Irodov 3.369: What Every Observer Agrees On – Amit & Deepali Sinha Foundation

Problem Irodov 3.369

An inertial frame $K'$ moves relative to an inertial frame $K$ with the constant velocity $\vect v$ . With $\beta = v/c$ , and with $\parallel$ and $\perp$ marking the components of a field along $\vect v$ and across it, the fields in the two frames are related by

\begin{gathered} \vect E'_\parallel = \vect E_\parallel, \qquad \vect B'_\parallel = \vect B_\parallel,\\ \vect E'_\perp = \frac{\vect E_\perp + \vect v\times\vect B}{\sqrt{1-\beta^2}}, \qquad \vect B'_\perp = \frac{\vect B_\perp - \vect v\times\vect E/c^2}{\sqrt{1-\beta^2}} \end{gathered}

Using these relations, show that the electromagnetic field has two quantities that are the same in every inertial frame:

the scalar product $\vect E\cdot\vect B$ ;
the combination $E^2 - c^2B^2$ .

Given

the relativistic transformation of $\vect E$ and $\vect B$ for a frame moving with $\vect v$

Find

that (a)

\vect E\cdot\vect B

and (b)

E^2 - c^2B^2

are the same in

K

and

K'

Solution

Step 1Split the fields and name the factor

Write $\gamma = 1/\sqrt{1-\beta^2}$ . The parallel and perpendicular parts are orthogonal, so a scalar product and a square each split into a parallel and a perpendicular term:

\vect E\cdot\vect B = E_\parallel B_\parallel + \vect E_\perp\cdot\vect B_\perp, \qquad E^2 = E_\parallel^2 + E_\perp^2, \qquad B^2 = B_\parallel^2 + B_\perp^2 \tag{1}

The same holds in $K'$ . The parallel parts are equal in the two frames, so everything turns on the perpendicular parts:

\vect E'_\perp = \gamma\,(\vect E_\perp + \vect v\times\vect B), \qquad \vect B'_\perp = \gamma\Bigl(\vect B_\perp - \frac{\vect v\times\vect E}{c^2}\Bigr) \tag{2}

Step 2Cross products in the plane normal to v

All the vectors in (2) lie in the plane normal to $\vect v$ , and in that plane a cross product with $\vect v$ only turns a vector through $90^\circ$ and multiplies its length by $v$ : $\vect v\times\vect B = \vect v\times\vect B_\perp$ and $\vect v\times\vect E = \vect v\times\vect E_\perp$ .

The plane normal to v, seen with v toward the viewer: E_⊥ and B_⊥, and the dashed vectors v· E and v· B, each its field turned through a right angle (lengths schematic). — The plane normal to $\vect v$ , seen with $\vect v$ toward the viewer: $\vect E_\perp$ and $\vect B_\perp$ , and the dashed vectors $\vect v\times\vect E$ and $\vect v\times\vect B$ , each its field turned through a right angle (lengths schematic).

A vector turned through $90^\circ$ is perpendicular to the vector it came from, and turning two vectors together leaves the angle between them unchanged:

\begin{aligned} \vect E_\perp\cdot(\vect v\times\vect E) &= 0, & \vect B_\perp\cdot(\vect v\times\vect B) &= 0,\\ (\vect v\times\vect B)\cdot(\vect v\times\vect E) &= v^2\,\vect B_\perp\cdot\vect E_\perp, & |\vect v\times\vect B| &= vB_\perp, \quad |\vect v\times\vect E| = vE_\perp \end{aligned} \tag{3}

One more relation pairs the two mixed products. By the cyclic rule for a triple product, both are the same triple product with opposite signs:

\vect E_\perp\cdot(\vect v\times\vect B) = \vect v\cdot(\vect B\times\vect E) = -\vect B_\perp\cdot(\vect v\times\vect E) \tag{4}

Step 3(a) The scalar product

Multiply the two relations (2) and expand into four terms:

\vect E'_\perp\cdot\vect B'_\perp = \gamma^2\Bigl[\vect E_\perp\cdot\vect B_\perp - \frac{\vect E_\perp\cdot(\vect v\times\vect E)}{c^2} + (\vect v\times\vect B)\cdot\vect B_\perp - \frac{(\vect v\times\vect B)\cdot(\vect v\times\vect E)}{c^2}\Bigr]

By (3) the two middle terms vanish and the last is $-\beta^2\,\vect E_\perp\cdot\vect B_\perp$ :

\vect E'_\perp\cdot\vect B'_\perp = \gamma^2(1-\beta^2)\,\vect E_\perp\cdot\vect B_\perp = \vect E_\perp\cdot\vect B_\perp

With the parallel parts unchanged, (1) in both frames gives

\vect E'\cdot\vect B' = \vect E\cdot\vect B

(5)

Step 4(b) The difference of squares

Square each relation (2), using the lengths in (3):

\begin{aligned} E'^2_\perp &= \gamma^2\bigl[E_\perp^2 + 2\,\vect E_\perp\cdot(\vect v\times\vect B) + v^2B_\perp^2\bigr]\\ c^2B'^2_\perp &= \gamma^2\bigl[c^2B_\perp^2 - 2\,\vect B_\perp\cdot(\vect v\times\vect E) + \beta^2E_\perp^2\bigr] \end{aligned}

Subtracting, the mixed terms combine into $2\gamma^2[\vect E_\perp\cdot(\vect v\times\vect B) + \vect B_\perp\cdot(\vect v\times\vect E)]$ , which is zero by (4). With $v^2B_\perp^2 = \beta^2c^2B_\perp^2$ :

E'^2_\perp - c^2B'^2_\perp = \gamma^2\bigl[(1-\beta^2)E_\perp^2 - (1-\beta^2)c^2B_\perp^2\bigr] = E_\perp^2 - c^2B_\perp^2

Adding the unchanged $E_\parallel^2 - c^2B_\parallel^2$ , (1) in both frames gives

E'^2 - c^2B'^2 = E^2 - c^2B^2

(6)

Check

A special case done directly: let $K$ hold only an electric field $\vect E$ normal to $\vect v$ . Then $\vect E' = \gamma\vect E$ and $\vect B' = -\gamma\,\vect v\times\vect E/c^2$ , of magnitude $\gamma\beta E/c$ and normal to $\vect E'$ . So

E'^2 - c^2B'^2 = \gamma^2E^2 - \gamma^2\beta^2E^2 = \gamma^2(1-\beta^2)\,E^2 = E^2

and $\vect E'\cdot\vect B' = 0 = \vect E\cdot\vect B$ , as (6) and (5) require. Units: $cB$ is $(\text{m/s})\cdot\text{T} = \text{V/m}$ , so $E^2$ and $c^2B^2$ can be subtracted.

Remark

The invariants settle questions without any transformation. If $\vect E\perp\vect B$ in one frame, they are perpendicular in all. If also $E > cB$ , there is a frame with no magnetic field at all, and if $E < cB$ , one with no electric field; when $E = cB$ , as in a light wave, neither field can be removed, in any frame. Every other quantity built from the six field components that all observers agree on is a function of these two.