Darwin Lagrangian

Derivation in vacuum

The relativistic interaction Lagrangian for a particle with charge q interacting with an electromagnetic field is $${\displaystyle L_{\text{int}}=-q\mathrm{\Phi }+\frac{q}{c}\mathbf{u}\cdot \mathbf{A},}$$ 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 $${\displaystyle {\mathrm{\nabla }}^2\mathbf{A}-\frac{1}{c^2}\frac{{\mathrm{\partial }}^2\mathbf{A}}{\mathrm{\partial }{t}^2}=-\frac{4\pi }{c}{\mathbf{J}}_t}$$ where the transverse current J_t 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 $${\displaystyle \mathbf{J}=q_2{\mathbf{v}}_2\delta \left(\mathbf{r}-{\mathbf{r}}_2\right),}$$ 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 $${\displaystyle \mathbf{k}\cdot {\mathbf{J}}_t(\mathbf{k})=0,}$$ which must be true if the divergence of the transverse current is zero. We see that ${\displaystyle {\mathbf{J}}_t(\mathbf{k})}$ 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 $${\displaystyle \mathbf{A}\left(\mathbf{k}\right)=\frac{4\pi }{c}\frac{q_2}{k^2}\left[\mathbf{1}-\hat{\mathbf{k}}\hat{\mathbf{k}}\right]\cdot {\mathbf{v}}_2{e}^{-i\mathbf{k}\cdot {\mathbf{r}}_2}}$$ where we have kept only the lowest order term in v/c.

The inverse Fourier transform of the vector potential is $${\displaystyle \mathbf{A}\left(\mathbf{r}\right)=\int \frac{d^{3}k}{{\left(2\pi \right)}^3}\mathbf{A}(\mathbf{k})e^{i\mathbf{k}\cdot {\mathbf{r}}_1}=\frac{q_2}{2c}\frac{1}{r}\left[\mathbf{1}+\hat{\mathbf{r}}\hat{\mathbf{r}}\right]\cdot {\mathbf{v}}_2}$$ where $${\displaystyle \mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2}$$ (see Common integrals in quantum field theory § Transverse potential with mass).

The Darwin interaction term in the Lagrangian is then $${\displaystyle L_{\text{D}}=\frac{q_1{q}_2}{r}\frac{1}{2c^2}{\mathbf{v}}_1\cdot \left[\mathbf{1}+\hat{\mathbf{r}}\hat{\mathbf{r}}\right]\cdot {\mathbf{v}}_2}$$ 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 $${\displaystyle \frac{d}{dt}\frac{\mathrm{\partial }}{\mathrm{\partial }{\mathbf{v}}_1}L\left({\mathbf{r}}_1,{\mathbf{v}}_1\right)={\mathrm{\nabla }}_{1}L\left({\mathbf{r}}_1,{\mathbf{v}}_1\right)}$$ $${\displaystyle \frac{d{\mathbf{p}}_1}{dt}={\mathrm{\nabla }}_{1}L\left({\mathbf{r}}_1,{\mathbf{v}}_1\right)}$$ 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 $${\displaystyle \frac{d}{dt}\left[\left(1+\frac{1}{2}\frac{v_1^2}{c^2}\right)m_1{\mathbf{v}}_1\right]=0}$$ $${\displaystyle {\mathbf{p}}_1=\left(1+\frac{1}{2}\frac{v_1^2}{c^2}\right)m_1{\mathbf{v}}_1}$$

Interacting particles

For interacting particles, the equation of motion becomes $${\displaystyle \frac{d}{dt}\left[\left(1+\frac{1}{2}\frac{v_1^2}{c^2}\right)m_1{\mathbf{v}}_1+\frac{q_1}{c}\mathbf{A}\left({\mathbf{r}}_1\right)\right]=-\mathrm{\nabla }\frac{q_1{q}_2}{r}+\mathrm{\nabla }\left[\frac{q_1{q}_2}{r}\frac{1}{2c^2}{\mathbf{v}}_1\cdot \left[\mathbf{1}+\hat{\mathbf{r}}\hat{\mathbf{r}}\right]\cdot {\mathbf{v}}_2\right]}$$ $${\displaystyle \frac{d{\mathbf{p}}_1}{dt}=\frac{q_1{q}_2}{r^2}\hat{\mathbf{r}}+\frac{q_1{q}_2}{r^2}\frac{1}{2c^2}\left\{{\mathbf{v}}_1\left(\hat{\mathbf{r}}\cdot {\mathbf{v}}_2\right)+{\mathbf{v}}_2\left(\hat{\mathbf{r}}\cdot {\mathbf{v}}_1\right)-\hat{\mathbf{r}}\left[{\mathbf{v}}_1\cdot \left(\mathbf{1}+3\hat{\mathbf{r}}\hat{\mathbf{r}}\right)\cdot {\mathbf{v}}_2\right]\right\}}$$ $${\displaystyle {\mathbf{p}}_1=\left(1+\frac{1}{2}\frac{v_1^2}{c^2}\right)m_1{\mathbf{v}}_1+\frac{q_1}{c}\mathbf{A}\left({\mathbf{r}}_1\right)}$$ $${\displaystyle \mathbf{A}\left({\mathbf{r}}_1\right)=\frac{q_2}{2c}\frac{1}{r}\left[\mathbf{1}+\hat{\mathbf{r}}\hat{\mathbf{r}}\right]\cdot {\mathbf{v}}_2}$$ $${\displaystyle \mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2}$$

