It is a standard exercise in relativistic electrodynamics to show that the electromagnetic field tensor $F_{\mu\nu}$, whose components equal the electric $E^i=cF^{i0}$ and magnetic $B_i=-\frac12\epsilon_{ijk}F^{jk}$ fields in the taken frame of reference, has two Lorentz invariant quantities, $$\frac12F^{\mu\nu}F_{\mu\nu}=\mathbf{B}^2-\mathbf{E}^2$$ and $$\frac14F_{\mu\nu} {}^\ast F^{\mu\nu}=\frac14\epsilon^{\mu\nu\alpha\beta }F_{\mu\nu}F_{\alpha\beta}=\mathbf{B}\cdot\mathbf{E}.$$

There is, however, a further Wikipedia article which states that these two quantities are fundamental, in the sense that any other invariant of this tensor must be a function of these two. While I find this plausible, I have never seen a proof of this fact, and it is absent from e.g. Jackson. Is there a simple proof of this fact? I'm particularly interested in higher-order invariants, but I would also like answers to include a proof that these are the only two bilinears.

To be more precise, I would like to see a proof that

Any function $I:F_{\mu\nu}\mapsto I(F)\in\mathbb{R}$ that takes electromagnetic field tensors to real scalars and which is Lorentz invariant (i.e. $I(\Lambda_{\mu}^\alpha \Lambda_{\nu}^\beta F_{\alpha\beta})= I(F_{\mu\nu})$ for all Lorentz transformations) must be a function $I(F)=I'(F^{\mu\nu}F_{\mu\nu},F_{\mu\nu}\ {}^\ast F^{\mu\nu})$ of the two fundamental invariants described above.

If there are multiple ways to arrive at this result, I would also appreciate comments on how they relate to each other.

  • $\begingroup$ @NijankowskiV. : The question is about high-order invariants (cubic, quartic, etc...). $\endgroup$
    – Trimok
    Nov 25, 2013 at 12:50
  • 2
    $\begingroup$ @EmilioPisanty : I wonder if there is a relation with the fact that the number of the independent Casimir operators depends only on the symmetry group, here the Poincaré group, and it does not depend on the representation. And we know that, there are only 2 independent Casimir for the Poincaré group... $\endgroup$
    – Trimok
    Nov 25, 2013 at 13:03
  • $\begingroup$ @Trimok While that may have something to it, any such explanation would have to either (a) take into account the two-tensor-ness and asymmetry of $F$, or (b) also produce strictly two independent invariants for tensors of any rank and symmetry. Whilst not implausible, (b) sound unlikely to me, whereas (a) sounds a bit removed from that. $\endgroup$ Nov 25, 2013 at 13:12
  • 4
    $\begingroup$ Perhaps they don't use the whole generality you seem to be looking for but have you checked arXiv:1309.4185? $\endgroup$
    – user34134
    Nov 25, 2013 at 16:48
  • $\begingroup$ Comments to the question (v3): (i) One could perhaps mention explicitly that the real scalars should also be gauge invariant (beside of being Lorentz invariant) to exclude e.g. $A_{\mu}A^{\mu}$. (ii) What about spacetime derivatives of $F_{\mu\nu}$? Should they be allowed? E.g. $F_{\mu\nu} \Box F^{\mu\nu}$, $\partial_{\lambda}F_{\mu\nu}\partial^{\lambda}F^{\mu\nu}$, etc? (iii) What about other dimensions than 4D? $\endgroup$
    – Qmechanic
    Nov 28, 2013 at 21:15

6 Answers 6


Here is the proof taken from Landau & Lifshitz' "Classical Theory of Fields":

Take the complex (3)-vector: $$ \mathbf{F} = \mathbf{E}+i\, \mathbf{B}. $$ Now consider the behavior of this vector under Lorentz transformations. It is easy to show that Lorentz boosts correspond to rotations through the imaginary angles, for example boost in $(x,t)$ plane: \begin{gather} F_x=F'_x,\\ F_y = F'_y \cosh \psi - i F'_z \sinh \psi = F'_y \cos i \psi - F'_z \sin i \psi. \\ F_z = F'_z \cos i \psi + F'_y \sin i \psi, \end{gather} where $\tanh \psi = \frac{v}c$, correspond to rotation of $\mathbf{F}$ through imaginary angle $i \psi$ in the $(y,z)$ plane.

Overall, the set of all Lorentz transformations (including also the purely spacial rotations) is equivalent to the set of all possible rotations through complex angles in three-dimensional space (where the six angles of rotation in four-space correspond to the three complex angles of rotation of the three-dimensional system).

