# Why is there a negative sign in the (non-relativistic) bivector formulation of the Lorentz force?

I'm currently trying to update my understanding of basic (Newtonian, non-relativistic) physics to use bivectors and Clifford products instead of pseudovectors and cross products. And I've come up against that most famous use of cross products, magnetic fields.

In the cross product formulation, the B-field is a pseudovector, and the force it exerts on a moving charged particle is $$\vec{F}_{mag} = q \vec{v} \times \vec{B}$$ .

In the bivector formulation, the B-field is a bivector, and the force it exerts on a moving charged particle is $$\vec{F}_{mag} = -q\vec{v} \vee \mathbf{B}$$ (where $$\vee$$ is the inner product).

Where does this extra negative sign come from? Why doesn't it show up in the cross product formulation?

• Isn’t it $\wedge$ rather than $\vee$? And isn’t it the exterior product? Aug 16, 2019 at 5:37
• @G.Smith $\wedge$ is the exterior/outer product (highest-grade part of the geometric product); $\vee$ is the interior/inner product (lowest-grade part of the geometric product). Some sources write the inner product with a dot (because for vectors it's equivalent to the dot product) but the first reference I found used the vee, so I've ended up sticking with that. Aug 16, 2019 at 5:39
• OK, thanks. That makes more sense! Aug 16, 2019 at 5:42
• I wonder whether you have considered an alternative (and more widely used) route here: adopt the field tensor and write the force as a contraction of this tensor with the four-current. The contraction here is an inner product, and tensor maths has much to recommend it. Aug 14, 2021 at 23:57

This is something I'm personally also trying to understand better, but here is something that might help. I was able to derive the minus sign by starting with the bivector and then deriving the usual $$\vec{B}$$ pseudovector field from that. Since I start by assuming the bivector is positive, I get the result that the pseudovector points in the "opposite sense", leading to a negative sign. If you turn my derivation backwards, assuming that the pseudovector is the positive one instead, you should get the formula you have.

First, I assume that there is a magnetic bivector $$\textbf{B}$$ which is written out with a right-handed bivector basis with scalar components $$B^i$$ (not to be confused with the pseudovector's components $$\vec{B}^i$$):

$$\textbf{B} = B^1e_{23} + B^2e_{31} + B^3e_{12}$$

This bivector is chosen to create a Lorentz force law which does not have a minus sign in it.

$$\vec{F}_{mag}=q\vec{v}\vee\textbf{B}$$

What will be shown is that this assumption leads to the cross-product-pseudovector version acquiring a minus sign,

$$\vec{F}_{mag}=-q\vec{v}\times\vec{B}$$

Which means that if you take $$\vec{B}$$ as the "starting point" instead, then the signs on both equations will flip and we'll get your version of the bivector force law that has a minus sign in it. I apologize for the "backwards" nature of this proof, but it helped me get a better handle on it because pseudovectors are imho less intuitive to work with than bivectors.

The key equation is this:

$$\omega\textbf{B}=-B^ie_i=\vec{B}^ie_i=\vec{B}$$

where $$\omega=e_1e_2e_3$$ is the pseudoscalar of 3-space. Let's work out the algebra to see how this is true.

$$\omega\textbf{B}=e_1e_2e_3(B^1e_{23} + B^2e_{31} + B^3e_{12}) \\ = e_1e_2e_3(B^1e_{23}) + e_1e_2e_3(B^2e_{31}) + e_1e_2e_3(B^3e_{12}) \\ = B^1e_1e_2e_3(e_{23}) + B^2e_1e_2e_3(e_{31}) + B^3e_1e_2e_3(e_{12})$$

We are free to permute the vectors making up $$\omega$$ as long as they stay in the same $$1\rightarrow2\rightarrow3$$ order, so

$$B^1e_1e_2e_3(e_{23}) + B^2e_1e_2e_3(e_{31}) + B^3e_1e_2e_3(e_{12}) \\=B^1e_1e_2e_3(e_{23}) + B^2e_2e_3e_1(e_{31}) + B^3e_3e_1e_2(e_{12})$$

$$e_ie_j(e_ie_j)=-e_ie_je_je_i=-e_ie_i=-1$$, so

$$B^1e_1e_2e_3(e_{23}) + B^2e_2e_3e_1(e_{31}) + B^3e_3e_1e_2(e_{12})\\=-B^1e_1 - B^2e_2 - B^3e_3$$

