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Both the elctromagnetic field strength tensor $F^{\mu\nu}$ and its dual $\tilde{F}^{\mu\nu}$ defined as $\tilde{F}^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\lambda\rho}F_{\lambda\rho}$ are examples of antisymmetric tensors of rank two with six independent components in (3+1) dimensional Minkowski spacetime. Both are 6-dimensional irreducible representations of proper Lorentz group ${\rm SO(3,1)}$. Let us define the objects $$ A^{\mu\nu}_{\pm}=F^{\mu\nu}\pm i \tilde{F}^{\mu\nu} $$

Questions

  1. What causes $A^{\mu\nu}_{\pm}$ to have three independent components rather than six? Do $A^{\mu\nu}_{+}$ and $A^{\mu\nu}_{-}$ satisfy additional constraints with further cut down their number of independent components?

  2. Can we attach physical meanings to the components of $A^{\mu\nu}_{\pm}$?

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    $\begingroup$ Where did you read that $A^{\mu\nu}_\pm$ have three independent components rather than six? $\endgroup$ – AccidentalFourierTransform Nov 17 '18 at 15:29
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    $\begingroup$ @SRS Surely you must realise your comment makes no sense. I asked why $A^{\mu\nu}_\pm$ has three independent components, and you responded that it does because it is an irrep of SO(3,1). But in the OP you ask why $A^{\mu\nu}_\pm$ has three independent components yourself, so you already know the answer to that question. $\endgroup$ – AccidentalFourierTransform Nov 17 '18 at 15:38
  • $\begingroup$ @SRS You didn't answer my previous comment. My point (which I had left for you to discover) was that your statement "Both are 6-dimensional irreducible representations of proper Lorentz group SO(3,1)." is false. An antisymmetrical tensor is a reducible representation of proper group: it is $(1,0)\oplus(0,1)$. It is irreducible for orthochronous group, i.e. the one also including space reflection. $\endgroup$ – Elio Fabri Nov 17 '18 at 17:14
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    $\begingroup$ @ElioFabri Sorry I overlooked the comment. The proper Lorentz group (as far as I know) is the one for which $\det \Lambda=+1$. It excludes time-reversal, parity and reflections. I am using Maggiore's QFT book where they show that for ${\rm SO(3,1)}$ $\textbf{16}$ dimensional representation (a tensor of rank 2) is reducible to $\textbf{16}=\textbf{1}\oplus \textbf{6}\oplus \textbf{9}$ where the representations on the RHS are irreducible. Maggiore page 21, Eq. 2.36. $\endgroup$ – SRS Nov 17 '18 at 21:53
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    $\begingroup$ Related question here. $\endgroup$ – knzhou Nov 19 '18 at 16:54
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The existing answers are good, but they kind of beat around the point and/or overcomplicate things, I think.

In short:

Do $A^{\mu\nu}_{+}$ and $A^{\mu\nu}_{-}$ satisfy additional constraints with further cut down their number of independent components?

Yes. Your $A^{\mu\nu}_{\pm}$ tensors have an additional symmetry that's not present in either of $F^{\mu\nu}$ or its dual, and that additional symmetry means that you need less components to reconstruct the full tensor.

So, what is this symmetry? It's called electric-magnetic duality, which is the fancy name for the idea that the replacements \begin{align} \mathbf E & \mapsto \mathbf B \\ \mathbf B & \mapsto -\mathbf E \end{align} will leave Maxwell's equations untouched, so it's a full symmetry of the theory. From the field-tensor expressions in md2perpe's answer it's easy to see that this symmetry can be rephrased as \begin{align} F^{\mu\nu} & \mapsto \tilde F^{\mu\nu} = +\frac{1}{2}\epsilon^{\mu\nu\lambda\rho} F_{\lambda\rho} \\ \tilde F^{\mu\nu} & \mapsto -F^{\mu\nu} = +\frac{1}{2}\epsilon^{\mu\nu\lambda\rho} \tilde F_{\lambda\rho} . \end{align} Since you've defined $$ A^{\mu\nu}_{\pm}=F^{\mu\nu}\pm i \tilde{F}^{\mu\nu}, $$ this means that under the duality transformation, your tensor will change to $$ A^{\mu\nu}_{\pm} = F^{\mu\nu}\pm i \tilde{F}^{\mu\nu} \mapsto \tilde{F}^{\mu\nu}\mp i F^{\mu\nu} =\mp i \left( F^{\mu\nu}\pm i \tilde{F}^{\mu\nu} \right) = \mp i A^{\mu\nu}_{\pm} , $$ i.e. it is an eigenvector of the transformation. Since the pure component-based aspect of the transformation requires that we send $$ A^{\mu\nu}_{\pm} \mapsto \frac{1}{2}\epsilon^{\mu\nu\lambda\rho} A_{\pm \ \lambda\rho} , $$ adding the demand that $A^{\mu\nu}_{\pm} \mapsto \mp i A^{\mu\nu}_{\pm}$ be an eigenvalue then implies that there needs to be a set of linear relationships between the tensor's components, $$ \frac{1}{2}\epsilon^{\mu\nu\lambda\rho} A_{\pm \ \lambda\rho} = \mp i A^{\mu\nu}_{\pm} , $$ and it is these linear relationships that are ultimately responsible for knocking out independent components from your tensor.

