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 In Cartesian coordinates , the Lagrangian of a non-relativistic classical particle in an electromagnetic field is (in SI Units ):
L = ∑ i 1 2 m x ˙ i 2 + ∑ i q x ˙ i A i − q φ {\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 Ai , i = 1, 2, 3 , are the components of the magnetic vector potential that may all explicitly depend on x i {\displaystyle x_{i}} and t {\displaystyle t} .
This Lagrangian, combined with Euler–Lagrange equation , produces the Lorentz force law
m x ¨ = q E + q x ˙ × B , {\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
p i = ∂ L ∂ x ˙ i = m x ˙ i + q A i {\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
P i ≡ m x ˙ i = p i − q A i {\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
H = { ∑ i x ˙ i p i } − L = ∑ i ( p i − q A i ) 2 2 m + q φ {\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,
A → A + ∇ f , φ → φ − f ˙ , {\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
L → L ′ = L + q d f d t , p → p ′ = p + q ∇ f , H → H ′ = H − q ∂ f ∂ t , {\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:
∂ H ′ ∂ x i | p i ′ = ∂ ∂ x i | p i ′ ( x ˙ i p i ′ − L ′ ) = − ∂ L ′ ∂ x i | p i ′ = − ∂ L ∂ x i | p i ′ − q ∂ ∂ x i | p i ′ d f d t = − d d t ( ∂ L ∂ x ˙ i | p i ′ + q ∂ f ∂ x i | p i ′ ) = − p ˙ i ′ {\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 The relativistic Lagrangian for a particle (rest mass m and charge q ) is given by:
L ( t ) = − m c 2 1 − x ˙ ( t ) 2 c 2 + q x ˙ ( t ) ⋅ A ( x ( t ) , t ) − q φ ( x ( t ) , t ) {\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
p ( t ) = ∂ L ∂ x ˙ = m x ˙ 1 − x ˙ 2 c 2 + q A {\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
x ˙ ( t ) = p − q A m 2 + 1 c 2 ( p − q A ) 2 {\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
H ( t ) = x ˙ ⋅ p − L = c m 2 c 2 + ( p − q A ) 2 + q φ {\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 )
p ˙ = − ∂ H ∂ x = q x ˙ ⋅ ( ∇ A ) − q ∇ φ = q ∇ ( x ˙ ⋅ A ) − q ∇ φ {\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
d d t ( m x ˙ 1 − x ˙ 2 c 2 ) = d d t ( p − q A ) = p ˙ − q ∂ A ∂ t − q ( x ˙ ⋅ ∇ ) A = q ∇ ( x ˙ ⋅ A ) − q ∇ φ − q ∂ A ∂ t − q ( x ˙ ⋅ ∇ ) A = q E + q x ˙ × B {\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 :
1 2 ∇ ( A ⋅ A ) = A ⋅ J A = A ⋅ ( ∇ A ) = ( A ⋅ ∇ ) A + A × ( ∇ × A ) . {\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 = γm ẋ (t ) = p - q A , is
H ( t ) = x ˙ ( t ) ⋅ P ( t ) + m c 2 γ + q φ ( x ( t ) , t ) = γ m c 2 + q φ ( x ( t ) , t ) = E + V {\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 = γmc 2 , plus the potential energy , V = eφ .