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The Orbital and Spin Momenta of Light

  1. Feb 15, 2014 #1
    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
  2. jcsd
  3. Feb 27, 2015 #2
    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,
    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,
    where I omit the summation symbol.
    After the integration by parts, we have
    J&=L+S+\epsilon_0 \int_V [(\nabla_j E_j )(r\times A) ]d^3 r,\\
    where we assume there is no charge.
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