In the context of Maxwell's equations, I was wondering whether there was any physical significance to the dual EM Field Tensor and/or its various derivations. It has components: $$\tilde{\textbf{F}} = \begin{bmatrix} 0 & B_1 & B_2 & B_3 \\ -B_1 & 0 & -E_3 & E_2 \\ -B_2 & E_3 & 0 & -E_1 \\ -B_3 & -E_2 & E_1 & 0 \end{bmatrix}.$$ I have seen two ways of deriving it, one which uses index lowering via the Minkowski Metric [1]: $$\tilde{F}_{\mu\nu} = \eta_{\mu\alpha}F^{\alpha\beta}\eta_{\beta\nu}$$ and another which uses the Levi-Civitia Tensor [2, P.111]: $$\tilde{F}_{\mu \nu} = \frac{1}{2}\epsilon_{\mu \nu \alpha \beta}F^{\alpha \beta}.$$ [I am also aware it can be reached via the substitution $ E_m \to -B_m$ and $B_m \to E_m$ however it seems as though this is just a consequence of the above definitions.]

Are both these definitions just formalisms so the dual EM tensor has the form it does so it can be used in certain physical equations (such as the simplification of Maxwell's equations: $\partial_\mu\tilde{F}_{\mu \nu}$) or is there any physical intuition to the way(s) in which it is defined?

  • $\begingroup$ Related: physics.stackexchange.com/q/331250/50583, physics.stackexchange.com/q/313091/50583 and possibly this answer of mine. $\endgroup$
    – ACuriousMind
    Aug 18, 2017 at 23:56
  • 4
    $\begingroup$ I don't think what you wrote about the definition of the dual by lowering indices is correct. A tensor with lowered indices is just the same tensor with lowered indices, and it is definitely not the same as the [Hodge dual][1], which is what you get when you contract with the Levi-Civita tensor, and is the correct definition of the dual electromagnetic tensor. Do you have a reference for that? [1]: en.wikipedia.org/wiki/Hodge_isomorphism $\endgroup$
    – user166321
    Aug 19, 2017 at 0:43
  • $\begingroup$ @Loopy see the updated question with citations added $\endgroup$ Aug 19, 2017 at 10:50
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    $\begingroup$ You must have misread something in that Wikipedia page, because it shows the covariant tensor and it clearly doesn't have E and B switched. $\endgroup$
    – Javier
    Aug 19, 2017 at 14:28

2 Answers 2


To understand the reason for defining the dual electromagnetic tensor, you need to understand electromagnetism in the language of differential forms. A highly recommended book for that is Gauge Fields, Knots and Gravity by Baez and Muniain.

When you read that book you will see that the electromagnetic tensor $F$ is a 2-form. In simple terms, an $n$-form can be described as a tensor with $n$ antisymmetric indices, and so it makes sense that $F_{\mu\nu}$ is a 2-form if it's antisymmetric in $\mu$ and $\nu$ (which it is).

In $d$ dimensions, there is an operation called the Hodge dual and denoted by a star $\star$, which takes an $n$-form $A$ to a $(d-n)$-form $\star A$. So, in the special case of the electromagnetic tensor $F$ in 2 dimensions, we have $n=2$ and $d=4$, and the dual $\star F$ of the electromagnetic tensor is another 2-form defined as follows:

$$(\star F)_{\mu\nu} \equiv \frac{1}{2}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}.$$

Now, in the language of differential forms, Maxwell's equations can be very expressed elegantly as:


$$d \star F = \star J,$$

where $J$ is the current, which is a 1-form. The first line gives you the first two Maxwell's equations in terms of $E$ and $B$, and the second line gives you the two other Maxwell's equations in terms of $E$ and $B$.

Crucially, it is impossible to get all 4 Maxwell's equations using $F$ alone; its dual $\star F$ must be used in the second equation.

I urge you to read the book I recommended above for more information; all of this, and much more, is explained very clearly there.

  • $\begingroup$ Technically, the source equations can be written neatly as $\delta F = j$, so you don't use the dual ⋆F, just the original form. $\endgroup$
    – DanielC
    Aug 19, 2017 at 18:53
  • $\begingroup$ Yes you do. That's just notation. Since $\delta$ is defined as $\delta \equiv \star d \star$ (up to a minus sign), we have $\delta F = \star d \star F$, and you are still using the dual $\star F$, you're just "hiding" it inside the definition of $\delta$... $\endgroup$
    – user166321
    Aug 19, 2017 at 19:19
  • $\begingroup$ Can't you write the second equation as $\partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta} =0$? I see this equivalent... $\endgroup$
    – FGSUZ
    Feb 16, 2018 at 19:05

Even though you've gotten an answer to this question, I thought it might also interest you that there is a whole underlying space which gives another perspective on the Hodge operator as in some sense a 90 degree rotation of the problem about some not-immediately-apparent axis.