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We have $$ F^{\mu\nu} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{pmatrix}, \quad \tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z & -E_y \\ B_y & -E_z & 0 & E_x \\ B_z & E_y & -E_x & 0 \\ \end{pmatrix}, $$ so $$ A_+^{\mu\nu} = F^{\mu\nu} + i \tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -E_x-iB_x & -E_y-iB_y & -E_z-iB_z \\ E_x+iB_x & 0 & -B_z+iE_z & B_y-iE_y \\ E_y+iB_y & B_z-iE_z & 0 & -B_x+iE_x \\ E_z+iB_z & -B_y+iE_y & B_x-iE_x & 0 \\ \end{pmatrix} \\ = \begin{pmatrix} 0 & -C_x & -C_y & -C_z \\ C_x & 0 & iC_z & -iC_y \\ C_y & -iC_z & 0 & iC_x \\ C_z & iC_y & -iC_x & 0 \\ \end{pmatrix}, $$ where $\vec{C} = \vec{E} + i \vec{B}.$

We see that $A_+^{\mu\nu}$ has only 3 complex independent components $C_x,\ C_y,\ C_z,$ but it still has 6 real independent components $E_x,\ E_y,\ E_z,\ B_x,\ B_y,\ B_z.$

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It's better for me to summarize my knowledge on the subject.

$\let\Lam=\Lambda \let\u=\uparrow \let\d=\downarrow \def\cL{{\cal L}} \def\R{{\Bbb R}^4} \def\bx{{\bf x}} \def\by{{\bf y}} \def\rT{{\mathrm T}} \def\qLq{\quad\Leftrightarrow\quad}$ Lorentz group $\cL$ is defined as the group of linear maps on $\R$ leaving invariant Minkowski metric: define $$x_\mu = g_{\mu\nu}\,x^\nu \qquad g_{\mu\nu} = \mathrm{diag}(1,-1,-1,-1).$$ If $\bx\in\R$ and $\by=\Lam\bx$ then $$\Lam \in \cL \qLq y^\mu y_\mu = x^\mu x_\mu.$$ Then, if $G=\|g_{\mu\nu}\|$ or, equivalently $$\Lam \in \cL \qLq \Lam^{-1} = G\,\Lam^\rT G.$$

From eq. (1) it can be shown $$\det\Lam = \pm 1 \tag2$$ and $$|\Lam^0{}_0| \ge 1 \tag3$$ both signs being allowed.

This shows that $\cL$ is the union of four unconnected components: $$\cL_+^\u,\quad \cL_+^\d,\quad \cL_-^\u,\quad \cL_-^\d.$$ $\cL_+^\u$ is a subgroup of $\cL$, known as proper orthochronous Lorentz group. The other three aren't, but their unions with $\cL_+^\u$ define different subgroups: proper, orthochronous, orthochorous (?). SO(3,1) is the proper group, which isn't connected. When it comes to representations it is $\cL_+^\u$ the first to be considered, since it is connected.

(Note that I was wrong in my previous comment when I identified the proper group with $\cL_+^\u$. My reference should have been to the latter.)

Why to bring up SL(2,C)? I already wrote about it, but let me repeat for completeness. The reason is $\cL_+^\u$ isn't simply connected and we are in need to also take into account its "two-valued" representations, just as for SO(3) when we are concerned with space rotations. SL(2,C) is to $\cL_+^\u$ exactly what SU(2) is to SO(3).

Irreps of SL(2,C) can be found via its Lie algebra. It can be shown that it is the direct product of two copies of su(2) (the Lie algebra of SU(2)). Then irreps of SL(2,C) are characterized by two quantum numbers, with integer or half-integer values: $(j,j')$. The order of this irrep is $(2j+1)(2j'+1)$. Note that when $j=j'$ we have a rep equivalent to a real one. If $j\ne j'$ a complex rep results.

