The Darwin Lagrangian (named after Charles Galton Darwin, grandson of the naturalist) describes the interaction to order ${\textstyle {v^{2))/{c^{2))}$ between two charged particles in a vacuum and is given by^{[1]: 596–598}

$${\displaystyle L=L_{\text{f))+L_{\text{int)),}$$

$${\displaystyle L_{\text{f))={\frac {1}{2))m_{1}v_{1}^{2}+{\frac {1}{8c^{2))}m_{1}v_{1}^{4}+{\frac {1}{2))m_{2}v_{2}^{2}+{\frac {1}{8c^{2))}m_{2}v_{2}^{4},}$$

$${\displaystyle L_{\text{int))=L_{\text{C))+L_{\text{D)),}$$

$${\displaystyle L_{\text{C))=-{\frac {q_{1}q_{2)){r)),}$$

$${\displaystyle L_{\text{D))={\frac {q_{1}q_{2)){r)){\frac {1}{2c^{2))}\mathbf {v} _{1}\cdot \left[\mathbf {1} +{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2}.}$$

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 ${\displaystyle {\hat {\mathbf {r} ))}$ is the unit vector in the direction of r.

The free Lagrangian 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^{[1]: 580–581}

$${\displaystyle L_{\text{int))=-q\Phi +{\frac {q}{c))\mathbf {u} \cdot \mathbf {A} ,}$$

The vector potential in the Coulomb gauge is described by^[1]: 242 (Gaussian units)

$${\displaystyle \nabla ^{2}\mathbf {A} -{\frac {1}{c^{2))}{\frac {\partial ^{2}\mathbf {A} }{\partial t^{2))}=-{\frac {4\pi }{c))\mathbf {J} _{t))$$

The current generated by the second particle is

$${\displaystyle \mathbf {J} =q_{2}\mathbf {v} _{2}\delta {\left(\mathbf {r} -\mathbf {r} _{2}\right)},}$$

which has a Fourier transform

$${\displaystyle \mathbf {J} \left(\mathbf {k} \right)\equiv \int d^{3}r\exp \left(-i\mathbf {k} \cdot \mathbf {r} \right)\mathbf {J} \left(\mathbf {r} \right)=q_{2}\mathbf {v} _{2}\exp \left(-i\mathbf {k} \cdot \mathbf {r} _{2}\right).}$$

The transverse component of the current is

$${\displaystyle \mathbf {J} _{t}(\mathbf {k} )=q_{2}\left[\mathbf {1} -{\hat {\mathbf {k} )){\hat {\mathbf {k} ))\right]\cdot \mathbf {v} _{2}e^{-i\mathbf {k} \cdot \mathbf {r} _{2)).}$$

It is easily verified that

$${\displaystyle \mathbf {k} \cdot \mathbf {J} _{t}(\mathbf {k} )=0,}$$

${\displaystyle \mathbf {J} _{t}(\mathbf {k} )}$

From the equation for the vector potential, the Fourier transform of the vector potential is

$${\displaystyle \mathbf {A} \left(\mathbf {k} \right)={\frac {4\pi }{c)){\frac {q_{2)){k^{2))}\left[\mathbf {1} -{\hat {\mathbf {k} )){\hat {\mathbf {k} ))\right]\cdot \mathbf {v} _{2}e^{-i\mathbf {k} \cdot \mathbf {r} _{2))}$$

The inverse Fourier transform of the vector potential is

$${\displaystyle \mathbf {A} \left(\mathbf {r} \right)=\int {\frac {d^{3}k}{\left(2\pi \right)^{3))}\;\mathbf {A} (\mathbf {k} )\;e^{i\mathbf {k} \cdot \mathbf {r} _{1))={\frac {q_{2)){2c)){\frac {1}{r))\left[\mathbf {1} +{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2))$$

$${\displaystyle \mathbf {r} =\mathbf {r} _{1}-\mathbf {r} _{2))$$

The Darwin interaction term in the Lagrangian is then

$${\displaystyle L_{\text{D))={\frac {q_{1}q_{2)){r)){\frac {1}{2c^{2))}\mathbf {v} _{1}\cdot \left[\mathbf {1} +{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2))$$

Lagrangian equations of motion

The equation of motion for one of the particles is

$${\displaystyle {\frac {d}{dt)){\frac {\partial }{\partial \mathbf {v} _{1))}L\left(\mathbf {r} _{1},\mathbf {v} _{1}\right)=\nabla _{1}L\left(\mathbf {r} _{1},\mathbf {v} _{1}\right)}$$

$${\displaystyle {\frac {d\mathbf {p} _{1)){dt))=\nabla _{1}L\left(\mathbf {r} _{1},\mathbf {v} _{1}\right)}$$

