5
$\begingroup$

I was wondering why we need the electromagnetic field tensor $F_{\mu\nu}$ to be a tensor and why can't we work with the electric and magnetic fields while dealing with the electromagnetic field Lagrangian density for flat spacetime. I kind of get the intuition behind it, because you cannot write the transformations as vector equations and need a second-order tensor, but where exactly does the math go wrong? I would prefer a purely mathematical reasoning, if possible.

$\endgroup$
3

3 Answers 3

12
$\begingroup$

The main reason why one needs a (2,0)-tensor is the fact that neither the electrical field $\mathbf{E}$ (3d-vector) nor the magnetic field $\mathbf{B}$ (3d-vector) can be fitted into a 4-vector, i.e. that transforms under fundamental representation of the Lorentz-group.

However, what one can do is the construction of two complex vectors $\mathbf{f^\pm}$

$$\mathbf{f^\pm} = \mathbf{E} \pm i\mathbf{B}$$

Being a (or two) complex vector(s) it transforms under complex rotations like a (of course) complex vector. The interesting fact is now that complex 3D-rotations can be seen as particular representations of the Lorentz-group since ${\cal{L}_+^\uparrow} \cong SO(3,\mathbb{C})$.

With a bit of knowledge of representation theory the anti-symmetric electromagnetic field tensor can be constructed from $\mathbf{f}^\pm$. However, in physics it is more useful to use a real tensor than a complex vector. Therefore the anti-symmetric field tensor is very common in physics whereas it complex counterpart $\mathbf{f}^\pm$ is mainly found in books on representation theory.

$\endgroup$
1
  • 1
    $\begingroup$ For the new students of this subject i want to add that the same construction (building a $\pm$ complex pair of the adeguate dimension) is carried out for all the other fields in relativistic field theory, and that those are the objects that transform well with the Lorentz/Poincare group $\endgroup$
    – LolloBoldo
    Jun 9, 2023 at 10:28
10
$\begingroup$

I think it's important to emphasize a point which was never emphasized to me when I was taking my courses. When a theory such as electromagnetism can be formulated in (roughly) equivalent ways at varying levels of sophistication, the more sophisticated forms should make your life easier, not harder.

Now, generally speaking this benefit only comes when you actually understand the added sophistication from both a physical and mathematical point of view, so for a time it may seem more complicated. But when that understanding has been achieved, there will be a moment where you suddenly see through the symbols and recognize everything as simpler than you first thought. The higher mathematics is there to serve you, not the other way around.


To put this in practice with electromagnetism, start from a relativistic standpoint and imagine that there was a force which is a linear function of the 4-velocity - that is, something of the form

$$\dot{\mathbf p} = qF \mathbf u$$

where $\mathbf p = m\mathbf u$ is the 4-momentum, $\mathbf u$ is the 4-velocity, and $q$ is some coupling constant. What are the properties of this $F$, which must be expressed as a matrix of some kind? Well, the first thing to point out is that because $\mathbf p\cdot \mathbf p = m^2c^4$ is constant, it follows that $\mathbf p \cdot \dot{\mathbf p} = 0$. But this requires that $$\mathbf u \cdot F\mathbf u = u^\mu\big(g_{\mu \alpha} F^\alpha_{\ \ \nu} \big)u^\nu \equiv F_{\mu\nu} u^\mu u^\nu =0$$ for any possible (timelike) 4-vector $\mathbf u$. The simplest way to satisfy this is to require that $F_{\mu\nu}$ be antisymmetric.

In any specific frame of reference, such a matrix acting on the 4-velocity can be broken up into a piece which depends on the ordinary 3-velocity $\vec v$ (in that frame) and a piece which does not. It's a straightforward exercise to show that

$$qF\mathbf u = q\gamma \pmatrix{\sum_{i=1}^3 F^0_{\ \ i} v^i \\ F^1_{\ \ 0} + \sum_{i=1}^3 F^1_{\ \ i} v^i \\ F^2_{\ \ 0} + \sum_{i=1}^3 F^2_{\ \ i} v^i \\ F^3_{\ \ 0} + \sum_{i=1}^3 F^3_{\ \ i} v^i }$$

This suggests that, for organizational purposes, we might define

$$F^i_{\ \ 0}\equiv E^i \qquad F^i_{\ \ j} = \mathcal B^i_{\ \ j}$$

where $\vec E$ is a 3-vector and $\mathcal B$ is a 3$\times$3 matrix with the property that $\mathcal B^i_{\ \ j} = -\mathcal B^j_{\ \ i}$ for each $i,j$. With this notation being defined, we have

$$F\mathbf u = q\gamma \pmatrix{\vec E \cdot \vec v \\ \vec E + \mathcal B\vec v}$$

