# Interpretation of the EM tensor as a rotation matrix

• A
In special relativity, the electromagnetic field is represented 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. Recalling the one-to-one correspondence between skew-symmetric matrices and special orthogonal [rotation] matrices established by Cayley’s transformation, one could think of this tensor as an infinitesimal rotation matrix. As Lorentz boosts can also be interpreted as rotations, I wonder if those two concepts might be related in some way.

Could this correspondence have any physical interpretation? Does it make any sense at all to associate a general rotation of space-time coordinates with a given field? I'd welcome any thoughts/insights on this subject coming from any more knowledgeable person.

jambaugh
Gold Member
In special relativity, the electromagnetic field is represented 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. Recalling the one-to-one correspondence between skew-symmetric matrices and special orthogonal [rotation] matrices established by Cayley’s transformation, one could think of this tensor as an infinitesimal rotation matrix. As Lorentz boosts can also be interpreted as rotations, I wonder if those two concepts might be related in some way.

Could this correspondence have any physical interpretation? Does it make any sense at all to associate a general rotation of space-time coordinates with a given field? I'd welcome any thoughts/insights on this subject coming from any more knowledgeable person.

Given that space-time 4-velocities and 4-momenta for a fixed mass particle have fixed 4-vector norms, all forces and accelerations on the particle will be Lorentz transformations which is to say 4-dim pseudo rotations. So yes, the electromagnet tensor IS the generator of a 4-(pseudo) rotation.
All conservative forces will take this form.

• Nugatory
Given that space-time 4-velocities and 4-momenta for a fixed mass particle have fixed 4-vector norms, all forces and accelerations on the particle will be Lorentz transformations which is to say 4-dim pseudo rotations. So yes, the electromagnet tensor IS the generator of a 4-(pseudo) rotation.
All conservative forces will take this form.

That is, including gravity? Isn't the Einstein tensor in general relativity symmetric?
Could you give another example of correlations between forces and Lorentz boosts?

That is, including gravity? Isn't the Einstein tensor in general relativity symmetric?

Yes it is symmetric, but GR does not use the concept of "force field" to define gravity - it does so via the concept of geodesic deviation. I don't really see a direct connection between the Einstein tensor and forces.

• dahemar4
Yes it is symmetric, but GR does not use the concept of "force field" to define gravity - it does so via the concept of geodesic deviation. I don't really see a direct connection between the Einstein tensor and forces.

I see. So, for instance, a "gravitational field tensor" (I don't know if this is used in some existing theory) would also be anti-symmetric?

jambaugh
Gold Member
Let me add that the gravitational force is not conservative in the same way. In general relativity one must look, not at the particle 4-momenta but a the full stress energy tensor. That is what is conserved. You must therefore look at how the (symmetric) stress energy tensor is transformed under a Lorentz pseudo-rotation. This is encoded in the Riemann's curvature tensor (which has both symmetric and anti-symmetric components, symmetric due to the entity it acts upon and anti-symmetric due to the components of action.) Think of the Riemann tensor as encoding two copies of such a pseudo-rotation tensor, one for each component of the symmetric stress energy tensor.

However in a somewhat frame-dependent way there is a corresponding anti-symmetric tensor to describe the gravitational effect on vector quantities, the Christoffel symbols which when contracted with the particle velocity (the particle's time direction) yield an anti-symmetric 2 tensor generating the pseudo-rotation of the particle's velocity as it travels. But this is frame dependent and frames here are arbitrary coordinate systems not just retilinear inertial frames. You can break the conservat[ive property (expressed in the old way) by choice of coordinates.]

The other conservative gauge forces likewise have a similar gauge charge dependent force tensor. In general:
$$\tfrac{d}{d\tau} P^\mu =q_a F^{a}_{[\mu\nu]} U^\nu$$
where the $q_a$ are components of whatever gauge charge the particle has and $P$ is the particle 4-momentum and $U$ its 4-velocity.

