In electrodynamics, minimal coupling is adequate to account for all electromagnetic interactions. Higher moments of particles are consequences of minimal coupling and non-zero spin.

Non-relativistic charged particle in an electromagnetic field edit

In Cartesian coordinates, the Lagrangian of a non-relativistic classical particle in an electromagnetic field is (in SI Units):

${\displaystyle {\mathcal {L}}=\sum _{i}{\tfrac {1}{2}}m{\dot {x}}_{i}^{2}+\sum _{i}q{\dot {x}}_{i}A_{i}-q\varphi }$ 

where q is the electric charge of the particle, φ is the electric scalar potential, and the A_i are the components of the magnetic vector potential that may all explicitly depend on ${\displaystyle x_{i}}$  and ${\displaystyle t}$ .

This Lagrangian, combined with Euler–Lagrange equation, produces the Lorentz force law

${\displaystyle m{\ddot {\mathbf {x} }}=q\mathbf {E} +q{\dot {\mathbf {x} }}\times \mathbf {B} \,,}$ 

and is called minimal coupling.

Note that the values of scalar potential and vector potential would change during a gauge transformation,^[2] and the Lagrangian itself will pick up extra terms as well, but the extra terms in the Lagrangian add up to a total time derivative of a scalar function, and therefore still produce the same Euler–Lagrange equation.

The canonical momenta are given by

${\displaystyle p_{i}={\frac {\partial {\mathcal {L}}}{\partial {\dot {x}}_{i}}}=m{\dot {x}}_{i}+qA_{i}}$ 

Note that canonical momenta are not gauge invariant, and are not physically measurable. However, the kinetic momenta

${\displaystyle P_{i}\equiv m{\dot {x}}_{i}=p_{i}-qA_{i}}$ 

are gauge invariant and physically measurable.

The Hamiltonian, as the Legendre transformation of the Lagrangian, is therefore

${\displaystyle {\mathcal {H}}=\left\{\sum _{i}{\dot {x}}_{i}p_{i}\right\}-{\mathcal {L}}=\sum _{i}{\frac {\left(p_{i}-qA_{i}\right)^{2}}{2m}}+q\varphi }$ 

This equation is used frequently in quantum mechanics.

Under a gauge transformation,

${\displaystyle \mathbf {A} \rightarrow \mathbf {A} +\nabla f\,,\quad \varphi \rightarrow \varphi -{\dot {f}}\,,}$ 

where f(r,t) is any scalar function of space and time, the aforementioned Lagrangian, canonical momenta and Hamiltonian transform like

${\displaystyle L\rightarrow L'=L+q{\frac {df}{dt}}\,,\quad \mathbf {p} \rightarrow \mathbf {p'} =\mathbf {p} +q\nabla f\,,\quad H\rightarrow H'=H-q{\frac {\partial f}{\partial t}}\,,}$ 

which still produces the same Hamilton's equation:

${\displaystyle {\begin{aligned}\left.{\frac {\partial H'}{\partial {x_{i}}}}\right|_{p'_{i}}&=\left.{\frac {\partial }{\partial {x_{i}}}}\right|_{p'_{i}}({\dot {x}}_{i}p'_{i}-L')=-\left.{\frac {\partial L'}{\partial {x_{i}}}}\right|_{p'_{i}}\\&=-\left.{\frac {\partial L}{\partial {x_{i}}}}\right|_{p'_{i}}-q\left.{\frac {\partial }{\partial {x_{i}}}}\right|_{p'_{i}}{\frac {df}{dt}}\\&=-{\frac {d}{dt}}\left(\left.{\frac {\partial L}{\partial {{\dot {x}}_{i}}}}\right|_{p'_{i}}+q\left.{\frac {\partial f}{\partial {x_{i}}}}\right|_{p'_{i}}\right)\\&=-{\dot {p}}'_{i}\end{aligned}}}$ 

In quantum mechanics, the wave function will also undergo a local U(1) group transformation^[3] during the gauge transformation, which implies that all physical results must be invariant under local U(1) transformations.

Relativistic charged particle in an electromagnetic field edit

The relativistic Lagrangian for a particle (rest mass m and charge q) is given by:

${\displaystyle {\mathcal {L}}(t)=-mc^{2}{\sqrt {1-{\frac {{{\dot {\mathbf {x} }}(t)}^{2}}{c^{2}}}}}+q{\dot {\mathbf {x} }}(t)\cdot \mathbf {A} \left(\mathbf {x} (t),t\right)-q\varphi \left(\mathbf {x} (t),t\right)}$ 

Thus the particle's canonical momentum is

${\displaystyle \mathbf {p} (t)={\frac {\partial {\mathcal {L}}}{\partial {\dot {\mathbf {x} }}}}={\frac {m{\dot {\mathbf {x} }}}{\sqrt {1-{\frac {{\dot {\mathbf {x} }}^{2}}{c^{2}}}}}}+q\mathbf {A} }$ 

that is, the sum of the kinetic momentum and the potential momentum.

