### Spatial cross products

The **cross product** of two three-vectors, $\vec{c},\vec{d}$ is usually thought of as another vector, $$\vec{a} = \vec{c} \times \vec{d}$$ such that $\vec{a}$ is perpendicular to both $\vec{c},\vec{d}$ and has magnitude $$a = c d \sin\theta$$ where $\theta$ is the angle between $\vec{c},\vec{d}$.

But, as we can see in figure 7.3, we can choose the angle, and hence the cross product, in two different ways.

Only by convention do we use the right-handed rule to decide between them (top of figure 7.3).

We would like to define an object that encompasses both aspects of the cross product.

Recall that the other way to calculate the components of a cross product of two vectors, $\vec{c},\vec{d}$, is to take the determinant of a seemingly contrived matrix, $$\vec{a} = \begin{vmatrix}\hat{x} & \hat{y} & \hat{z}\\c_x & c_y & c_z\\d_x & d_y & d_z\end{vmatrix}$$ which leads to the formulae for the components of the product as a three-vector, $$\begin{align*}a_x &= c_y d_z - c_z d_y\\a_y &= c_z d_x - c_x d_z\\a_z &= c_x d_y - c_y d_x\end{align*}$$

This is an equivalent definition of the (spatial) cross product, and as we have seen before, displays a cyclic relationship in the components, $$a_m = c_n d_k - c_k d_n \text{ where } (m,n,k) \text{ is a cycle of } (1,2,3)$$

Nb. We have flipped from using $x,y,z$ as indices to $1,2,3$. Also, recall that we only use Latin subscripts for three dimensional tensors.

If we define a spatial rank-two tensor by the rule, $$A_{n k} = c_n d_k - c_k d_n$$ then it is immediately anti-symmetric, $$A_{k n} = c_k d_n - c_n d_k = -(c_n d_k - c_k d_n) = -A_{n k}$$ and we can write the components of the cross product in terms of this tensor, $$a_m = A_{n k} \text{ where } (m,n,k) \text{ is a cycle of } (1,2,3)$$

That is, $$\begin{align*}a_1 &= A_{23} &(= -A_{32})\\a_2 &= A_{31} &(= -A_{13})\\a_3 &= A_{12} &(= -A_{23})\end{align*}$$

Thus, using the pattern of a cross product, we can convert any (spatial) vector, $\vec{a} = (a_1,a_2,a_3)$ into a corresponding (spatial) anti-symmetric rank-two tensor, $$A_{m n} \rightarrow \begin{bmatrix}0 & a_3 & -a_2\\-a_3 & 0 & a_1\\a_2 & -a_1 & 0\end{bmatrix}$$

Nb. The notion of a cross product does generalise to higher dimensions, but only in three dimensions do we have the equivalence between vectors and anti-symmetric rank-two tensors.

### Magnetic field as a cross product

For the moment, suppose again that we are only dealing with the magnetic component of the Lorentz force, $$\vec{F} = q\vec{v} \times \vec{B}$$ which we can write in index notation as $$F_m = q\left[\vec{v} \times \vec{B}\right]_m, \text{ for }m = 1,2,3$$

Historically, the rank-two anti-symmetric tensor associated with $\vec{B} = (B_1,B_2,B_3)$ is the negative of the what we might expect to be the conventional option, $$\begin{align*}B_1 &= -B_{2 3} &(= B_{3 2})\\B_2 &= -B_{3 1} &(= B_{1 3})\\B_3 &= -B_{1 2} &(= B_{2 1})\end{align*}$$ which we can summarise as $$B_m = -B_{n k} (= B_{k n}) \text{ where } (m,n,k) \text{ is a cycle of } (1,2,3)$$

Nb. The minus-sign is related to the fact that $\vec{B}$ appears on the right of the cross product in the force law.

The corresponding anti-symmetric tensor is then $$B_{n k} \rightarrow \begin{bmatrix}0 & -B_3 & B_2\\B_3 & 0 & -B_1\\-B_2 & B_1 & 0\end{bmatrix}$$

Now consider the $x$-component of $\vec{v} \times \vec{B}$, $$\begin{align*}\left[\vec{v} \times \vec{B}\right]_1 &= v_2 B_3 - v_3 B_2\\&= (B_3) v_2 + (-B_2) v_3\\&= (B_{2 1}) v_2 + (B_{3 1}) v_3\\&= B_{1 1} v_1 + B_{2 1} v_2 + B_{3 1} v_3 & \text{ since } B_{1 1} = 0\\&= B_{1 n} v_n\end{align*}$$

The other components follow a similar pattern, leading to a new formula for the magnetic part of the Lorentz force, $$F_m = q\left[\vec{v} \times \vec{B}\right]_m = B_{m n} v_n = - v_n B_{n m}$$

If we let $\mathbf{B}$ be the matrix associated with the tensor $B_{n k}$, then $$\mathbf{B} \vec{v} = \vec{B} \times \vec{v}$$ so, in three-vector terms, the magnetic part of the Lorentz force could be written as $$\vec{F} = - q \mathbf{B} \vec{v}$$

#### Example: the speed of a particle in a static, purely magnetic field is constant

We have already seen, that for any anti-symmetric rank-two tensor, $T^{\mu \nu}$ and any (covariant) four-vector, $X_\mu$, $$T^{\mu \nu} X_\mu X_\nu = 0$$

This holds for three-vectors as well - and for the same reasons. The only difference is that covariant and contravariant components of a three-vector are the same, hence we use (Latin) subscripts for all objects.

Thus, it follows that $$F_n v_n = B_{n k} v_n v_k = 0$$

But, $$\begin{align*}F_n v_n &= m \frac{\mathrm{d} v_n}{\mathrm{d} t} v_n\\&= \frac{m}{2} \frac{\mathrm{d} }{\mathrm{d} t} \left(v_n v_n\right) & \text{ summed over } n\end{align*}$$ and so $$\frac{\mathrm{d} }{\mathrm{d} t} \left(v_n v_n\right) = 0$$

Hence, the speed $v = \sqrt{v_n v_n}$ is constant.

Nb. This is just saying that the magnetic portion of the Lorentz force is perpendicular to the velocity and hence doesn't change the velocity - something which is apparent from the Newtonian form of the equation, since $\vec{v} \times \vec{B} \cdot \vec{v} = 0$ by definition.