Note I'm being a bit simplistic here and there's no magic mystery in this. It is simply expressing the fact that in SR a force is a pseudo-rotation generator provided it does not change the mass of the particle it acts upon. In the case of gravitation, when its momentum becomes its charge this becomes a Christoffel symbol $q_a\to P_\lambda, F^{a}_{[\mu,\nu]} \to \Gamma^\lambda_{\mu\nu}$.
[EDIT: This isn't quite right, the Christoffel coefficient indices are not in standard notation order here, there's actually a problem with the general force too. To avoid having to pull out the metric we should express force tensor with one raise and one lowered index.]

Now you can interpret this force as geodesic motion but that is one interpretation. You should keep in mind that the whole curved space-time paradigm of GR is a model not the theory itself. The theory only states how objects behave empirically. Whether we attribute all or part of that to geometry is a question of style. This ambiguity is the relativity in GR and is the deepest application of the equivalence principle.

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• dahemar4
jambaugh
Gold Member
Here's my revised equations:
$$\dot{P}^\mu = q^a {F_a}^\mu_\nu U^\nu$$
$$q^a \to P^\lambda, \quad {F_a}^\mu_\nu \to \Gamma^\mu_{\lambda\nu} = \Gamma^\mu_{\nu\lambda}$$

The anti-symmetry in a lorentz transformation only manifests when we use the metric to bring both indices to the same level:
$$\Lambda_{\mu \nu} = g_{\mu\alpha}\Lambda^\alpha_\nu= -\Lambda_{\nu\mu}$$
It is the $\Lambda^\alpha_\nu$ components which are in the form of an operator's components. $dV^\mu = \Lambda^\mu_\nu V^\nu d\tau$
Anti-symmetry in the operator form will only manifest when we have a standard diagional metric, (rectilinear coordinates).

• dahemar4
Thank you for your detailed explanation, I think the point with gravitation is clear now.

I also have understood that a force is a generator of Lorentz transformations provided it does not change the mass of the particle it acts upon, but I've still got some doubts: according to Wikipedia, the generator for a general Lorentz transformation can be written as follows: where
The axis-angle vector θ and rapidity vector ζ are altogether six continuous variables which make up the group parameters (in this particular representation), and J and K are the corresponding six generators of the group. Physically, the generators of the Lorentz group are operators that correspond to important symmetries in spacetime: J are the rotation generators which correspond to angular momentum, and K are the boost generators which correspond to the motion of the system in spacetime.

My first doubt is: from what we just said, shouldn't this matrix be anti-symmetric? On the other hand, if the electromagnetic field tensor is also a generator, then, are ζi and θi in some way related to the magnetic and electric field components, Ei and Bi?

jambaugh
Gold Member
The matrix should be anti-self adjoint with respect to the metric which in this case is not definite. This is in part why we use the raised vs lowered index notation for components. But using matrix notation to map column vectors to column vectors you should think of the components of that matrix as having one raised and one lowered index. In order to see the proper anti-symmetry you need to contract with the metric. I'll again use a matrix but this time not as an operator it has two lowered indices and so when we matrix multiply we'll be lowering one operator index. How that manfests is as so:

Your operator contracted with metric yields:
$$\left(\begin{array}{cccc} -1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1 \end{array}\right)\left[\begin{array}{cccc}0&\zeta_x&\zeta_y&\zeta_z\\ \zeta_x&0&-\theta_z&\theta_y\\ \zeta_y&\theta_z&0&-\theta_x\\ \zeta_z&-\theta_y&\theta_x&0 \end{array}\right)= \left( \begin{array}{cccc}0&-\zeta_x&-\zeta_y&-\zeta_z\\ \zeta_x&0&-\theta_z&\theta_y\\ \zeta_y&\theta_z&0&-\theta_x\\ \zeta_z&-\theta_y&\theta_x&0\end{array}\right)$$

I'm using here a made up notation where a parenthesis side of a matrix indicates its index is lowered (covariant component) and a square bracket side indicates a raised index (contravariant component).

In order to swap indices in a non-frame-dependent way they must be at the same level, indicating you are swapping basis vectors within the same space.

In index notation see the last equation in my last post.

• dahemar4