What is the intepretation of the electromagnetic tensor?

Question

Let $A$ be the four-potential, then we know that we can form the electromagnetic tensor as $F=dA$. This is usually done as a way to have a better writing of Maxwell's equations. So, to simplify the equations and make then covariant we simply notice that we can join electric and magnetic potentials in a single one and then take the derivative and then we find Maxwell's equations from $d^2A = 0$ and all of that.

That's all fine, but $F$ is a differential $2$-form and such objects are highly geometrical. What is then the interpretation of the electromagnetic tensor? I know that $2$-forms represents "objects that perform $2$d measures", but in this case what does $F$ measures?

Until now all approaches I've seem to introduce this tensor were mainly to rewrite something in a better way.

Answer 1

If you've studied electric and magnetic fields before then you know they are related to forces on charged particles. However, if you look at how the electromagnetic field $F$ interacts with charges to produce forces you see the metric get involved. So the metric is essential to the field as a thing that exert forces, which is the actual definition.

There is a branch of mathematics that has higher rank objects and metric all rolled together, it is called Clifford Algebra. And as far as I know it is the only known way to do relativistic quantum mechanics, so it is unavoidable, you'll have to learn it eventually.

And then there is a particular applied version called geometric algebra (or geometric calculus) where you make additional products out of the original one in a way designed to respect both the geometry and the metric.

In that setting all you need to start is an orthonormal basis e.g. four vectors, $e_0=\hat t,$ $e_1=\hat x,$ $e_2=\hat y,$ and $e_3=\hat z.$ Then you assert that $e_ie_j+e_je_i=\pm 2\delta_{ij},$ where $\delta_{ij}$ is the Kronecker delta and $-e_0e_0$=$-1$=$e_1e_1$=$e_2e_2$=$e_3e_3.$ So orthogonal vectors anticommute. Then you can make an associative and distributive algebra by taking the space of linear combinations of products of vectors. So for instance $3e_0e_0+4e_2+5e_1e_3+6e_1e_2e_3+7e_0e_1e_2e_3$ is a linear combination of products of vectors. As an algebra it is just like the Dirac algebra (or you could start with objects like $e_0e_1,$ $e_0e_2,$ and $e_0e_3$ and note they square to 1 and anticommute so linear combinations of products of then are identical ... as an algebra ... to the Pauli algebra) but you don't need to pick a representation because there is no law that says if you have an algebra you must pick a representation.

Now you can define any linear combination of the original vectors as a vector $v=\alpha e_0+\beta e_1 +\gamma e_2+\delta e_3.$ And to make it more than a Clifford Algebra we can define additional products, for instance for vectors $v$ and $w$ you can define $v\cdot w=\frac{1}{2}(vw+wv)$ which ends up a scalar and $v\wedge w=\frac{1}{2}(vw-wv)$ which is a bivector (linear combination of products of two orthogonal vectors). And both operations can be generalized.

The generalization comes from a linear involution that sends every vector to its negative $v\mapsto \overline v=-v$ and sends a product via $\overline{AB}=\overline A \quad \overline B$ so in particular it sends a product of an odd number of vectors to it's negative and sends a product of an even number of vectors to itself. Then for vectors $a$ and a general object $A$ of the algebra (called a multivector since it is a linear combination of products of vectors) you can define

$$a\rfloor A =\frac{1}{2}\left(aA-\overline Aa\right)$$ and

$$a\wedge A =\frac{1}{2}\left(aA+\overline Aa\right).$$

So really it is just the symmetric (or antisymmetric) product on odd (or even) objects. And then we can extend it to all objects by saying it is linear and via

$$(A\wedge B)\rfloor C=A\rfloor(B\rfloor C)$$ and

$$(A\wedge B)\wedge C=A\wedge(B\wedge C).$$

The wedge ends up being associative, just like the original product and for orthogonal vectors $v$ and $w$ you get $vw=v\wedge w$ so knowing those rules for wedge products tells us how they work on every object.