Finally, observe that in 3 dimensions, the action of an antisymmetric matrix on a vector can be re-expressed as as a cross product: $$\pmatrix{0 & \mathcal B^1_{\ \ 2} & \mathcal B^1_{\ \ 3} \\ -\mathcal B^1_{\ \ 2} & 0 & \mathcal B^2_{\ \ 3}\\ -\mathcal B^1_{\ \ 3} & -\mathcal B^2_{\ \ 3} & 0}\pmatrix{v^1\\v^2\\v^3} = \pmatrix{\mathcal B^1_{\ \ 2}v^2 + \mathcal B^1_{\ \ 3} v^3 \\ - \mathcal B^2_{\ \ 1} v^1 + \mathcal B^2_{\ \ 3}v^3\\ -\mathcal B^1_{\ \ 3} v^1 - \mathcal B^2_{\ \ 3} v^2 } = \vec v \times \vec B $$ where we have defined the 3-vector $\vec B \equiv (-\mathcal B^2_{\ \ 3}, \mathcal B^1_{\ \ 3}, -\mathcal B^1_{\ \ 2})$. So finally we have obtained

$$F\mathbf u = \gamma \pmatrix{q\vec E \cdot \vec v\\ q\big(\vec E + \vec v\times \vec B\big)}$$


The main lessons from this go as follows.

  1. The fact that the force is linear in $\mathbf u$ and preserves the rest mass of the particle is sufficient to show that in any specific reference frame, it can be decomposed into timelike and spacelike parts - or put differently, into parts proportional to $\vec v$ and parts which are not.
  2. This splitting is obviously frame-dependent, because $\vec v$ changes from frame to frame.
  3. The $\vec v$-dependent part is the action of an antisymmetric matrix $\mathcal B$ on $\vec v$. The fact that this can be expressed as the cross product of $\vec v$ with a vector $\vec B$ is an accident of 3-dimensions, essentially deriving from the fact that an $n\times n$ antisymmetric matrix has $n(n-1)/2$ independent components.
  4. If we want Lorentz-invariant quantities, we should obtain them at the level of $F$. Examples include $F^2 \equiv F_{\mu\nu}F^{\mu \nu} = 2(B^2-E^2/c^2)$ and $\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta} \propto \vec E \cdot \vec B$.

If we wanted to go further and endow this $F$ with dynamics of its own, we could start to think about the simplest, Lorentz-covariant differential equations we could write down. In what may or may not be surprising, the Maxwell equations take the rather straightforward form $$\nabla_\mu F^{\mu\nu} = \mu_0 J^\nu \qquad \epsilon^{\mu \alpha\beta\gamma}\nabla_\alpha F_{\beta\gamma} = 0$$ where indices have been freely lowered and raised with the metric. The latter can be re-interpreted by defining the dual tensor $$G^{\alpha\mu} = \frac{1}{2}\epsilon^{\alpha \mu \beta \gamma}F_{\beta \gamma}$$ in which case the second equation becomes $\nabla_\mu G^{\mu\nu} = 0$. These equations can be discussed and motivated at various levels as well, but that's beyond the scope of this answer.

$\endgroup$
1
  • $\begingroup$ I have never seen the electromagnetic tensor been explained/motivated so clearly! Very nice $\endgroup$ Oct 13, 2023 at 10:00
7
$\begingroup$

You can if you like work with a pair of 3-vector fields and you can fully and correctly describe classical electromagnetism in all its special relativistic correctness. Everything will be there, nothing missed out, and all relativistic predictions are available.

It also possible for a left-handed player to play tennis with the racket in their right hand but they will not play very well.

By realising that $\bf E$ and $\bf B$ are parts of a tensor one gains insight and easier methods of deduction. One immediately obtains a couple of scalar invariants and one can write down the transformation laws with almost no thought needed. (To derive them using vector $\bf E$ and $\bf B$ can be done but it takes a couple of pages of algebra and careful reasoning.) One can also manipulate stress-energy quantities without needing to know which part is electric, which magnetic.

If the question was, why do we need a second-rank object; why can't we just use first-rank tensors? Then the answer is that the field equations (Maxwell equations) and Lorentz force equation are what they are. One can argue that they are the simplest equations that allow a field that does not change the rest mass of objects it exchanges energy-momentum with and that allow a conserved scalar charge, all the while respecting Lorentz covariance. If you try for a scalar field or a 4-vector field you don't get these properties except possibly by much less simple field equations.

$\endgroup$
6
  • $\begingroup$ You are essentially saying "it is what it is" when in fact, we have good reasons for it. $\endgroup$ Jun 9, 2023 at 13:54
  • $\begingroup$ I thought I gave plenty of reasons, including, for example, the whole of my third and fourth paragraphs, especially the latter. $\endgroup$ Jun 9, 2023 at 15:05
  • $\begingroup$ Those are not. You even specifically said "are what they are". That is not a way of writing if you are interested in giving reasons. $\endgroup$ Jun 9, 2023 at 15:25
  • $\begingroup$ In your remark "in fact, we have good reasons for it" what reasons do you mean (in addition to the ones I mention of course)? $\endgroup$ Jun 9, 2023 at 15:59
  • $\begingroup$ Check something like Frederic Thomas's answer above. e.g. we worked out all the possible SR invariant wave equations, and discovered that the only way to write down a gauge invariant spin 1 field respecting SR invariances is the Maxwell equations, and that results in almost-a-4vector potentials or the Faraday rank 2 tensor. There are a myriad of other similar answers, like the discovery of how E and B fields transform, i.e. not at all like a 4-vector, and thus necessitating a different interpretation. Not what you wrote. $\endgroup$ Jun 9, 2023 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.