# Is there a more clever way to apply the cross product of two vectors to magnetism?

I am just beginning to learn magnetism and my book used two ways to define the force caused by the magnetic field, brushing over the latter. The first:

$$F = q v B \sin (\theta).$$

And:

$$\vec{F} = q(\vec{v} \times \vec{B})$$

where the $\times$ is the cross product. After looking up the cross product, I found that it was defined as:

$\vec{a} \times \vec{b} = |\vec{a}||\vec{b}| \sin (\theta) \vec{n}$ where $\vec{n}$ is the unit vector found via the "right hand rule."

While I have no doubt the right hand rule is a useful tool, I wonder if there is a more "on paper" way to find the direction of the force a magnetic field applies to a charged particle. Basically, shouldn't there be a way to avoid the right-hand rule and to do it all via equations?

• -1: This is trivially answered by the wikipedia article on cross product
– user2963
Mar 7 '12 at 2:16
• @zephyr i'd rather sugges the wiki page on oriented coordinate systems: en.wikipedia.org/wiki/… Mar 7 '12 at 3:15
• @zephyr I guess I thought that there might be a quicker way than the jumble of variables wikipedia had. I guess the jumble is inherent in matrices. (I am pre-calculus). Mar 7 '12 at 3:21

Any equations you set up will be uglier than the cross product (they will basically by expansions of $\times$)

Remember, vectors only can define a direction. The issue with the magnetic field is that it is heavily related to planes, and is a perpendicular force. We can't assign a magnetic field analogous to the electric field (as in: a field which shows the direction of force), as it depends upon the direction of velocity.

The cross product enables us to define a plane with a vector perpendicular to it. Aside from that, it gives perpendicular stuff. This clearly points to the use of the cross product.

Besides, the cross product isn't that dirty. It makes stuff simpler, if anything.

If you want to do it all via equations, use this:

$$\vec{a}\times\vec{b}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \\ \end{vmatrix}$$ These rules may help you avoid the determinant form for simpler cross products: $$\hat{i}\times\hat{j}=\hat{k};\:\hat{j}\times\hat{k}=\hat{i};\:\hat{k}\times\hat{i}=\hat{j};\:\vec{a}\times\vec{a}=0;\:\vec{a}\times\vec{b}=-\vec{b}\times\vec{a}$$

Over here, the right hand rule is implied by our choice of a right handed system of $\hat{i} , \hat{j} , \hat{k}$ or $x,y,z$. So you don't need to touch it, it is already inherent in the coordinate system.

To express mathematically an observed vector, that is perpendicular to two other vectors, we use the cross product to define that observation. As there are two possible directions to the resulting vector, the right hand rule is used to define which way is positive. As Manishearth said, it’s already embedded in the directions of the axes used to locate the vectors.

One problem: I have seen different methods of applying the right hand rule (some difficult to remember). The one I like applies to all vector products: Point your right-hand fingers in the direction of the first named vector. Hold your hand so that you can curl your fingers in the direction of the second named vector. Your thumb gives the direction of the product. (For a “right-handed” system you go from (x) to (y) and get (z).

You are right that the cross product notation is hiding something. A cross product between vectors does not give a vector but a pseudovector, but is really an antisymmetric tensor. Only in 3D this behaves like a vector, except that it does not change sign under coordinate inversion. However, the magnetic part of the Lorentz force, like the electric part, is a vector, because the cross product of a vector and a pseudovector gives a vector.

When you use the antisymmetric notation you will not need the right hand screw rule.

You can avoid cross products all together by writing the magnetic field as:

$$B_{ij}=-\frac{\mu_0}{4\pi}\int\big[ \frac{(r_i-r'_i)J_j({\bf r'},t_r)-(r_j-r'_j)J_i({\bf r'},t_r)} {|{\bf r}-{\bf r'}|^3}+ \frac{(r_i-r'_i)\partial J_j({\bf r'},t_r)/\partial t-(r_j-r'_j)\partial J_i({\bf r'},t_r)/\partial t} {c|{\bf r}-{\bf r'}|^2} \big]d^3{\bf r}'$$

That is a standard sourced-from-currents-only formulation with the cross product replaced by an antisymmetric dyad:

$${\bf u}\times{\bf v}\rightarrow {\bf u}{\bf v}-{\bf v}{\bf u}$$

So for instance, for an current flowing on the $$z$$-axis:

$${\bf J} = I\delta(z){\bf\hat z}$$

at some position on the $$x$$-axis:

$${\bf r} = R{\bf x}$$

the usual vector formulation:

$${\bf B} = B_y(R){\hat y}$$

is replaced with:

$${\bf B}=-B_y({\bf \hat x\hat z}-{\bf \hat z\hat x})= \left(\begin{array}{ccc} 0&0&-B_y\\0&0&0\\B_y&0&0\end{array}\right)$$

In general, the Lorentz Force Law is:

$$F_i =q(E_i + B_{ij}v_j)$$

The direction is no-longer ambiguous, you just have to carry around a rank-2 tensor that is mostly zeros. Ppl prefer axial-vectors:

$$B_i= \frac 1 2 \epsilon_{ijk}B_{jk}$$

The tensor method is "nice" in that explicitly shows that the magnetic field's geometric nature differs from that the electric field (e.g.,$$B_{ij}$$ is explicitly even under parity transformations, while $$E_i$$ is regular (odd) vector).

There deeper ways to address this with so-called geometric algebra and Clifford algebras.

• I suspect that if someone is "just beginning to learn magnetism", introducing dyads & Clifford algebras is not the way to go — unless you want to convey that the other options are even more complicated and they should be happy with the cross product. :-) Jul 5 '21 at 19:03
• The question as posed was, "Shouldn't there be a way to avoid the right hand rule with and to do it all via equations." The answer is yes, I showed how to do it. While an answer the just uses the cross product with the RHR gets upvoted, so idk what the OP wants. Usually ppl come here to learn.
– JEB
Jul 6 '21 at 3:16