The only invariant of a vector with respect to rotation is its square: $\mathbf{F}^2 = E^2 - B^2 + 2 i (\mathbf{E}\cdot \mathbf{B})$ thus the real quantities $E^2-B^2$ and $(\mathbf{E}\cdot \mathbf{B})$ are the only two independent invariants of the tensor $F_{\mu\nu}$.

So in essence, we reduce the problem of invariant of $F_{\mu\nu}$ under Lorentz tranform to invariants of a 3-vector under rotations which is square of a vector (and only it). So any invariant $I(F)$ has to be the function of $\Re(F^2)$ and $\Im(F^2)$.


Here is another proof.

Let us assume that there is another invariant $I_3$ functionally independent of $I_1= E^2-B^2$ and $I_2=\mathbf{B}\cdot\mathbf{E}$. This would mean that

  1. There are pairs of vectors $(\mathbf{E},\mathbf{B})$ and $(\mathbf{E}',\mathbf{B}')$, which have the same $I_1$ and $I_2$ but which cannot be turned into one another by some Lorentz transformation (because they have different values of invariant $I_3$).

  2. If a Lorentz transformation changes a pair $(\mathbf{E},\mathbf{B})$ into $(\mathbf{E}',\mathbf{B}')$, then there is another pair $(\mathbf{E}'',\mathbf{B}'')$ with the same $I_1$ and $I_2$ (but different $I_3$) which no Lorentz transformation can transform into $(\mathbf{E}',\mathbf{B}')$.

It is easy to prove that both 1 and 2 are false. Let's disprove (2). To do that, let us choose the unique particular form of $(\mathbf{E}',\mathbf{B}')$ where $\mathbf{E}'$ and $\mathbf{B}'$ are both parallel to the $x$ axis (and $E_x\ge 0$). This can always be done in case when at least one of $I_1$ or $I_2$ is nonzero with a combination of boost along the mutually orthogonal to $\bf E$ and $\bf B$ direction with the speed $\bf v$ satisfying $$ \frac{\mathbf{v}/c}{1+v^2/c^2}=\frac{[\mathbf{E}\times \mathbf{B}]}{E^2+B^2} $$ and spacial rotation (note, that such a transformation is not unique). Since such transformation exist for all pairs of $(\mathbf{E},\mathbf{B})$ and a pair $(\mathbf{E}',\mathbf{B}')$ is uniquely defined by $I_1$ and $I_2$ we proved that 2 is false. So we have a contradiction and no independent invariant $I_3$ exists.

Note: A special case of $I_1=0$, $I_2=0$ has to be considered separately, but poses no special problems.


A (constructive) proof based on The invariants of the electromagnetic field (arxiv, 2014)

We present a constructive proof that all gauge invariant Lorentz scalars in Electrodynamics can be expressed as a function of the quadratic ones.


Assuming a generalised matrix notation for the tensors in electrodynamics.

A convenient way of classifying all the scalars and pseudoscalars is by writing an invariant of order $n$ (even or odd) in the field strength as:

$$I^{(n)} = F^{\alpha\beta} \cdots F^{\kappa\lambda} I_{\alpha\beta \ldots \kappa\lambda} \space\space\space (n \space \textrm{factors)}$$

where $I_{\alpha\beta \ldots \kappa\lambda}$ is constructed from the only tensor and pseudotensor that are invariant under the proper Lorentz transformations: $\eta_{\mu\nu}$ and $\epsilon_{\alpha\beta\mu\nu}$.

Now there are 3 cases:


The $I_{\alpha\beta \ldots \kappa\lambda}$ does not contain the $\epsilon_{\alpha\beta\mu\nu}$ tensor.

Then the invariants have the generic form:

$$I^{(n)} = Tr(F^q)Tr(F^p) \cdots Tr(F^r)$$

with $p+q+\cdots+r=n$

The anti-symmetry of $F$ implies that $Tr(F^q)=0$ when $q$ is odd.

For even $p$, the parity conservation and the recurrence relation:

$$Tr(F^p) = −\frac{F}{2}Tr(F^{p−2}) + \frac{G}{16}Tr(F^{p−4})$$

implies that the all invariants of this form are reduced to the quadratic invariants (and functions of them).


The $I_{\alpha\beta \ldots \kappa\lambda}$ contains $\epsilon_{\alpha\beta\mu\nu}$ tensor even number of times.

