# Darwin Lagrangian

The Darwin Lagrangian (named after Charles Galton Darwin, grandson of the naturalist) describes the interaction to order ${\frac {v^{2}}{c^{2}}}$ between two charged particles in a vacuum and is given by

$L=L_{\text{f}}+L_{\text{int}},$ where the free particle Lagrangian is

$L_{\text{f}}={\frac {1}{2}}m_{1}v_{1}^{2}+{\frac {1}{8c^{2}}}m_{1}v_{1}^{4}+{\frac {1}{2}}m_{2}v_{2}^{2}+{\frac {1}{8c^{2}}}m_{2}v_{2}^{4},$ and the interaction Lagrangian is

$L_{\text{int}}=L_{\text{C}}+L_{\text{D}},$ where the Coulomb interaction is

$L_{\text{C}}=-{\frac {q_{1}q_{2}}{r}},$ and the Darwin interaction is

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

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

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 (Gaussian units)

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

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

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

which has a Fourier transform

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

The transverse component of the current is

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

It is easily verified that

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

which must be true if the divergence of the transverse current is zero. We see that

$\mathbf {J} _{t}\left(\mathbf {k} \right)$

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

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

where we have kept only the lowest order term in v/c.

The inverse Fourier transform of the vector potential is

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

where

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

The Darwin interaction term in the Lagrangian is then

 $L_{\rm {D}}={q_{1}q_{2} \over r}{1 \over 2c^{2}}\mathbf {v} _{1}\cdot \left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\cdot \mathbf {v} _{2}$ 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

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

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

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

### Interacting particles

For interacting particles, the equation of motion becomes

${d \over dt}\left[\left(1+{1 \over 2}{v_{1}^{2} \over c^{2}}\right)m_{1}\mathbf {v} _{1}+{q_{1} \over c}\mathbf {A} \left(\mathbf {r} _{1}\right)\right]=-\nabla {q_{1}q_{2} \over r}+\nabla \left[{q_{1}q_{2} \over r}{1 \over 2c^{2}}\mathbf {v} _{1}\cdot \left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\cdot \mathbf {v} _{2}\right]$
 ${d\mathbf {p} _{1} \over dt}={q_{1}q_{2} \over r^{2}}{\hat {\mathbf {r} }}+{q_{1}q_{2} \over r^{2}}{1 \over 2c^{2}}\left\{\mathbf {v} _{1}\left({{\hat {\mathbf {r} }}\cdot \mathbf {v} _{2}}\right)+\mathbf {v} _{2}\left({{\hat {\mathbf {r} }}\cdot \mathbf {v} _{1}}\right)-{\hat {\mathbf {r} }}\left[\mathbf {v} _{1}\cdot \left(\mathbf {1} +3{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)\cdot \mathbf {v} _{2}\right]\right\}$ $\mathbf {p} _{1}=\left(1+{1 \over 2}{v_{1}^{2} \over c^{2}}\right)m_{1}\mathbf {v} _{1}+{q_{1} \over c}\mathbf {A} \left(\mathbf {r} _{1}\right)$
$\mathbf {A} \left(\mathbf {r} _{1}\right)={q_{2} \over 2c}{1 \over r}\left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\cdot \mathbf {v} _{2}$
$\mathbf {r} =\mathbf {r} _{1}-\mathbf {r} _{2}$

## 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

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

The Hamiltonian becomes

 $H\left(\mathbf {r} _{1},\mathbf {p} _{1},\mathbf {r} _{2},\mathbf {p} _{2}\right)=\left(1-{1 \over 4}{p_{1}^{2} \over m_{1}^{2}c^{2}}\right){p_{1}^{2} \over 2m_{1}}\;+\;\left(1-{1 \over 4}{p_{2}^{2} \over m_{2}^{2}c^{2}}\right){p_{2}^{2} \over 2m_{2}}\;+\;{q_{1}q_{2} \over r}\;-\;{q_{1}q_{2} \over r}{1 \over 2m_{1}m_{2}c^{2}}\mathbf {p} _{1}\cdot \left[\mathbf {1} +\mathbf {\hat {r}} \mathbf {\hat {r}} \right]\cdot \mathbf {p} _{2}.$ ## Hamiltonian equations of motion

The Hamiltonian equations of motion are

$\mathbf {v} _{1}={\partial H \over \partial \mathbf {p} _{1}}$

and

${d\mathbf {p} _{1} \over dt}=-\nabla _{1}H$

which yield

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

and

 ${d\mathbf {p} _{1} \over dt}={q_{1}q_{2} \over r^{2}}{\hat {\mathbf {r} }}\;+\;{q_{1}q_{2} \over r^{2}}{1 \over 2m_{1}m_{2}c^{2}}\left\{\mathbf {p} _{1}\left({{\hat {\mathbf {r} }}\cdot \mathbf {p} _{2}}\right)+\mathbf {p} _{2}\left({{\hat {\mathbf {r} }}\cdot \mathbf {p} _{1}}\right)-{\hat {\mathbf {r} }}\left[\mathbf {p} _{1}\cdot \left(\mathbf {1} +3{\hat {\mathbf {r} }}{\hat {\mathbf {r} }}\right)\cdot \mathbf {p} _{2}\right]\right\}$ Note that the quantum mechanical Breit equation originally used the Darwin Lagrangian with the Darwin Hamiltonian as its classical starting point though the Breit equation would be better vindicated by the Wheeler–Feynman absorber theory and better yet quantum electrodynamics.