Hi everybody, In most classical or quantum optics texts an angular momentum is considered for the EM radiation as the following: [itex] J = ε_0 ∫_V r × [E(r, t) × B(r, t)] d^3 r [/itex] Then it is claimed that: "Using the usual formula for a double vector product and integrating by parts, bearing in mind the assumption that the fields are zero at the surface of volume V introduced for the mode expansion, one finds that JR can be written as a sum of two terms: [itex]J = L + S , [/itex] given by [itex]L = ε_0 ∑_{j=(x,y,z)} ∫d^3r Ej (r, t)(r × ∇)Aj (r, t) ,[/itex] [itex]S = ε_0 ∫d^3r E(r, t) × A(r, t) [/itex] " Would you please help me derive them? Thank you
Hi kderakhshani, The expressions of L and S include the vector potential. So, first, let's use the vector potential, [tex]J=\epsilon_0 \int_Vr\times[E\times [\nabla \times A]]d^3 r.[/tex] We can proceed by using the formula [itex]a\times(b\times c)=b(a\cdot c)-c(a\cdot b)[/itex]. There are two possibilities, [itex]a=r[/itex] or [itex]a=E[/itex]. After some consideration, it seems that the former is the right way to go, [tex] \begin{align} J&=\epsilon_0 \int_Vr\times[ E_j \nabla A_j -E_j \nabla_j A ]d^3 r,\\ &=L+\epsilon_0 \int_Vr\times[ -E_j \nabla_j A ]d^3 r,\\ &=L+\epsilon_0 \int_V [-E_j \nabla_j (r\times A)+E\times A ]d^3 r,\\ &=L+S+\epsilon_0 \int_V [-E_j \nabla_j (r\times A) ]d^3 r, \end{align} [/tex] where I omit the summation symbol. After the integration by parts, we have [tex] \begin{align} J&=L+S+\epsilon_0 \int_V [(\nabla_j E_j )(r\times A) ]d^3 r,\\ &=L+S+0, \end{align} [/tex] where we assume there is no charge.