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 $${\displaystyle H\left({\mathbf{r}}_1,{\mathbf{p}}_1,{\mathbf{r}}_2,{\mathbf{p}}_2\right)=\left(1-\frac{1}{4}\frac{p_1^2}{m_1^2{c}^2}\right)\frac{p_1^2}{2m_1}+\left(1-\frac{1}{4}\frac{p_2^2}{m_2^2{c}^2}\right)\frac{p_2^2}{2m_2}+\frac{q_1{q}_2}{r}-\frac{q_1{q}_2}{r}\frac{1}{2m_1{m}_2{c}^2}{\mathbf{p}}_1\cdot \left[\mathbf{1}+\hat{\mathbf{r}}\hat{\mathbf{r}}\right]\cdot {\mathbf{p}}_2.}$$

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 ${\displaystyle \mathbf{p}=m\mathbf{v}}$ in the last term and reverse its sign.

Equations of motion

The Hamiltonian equations of motion are $${\displaystyle {\mathbf{v}}_1=\frac{\mathrm{\partial }H}{\mathrm{\partial }{\mathbf{p}}_1}}$$ and $${\displaystyle \frac{d{\mathbf{p}}_1}{dt}=-{\mathrm{\nabla }}_{1}H,}$$ which yield $${\displaystyle {\mathbf{v}}_1=\left(1-\frac{1}{2}\frac{p_1^2}{m_1^2{c}^2}\right)\frac{{\mathbf{p}}_1}{m_1}-\frac{q_1{q}_2}{2m_1{m}_2{c}^2}\frac{1}{r}\left[\mathbf{1}+\hat{\mathbf{r}}\hat{\mathbf{r}}\right]\cdot {\mathbf{p}}_2}$$ and $${\displaystyle \frac{d{\mathbf{p}}_1}{dt}=\frac{q_1{q}_2}{r^2}\hat{\mathbf{r}}+\frac{q_1{q}_2}{r^2}\frac{1}{2m_1{m}_2{c}^2}\left\{{\mathbf{p}}_1\left(\hat{\mathbf{r}}\cdot {\mathbf{p}}_2\right)+{\mathbf{p}}_2\left(\hat{\mathbf{r}}\cdot {\mathbf{p}}_1\right)-\hat{\mathbf{r}}\left[{\mathbf{p}}_1\cdot \left(\mathbf{1}+3\hat{\mathbf{r}}\hat{\mathbf{r}}\right)\cdot {\mathbf{p}}_2\right]\right\}}$$

Quantum electrodynamics

The structure of the Darwin interaction can also be clearly seen in quantum electrodynamics and due to the exchange of photons in lowest order of perturbation theory. When the photon has four-momentum p^μ = ħk^μ with wave vector k^μ = (ω /c, k), its propagator in the Coulomb gauge has two components.

${\displaystyle D_{00}(k)=\frac{1}{{\mathbf{k}}^2}}$

gives the Coulomb interaction between two charged particles, while

${\displaystyle D_{ij}(k)=\frac{1}{{\omega }^2-c^2{\mathbf{k}}^2}\left({\delta }_{ij}-\frac{k_i{k}_j}{{\mathbf{k}}^2}\right)}$

describes the exchange of a transverse photon. It has a polarization vector ${\displaystyle {\mathbf{e}}_{\lambda }}$ and couples to a particle with charge ${\displaystyle q}$ and three-momentum ${\displaystyle \mathbf{p}}$ with a strength ${\displaystyle -q\sqrt{4\pi }{\mathbf{e}}_{\lambda }\cdot \mathbf{p}/m.}$ Since ${\displaystyle {\mathbf{e}}_{\lambda }\cdot \mathbf{k}=0}$ in this gauge, it doesn't matter if one uses the particle momentum before or after the photon couples to it.

In the exchange of the photon between the two particles one can ignore the frequency ${\displaystyle \omega }$ compared with ${\displaystyle c\mathbf{k}}$ in the propagator working to the accuracy in ${\displaystyle v^2/c^2}$ that is needed here. The two parts of the propagator then give together the effective Hamiltonian

${\displaystyle H_{int}(\mathbf{k})=\frac{4\pi q_1{q}_2}{{\mathbf{k}}^2}-\frac{4\pi q_1{q}_2}{m_1{m}_2{c}^2{\mathbf{k}}^2}{\mathbf{p}}_1\cdot \left(\mathbf{1}-\hat{\mathbf{k}}\hat{\mathbf{k}}\right)\cdot {\mathbf{p}}_2}$

for their interaction in k-space. This is now identical with the classical result and there is no trace of the quantum effects used in this derivation.

A similar calculation can be done when the photon couples to Dirac particles with spin s = 1/2 and used for a derivation of the Breit equation. It gives the same Darwin interaction but also additional terms involving the spin degrees of freedom and depending on the Planck constant.