Free particle

The equation of motion for a free particle neglecting interactions between the two particles is

$${\displaystyle {\frac {d}{dt))\left[\left(1+{\frac {1}{2)){\frac {v_{1}^{2)){c^{2))}\right)m_{1}\mathbf {v} _{1}\right]=0}$$

$${\displaystyle \mathbf {p} _{1}=\left(1+{\frac {1}{2)){\frac {v_{1}^{2)){c^{2))}\right)m_{1}\mathbf {v} _{1))$$

Interacting particles

For interacting particles, the equation of motion becomes

$${\displaystyle {\frac {d}{dt))\left[\left(1+{\frac {1}{2)){\frac {v_{1}^{2)){c^{2))}\right)m_{1}\mathbf {v} _{1}+{\frac {q_{1)){c))\mathbf {A} \left(\mathbf {r} _{1}\right)\right]=-\nabla {\frac {q_{1}q_{2)){r))+\nabla \left[{\frac {q_{1}q_{2)){r)){\frac {1}{2c^{2))}\mathbf {v} _{1}\cdot \left[\mathbf {1} +{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2}\right]}$$

$${\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\))$$

$${\displaystyle \mathbf {p} _{1}=\left(1+{\frac {1}{2)){\frac {v_{1}^{2)){c^{2))}\right)m_{1}\mathbf {v} _{1}+{\frac {q_{1)){c))\mathbf {A} \left(\mathbf {r} _{1}\right)}$$

$${\displaystyle \mathbf {A} \left(\mathbf {r} _{1}\right)={\frac {q_{2)){2c)){\frac {1}{r))\left[\mathbf {1} +{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {v} _{2))$$

$${\displaystyle \mathbf {r} =\mathbf {r} _{1}-\mathbf {r} _{2))$$

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

$${\displaystyle H=\mathbf {p} _{1}\cdot \mathbf {v} _{1}+\mathbf {p} _{2}\cdot \mathbf {v} _{2}-L.}$$

The Hamiltonian becomes

$${\displaystyle H\left(\mathbf {r} _{1},\mathbf {p} _{1},\mathbf {r} _{2},\mathbf {p} _{2}\right)=\left(1-{\frac {1}{4)){\frac {p_{1}^{2)){m_{1}^{2}c^{2))}\right){\frac {p_{1}^{2)){2m_{1))}\;+\;\left(1-{\frac {1}{4)){\frac {p_{2}^{2)){m_{2}^{2}c^{2))}\right){\frac {p_{2}^{2)){2m_{2))}\;+\;{\frac {q_{1}q_{2)){r))\;-\;{\frac {q_{1}q_{2)){r)){\frac {1}{2m_{1}m_{2}c^{2))}\mathbf {p} _{1}\cdot \left[\mathbf {1} +\mathbf {\hat {r)) \mathbf {\hat {r)) \right]\cdot \mathbf {p} _{2}.}$$

Hamiltonian equations of motion

The Hamiltonian equations of motion are

$${\displaystyle \mathbf {v} _{1}={\frac {\partial H}{\partial \mathbf {p} _{1))))$$

$${\displaystyle {\frac {d\mathbf {p} _{1)){dt))=-\nabla _{1}H}$$

$${\displaystyle \mathbf {v} _{1}=\left(1-{\frac {1}{2)){\frac {p_{1}^{2)){m_{1}^{2}c^{2))}\right){\frac {\mathbf {p} _{1)){m_{1))}-{\frac {q_{1}q_{2)){2m_{1}m_{2}c^{2))}{\frac {1}{r))\left[\mathbf {1} +{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right]\cdot \mathbf {p} _{2))$$

$${\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}{2m_{1}m_{2}c^{2))}\left\{\mathbf {p} _{1}\left(((\hat {\mathbf {r} ))\cdot \mathbf {p} _{2))\right)+\mathbf {p} _{2}\left(((\hat {\mathbf {r} ))\cdot \mathbf {p} _{1))\right)-{\hat {\mathbf {r} ))\left[\mathbf {p} _{1}\cdot \left(\mathbf {1} +3{\hat {\mathbf {r} )){\hat {\mathbf {r} ))\right)\cdot \mathbf {p} _{2}\right]\right\))$$