The Darwin Lagrangian (named after Charles Galton Darwin, grandson of the naturalist) describes the interaction to order v 2 / c 2 {\textstyle {v^{2}}/{c^{2}}} between two charged particles in a vacuum where c is the speed of light. It was derived before the advent of quantum mechanics and resulted from a more detailed investigation of the classical, electromagnetic interactions of the electrons in an atom. From the Bohr model it was known that they should be moving with velocities approaching the speed of light.
Formulation The full Lagrangian for two interacting particles is
L = L f + L int , {\displaystyle L=L_{\text{f}}+L_{\text{int}},}
where the free particle part is
L f = 1 2 m 1 v 1 2 + 1 8 c 2 m 1 v 1 4 + 1 2 m 2 v 2 2 + 1 8 c 2 m 2 v 2 4 , {\displaystyle L_{\text{f}}={\frac {1}{2}}m_{1}v_{1}^{2}+{\frac {1}{8c^{2}}}m_{1}v_{1}^{4}+{\frac {1}{2}}m_{2}v_{2}^{2}+{\frac {1}{8c^{2}}}m_{2}v_{2}^{4},}
The interaction is described by
L int = L C + L D , {\displaystyle L_{\text{int}}=L_{\text{C}}+L_{\text{D}},}
where the Coulomb interaction in Gaussian units is
L C = − q 1 q 2 r , {\displaystyle L_{\text{C}}=-{\frac {q_{1}q_{2}}{r}},}
while the Darwin interaction is
L D = q 1 q 2 r 1 2 c 2 v 1 ⋅ [ 1 + r ^ r ^ ] ⋅ v 2 . {\displaystyle L_{\text{D}}={\frac {q_{1}q_{2}}{r}}{\frac {1}{2c^{2}}}\mathbf {v} _{1}\cdot \left[\mathbf {1} +{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right]\cdot \mathbf {v} _{2}.}
Here q1 and q2 are the charges on particles 1 and 2 respectively, m1 and m2 are the masses of the particles, v1 and v2 are the velocities of the particles, c is the speed of light, r is the vector between the two particles, and r ^ {\displaystyle {\hat {\mathbf {r} }}} is the unit vector in the direction of r. The first part is the Taylor expansion of free Lagrangian of two relativistic particles to second order in v. The Darwin interaction term is due to one particle reacting to the magnetic field generated by the other particle. If higher-order terms in v/c are retained, then the field degrees of freedom must be taken into account, and the interaction can no longer be taken to be instantaneous between the particles. In that case retardation effects must be accounted for.
Derivation in vacuum The relativistic interaction Lagrangian for a particle with charge q interacting with an electromagnetic field is
L int = − q Φ + q c u ⋅ A , {\displaystyle L_{\text{int}}=-q\Phi +{\frac {q}{c}}\mathbf {u} \cdot \mathbf {A} ,}
where u is the relativistic velocity of the particle. The first term on the right generates the Coulomb interaction. The second term generates the Darwin interaction. The vector potential in the Coulomb gauge is described by
∇ 2 A − 1 c 2 ∂ 2 A ∂ t 2 = − 4 π c J t {\displaystyle \nabla ^{2}\mathbf {A} -{\frac {1}{c^{2}}}{\frac {\partial ^{2}\mathbf {A} }{\partial t^{2}}}=-{\frac {4\pi }{c}}\mathbf {J} _{t}}
where the transverse current Jt is the solenoidal current (see Helmholtz decomposition) generated by a second particle. The divergence of the transverse current is zero. The current generated by the second particle is
