Editing Lorentz force (section)

== Lorentz force and analytical mechanics ==

{{see also|Magnetic vector potential#Interpretation as Potential Momentum}}
The [[Lagrangian_mechanics#Electromagnetism|Lagrangian]] for a charged particle of mass {{math|''m''}} and charge {{math|''q''}} in an electromagnetic field equivalently describes the dynamics of the particle in terms of its ''energy'', rather than the force exerted on it. The classical expression is given by:<ref name="Kibble">{{cite book | last=Kibble | first=T. W. B. | last2=Berkshire | first2=Frank H. | title=Classical Mechanics | publisher=World Scientific Publishing Company | publication-place=London : River Edge, NJ | date=2004 | isbn=1-86094-424-8 | oclc=54415965 | chapter=10.5 Charged Particle in an Electromagnetic Field}}</ref>
<math display="block">L = \frac{m}{2} \mathbf{\dot{r} }\cdot\mathbf{\dot{r} } + q \mathbf{A}\cdot\mathbf{\dot{r} }-q\phi</math>
where {{math|'''A'''}} and {{math|''ϕ''}} are the potential fields as above. The quantity <math>V = q(\phi - \mathbf{A}\cdot \mathbf{\dot{r}})</math> can be identified as a generalized, velocity-dependent potential energy and, accordingly, <math>\mathbf{F}</math> as a [[Conservative_force#Non-conservative_force|non-conservative force]].<ref>{{cite journal | last=Semon | first=Mark D. | last2=Taylor | first2=John R. | title=Thoughts on the magnetic vector potential | journal=American Journal of Physics | volume=64 | issue=11 | date=1996 | issn=0002-9505 | doi=10.1119/1.18400 | pages=1361–1369}}</ref> Using the Lagrangian, the equation for the Lorentz force given above can be obtained again.

{{math proof|title=Derivation of Lorentz force from classical Lagrangian (SI units)| proof =
For an {{math|1='''A'''}} field, a particle moving with velocity {{math|1='''v''' = '''ṙ'''}} has [[Momentum#In_electromagnetics|potential momentum]] <math>q\mathbf{A}(\mathbf{r}, t)</math>, so its potential energy is <math>q\mathbf{A}(\mathbf{r},t)\cdot\mathbf{\dot{r}}</math>. For a ''ϕ'' field, the particle's potential energy is <math>q\phi(\mathbf{r},t)</math>.

The total [[potential energy]] is then:
<math display="block">V = q\phi - q\mathbf{A}\cdot\mathbf{\dot{r}}</math>
and the [[kinetic energy]] is:
<math display="block">T = \frac{m}{2} \mathbf{\dot{r}}\cdot\mathbf{\dot{r}}</math>
hence the Lagrangian:
<math display="block">\begin{align}
L &= T - V \\[1ex]
&= \frac{m}{2} \mathbf{\dot{r} } \cdot \mathbf{\dot{r} } + q \mathbf{A} \cdot \mathbf{\dot{r} } - q\phi \\[1ex]
&= \frac{m}{2} \left(\dot{x}^2 + \dot{y}^2 + \dot{z}^2\right) + q \left(\dot{x} A_x + \dot{y} A_y + \dot{z} A_z\right) - q\phi
\end{align}</math>

