I have an intuition that magnetic fields are bivectors. While trying to derive some equations with bivectors, I arrived at this formula $$F=\frac{q_{m_a}q_{m_b}}{r^2}\hat r$$ So we have, $$F=q_mB$$ Now, what is pole strength? It maps bivectors to vectors?

Edit Note: I want to generalize magnetism to n-dimensions

Edit 2: Do I need to consider magnetic field as pseudovectors i.e. grade-(n-1) multivector? For 3d bivector, For 4d trivector such that? But this means magnetic field is a vector in 2d? Am I right?


2 Answers 2


The Geometric Algebra approach to electromagnetism (see here) finds that the electromagnetic field is a bivector in 4D spacetime. We can shift our point of view to a specific 3D reference frame by multiplying everything by the time vector (called a spacetime split), which splits the electromagnetic field into a vector electric field and a trivector magnetic field.

We first express 4D spacetime vectors as a combination of 4 unit basis vectors in each direction: $e_t$, $e_x$, $e_y$, $e_z$. (I'm not using the proper notation here, to shorten the exposition. See the paper linked above for a more rigorous and systematic explanation.)

The electromagnetic bivector then has six components: $e_xe_t$, $e_ye_t$, $e_ze_t$, $e_ye_z$, $e_xe_z$, $e_xe_y$. Each of these is a geometric product of two of the unit vectors.

We multiply the whole thing by the constant $e_t$. Since $e_t^2=1$, the $e_t$ component disappears on the first three, and we get: $e_x$, $e_y$, $e_z$, and $e_ye_ze_t$, $e_xe_ze_t$, $e_xe_ye_t$. The vector part with components $e_x$, $e_y$, $e_z$ is the electric field.

Because Gibbs' vector algebra can't cope with trivectors, we convert the other three components to a vector by taking the dual. This is equivalent to the Hodge star operation used in differential geometry, and can be done in Geometric Algebra by dividing by the constant pseudoscalar $I=e_xe_ye_ze_t$. Since $e_x^2=e_y^2=e_z^2=-1$, this has the effect of cancelling any components that were there, and inserting any components that weren't. So $(e_ye_ze_t)(e_xe_ye_ze_t)^{-1}=e_x$, and so on.

The electromagnetic field is thus written $Fe_t=E+BI$. $F$ is the bivector field, $e_t$ is the unit vector in the time direction that expresses everything in a particular inertial reference frame, $E$ is the electric field 3-vector, $B$ is the magnetic field 3-vector, and $I$ is the pseudoscalar used to convert a trivector $BI$ to a vector in 3D.

The expression $E+BI$ expresses the electromagnetic field as a 'Complex' vector with Real and Imaginary components. This is the basis of the Riemann-Silberstein vector. Although $I$ acts somewhat like the unit imaginary in that $I^2=-1$, it has a geometric meaning that means that when you change to a different reference frame, it rotates $E$ and $B$ components into each other, maintaining reference-frame independence. The original Riemann-Silberstein approach only worked in a particular reference frame, because it treated $i$ as an imaginary scalar, that doesn't transform when we switch coordinates.

Although in our universe we only have electric charges and currents, the equations have a duality symmetry in which we can multiply $E+BI$ by $e^{I\theta}$ and get another solution, switching and mixing the roles of $E$ and $B$. The electric current is a 4-vector $J_e$ (which we can do the spacetime split on to get a scalar electric charge and 3-vector electric current), the magnetic equivalent is the dual to this which is a trivector. For convenience, we can use $I$ to express the current $J=J_e+J_mI$, where $J_m$ is a 4-vector, which can be split into a magnetic monopole charge, and a magnetic current.

The duality symmetry only applies in a vaccuum, in our universe, because the coupling to matter breaks the symmetry. We only have electric charges, the imaginary component of $J=J_e+J_mI$ is always zero. But according to the paper I linked above, the duality symmetry is part of the larger electroweak symmetry that exists prior to the separation of electromagnetism and the weak nuclear force.

In summary: yes, in the underlying physics the electromagnetic field is a 4D bivector, but in order to squeeze it into the straitjacket of Gibbs' 3D vector algebra we have to split it into vector electric and trivector magnetic components, and then dualise the trivector magnetic component to turn it into a 3D vector field. These manipulations have nothing to do with the underlying physics - they are purely done so that we can use 3-vectors.

Similarly, the unified electric-magnetic current is a 'Complex' mix of vector and trivector components. To work with 3D Gibbs' vectors we again split it up and dualise it artificially.

If you're interested in an alternate geometric viewpoint on vectors, bivectors, spinors, and so on, then the Space-Time Algebra/Geometric Algebra approach is worth looking at. It doesn't do anything the mathematicians didn't already know (they're just Real Clifford algebras), but it gives a geometric Real-number interpretation that physicists find a lot easier to work with, and provides a lot of interesting insights into what the somewhat opaque mathematical abstractions really mean. For example, all the weird imaginary numbers in quantum physics get replaced by Real geometric entities like bivectors and pseudoscalars (plane elements and volume elements). Spinors are just the even subalgebra of a Geometric Algebra, and have a comparatively simple and intuitive geometric meaning. It takes a while to get up to speed on the mindset - you effectively have to go back and relearn vectors from scratch - but I've found it well worth the effort.

  • $\begingroup$ Can this be progressed to 2d and 4d? $\endgroup$
    – JRBros
    Aug 8, 2022 at 2:39

The magnetic monopole charge is a pseudoscalar. (See https://en.wikipedia.org/wiki/Pseudoscalar )


The magnetic field [magnetic induction] B in Gauss's Law for magnetism (and in Faraday's Law] is a bivector, like the electric field (call it D) is in Gauss' Law and Faraday's Law.
(These are associated with fluxes.)

However, the magnetic field (call it H) in the force law (and in Ampere-Maxwell) is a [pseudo-]vector, like the electric field E is in the force law (and in Faraday).
(These are associated with circulation.)

Of course, force is a vector.

(Depending on your starting point, maybe the force law should be written as an equation of one-forms).

So, it might be better to think of it as $$F=q_m H,$$ the product of a pseudoscalar and a pseudovector.

  • $\begingroup$ So u mean the force equation cannot be denoted in form of bivector? $\endgroup$
    – JRBros
    Aug 7, 2022 at 4:06
  • 1
    $\begingroup$ @JRBros For the force on a magnetic monopole, I think you would need the 3D-Hodge dual of the magnetic-field-bivector. For the force on an electric charge, you would use the magnetic-field-bivector contracted with the electric-charge's velocity-vector. $\endgroup$
    – robphy
    Aug 7, 2022 at 4:09
  • $\begingroup$ What is the 3d hodge star operator? And how should i use it? $\endgroup$
    – JRBros
    Aug 7, 2022 at 4:12
  • 1
    $\begingroup$ @JRBros The 3D-Hodge dual is how we associate a vector with the cross-product of two vectors. ( en.wikipedia.org/wiki/Hodge_star_operator#Three_dimensions ) For magnetism, it is how B and H are related (a constitutive relation using the volume element and something to describe a medium [possibly vacuum]). $\endgroup$
    – robphy
    Aug 7, 2022 at 4:15
  • $\begingroup$ @JRBros Possibly useful: physics.stackexchange.com/questions/692414/… $\endgroup$
    – robphy
    Aug 7, 2022 at 4:19

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