So know we have objects that are multilinear, and take the metric into account. We are forced to use them for relativistic quantum mechanics but can also use them for electromagnetism and/or classical mechanics if we want to. They are exactly what you need to geometrically relate the electromagnetic field to forces charged particles feel.

And note that you have two associative products (the original products that took the metric into account and the wedge product that ignores it) and if you take the objects (the linear combinations of products of vectors) the objects can be written on a basis of wedge products of orthogonal vectors or on a basis of products of orthogonal vectors. And if you only used the wedge product (of the two associative products available) then you get something that as an algebra looks just like forms so any form you imagine can be instead a multivector.

So if you start with multivectors you can get things just like forms but you have additional products. If you started with forms you'd have to add a metric and then the force would be a combination of the two. But if you started with geometric objects that had lengths and angles with each other then the objects already have the metric geometry as part of themselves.

In this case, given some vectors in the space of vectors $V$ such as $u,$ $v,$ and $w$ all from $V$ you can get the plane spanned by two of them e.g. $v\wedge w$ is a plane that contains $\{x\in V: x\wedge (v\wedge w)=0\}$ or the 3-volume spanned by them e.g. $u\wedge v\wedge w$ is a 3-volume that contains $\{x\in V: x\wedge (u\wedge v\wedge w)=0\}.$

But forms could do that. But with multivectors you can also do rotations. And that's excellent because that is what forces do.

A classical particle has a world line that world line has a tangent. If you multiply that future pointing tangent by the rest mass and you literally get the energy-momentum vector for the particle. When you apply a force and wait the particle has the same rest mass and it still has a unit future pointing tangent to its world line. So the net effect of the force was to rotate the energy-momentum vector to a new one.

In an infinitesimally small time interval a force does an infinitesimally small rotation of the energy-momentum vector. So forced are rightly little rotations or linear combinations of planes, so a bivector. And the space of bivectors is 6d as a linear space it has a basis $\{e_0e_1, e_0e_2, e_0e_3, e_1e_2, e_1e_3, e_2e_3\}$ where the first three increase the speed of the particle and the last three change the direction of the particle. So the first three are the basis for the electric field and the last three are the basis for the magnetic field. But all together they are just the basis for an infinitesimal rotation in spacetime, which is what a force is.

And when you look at the algebra is it can't tell what the basis was so you could pick any four mutually orthogonal vectors that have a length of $\pm 1$ and they work just as well. That is why an electromagnetic field, which is a bivector, is broken up into two vector fields differently by two observers but geometrically it just does an infinitesimal rotation in spacetime, a geometrically meaningful thing.

So what is the electromagnetic field $F$? It is a bivector such that for a particle with a unit tangent $u$ to it's world line and a charge $q$ the change in momentum is $d mu /dt=qu\rfloor F$ and the field $F$ is the generator of the infinitesimal rotation. You'll also note that the force is just the electric force in the frame of the particle that feels it but the field needs six components to correctly give the forces for every possible particle moving in every possible way.

There is also a nice geometric interpretation of $u\rfloor (v\wedge w)$ for vectors $u, v,$ and $w$ namely it is the orthogonal complement in the plane $v\wedge w$ of the projection of $u$ into the plane $v\wedge w.$

This might seem like a lot of work. Keep in mind you will have to learn the Dirac algebra someday anyway, so that work that has to be done no matter what. And you can learn to interpret it geometrically or not, most people don't, but you wanted to understand the geometry of forces and this is it.

And it's a bit your instructors' fault if you are a physicist since you could have learned this from the get go and then the Gibbs vector algebra with all the dot products, cross products and such that don't generalize to 4d could have used this stuff all along. So you could have just this one thing and used it for everything. It's just the geometry of space and the geometry of spacetime, with the metric as well as the spanning.

