_a E_b + E_a \delta^0_b + \epsilon_{0dab} B^d
\end{align*}from which you recover the familiar-looking matrix elements.

vanhees71, Ibix and wnvl2
carpinus
I'm pondering over the meaning of the quantity ## ϵ_{abcd} ## occurring in (4.2.22).
In his text Wald says ## ϵ_{abcd} ## is (a totally antisymmetric) tensor. Then according to (4.2.22) ## B_a ## should be a tensor (4-vector) as well. However it seems to me ## ϵ_{abcd} ## is considered to be the Levi-Civita-Symbol.

From (B.2.17) ## \epsilon = \sqrt {|g|} dx^1 \wedge .. \wedge dx^n ## I think it follows that ## ϵ_{abcd} ## generally is a function of the position x in spacetime. I think even for a flat spacetime the metric may vary with x when expressed in curvilinear coordinates (e.g. spherical coordinates). On the other hand, (B.2.14) ## ϵ_{\mu_1,..,\mu_n} = (-1)^p ## , with p the signature of the permutation ## (1,..,n) → (\mu_1,..,\mu_n) ##, looks more like the Levi-Civita-Symbol

wnvl2
Epsilon is called a tensor density of weight 1.

Staff Emeritus
I am curious: is this true only for cartesian coordinates or for general coordinates?
thank you for the link.

In general, there isn't any such thing as "cartesian coordiantes" in GR.

Instead, focus on an "orthonormal set of basis vectors". The basis vectors exist in the tangent space at some point on the four dimensional manifold of space-time. The definition of orthonormal has two parts - the orthogonality of basis vectors, meaning their inner product is zero for two basis vectors which are not the same, and the normal part, which states that the inner product of a basis vector itself has a magnitude of 1, though in general it will have a + or - sign, the sign depending on the sign convention and whether the vector in question is timelike or spacelike. There is generally one timelike vector and three spacelike vectors in an orthonormal basis in the 4 dimensional manifold of GR, because the manifold has a Lorentzian signature.

You should be able to prove that the operation specified by Wald using the Levi-Civita symbols, which is sometimes called the hodges dual, is equivalent to picking out specific components of the Farday tensor in an orthonormal basis. Furthermore, you should be able to demonstrate by choosing a basis that it's not true in a general basis.

Basically, once you revise your statement from the ill-defined notion of "cartesian coordinates" to a statement about basis vectors, it's subject to mathematical proof.

carpinus
thanks to everyone