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How to prove that $\mathbf{E}^2-\mathbf{B}^2$ and $\mathbf{E}\cdot\mathbf{B}$ are the only two independent Lorentz invariant quantities that are constructed by $\mathbf{E}$ and $\mathbf{B}$?

It's easy to prove they are Lorentz invariant quantities and independent from each other because they are $F_{\mu\nu}F^{\mu\nu}$ and $\epsilon_{abcd}F^{ab}F^{cd}$ up to a constant. But how to prove they are the unique two independent Lorentz invariant quantities? i.e. Any other Lorentz invariant quantities constructed by $\mathbf{E}$, $\mathbf{B}$ or $F_{\mu\nu}$ can be represented as a function of $\mathbf{E}^2-\mathbf{B}^2$ and $\mathbf{E}\cdot\mathbf{B}$.

Qmechanic
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346699
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2 Answers2

1

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.

Misha
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$\vec{E}$ and $\vec{B}$ are vectors, and geometrically there are just 3 scalar quadratic combinations of these fields -- $E^2$, $B^2$ and $(\vec{E}, \vec{B})$. The last one is Lorentz invariant by itself, while in order to get the invariant one from the rest, one has to subrtact one from the other. There are also cubic combinations $\epsilon_{ijk}E^i E^j B^k$, $\epsilon_{ijk}E^i B^j B^k$, etc., but all of them vanish due to antisymmetry of $\epsilon$-tensor.

Andrey Feldman
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