Fundamental invariants of the electromagnetic field

Question

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.

Answer 1

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 $\Im(F^2)$.

Answer 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.

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

Answer 3

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.

Summary

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:

A.

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).

B.

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.

C.

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)

Answer 4

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.

Answer 5

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 $F\,{}^*F$

Answer 6

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).