You can do the vector calculus too in addition to the geometric algebra with a $\nabla F=(e_0\partial_0-e_1\partial_1-e_2\partial_2-e_3\partial_3)F$ and $\nabla \rfloor F=(e_0\rfloor\partial_0-e_1\rfloor\partial_1-e_2\rfloor\partial_2-e_3\rfloor\partial_3)F$ and $\nabla \wedge F=(e_0\wedge\partial_0-e_1\wedge \partial_1-e_2\wedge\partial_2-e_3\wedge\partial_3)F$ where the $\partial_i$ first acts as $\partial F /\partial x_i$ on $F$ and then next you have the vector act on it from the left. Unlike the Gibbs curl it generalizes to higher dimensions. And unlike the exterior derivative $\nabla F$ is reversible, i.e. given the derivative $\nabla F=J$ (yes that is just one equation and it is the Maxwell equation) you can find $F$ (given appropriate boundary conditions).

When you include the calculus it is called geometric calculus (instead of geometric algebra) and you can define the derivative $\nabla F$ through $\nabla F@p =\lim_{V\rightarrow \{p\}} \frac{\int_{S=\partial V}dSnF }{ \int_V dV}$ thus making the fundamental theorem trivial and it is a generalization to the generalized stokes theorem since $F$ could be a multivector valued function not just scalar valued. The divergence and curl can then be the grade lowering and grade raising part of the derivative.

So if you just have the one metric product it is called Clifford Algebra, and you have to learn that to do relativistic quantum mechanics. Throw in the other products too and you can see the geometry and easily do rotations and surfaces and spanning and projections, everything you use vector algebra for but you can do it to multivectors too in any dimension space and any dimension object (not just vectors, which is essential in 4d since planes have to be planes since there isn't a vector normal to a 2d plane anymore). Though integration and you can use the fundamental theorem to define a derivative and this is geometric calculus.

At this point you can do anything that forms do, plus more. And it is also unavoidable, the Feynman slash is just making vectors out of things. The Dirac operator is just the derivative (which is the same operator regardless of dimension of what it acts on). And now your derivative is a reversible operation with boundary conditions.

It is possible to ignore all the geometry and just have a long list of algebraic identities that are just exactly like the geometric identities I listed. You can do the calculations. And if that helps with a shut-up-and-calculate teaching philosophy then not learning the geometric meaning can be part of your plan.

But don't assume your instructor has such a detailed plan. In all likelihood then never learned geometric algebra or geometric calculus themselves and that's why it isn't covered. Or it's a political consequence of no clear agreed upon prerequisite structure and agreement about when and where to learn it and how to handle transfer into or out of your school.

Plus you do have to learn ungeometrical ways to do things if you want to read old papers or talk or communicate with people that don't know the geometry behind their algebra and/or calculus. But you can see those things sitting inside geometric calculus.

For instance in 3d, compare $v\times w$ to $\pm(v\wedge w)(\hat x \hat y \hat z)$=$\pm(v\wedge w)\rfloor(\hat x \hat y \hat z).$ Anything you can do by ignoring geometry you can do my being aware of geometry. So being geometry aware is just being more empowered.

Oh, also. Many things that come up in quantum mechanics actually come up in geometric calculus. People mistaken say it is an effect of quantum mechanics because they put off using Clifford Algebra until quantum mechanics. If they had used geometric calculus all along they would know it is just a geometry thing.

So if you use different math for different subjects you can't tell what is a consequence of the math versus the subject. If you use geometric calculus for every subject you can tell what the subject it telling you.

And that is the point of Physics. To understand the universe. To learn what nature is telling us by looking at our best v theories b about how nature works. Avoiding geometric calculus is like signing up for ignorance.

I know it's hard to relearn all of vector algebra, vector calculus, and forms and then learn more on top all by reading a single post. But I've laid out what you should learn if you truly want to understand. Because we know what electromagnetic field is, its a thing that produces the force on charged particles, so something that rotates the energy momentum vectors so a linear combination of basic planes in 4d with a built in metric so you can do rotations. So you need to learn the geometry of spacetime and rotations in spacetime.

So you need geometric calculus to truly understand an electromagnetic field.

