Dot and cross products

How much of one arrow lies along another, and which way does a magnetic force push?

The dot product multiplies two arrows into a single number: $\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$, which also equals the two lengths times the cosine of the angle between them. With an arrow of length 1 along a chosen direction, the dot product is how much of the other arrow lies along that direction: its projection. Arrows at right angles have a dot product of zero.

For a boost along x, the part of an electric field along the motion is its dot product with an arrow of length 1 along x. A field of 5 units pointing 3 along x and 4 along y has 3 along the boost; that part is the same in both frames, and only the rest changes.

The cross product $\mathbf{a}\times\mathbf{b}$ is an arrow at right angles to both a and b. Its length is the area of the parallelogram they span, the two lengths times the sine of the angle between them, and its direction follows the right-hand rule: curl the fingers of the right hand from a toward b, and the thumb points along the product. Swapping the order reverses the arrow.

The magnetic force on a charge q moving with velocity v through a field B is $q\,\mathbf{v}\times\mathbf{B}$. It is at right angles to the motion, so it changes the charge's direction and not its speed. For a positive charge moving along +x through a field along +z, $\mathbf{v}\times\mathbf{B}$ points along −y. In the frame moving with the charge the same push is electric: the relativity paper's §6 gives that frame an electric field along −y.

One combination of the two fields, $\mathbf{E}\cdot\mathbf{B}$, comes out the same in every frame. That is a modern check on the field transformation, useful for catching an error in a calculation. It is not a premise of the 1905 argument, which derives the transformation from the equations of electrodynamics.

Worked example: A projection, a force direction and a quantity that stays put

1. Projection: $\mathbf{E} = (3, 4, 0)$ and the boost direction $(1, 0, 0)$ give $\mathbf{E}\cdot(1, 0, 0) = 3$, the part along the boost. The field's length is $\sqrt{3^2 + 4^2} = 5$.
2. Direction: with v along +x and B along +z, $(1, 0, 0)\times(0, 0, 1) = (0, -1, 0)$, so a positive charge is pushed along −y.
3. Invariance, measuring fields so that c = 1: take $\mathbf{E} = (0, 0.6, 0.8)$ and $\mathbf{B} = (0, 0, 1)$, so $\mathbf{E}\cdot\mathbf{B} = 0.8$.
4. Seen from a frame moving at 0.6 along x, where γ = 1.25 (the factor 1/√(1 − v²/c²), which the relativity paper prints as β), the §6 rules give $\mathbf{E}' = (0, 0, 1)$ and $\mathbf{B}' = (0, 0.6, 0.8)$. Both fields changed, and $\mathbf{E}'\cdot\mathbf{B}' = 1 \times 0.8 = 0.8$ did not.