Background I'm familiar enough with the Faraday tensor $F_{\alpha\beta}$ to know that it's is a 2-form. Hence, at each point $P$ in spacetime $V$, it's a multilinear map $$F: T_PV\times T_pV\to\mathbb{R}$$ However, I've never seen it used in this manner, taking two tangent vectors to the real numbers. The closest I have seen to using the tensor as a map is in the Lorentz force law, but only one "slot" is taken up, so we are left with a covector representing the force.

$$f_\alpha = \frac{dp_\alpha}{d\tau}=qF_{\alpha\beta}U^\beta$$

My own thought process:

Using Lorentz-Heaviside natural units, elements of the tangent space (e.g. 4-velocities) do not have units. At the same time, the tensor itself has field strength units of energy squared. Then, $F(\textbf{u},\textbf{v})=u^\alpha v^\beta F_{\mu\nu}$ will then have units of energy squared as well. While this cannot represent the force, it could indeed represent the power. Specifically, if a particle is moving in a direction $\textbf{u}$ through a region with non-zero fields, then we can measure the change in power along a direction $\textbf{v}$.

This works with the Lorentz force law as well, because

$$u^\mu v^\nu F_{\mu\nu} = v^\nu F_{\mu\nu}u^\mu = \frac{1}{q}v^\nu f_\nu$$

In this form, the scalar reminds me of the formula $P=\vec F\cdot\vec v$ from classical mechanics.

My question I would like to know whether there is a way to view the Faraday tensor as a map that takes two tangent vectors. Given two tangent vectors $u,v$, wis the real number $F(u,v)$ some form of power? If so, how should I interpret this?

  • $\begingroup$ Nobody calls that the "Faraday tensor." $\endgroup$ – Buzz Jul 18 '19 at 0:09
  • 2
    $\begingroup$ MTW calls it Faraday. $\endgroup$ – Puk Jul 18 '19 at 0:51

Of course there is. Each time a flux is computed tangent vectors are put into the corresponding components of the electromagnetic field tensor. For instance one wants to know the flux of magnetic flux density $\mathbf{B}$ through a surface $S$ in $\mathbb{R}^3$, in order to compute the integral $\int_S \mathbf{B} d\mathbf{f}$ one actually puts tangent vectors which are tangent to the surface $S$ into the 2-form and evaluates it at these tangent vectors. I are going to be more explicit: The first step is to provide a parametrisation of the surface, i.e. a surjective and smooth map $g$ which maps an area in $\mathbb{R}^2$ on $S$: $g: (x,y)\in \mathbb{R} \rightarrow S \in \mathbb{R}^3$. The parametrisation allows that a grid on $\mathbb{R}^2$ can be mapped onto a grid expanded over the 3D surface $S$ and at each of the points of the surface grid the smooth tangent vectors $(\xi_{1,i},\xi_{2,j})$ can be easily obtained by ($\Delta x_i$ and $\Delta y_j$ are the x- respectively y-intervals of the grid in $\mathbb{R}^2$ where $i$ and $j$ the indices which enumerate the grid points)

$$ \xi_{1,i} = (\frac{\partial g_x}{\partial x}, \frac{\partial g_y}{\partial x}, \frac{\partial g_z}{\partial x})|_{g(x_i),g(y_j)}\,\,\Delta x_i$$

$$\xi_{2,j} = (\frac{\partial g_x}{\partial y}, \frac{\partial g_y}{\partial y}, \frac{\partial g_z}{\partial y})|_{g(x_i),g(y_j)}\,\,\Delta y_j$$

The 2-form is evaluated on the tangent vectors $\xi_{1,i}$ and $\xi_{2,j}$ at all the grid points $(x_i,y_j)$ of the surface and for the evaluation of the integral Riemann sums have to built up from the resulting values $\sum_i \sum_j F(\xi_{1,i},\xi_{2,j})$. If everything is done correctly, one ends up with an approximate integral where even the Jacobi determinants are included automatically (2-forms are made like to produce the Jacobi determinant automatically!) and the only missing step is to take the limit $ \Delta x_i , \Delta y_j \rightarrow 0$ to obtain the desired field flux through the surface. This explanation is quite rough and should be backed up with an expert text on integration of differential forms which explains each step on sound mathematical ground. But grosso modo this is how forms are evaluated on tangent vectors. Of course this is just a special but very frequent case, other evaluations of the electromagnetic field tensor $F$ on other vectors are also possible.

  • $\begingroup$ I'm a bit late to get back, but I wanted to thank you for your answer! I hadn't actually thought of things in this manner, so I liked your take. $\endgroup$ – Noah M Oct 7 '19 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.