Spin space

So there exists a more fundamental theory out of which the Minkowski space naturally emerges, and it can be called "spinor algebra" or "spin space" or what have you. The basic idea is that we take the Pauli matrices $\sigma_{1,2,3}$ that you might be familiar with from quantum mechanics, adjoin the identity matrix as $\sigma_w$, and then take the components of a worldvector $v^\mu$ in Minkowski space and form the 2x2 Hermitian matrix $V = \sigma_w v^w + \sigma_x v^x + \sigma_y v^y + \sigma_z v^z.$ (As usual, $w=ct.$)

This seems awfully arbitrary at first glance but it becomes less arbitrary when one realizes that $\det V = v_\mu~v^\mu$ in the $(+~-~-~-)$ metric. Since the definition of Lorentz transform is "anything which preserves the metric" we now have recast this as "anything which maps these Hermitian matrices to other Hermitian matrices with the same determinant", and we find out that the Lorentz transforms naturally fall into four separate populations: the "main one" is continuous with the identity transform and takes the form $V\mapsto L V L^\dagger$ where $L$ is a 2x2 complex matrix with determinant 1; the other populations come by composing these with the two parity transforms $V \mapsto -V$ and $V \mapsto (\det V)V^{-1}.$ So this is the connection between the Lorentz group and the group $\text{SL}(2, \mathbb C),$ seen directly. We also get a very nice theory benefit that the subgroup of rotations is precisely the case where $L$ is unitary.

If we look carefully at what $L$ embodies a rotation by $\theta$ about the $z$-axis we chose above, we can discover that it is in fact $L = \exp(i\sigma_z~\theta/2) = \cos(\theta/2) ~I+ i\sin(\theta/2)\sigma_z,$ having the property that after a rotation by $2\pi$ we do not end at $I$ but rather at $-I.$ Of course the fact that the transform is $V\mapsto L~V~L^\dagger$ means that the two minus signs cancel out, but not on the natural vectors of this matrix's space. So for example if we have a null vector $\det V = 0$ we know that that's a projection and it can be written as an outer product, $V = v~v^\dagger,$ but a 360 degree rotation maps $v \mapsto - v$ and therefore $v$ is a spinor, this property of "after a 360 degree rotation we get back the negative of what we started with" being the defining property of spinors. So the natural vector space which these $V$ matrices live atop is this spin-space and its vectors are spinors with respect to the group elements which rotate our 3D space. We can see that we need 4 complex components that we can call $v^{00}, v^{01}, v^{10}, v^{11},$ to describe our 4-vector, but they are related by the Hermitian restriction.

To be able to express this restriction in our formalism we have to imagine that $\psi^a$ lives in a spinor space $\mathcal S^a$ and has a canonical conjugate $\bar\psi^{\bar a}$ living in the spinor space $\mathcal S^\bar a.$ This canonical conjugation relation distributes over products and sums the straightforward way, and over scalar multiples by $\bar{\alpha~\psi^a} = \alpha^*~\bar\psi^{\bar a}.$ And then we can say that our real 4-vectors are these 2-spinors $v^{a\bar a}$ with a pair of a normal and barred index, and they are real in the sense that $v^{a\bar a} = \bar v^{a\bar a}.$ Note that the barred spaces and the unbarred ones cannot be easily confused with each other (it leads to pretty immediate type-errors) so we can just allow barred and unbarred indices to commute freely with each other in spinor expressions.

We must therefore represent an index of the worldvector space $v^\alpha$ with a pair of related indices $v^{a\bar a},$ where the bar indicates that on conjugation this symbol $\bar a$ is supposed to correspond to the symbol $a$. In other words, we define the operation $\bar{p^A} = \bar p^{\bar A}$ as a complex-conjugation operation on the spin space $\bar{\alpha p^a + \beta q^a} = \bar p^{\bar A}$with Without loss of generality we can then commute barred indices with unbarred ones, one just belongs to the "conjugate space".