J = q 2 v 2 δ ( r − r 2 ) , {\displaystyle \mathbf {J} =q_{2}\mathbf {v} _{2}\delta {\left(\mathbf {r} -\mathbf {r} _{2}\right)},}
which has a Fourier transform
J ( k ) ≡ ∫ d 3 r exp ( − i k ⋅ r ) J ( r ) = q 2 v 2 exp ( − i k ⋅ r 2 ) . {\displaystyle \mathbf {J} \left(\mathbf {k} \right)\equiv \int d^{3}r\exp \left(-i\mathbf {k} \cdot \mathbf {r} \right)\mathbf {J} \left(\mathbf {r} \right)=q_{2}\mathbf {v} _{2}\exp \left(-i\mathbf {k} \cdot \mathbf {r} _{2}\right).}
The transverse component of the current is
J t ( k ) = q 2 [ 1 − k ^ k ^ ] ⋅ v 2 e − i k ⋅ r 2 . {\displaystyle \mathbf {J} _{t}(\mathbf {k} )=q_{2}\left[\mathbf {1} -{\hat {\mathbf {k} }}{\hat {\mathbf {k} }}\right]\cdot \mathbf {v} _{2}e^{-i\mathbf {k} \cdot \mathbf {r} _{2}}.}
It is easily verified that
k ⋅ J t ( k ) = 0 , {\displaystyle \mathbf {k} \cdot \mathbf {J} _{t}(\mathbf {k} )=0,}
which must be true if the divergence of the transverse current is zero. We see that J t ( k ) {\displaystyle \mathbf {J} _{t}(\mathbf {k} )} is the component of the Fourier transformed current perpendicular to k. From the equation for the vector potential, the Fourier transform of the vector potential is
A ( k ) = 4 π c q 2 k 2 [ 1 − k ^ k ^ ] ⋅ v 2 e − i k ⋅ r 2 {\displaystyle \mathbf {A} \left(\mathbf {k} \right)={\frac {4\pi }{c}}{\frac {q_{2}}{k^{2}}}\left[\mathbf {1} -{\hat {\mathbf {k} }}{\hat {\mathbf {k} }}\right]\cdot \mathbf {v} _{2}e^{-i\mathbf {k} \cdot \mathbf {r} _{2}}}
where we have kept only the lowest order term in v/c. The inverse Fourier transform of the vector potential is
A ( r ) = ∫ d 3 k ( 2 π ) 3 A ( k ) e i k ⋅ r 1 = q 2 2 c 1 r [ 1 + r ^ r ^ ] ⋅ v 2 {\displaystyle \mathbf {A} \left(\mathbf {r} \right)=\int {\frac {d^{3}k}{\left(2\pi \right)^{3}}}\;\mathbf {A} (\mathbf {k} )\;e^{i\mathbf {k} \cdot \mathbf {r} _{1}}={\frac {q_{2}}{2c}}{\frac {1}{r}}\left[\mathbf {1} +{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right]\cdot \mathbf {v} _{2}}
where
r = r 1 − r 2 {\displaystyle \mathbf {r} =\mathbf {r} _{1}-\mathbf {r} _{2}}
(see Common integrals in quantum field theory § Transverse potential with mass). The Darwin interaction term in the Lagrangian is then
L D = q 1 q 2 r 1 2 c 2 v 1 ⋅ [ 1 + r ^ r ^ ] ⋅ v 2 {\displaystyle L_{\text{D}}={\frac {q_{1}q_{2}}{r}}{\frac {1}{2c^{2}}}\mathbf {v} _{1}\cdot \left[\mathbf {1} +{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right]\cdot \mathbf {v} _{2}}
where again we kept only the lowest order term in v/c.
Lagrangian equations of motion The equation of motion for one of the particles is
d d t ∂ ∂ v 1 L ( r 1 , v 1 ) = ∇ 1 L ( r 1 , v 1 ) {\displaystyle {\frac {d}{dt}}{\frac {\partial }{\partial \mathbf {v} _{1}}}L\left(\mathbf {r} _{1},\mathbf {v} _{1}\right)=\nabla _{1}L\left(\mathbf {r} _{1},\mathbf {v} _{1}\right)}
d p 1 d t = ∇ 1 L ( r 1 , v 1 ) {\displaystyle {\frac {d\mathbf {p} _{1}}{dt}}=\nabla _{1}L\left(\mathbf {r} _{1},\mathbf {v} _{1}\right)}
where p1 is the momentum of the particle.