In this case the epsilon anti-symmetric tensors can be reduced according to:

$$\boxed{\;\;\epsilon_{\mu\nu\rho\sigma}\epsilon_{\pi\delta\kappa\lambda } =\det\left[\begin{array}{cccc} \eta_{\mu\pi} & \eta_{\mu\delta} &\eta_{\mu\kappa} &\eta_{\mu\lambda}\\ \eta_{\nu\pi} &\eta_{\nu\delta} &\eta_{\nu\kappa} & \eta_{\nu\lambda} \\ \eta_{\rho\pi} & \eta_{\rho\delta} &\eta_{\rho\kappa} & \eta_{\rho\lambda}\\ \eta_{\sigma\pi} & \eta_{\sigma\delta} &\eta_{\sigma\kappa} &\eta_{\sigma\lambda}\end{array}\right]\;\;}$$

and then handled as in case A.


The $I_{\alpha\beta \ldots \kappa\lambda}$ contains $\epsilon_{\alpha\beta\mu\nu}$ tensor odd number of times.

Similarly to case B. above all but one epsilon factor can be reduced which leads to the generic form:

$$I_{\alpha\beta \ldots \kappa\lambda\mu\nu\pi\delta} = \eta_{\alpha\beta}\cdots\eta_{\kappa\lambda}\epsilon_{\mu\nu\pi\delta} \space\space\space(n-2 \space\space \text{factors})$$

The only invariant with one epsilon tensor reduces to the factor of the generic form:

$$I^{(q+r)} = (F^q)^{\kappa\lambda}\epsilon_{\kappa\lambda\pi\delta}(F^r)^{\pi\delta}$$

which with similar recurrence relations to those of part A reduces to quadratic invariants.

(refer to the paper for details and the recurrence relations)


This is a copy of my answer to another question which was marked as a duplicate of this one.

I must admit that I am not very familiar with Lorentz group, but this kind of questions is definitely for the group theory. From wikipedia I conclude that electromagnetic field tensor transforms under $(1,0)\oplus(0,1)$ representation. The general idea is to find, how many invariants (i.e., $(0,0)$) may be formed from two values which transform under $(1,0)\oplus(0,1)$. So, we need to find the result of the direct producs $\left[ (1,0)\oplus(0,1) \right] \otimes \left[ (1,0)\oplus(0,1) \right]$.

Based on explanation given here I conclude that it is equal to $$ \left[ (1,0)\oplus(0,1) \right] \otimes \left[ (1,0)\oplus(0,1) \right] = $$ $$ [(1,0)\otimes(1,0)]\oplus[(0,1)\otimes(0,1)] \oplus 2\cdot[(1,0)\otimes(1,0)]= $$ $$ [(0,0)\oplus(1,0)\oplus(2,0)] \oplus [(0,0)\oplus(0,1)\oplus(0,2)] \oplus 2 \cdot(1,1) = $$ $$ 2\cdot (0,0) \oplus[(1,0)\oplus(0,1)] \oplus[(2,0)\oplus(0,2)] \oplus 2 \cdot(1,1) $$

The number of scalars ($(0,0)$ representation) in the product is 2. So, we may construct only two scalars out of product of two electromagnetic field tensors.


I think the point is that the only invariant tensors (under proper Lorentz transformations) are $\epsilon^{\mu\nu\alpha\beta}$ and $\eta^{\mu\nu}$, so any invariant will contain some number of powers of $F_{\mu\nu}$ where the indices are contracted (raised) with these two invariant tensors. Because of antisymmetry and symmetry, $\eta$ can only act once on $F$ and $\epsilon$ can't act on the same thing $F$ more than twice. So that reduces the possibilities to powers of $F^2$ and $F\,{}^*F$


Let us calculate the characteristic polynomial of the tensor:

$${\mathbf{F}^a}_b = \begin{pmatrix} 0 & E_x & E_y & E_z \\ E_x & 0 & -B_z & B_y \\E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix}$$

which turns out to be:

$$p_{\mathbf{F}}(\lambda) = \lambda^4 - (\boldsymbol{E}^2-\boldsymbol{B}^2) \lambda^2 - (\boldsymbol{E}\cdot\boldsymbol{B})^2$$

It is a well know result that that finding the invariants of a tensor can be reduced to finding the invariants that are polynomial functions of its coordinates, see Zheng (1994).

  • $\begingroup$ This places a lot of weight on the cited reference. Can you expand a bit on the nature of the result reported there? $\endgroup$ Jul 6, 2022 at 22:16
  • $\begingroup$ Esentially, for every invariant $f(F)$ we can found a polinomial invariant $p(F)$ such that: $f(F) = \phi\circ p(F)$, so non-polynomial invariants can be skipped under reasonable conditions. $\endgroup$
    – Davius
    Jul 6, 2022 at 22:47

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