This expression is a vector, "$$-B^ie_i$$", whose vector-components (coefficients tacked onto basis vectors $$e_i$$) are expressed in terms of the bivector's components ($$B^i$$, which are just scalars). You'll note the minus sign in front. You have two options on how you interpret this. For now, we will define the "usual" pseudovector field $$\vec{B}$$ to include this minus sign (relative to the original bivector components).

$$\vec{B}\equiv\vec{B}^ie_i\equiv-B^ie_i$$

To be totally explicit, we're saying that the (scalar) components of the geometrical object "pseudovector $$\vec{B}$$", which we label $$\vec{B}^i$$, are each defined to have the numerical value found by evaluating the scalar expression "$$-1\cdot B^i$$". Where $$B^i$$ is, as always, "the coefficients of the $$e_{23}$$, $$e_{31}$$, and $$e_{12}$$ that add up to make up the geometrical object that is the bivector $$\textbf{B}$$".

So, now we've seen that $$\omega\textbf{B}=\vec{B}$$ ... in essence we can "convert" from $$\textbf{B}$$ to $$\vec{B}$$ with this formula. If we have a bivector, we can find its corresponding pseudovector. What about the other way around? Well, let's do the algebra.

$$\omega\vec{B}=e_1e_2e_3(\vec{B}^1e_{1} + \vec{B}^2e_{2} + \vec{B}^3e_{3}) \\=e_1e_2e_3(\vec{B}^1e_{1}) + e_1e_2e_3(\vec{B}^2e_{2}) + e_1e_2e_3(\vec{B}^3e_{3}) \\=\vec{B}^1e_1e_2e_3(e_{1}) + \vec{B}^2e_1e_2e_3(e_{2}) + \vec{B}^3e_1e_2e_3(e_{3}) \\=\vec{B}^1e_2e_3e_1(e_{1}) + \vec{B}^2e_3e_1e_2(e_{2}) + \vec{B}^3e_1e_2e_3(e_{3}) \\=\vec{B}^1e_2e_3 + \vec{B}^2e_3e_1 + \vec{B}^3e_1e_2 \\ (\vec{B}^i\equiv-B^i) \\ =-B^1e_2e_3 - B^2e_3e_1 - B^3e_1e_2 \\=-\textbf{B}$$

In summary,

$$\omega\vec{B}=-\textbf{B} \\ \omega\textbf{B}=\vec{B}$$

This relationship is the source of the "opposite sense" between the bivector field and the pseudovector one. Now that we have this fact, we are almost ready to recover your equation in the form you have it.

Continuing to stick with the assumption that $$\textbf{B}$$ is the "default" which leads to the following force law

$$\vec{F}_{mag}=q\vec{v}\vee\textbf{B}$$

Using the equation we found above, $$\omega\vec{B}=-\textbf{B}$$, we find that $$-\omega\vec{B}=\textbf{B}$$, so let's substitute that into our force law:

$$\vec{F}_{mag}=q\vec{v}\vee(-\omega\vec{B})$$

We can move that minus sign out front:

$$\vec{F}_{mag}=-q\vec{v}\vee(\omega\vec{B})$$

And re-use our definitions for "converting":

$$\omega\vec{B}=-\textbf{B} \\ \vec{F}_{mag}=-q\vec{v}\vee(-\textbf{B})$$

Seems redundant? Well, now we can finally make our conclusion. Remember how I did this all "backwards", starting by assuming the bivector $$\textbf{B}$$ is positive. However, by the definition of $$\vec{B}$$, you can see it's just as valid to assume that $$\vec{B}$$'s components are positive, and $$\textbf{B}$$ is the negative one:

$$\vec{B}\equiv\vec{B}^ie_i=-B^ie_i \\ \therefore B^ie_i=-\vec{B}^ie_i$$

So if we apply that to our "redundant" force law we get

$$\vec{F}_{mag}=-q\vec{v}\vee(\textbf{B})$$

instead, which was the result to be shown. In short, the reason this "extra minus sign" shows up is because assuming that the cross-product is the "default" law with a positive sign entails that the bivector formula must have a minus sign to "make up for" the minus sign that appears when you "convert" between the axial vector/pseudovector $$\vec{B}$$ and the corresponding bivector $$\textbf{B}$$.

I worry that my notation and especially the sign-flipping might end up confusing more than it clears up, but all the same, at the very least I've shown you some of the reason why there is always a minus sign difference between the bivector and pseudovector versions of the magnetic force law. Hope it helps!