In the radiation zone, both fields are transverse to

x

and obey

E_{0} = - c x \times B_{0}, ∣ E_{0} ∣ = c ∣ B_{0} ∣.

S = \frac{1}{μ _{0}} E \times B = \frac{c}{μ _{0}} ∣ B_{0} ∣^{2} sin^{2} (ω t - k \cdot x) x,

⟨ S_{r} ⟩ = \frac{c}{2 μ _{0}} ∣ B_{0} ∣^{2} .

An absorbing shell receives the incident momentum rather than reversing it. Hence radiation pressure on a perfectly absorbing surface gives

p (x) = \frac{⟨ S _{r} ⟩}{c} = \frac{∣ B _{0} ∣ ^{2}}{2 μ _{0}} .

At

r = R

,

∣ B_{0} ∣^{2} = \frac{μ _{0}^{2} ω ^{4}}{16 π ^{2} R ^{2} c ^{2}} ∣ x \times p_{0} ∣^{2},

and therefore

p (x) = \frac{μ _{0} ω ^{4}}{32 π ^{2} R ^{2} c ^{2}} ∣ x \times p_{0} ∣^{2} .

If

ϑ

is the angle between

x

and

p_{0}

, this is

p (ϑ) = \frac{μ _{0} ω ^{4} ∣ p _{0} ∣ ^{2}}{32 π ^{2} R ^{2} c ^{2}} sin^{2} ϑ .

μ_{0} ω^{4} ∣ p_{0} ∣^{2} / (48 π^{2} R^{2} c^{2})

; the vector force integrates to zero by the inversion symmetry of the electric dipole radiation pattern.

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 Incoming links: Radiation pressure on a perfectly absorbing surface