This material generalizes the Bohr model for a particle in a central force field when relativistic effects are taken into account. The amazing result is that angular momentum p_θ is quantized in exactly the same manner as for the non-relativistic case; i.e.,

Δp_θ = h
and hence
p_θ = lh

where h is Planck's constant divided by 2π and l is an integer.

Consider a particle in a central force field with a potential energy function V(r). The particle is in a circular orbit of radius r and has a velocity v. The relative velocity is β=(v/c) where c is the speed of light. The kinetic energy is a function of the relative velocity β; i.e.,

K(β) = m₀c²[(1−β²)^−½−1]

The total energy E of the particle is then

E = K(β) + V(r)

By Bohr's analysis

ΔE = (dE/dp_θ)Δp_θ = h(v/r)

But (dE/dp_θ) = (dE/dβ)/(dp_θ/dβ)

Thus

dE/dβ = K'(β) + V'(r)(dr/dβ)
and
dp_θ/dβ = d(mvr)/dβ = (d(mv)/dβ)r + mv(dr/dβ)

Since

K(β)=m₀c²[(1−β²)^−½−1]
K'(β) = m₀c²((−½)(−2β(1−β²)^−3/2 = m₀c²β(1−β²)^−3/2

On the other hand,

d(mv/dβ = cd(mβ)/dβ

Since

d(mβ)/dβ = d[m₀β/(1−β²)^½)]
= m₀/(1−β²)^½) + (−½)m₀β(−β)/(1−β²)^3/2)
= m₀[1/(1−β²)^½) + β²/(1−β²)^3/2]
= m₀[(1−β²+β²)/(1−β²)^3/2) = m₀(1−β²)^−3/2)

This means that

d(mv)/dβ = m₀c(1−β²)^−3/2

In a circular orbit mv²/r=V'(r) so mv=rV'(r)/v. Therefore

dp_θ/dβ = (rV'(r)/v)(dr/dβ) + rm₀c(1−β²)^−3/2
which, by factoring out a (r/v),
is equal to
dp_θ/dβ = (r/v)[V'(r)/v)(dr/dβ) + vm₀c(1−β²)^−3/2]

Thus

ΔE = [V'(r)/v)(dr/dβ)
+ βm₀c²(1−β²)^−3/2]/{(r/v)[V'(r)/v)(dr/dβ)
+ βm₀c²(1−β²)^−3/2]}Δp_θ
= h(v/r)

This reduces to the amazing result

Δp_θ = h

Thus the quantization of angular momentum in the relativistic case is independent of the potential energy function V(r) and is exactly the same as the quantization for the non-relativistic case.