# Can we think of the EM tensor as an infinitesimal generator of Lorentz transformations?

I'm asking this question because I'm feeling a bit confused about how Lorentz transformations relate to the electromagnetic tensor, and hope someone can help me clear out my possible misunderstandings. Please excuse me if the answer is obvious.

In special relativity, the EM field is described by the tensor

$$F^{\mu\nu} = \begin{pmatrix}0 & -E_{x} & -E_{y} & -E_{z}\\ E_{x} & 0 & -B_{z} & B_{y}\\ E_{y} & B_{z} & 0 & -B_{x}\\ E_{z} & -B_{y} & B_{x} & 0 \end{pmatrix}$$

which is an anti-symmetric matrix. Then, recalling the one-to-one correspondence between skew-symmetric matrices and orthogonal matrices established by Cayley’s transformation, one could view this tensor as an infinitesimal rotation matrix, that is, a generator of 4-dim pseudo-rotations. This seems at first natural: given that space-time 4-velocities and 4-momenta for a fixed mass particle have fixed 4-vector norms, all forces (including EM) and accelerations on the particle will be Lorentz transformations. However, this page is the unique reference I've found which states such relationship (and I don't fully understand the discussion which follows, which I find somewhat disconcerting).

• Is this line of reasoning correct?

On the other hand, according to Wikipedia, a general Lorentz transformation can be written as an exponential,

$$\mathbf \Lambda(\mathbf ζ,\mathbf θ) = e^{-\mathbf ζ \cdot \mathbf K + \mathbf θ \cdot \mathbf J}$$

where (I'm quoting) $\mathbf J$ are the rotation generators which correspond to angular momentum, $\mathbf K$ are the boost generators which correspond to the motion of the system in spacetime, and the axis-angle vector $\mathbf θ$ and rapidity vector $\mathbf ζ$ are altogether six continuous variables which make up the group parameters in this particular representation (here the group is the Lie group $SO^+(3,1)$). Then, the generator for a general Lorentz transformation can be written as $$-\mathbf ζ \cdot \mathbf K + \mathbf θ \cdot \mathbf J = -ζ_xK_x - ζ_yK_y - ζ_zK_z + θ_xJ_x + θ_yJ_y +θ_zJ_z = \begin{pmatrix}0&-\zeta_x&-\zeta_y&-\zeta_z\\ \zeta_x&0&-\theta_z&\theta_y\\ \zeta_y&\theta_z&0&-\theta_x\\ \zeta_z&-\theta_y&\theta_x&0\end{pmatrix}.$$

• How does this matrix relate with the EM tensor? By comparison between the two matrices, it would appear that the components of the electric and magnetic field ($\mathbf E$ and $\mathbf B$) should be linked, respectively, with $\mathbf ζ$ and $\mathbf θ$. I'm missing what the physical interpretation of this would be.
• There could be something interesting here, but a priori, there's no reason for these things to be linked. The only thing they have in common is that they're both antisymmetric rank 2 tensors. There are a lot of those. – knzhou Nov 14 '16 at 2:00
• @knzhou Well, if I understood correctly the Wikipedia article, the second tensor is a general way to write the generators for a general Lorentz transformation, and I'm asking if the EM tensor can be thought of as a generator... – David Herrero Martí Nov 14 '16 at 2:09

Physically, the only thing that the electromagnetic field tensor and a Lorentz transformation generator have in common is that they both happen to be antisymmetric rank 2 tensors. The link doesn't go any farther than that.

However, this coincidence does lead to a few analogies. For example, if you know about Lorentz transformations, then you know that an antisymmetric rank 2 tensor contains two three-vectors inside it, namely $$\boldsymbol{\zeta}$$ and $$\mathbf{K}$$. Then if somebody tells you the electromagnetic field is the same kind of tensor, you'll automatically know that it can be broken down into two three-vectors, namely the electric and magnetic fields. But this is a purely mathematical analogy.