This also allows you to take an interesting "metric square root": if we imagine that the inner product $g_{\alpha\beta} u^\alpha v^\beta$ is embodied by a spin-space tensor $\epsilon_{ab}\epsilon_{\bar a\bar b}~u^{a\bar a}v^{b\bar b}$ we can find that a natural choice resulting in our $(+~-~-~-)$ metric would simply be written $\epsilon_{ab} = \begin{bmatrix}0&1\\-1&0\end{bmatrix}$. In other words we naturally find that if we give the spin-space an orientation tensor, it magically becomes a metric on the world-space.

It also naturally orients Minkowski space at the same time, with $$\epsilon_{\alpha\beta\gamma\delta} = \epsilon_{a\bar a b\bar b g\bar g d\bar d} = i\epsilon_{ag}\epsilon_{bd}\epsilon_{\bar a\bar d}\epsilon_{\bar b\bar g} - i \epsilon_{ad}\epsilon_{bg}\epsilon_{\bar a\bar g}\epsilon_{\bar b\bar d}.$$

What antisymmetric tensors look like in spin-space.

Your antisymmetric valence-[0, 2] world-tensor obeys $F_{\alpha\beta} = -F_{\beta\alpha},$ which in spin-space looks like $$F_{a\bar ab\bar b} = -F_{b\bar ba\bar a}.$$

Now commute the barred indices to the end and use this to replace it with the sum of two halves of itself, $$F_{ab\bar a\bar b} = \frac12 F_{ab\bar a\bar b} - \frac12 F_{ba\bar b\bar a}.$$ Adding and subtracting the "backwards" mixture $\pm F_{ab \bar b \bar a}$ to this expression with the associated symmetry; one has: $$F_{ab\bar a \bar b} = \frac12 (F_{ab\bar a\bar b} - F_{ab\bar b \bar a}) + \frac12 (F_{ab\bar b\bar a} - F_{ba\bar b \bar a}).$$ Now a theorem in the spin space is that if a pair of normal (either both unconjugated or both conjugated) indices is antisymmetric it can be "boiled down" to a simple $\epsilon_{ab}$ term, and here we see that appearing in two places, $$F_{ab\bar a \bar b} = \varphi_{ab}~\epsilon_{\bar a \bar b} + \epsilon_{ab}~\bar \varphi_{\bar a \bar b},$$where the requirement that these be conjugate 2-spinors comes from the fact that $F$ is real. One can also work out that the 2-spinor $\varphi_{ab}$ is symmetric and therefore it has 3 complex degrees of freedom, or 6 real ones.

Applying the orientation we can write, $$\epsilon_{\alpha\beta\gamma\delta}~F^{\gamma\delta} = (i\epsilon_{ag}\epsilon_{bd}\epsilon_{\bar a\bar d}\epsilon_{\bar b\bar g} - i \epsilon_{ad}\epsilon_{bg}\epsilon_{\bar a\bar g}\epsilon_{\bar b\bar d})(\varphi^{gd}~\epsilon^{\bar g \bar d} + \epsilon^{gd}~\bar \varphi^{\bar g \bar d}).$$ Here the raising of indices has been done with $\epsilon^{ab}$ defined so as to make $\epsilon^{ab}~\epsilon_{cb} = \delta^a_c,$ and this will in turn collapse those expressions to: $$\epsilon_{\alpha\beta\gamma\delta}~F^{\gamma\delta} = -2i\varphi_{ab}~\epsilon_{\bar a \bar b} - 2i\epsilon_{ab}\bar\varphi_{\bar a\bar b}.$$ In other words the electromagnetic field is a [0 2]-spinor, and there is an operation which we could call Hodge rotation, $\varphi_{ab} \mapsto e^{-i\theta} \varphi_{ab},$ and the Hodge dual is just the 90-degree Hodge rotation. The fact that it is important is really saying that we need all parts of this 2-spinor, not just the "real part" as naturally seen in the world-tensor, to describe its evolution and scalar invariants.

It also helps to understand this by forming the Lorentz invariants $\varphi_{ab}~\varphi^{ab}.$ If you look carefully the product $F_{ab}F^{ab}$ is only going to give you the real part of this, which is $E^2 - B^2.$ But there's another Lorentz invariant out there, which is the imaginary part $E \cdot B.$ And this Lorentz invariant only comes naturally out of $F_{ab} \tilde F^{ab}$. So you need this orientation tensor to properly show you this extra Lorentz invariant that was "hidden" from you before.


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