17A (Electromagnetism)

For a boost of speed

v

in the

x

direction, with

γ = \frac{1}{1 - v ^{2} / c ^{2}},

E_{x}^{'} = E_{x}, B_{x}^{'} = B_{x}, E_{y}^{'} = γ (E_{y} - v B_{z}), B_{y}^{'} = γ (B_{y} + v E_{z} / c^{2}), E_{z}^{'} = γ (E_{z} + v B_{y}), B_{z}^{'} = γ (B_{z} - v E_{y} / c^{2}) .

Expanding

E_{y}^{'} B_{y}^{'} + E_{z}^{'} B_{z}^{'}

, the mixed terms cancel and the remaining transverse terms acquire the factor

γ^{2} (1 - v^{2} / c^{2}) = 1

. Together with the unchanged longitudinal product, this proves the Lorentz invariant

E^{'} \cdot B^{'} = E \cdot B .

An independent invariant is

E^{2} - c^{2} B^{2} .

If only

E_{y} = E

and

B_{y} = B

are nonzero, then

E^{'} = (0, γ E, γ v B), B^{'} = (0, γ B, - γ v E / c^{2}),

and direct subtraction gives

E^{'2} - c^{2} B^{'2} = γ^{2} (1 - v^{2} / c^{2}) (E^{2} - c^{2} B^{2}) = E^{2} - c^{2} B^{2} .

Now put

c B = λ E

and

β = v / c

. Invariance of the dot product gives

E^{'} \cdot B^{'} = λ E^{2} / c

, while the displayed components give

E^{'2} = γ^{2} E^{2} (1 + λ^{2} β^{2}), B^{'2} = \frac{γ ^{2} E ^{2}}{c ^{2}} (λ^{2} + β^{2}) .

Hence

cos θ = \frac{λ ( 1 - β ^{2} )}{s q r t ( 1 + λ ^{2} β ^{2} ) ( λ ^{2} + β ^{2} )} .

β = 0

this equals

sgn λ

, and as

∣ β ∣ ↑ 1

it tends continuously to zero with the same sign. Thus boosts realize every

0 \leq θ < π /2

when

λ > 0

, and every

π /2 < θ \leq π

when

λ < 0

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