Examples:

  • $(0,0)$ is a trivial rep, the one of Lorentz scalars.
  • $(1/2,0)$ is 2D complex - these are spinors.
  • $(1/2,0)\oplus(0,1/2)$ is a complex 4D rep, the one of Dirac bispinors. Why to introduce a sum of irreps I'll explain presently.
  • $(1/2,1/2)$ is real 4D. These are Lorentz 4-vectors.
  • $(1,0)$ is complex 3D. Here we are! $A^{\mu\nu}_+$ trasforms according this rep. $A^{\mu\nu}_-$ according the conjugate $(0,1)$ (or maybe the reverse? I didn't check).
  • $(1,1)$ is real 9D, the symmetrical traceless tensors. Traceless because the trace is a scalar by itself, so it must be subtracted away to have an irrep.

I'm left with an answer still waiting: why $(1/2,0)\oplus(0,1/2)$? And what about $(1,0)\oplus(0,1)$? Remember that $\cL_+^\u$ doesn't contain space inversion (whose determinant is $-1$). In order to include it we have to examine $\cL^\u$, the orthochronous group. Since $\cL_+^\u$ is a subgroup of $\cL^\u$, many (not all) irreps of the latter do reduce when viewed as reps of the former. More exactly, the reduction will be into two of halved dimension.

It happens that irreps $(j,j)$ of $\cL_+^\u$ are also reps of $\cL^\u$, whereas it isn't so if $j\ne j'$. In that case we must take the direct sum $(j,j')\oplus(j',j)$. Here is why bispinors: Dirac theory is also invariant under space reflections. And the direct sum $(1,0)\oplus(0,1)$ gives a 6D irrep of $\cL^\u$, $(\vec E + i \vec B, \vec E - i \vec B)$. Rearranging these 6 components you get an antisymmetric tensor. These are irreducible under $\cL^\u$, reducible under $\cL_+^\u$.

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  1. Minkowski signature:

    • The real antisymmetric EM field strength $F_{\mu\nu}$ transforms in a real 6-dimensional irreducible representation ${\bf 6}_{\mathbb{R}}$ of the Lorentz group$^1$ $O(1,3;\mathbb{R})$.

    • The complexified antisymmetric EM field strength $F^{\mathbb{C}}_{\mu\nu}$ transforms in a complex 6-dimensional reducible representation $$(0,1)~\oplus~ (1,0) \tag{1A}$$ of the complexified proper Lorentz group$^1$ $$SO(1,3;\mathbb{C})~\cong~[SL(2,\mathbb{C}) \times SL(2,\mathbb{C})]/\mathbb{Z}_2,\tag{1B}$$ cf. e.g. this and this Phys.SE posts. The split (1A) does not make sense over real vector spaces.

    • The selfdual and anti-selfdual complexified antisymmetric EM field strengths
      $$0~=~F^{\mathbb{\mp}}_{\mu\nu}~:=~ F^{\mathbb{C}}_{\mu\nu}\mp i (\star F^{\mathbb{C}})_{\mu\nu} \tag{1C}$$ transform in the complex 3-dimensional irreducible representations $(0,1)$ and $(1,0)$ of the Lie group (1B). The selfdual and anti-selfdual conditions (1C) each reduce the complex dimension from 6 to 3, cf. OP's title question.

  2. Euclidean signature:

    • The real antisymmetric EM field strength $F_{\mu\nu}$ transforms in a real 6-dimensional irreducible representation ${\bf 6}_{\mathbb{R}}$ of the real orthogonal group$^1$ $O(4;\mathbb{R})$.

    • The real antisymmetric EM field strength $F_{\mu\nu}$ transforms in an real 6-dimensional reducible representation $$(0,1)~\oplus~ (1,0) \tag{2A}$$ of the real special orthogonal group$^1$ $$SO(4;\mathbb{R})~\cong~[SU(2) \times SU(2)]/\mathbb{Z}_2.\tag{2B}$$ The split (2A) makes sense over real vector spaces.

    • The selfdual and anti-selfdual real antisymmetric EM field strengths
      $$0~=~F^{\mathbb{\mp}}_{\mu\nu}~:=~ F_{\mu\nu}\mp (\star F)_{\mu\nu} \tag{2C}$$ transform in the real 3-dimensional irreducible representations $(0,1)$ and $(1,0)$ of the Lie group (2B). The selfdual and anti-selfdual conditions (2C) each reduce the real dimension from 6 to 3, cf. OP's title question.

  3. Under a Wick-rotation, the electric field ${\bf E}$ and the Hodge $\star$ operation are multiplied with an imaginary unit $i$.

  4. See also this related Phys.SE post.

References:

  1. M. Maggiore, A modern intro to QFT, 2005; Eq. (2.36) p. 21, Exercise 2.5 p. 42 & its solution p. 268-269.

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$^1$ Since all pertinent spins here are integers, we don't have to go to the double cover of the Lorentz group.

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