Answer 2

Here is a more intuitive way to look at the electromagnetic field tensor: The electromagnetic field tensor is correlated to the Lorentz force law $$\frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = e (\mathbf{E} + \mathbf{v} \times \mathbf{B}).$$ To problem with this equation is that it is not a "geometric equation", i.e. neither the righthand nor the lefthand side are frame independent.

Let $p^\alpha$ be the particles 4-momentum vector, the time shall be measured with the particles own clock. The frame-independent equation has the form $$\frac{\mathrm{d}p^\alpha}{\mathrm{d}\tau} = \frac{1}{\sqrt{1 - v^2}}e(\mathbf{E} + \mathbf{v} \times \mathbf{B}) = e(u^0\mathbf{E} + \mathbf{u} \times \mathbf{B})$$ The above equation is linear in the coordinates of the particle's 4-velocity $\mathbf{u}$, from where conclude that the lefthand side corresponds to a tensor quantity $\mathbf{F}$, named electromagnetic field tensor or Faraday tensor. $\mathbf{F}$ eats the particle's velocity vector $\mathbf{u}$ and puts out the electromagnetic 4-force vector.

One might wonder why $\mathbf{F}$ just takes one argument since a two-form usually eats to vectors and puts out a scalar. The answer to this is quite simple, just plug in another arbitrary 1-form $\mathbf{e}$ into $\mathbf{F}$ so that the output becomes a scalar: $$\mathbf{F}(\mathbf{e},\mathbf{u}) = \langle \mathbf{e}, \mathbf{F}(\mathbf{u})\rangle = \text{scalar}.$$ The Faraday tensor demonstrates the interrelation between the electric and magnetic field; neither one is frame-independent by itself, but by merging them into a tensor makes them frame-independent.

Answer 3

Here's how I think about it. The Lorentz force law for a charged particle of mass $m$ and charge $q$ takes the coordinate-free form

$$m \frac{dU}{d\tau} = q F U,$$

where $F$ (the electromagnetic tensor) is thought of as a linear operator and $U$ (the four-velocity) as a vector. Now in Euclidean space, multiplication by an antisymmetric matrix represents "$\vec{a}\, \times \ $_" operator, where $\vec{a}$ is some fixed 3-vector. The equation $\frac{d\vec{v}}{dt} = \vec{\omega} \times \vec{v}$ is the equation of motion for a vector precessing around the axis $\hat{\omega}$ with angular velocity $\omega$. So by analogy, the Lorentz force law has the interpretation that the electromagnetic tensor tells you what plane through spacetime the four-velocity is "rotating" through at a given instant, and at what "angular velocity".

To interpret Maxwell's equations, it's more natural to intepret $F$ as a differential 2-form rather than a linear operator. If we let $V$ be an arbitrary 3D region of spacetime with 2D boundary $\partial V$, then the integral forms of the two sourced Maxwell's equations can be combined into a single differential form equation $$\iint_{\partial V} \ast F = Q.$$

(That $\iint$ should really be an \oiint, but annoyingly, MathJax doesn't recognize that LaTeX command.) Here, $$Q := \iiint_V J$$ (where $J$ is the current 3-form) represents the total charge contained within the region $V$, but we need to interpret that statement carefully. If $V$ happens to lie within a single constant-time slice (in some Lorentz frame), then $Q$ is just the ordinary charge $\iiint_V \rho\, dV$. But in general, it's a Lorentz-covariant generalization of this quantity. For example, if V is instead the region that a 2D surface $D$ traces out over some finite time $\Delta t$ (in some Lorentz frame), then $Q$ represents the total charge that passes through that surface over that time period.

Anyway, the equation above tells us that integrating $\ast F$ over any closed 2D surface gives the charge enclosed within any 3D region with that surface as a boundary - a straightforward generalization of the integral form of Gauss's law. So the electromagnetic tensor (or at least its Hodge dual) really isn't so different from just its electric field component. It "tracks" the charge distribution over spacetime in a nonlocal way such that you can integrate it over closed surfaces in order to quantify the enclosed charge.