A more physical result comes from the equation of motion $$\frac{d u_\mu}{d\tau} = (q/m) F_{\mu\nu} u^\nu.$$ where $$u^\mu$$ is the four-velocity; you can expand this in components to verify it's just the Lorentz force law. Now, comparing this with an infinitesimal (active) Lorentz transformation $$\Delta u_\mu = \Lambda_{\mu\nu} u^\nu$$ we see that the Lorentz force is equivalent to an active Lorentz transformation acting on the four-velocity, with generator $$(q/m) F_{\mu\nu}$$.

We can do some quick sanity checks:

• Magnetic fields cause rotations. If we start with a nonzero three-velocity and apply a magnetic field, the velocity spins around.
• Electric fields cause boosts. If we apply an electric field, the three-velocity grows in the direction of the field, just like it does in the direction of a boost.

Two caveats to this result:

• As stated in the link you gave, this result doesn't allow us to think of electromagnetism as a geometric phenomenon, because different particles have different values of $$q/m$$ and hence are acted on by different Lorentz transformations. It's just a nice heuristic.
• Be careful to distinguish between active and passive Lorentz transformations. Most of the ones you'll run into are passive (i.e. used to switch between coordinate systems), but as ACM points out, such transformations are described by matrices, not tensors. Above, I'm considering active rotations and boosts, and everything is taking place in a single coordinate system.
• Thank you for your clear answer, it solved some of my doubts! However, I'm afraid @ACuriousMind is right: I mixed up some of the signs in my generator matrix. The correct one seems to be $\begin{pmatrix}0&\zeta_x&\zeta_y&\zeta_z\\ \zeta_x&0&-\theta_z&\theta_y\\ \zeta_y&\theta_z&0&-\theta_x\\ \zeta_z&-\theta_y&\theta_x&0\end{pmatrix}.$ Is it possible that we should compare it to the mixed-variance form of the EM tensor ($F^{\mu}_{\nu}$) instead? – David Herrero Martí Nov 14 '16 at 2:57
• @DavidHerreroMartí Yes, you need to make sure the indices on both line up. The usual Lorentz transformation has mixed indices. You want to lower the upper index to properly compare it with $F_{\mu\nu}$. If you do this, you end up with antisymmetric components, as expected. – knzhou Nov 14 '16 at 3:00
• This is a great analogy. Nice find! – Javier Nov 14 '16 at 3:18
• The field strength does not generate a Lorentz transformation, the signs are wrong - the boost generators are not anti-symmetric. I therefore don't understand why you're saying that the Lorentz force is "equivalent to a Lorentz transformation". – ACuriousMind Nov 14 '16 at 3:27
• @ACuriousMind Your components are in mixed indices. If you just lower an index, you get a sign flip and the signs work out fine. – knzhou Nov 14 '16 at 3:37

The electromagnetic field strength tensor is not a Lorentz generator.

First, even when written in matrix form, the signs are wrong. The boost generators are of the form $$\begin{pmatrix} 0 & v_x & v_y & v_z \\ v_x & 0 & 0 & 0 \\ v_y & 0 & 0 & 0 \\ v_z & 0 & 0 & 0 \end{pmatrix},$$ which are not antisymmetric.

Second, the EM tensor is not a matrix, it's a 2-form $F = F_{\mu\nu}\mathrm{d}x^\mu \wedge\mathrm{d}x^\mu$, while the Lorentz generators are actual matrices, not coefficients of a form. Writing $F_{\mu\nu}$ as a matrix does not reflect its geometric nature. It does not generate Lorentz transformations, it itself transforms under them as an ordinary 2-tensor.

• Thank you for your answer. Sorry, you're right, I got the signs wrong. However, the relationship I was referring to seems to stand (look at the comments in knzhou's answer). I'm afraid I don't know the difference between a matrix and a 2-form (I'll look it up), but I fail to see why this implies that the EM tensor does not generate Lorentz transformations. – David Herrero Martí Nov 14 '16 at 3:08
• I don't agree with ACM in his last sentence. The EM tensor is an active local Lorentz generator, since it has the exact same effect on a charged particle as a regular active LT. Also, the Lorentz parameters of a regular Lorentz transformation do transform as a 2-tensor under successive Lorentz transformations. It even lead to the Thomas precession. – Cham Dec 21 '18 at 19:34