Free particle The equation of motion for a free particle neglecting interactions between the two particles is
d d t [ ( 1 + 1 2 v 1 2 c 2 ) m 1 v 1 ] = 0 {\displaystyle {\frac {d}{dt}}\left[\left(1+{\frac {1}{2}}{\frac {v_{1}^{2}}{c^{2}}}\right)m_{1}\mathbf {v} _{1}\right]=0}
p 1 = ( 1 + 1 2 v 1 2 c 2 ) m 1 v 1 {\displaystyle \mathbf {p} _{1}=\left(1+{\frac {1}{2}}{\frac {v_{1}^{2}}{c^{2}}}\right)m_{1}\mathbf {v} _{1}}
Interacting particles For interacting particles, the equation of motion becomes
d d t [ ( 1 + 1 2 v 1 2 c 2 ) m 1 v 1 + q 1 c A ( r 1 ) ] = − ∇ q 1 q 2 r + ∇ [ q 1 q 2 r 1 2 c 2 v 1 ⋅ [ 1 + r ^ r ^ ] ⋅ v 2 ] {\displaystyle {\frac {d}{dt}}\left[\left(1+{\frac {1}{2}}{\frac {v_{1}^{2}}{c^{2}}}\right)m_{1}\mathbf {v} _{1}+{\frac {q_{1}}{c}}\mathbf {A} \left(\mathbf {r} _{1}\right)\right]=-\nabla {\frac {q_{1}q_{2}}{r}}+\nabla \left[{\frac {q_{1}q_{2}}{r}}{\frac {1}{2c^{2}}}\mathbf {v} _{1}\cdot \left[\mathbf {1} +{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right]\cdot \mathbf {v} _{2}\right]}
d p 1 d t = q 1 q 2 r 2 r ^ + q 1 q 2 r 2 1 2 c 2 { v 1 ( r ^ ⋅ v 2 ) + v 2 ( r ^ ⋅ v 1 ) − r ^ [ v 1 ⋅ ( 1 + 3 r ^ r ^ ) ⋅ v 2 ] } {\displaystyle {\frac {d\mathbf {p} _{1}}{dt}}={\frac {q_{1}q_{2}}{r^{2}}}{\hat {\mathbf {r} }}+{\frac {q_{1}q_{2}}{r^{2}}}{\frac {1}{2c^{2}}}\left\{\mathbf {v} _{1}\left({{\hat {\mathbf {r} }}\cdot \mathbf {v} _{2}}\right)+\mathbf {v} _{2}\left({{\hat {\mathbf {r} }}\cdot \mathbf {v} _{1}}\right)-{\hat {\mathbf {r} }}\left[\mathbf {v} _{1}\cdot \left(\mathbf {1} +3{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)\cdot \mathbf {v} _{2}\right]\right\}}
p 1 = ( 1 + 1 2 v 1 2 c 2 ) m 1 v 1 + q 1 c A ( r 1 ) {\displaystyle \mathbf {p} _{1}=\left(1+{\frac {1}{2}}{\frac {v_{1}^{2}}{c^{2}}}\right)m_{1}\mathbf {v} _{1}+{\frac {q_{1}}{c}}\mathbf {A} \left(\mathbf {r} _{1}\right)}
A ( r 1 ) = q 2 2 c 1 r [ 1 + r ^ r ^ ] ⋅ v 2 {\displaystyle \mathbf {A} \left(\mathbf {r} _{1}\right)={\frac {q_{2}}{2c}}{\frac {1}{r}}\left[\mathbf {1} +{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right]\cdot \mathbf {v} _{2}}
r = r 1 − r 2 {\displaystyle \mathbf {r} =\mathbf {r} _{1}-\mathbf {r} _{2}}
Hamiltonian for two particles in a vacuum The Darwin Hamiltonian for two particles in a vacuum is related to the Lagrangian by a Legendre transformation
H = p 1 ⋅ v 1 + p 2 ⋅ v 2 − L . {\displaystyle H=\mathbf {p} _{1}\cdot \mathbf {v} _{1}+\mathbf {p} _{2}\cdot \mathbf {v} _{2}-L.}
The Hamiltonian becomes
H ( r 1 , p 1 , r 2 , p 2 ) = ( 1 − 1 4 p 1 2 m 1 2 c 2 ) p 1 2 2 m 1 + ( 1 − 1 4 p 2 2 m 2 2 c 2 ) p 2 2 2 m 2 + q 1 q 2 r − q 1 q 2 r 1 2 m 1 m 2 c 2 p 1 ⋅ [ 1 + r ^ r ^ ] ⋅ p 2 . {\displaystyle H\left(\mathbf {r} _{1},\mathbf {p} _{1},\mathbf {r} _{2},\mathbf {p} _{2}\right)=\left(1-{\frac {1}{4}}{\frac {p_{1}^{2}}{m_{1}^{2}c^{2}}}\right){\frac {p_{1}^{2}}{2m_{1}}}\;+\;\left(1-{\frac {1}{4}}{\frac {p_{2}^{2}}{m_{2}^{2}c^{2}}}\right){\frac {p_{2}^{2}}{2m_{2}}}\;+\;{\frac {q_{1}q_{2}}{r}}\;-\;{\frac {q_{1}q_{2}}{r}}{\frac {1}{2m_{1}m_{2}c^{2}}}\mathbf {p} _{1}\cdot \left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\cdot \mathbf {p} _{2}.}
This Hamiltonian gives the interaction energy between the two particles. It has recently been argued that when expressed in terms of particle velocities, one should simply set p = m v {\displaystyle \mathbf {p} =m\mathbf {v} } in the last term and reverse its sign.
Equations of motion The Hamiltonian equations of motion are
v 1 = ∂ H ∂ p 1 {\displaystyle \mathbf {v} _{1}={\frac {\partial H}{\partial \mathbf {p} _{1}}}}
and