2
$\begingroup$

When formulating classical electromagnetism using differential forms, Maxwell's equations can be written as $$ \begin{align} dD &= \rho \\ dB &= 0 \\ dE &= -\dot{B} \\ dH &= \dot{D} + j \end{align} $$ where $d$ is the exterior derivative, and the symbols appearing there are summarized in the following table:

enter image description here

where $D = \star E$ and $H = \star B$, where $\star$ is the Hodge star encoding the permittivity/permeability of the medium (i.e. defining the metric for the fields).

I have two questions about this:

  • Is there a standard convention for the names of $E$, $H$, $D$, and $B$? Here I called the 1-forms "field" and the 2-forms "induction", I'm not sure if this naming convention is standard or not.
  • Whereas the electric field integrated along a curve yields the change in potential energy for a unit electric charge moving along that path, reflected in the units Volts/meter, what is the equivalent interpretation of $H$? While I know that certain magnetostatics problems can be solved using a scalar potential such that $H = d\Phi$, in general I don't think that interpreting it as "change in potential for a unit magnetic charge" is particularly useful, and furthermore I struggle to understand why its units are Amp/meter.
$\endgroup$
7
  • $\begingroup$ I use idiosyncratic terminology where a field is short for a tangent field and a cofield is short for a cotangent field. Since a 1-form is a cotangent field, then I would call this a cofield. Similarly, a 3-form would be a 3-cofield. I prefer this language as its closer to the field terminology of physics rather than the more mathematically motivated notion of a form. $\endgroup$ Feb 3, 2022 at 2:33
  • $\begingroup$ @MoziburUllah I can appreciate that, but I am hoping to convince some experimentalists that differential forms are the "right" way to think about these fields, so was hoping to make contact with more "standard" terminology $\endgroup$
    – Kai
    Feb 4, 2022 at 15:51
  • $\begingroup$ Sure. There's a book on the foundations of electromagnetism by Hehl & Okulov that uses differential forms but I can't remember their terminology off-hand. Personally, I would say that differential forms allow us to generalise electromagnetism to any manifold - ie curved and higher dimensional - shows that itvis the right way to go. $\endgroup$ Feb 4, 2022 at 22:50
  • $\begingroup$ @MoziburUllah agreed with you, I think also the distinction between 1-forms and 2-forms is really crucial, taking a vector calculus approach completely destroys this distinction (or hides it in "vectors" vs "pseudovectors" which imo are unintuitive). I will check out that reference, I haven't heard of it before $\endgroup$
    – Kai
    Feb 5, 2022 at 2:39
  • 1
    $\begingroup$ Yes, you're right , it is Obukhov. I also agree with what you say about pseudo-vectors. It becomes even more apparent in higher dimensions. For example, something as basic as angular momentum in 4d can't be represented by a pseudo-vector. $\endgroup$ Feb 5, 2022 at 4:06

2 Answers 2

1
$\begingroup$

This isn’t a complete answer, but consider the source of asymmetry in Maxwell’s Equations between electric and magnetic fields: There are apparently no magnetic monopoles. The sources for the fields are $\rho$ (electric charge density) and $j$ (electric current density). It’s a standard exercise to reformulate Maxwell’s Equations to include magnetic charges and currents, and in that case, things are much more symmetric. So when interpreting the fields, they aren’t fundamentally different from one another. Rather, there’s just don’t seem to be any magnetic charges around.

$\endgroup$
1
  • $\begingroup$ I totally agree with you, but I'm wondering why these units (Amps/meter) were chosen for the SI system for this field. I'm not sure how an experimentalist would think about $H$ or its units. $\endgroup$
    – Kai
    Feb 4, 2022 at 15:52
1
$\begingroup$

This doesn't answer your question about potential, but it might help making sense of the units A/m.

Consider a sheet of current having linear current density $j$. First we have to make sure that we understand what linear current density is. First consider "regular" current density (A/m${}^2$): the current passing through a two-dimensional sheet divided by the area of the sheet for a vanishingly small area. The current is perpendicular to the sheet. By analogy, linear current density $j$ (A/m) is the current passing through a line segment divided by the length of the line segment. The current is perpendicular to the line segment.

Imagine a current sheet having linear current density $j$. Draw an amperian loop perpendicular to $j$ having a vanishingly small radius. The field at that radius is ${\bf H}=j$. (Left as an exercise for the reader; consider the current passing through the loop.) So if it help visualization, you can imagine an $\bf H$ field as being "the same as" an imaginary current density at that point.

Similarly, by the way, $\bf D$ can be imagined as an imaginary charge density at the point of interest, ${\bf D}=\sigma$. (Again left as an exercise for the reader).

$\endgroup$

Your Answer

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

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