Lagrange's equations are
<math display="block">\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x}</math>
(same for {{math|''y''}} and {{math|''z''}}). So calculating the partial derivatives:
<math display="block">\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{x} } &= m\ddot{x} + q\frac{\mathrm{d} A_x}{\mathrm{d}t} \\
& = m\ddot{x} + q \left[\frac{\partial A_x}{\partial t} + \frac{\partial A_x}{\partial x}\frac{dx}{dt} + \frac{\partial A_x}{\partial y}\frac{dy}{dt} + \frac{\partial A_x}{\partial z}\frac{dz}{dt}\right] \\[1ex]
& = m\ddot{x} + q\left[\frac{\partial A_x}{\partial t} + \frac{\partial A_x}{\partial x}\dot{x} + \frac{\partial A_x}{\partial y}\dot{y} + \frac{\partial A_x}{\partial z}\dot{z}\right]\\
\end{align}</math>
<math display="block">\frac{\partial L}{\partial x}= -q\frac{\partial \phi}{\partial x}+ q\left(\frac{\partial A_x}{\partial x}\dot{x}+\frac{\partial A_y}{\partial x}\dot{y}+\frac{\partial A_z}{\partial x}\dot{z}\right)</math>
equating and simplifying:
<math display="block">m\ddot{x}+ q\left(\frac{\partial A_x}{\partial t}+\frac{\partial A_x}{\partial x}\dot{x}+\frac{\partial A_x}{\partial y}\dot{y}+\frac{\partial A_x}{\partial z}\dot{z}\right)= -q\frac{\partial \phi}{\partial x}+ q\left(\frac{\partial A_x}{\partial x}\dot{x}+\frac{\partial A_y}{\partial x}\dot{y}+\frac{\partial A_z}{\partial x}\dot{z}\right)</math>
<math display="block">\begin{align}
F_x & = -q\left(\frac{\partial \phi}{\partial x}+\frac{\partial A_x}{\partial t}\right) + q\left[\dot{y}\left(\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}\right)+\dot{z}\left(\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}\right)\right] \\[1ex]
& = qE_x + q[\dot{y}(\nabla\times\mathbf{A})_z-\dot{z}(\nabla\times\mathbf{A})_y] \\[1ex]
& = qE_x + q[\mathbf{\dot{r}}\times(\nabla\times\mathbf{A})]_x \\[1ex]
& = qE_x + q(\mathbf{\dot{r}}\times\mathbf{B})_x
\end{align}</math>
and similarly for the {{math|''y''}} and {{math|''z''}} directions. Hence the force equation is:
<math display="block">\mathbf{F}= q(\mathbf{E} + \mathbf{\dot{r}}\times\mathbf{B})</math>
}}

The relativistic Lagrangian is
<math display="block">L = -mc^2\sqrt{1-\left(\frac{\dot{\mathbf{r} } }{c}\right)^2} + q \mathbf{A}(\mathbf{r}) \cdot \dot{\mathbf{r} } - q \phi(\mathbf{r}) </math>

The action is the relativistic [[arclength]] of the path of the particle in [[spacetime]], minus the potential energy contribution, plus an extra contribution which [[Quantum Mechanics|quantum mechanically]] is an extra [[phase (waves)|phase]] a charged particle gets when it is moving along a vector potential.

{{math proof
|title=Derivation of Lorentz force from relativistic Lagrangian (SI units)
|proof=
The equations of motion derived by [[calculus of variations|extremizing]] the action (see [[matrix calculus]] for the notation):
<math display="block"> \frac{\mathrm{d}\mathbf{P}}{\mathrm{d}t} =\frac{\partial L}{\partial \mathbf{r}} = q {\partial \mathbf{A} \over \partial \mathbf{r}}\cdot \dot{\mathbf{r}} - q {\partial \phi \over \partial \mathbf{r} }</math>
<math display="block">\mathbf{P} -q\mathbf{A} = \frac{m\dot{\mathbf{r}}}{\sqrt{1-\left(\frac{\dot{\mathbf{r}}}{c}\right)^2}}</math>
are the same as [[Hamiltonian mechanics|Hamilton's equations of motion]]:
<math display="block"> \frac{\mathrm{d}\mathbf{r} }{\mathrm{d}t} = \frac{\partial}{\partial \mathbf{p} } \left ( \sqrt{(\mathbf{P}-q\mathbf{A})^2 + (mc^2)^2} + q\phi \right ) </math>
<math display="block"> \frac{\mathrm{d}\mathbf{p} }{\mathrm{d}t} = -\frac{\partial}{\partial \mathbf{r}} \left ( \sqrt{(\mathbf{P}-q\mathbf{A})^2 + (mc^2)^2} + q\phi \right ) </math>
both are equivalent to the noncanonical form:
<math display="block"> \frac{\mathrm{d} }{\mathrm{d}t} {m\dot{\mathbf{r} } \over \sqrt{1-\left(\frac{\dot{\mathbf{r} } }{c}\right)^2} } = q\left ( \mathbf{E} + \dot\mathbf{r} \times \mathbf{B} \right ) . </math>
This formula is the Lorentz force, representing the rate at which the EM field adds relativistic momentum to the particle.
}}