Solving for the velocity, we get

${\displaystyle {\dot {\mathbf {x} }}(t)={\frac {\mathbf {p} -q\mathbf {A} }{\sqrt {m^{2}+{\frac {1}{c^{2}}}{\left(\mathbf {p} -q\mathbf {A} \right)}^{2}}}}}$ 

So the Hamiltonian is

${\displaystyle {\mathcal {H}}(t)={\dot {\mathbf {x} }}\cdot \mathbf {p} -{\mathcal {L}}=c{\sqrt {m^{2}c^{2}+{\left(\mathbf {p} -q\mathbf {A} \right)}^{2}}}+q\varphi }$ 

This results in the force equation (equivalent to the Euler–Lagrange equation)

${\displaystyle {\dot {\mathbf {p} }}=-{\frac {\partial {\mathcal {H}}}{\partial \mathbf {x} }}=q{\dot {\mathbf {x} }}\cdot ({\boldsymbol {\nabla }}\mathbf {A} )-q{\boldsymbol {\nabla }}\varphi =q{\boldsymbol {\nabla }}({\dot {\mathbf {x} }}\cdot \mathbf {A} )-q{\boldsymbol {\nabla }}\varphi }$ 

from which one can derive

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}\left({\frac {m{\dot {\mathbf {x} }}}{\sqrt {1-{\frac {{\dot {\mathbf {x} }}^{2}}{c^{2}}}}}}\right)&={\frac {\mathrm {d} }{\mathrm {d} t}}(\mathbf {p} -q\mathbf {A} )={\dot {\mathbf {p} }}-q{\frac {\partial A}{\partial t}}-q({\dot {\mathbf {x} }}\cdot \nabla )\mathbf {A} \\&=q{\boldsymbol {\nabla }}({\dot {\mathbf {x} }}\cdot \mathbf {A} )-q{\boldsymbol {\nabla }}\varphi -q{\frac {\partial A}{\partial t}}-q({\dot {\mathbf {x} }}\cdot \nabla )\mathbf {A} \\&=q\mathbf {E} +q{\dot {\mathbf {x} }}\times \mathbf {B} \end{aligned}}}$ 

The above derivation makes use of the vector calculus identity:

${\displaystyle {\tfrac {1}{2}}\nabla \left(\mathbf {A} \cdot \mathbf {A} \right)\ =\ \mathbf {A} \cdot \mathbf {J} _{\mathbf {A} }\ =\ \mathbf {A} \cdot (\nabla \mathbf {A} )\ =\ (\mathbf {A} {\cdot }\nabla )\mathbf {A} \,+\,\mathbf {A} {\times }(\nabla {\times }\mathbf {A} ).}$ 

An equivalent expression for the Hamiltonian as function of the relativistic (kinetic) momentum, P = γmẋ(t) = p - qA, is

${\displaystyle {\mathcal {H}}(t)={\dot {\mathbf {x} }}(t)\cdot \mathbf {P} (t)+{\frac {mc^{2}}{\gamma }}+q\varphi (\mathbf {x} (t),t)=\gamma mc^{2}+q\varphi (\mathbf {x} (t),t)=E+V}$ 

This has the advantage that kinetic momentum P can be measured experimentally whereas canonical momentum p cannot. Notice that the Hamiltonian (total energy) can be viewed as the sum of the relativistic energy (kinetic+rest), E = γmc², plus the potential energy, V = eφ.

In studies of cosmological inflation, minimal coupling of a scalar field usually refers to minimal coupling to gravity. This means that the action for the inflaton field ${\displaystyle \varphi }$  is not coupled to the scalar curvature. Its only coupling to gravity is the coupling to the Lorentz invariant measure ${\displaystyle {\sqrt {g}}\,d^{4}x}$  constructed from the metric (in Planck units):

${\displaystyle S=\int d^{4}x\,{\sqrt {g}}\,\left(-{\frac {1}{2}}R+{\frac {1}{2}}\nabla _{\mu }\varphi \nabla ^{\mu }\varphi -V(\varphi )\right)}$ 

where ${\displaystyle g:=\det g_{\mu \nu }}$ , and utilizing the gauge covariant derivative.