This tutorial gives a step-by-step derivation of the Lorentz force law from the Lagrangian of a charged particle in electromagnetic potentials. Along the way, we compute the Euler-Lagrange terms carefully, explain the total derivative of the vector potential, and verify the vector identity used in the derivation.
For a charged particle of mass $m$ and charge $q$ moving in electromagnetic fields described by a vector potential $\vec{A}(\vec{x},t)$ and a scalar potential $\phi(\vec{x},t)$, the Lagrangian is
\[L = \frac{1}{2}m\dot{\vec{x}}\cdot\dot{\vec{x}} + q\dot{\vec{x}}\cdot\vec{A} – q\phi .\]
Here $\dot{\vec{x}}$ is the particle velocity, and the dot denotes a time derivative.
The derivative of the Lagrangian with respect to the velocity is
\[\frac{\partial L}{\partial \dot{\vec{x}}} = m\dot{\vec{x}} + q\vec{A}.\]
This quantity is the canonical momentum. It differs from the mechanical momentum $m\dot{\vec{x}}$ by the electromagnetic term $q\vec{A}$.
We will use the identity
\[\nabla(\vec{U}\cdot\vec{V}) = (\vec{U}\cdot\nabla)\vec{V} + (\vec{V}\cdot\nabla)\vec{U} + \vec{U}\times(\nabla\times\vec{V}) + \vec{V}\times(\nabla\times\vec{U}).\]
In this derivation, take $\vec{U}=\dot{\vec{x}}$ and $\vec{V}=\vec{A}$. When taking $\nabla$ with respect to $\vec{x}$, the velocity $\dot{\vec{x}}$ is treated as independent of position, so
\[(\vec{A}\cdot\nabla)\dot{\vec{x}}=0,
\qquad
\nabla\times\dot{\vec{x}}=0.\]
Therefore the identity reduces to
\[\nabla(\dot{\vec{x}}\cdot\vec{A}) = (\dot{\vec{x}}\cdot\nabla)\vec{A} + \dot{\vec{x}}\times(\nabla\times\vec{A}).\]
Now compute the partial derivative of $L$ with respect to $\vec{x}$. The kinetic term has no explicit dependence on $\vec{x}$, so only the potential terms contribute:
\[\frac{\partial L}{\partial \vec{x}}
= \nabla\left(q\dot{\vec{x}}\cdot\vec{A} – q\phi\right).\]
The scalar potential term gives
\[\nabla(-q\phi) = -q\nabla\phi.\]
Using the reduced vector identity above, the vector-potential term gives
\[\nabla(q\dot{\vec{x}}\cdot\vec{A})
= q(\dot{\vec{x}}\cdot\nabla)\vec{A} + q\dot{\vec{x}}\times(\nabla\times\vec{A}).\]
Combining the two terms,
\[\frac{\partial L}{\partial \vec{x}}
= -q\nabla\phi + q(\dot{\vec{x}}\cdot\nabla)\vec{A} + q\dot{\vec{x}}\times(\nabla\times\vec{A}).\]
The Euler-Lagrange equation is
\[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\vec{x}}}\right) – \frac{\partial L}{\partial \vec{x}} = 0.\]
From Step 2,
\[\frac{d}{dt}\left(m\dot{\vec{x}}+q\vec{A}\right)
= m\ddot{\vec{x}} + q\frac{d\vec{A}}{dt}.\]
Because $\vec{A}$ depends on both $t$ and the particle position $\vec{x}(t)$, its total time derivative is
\[\frac{d\vec{A}}{dt}
= \frac{\partial\vec{A}}{\partial t} + (\dot{\vec{x}}\cdot\nabla)\vec{A}.\]
Therefore,
\[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\vec{x}}}\right)
= m\ddot{\vec{x}} + q\left(\frac{\partial\vec{A}}{\partial t} + (\dot{\vec{x}}\cdot\nabla)\vec{A}\right).\]
Substitute the two computed terms into the Euler-Lagrange equation:
\[m\ddot{\vec{x}} + q\left(\frac{\partial\vec{A}}{\partial t} + (\dot{\vec{x}}\cdot\nabla)\vec{A}\right)
–
\left[-q\nabla\phi + q(\dot{\vec{x}}\cdot\nabla)\vec{A} + q\dot{\vec{x}}\times(\nabla\times\vec{A})\right]
=0.\]
Expanding and cancelling the repeated $(\dot{\vec{x}}\cdot\nabla)\vec{A}$ terms gives
\[m\ddot{\vec{x}} + q\frac{\partial\vec{A}}{\partial t} + q\nabla\phi – q\dot{\vec{x}}\times(\nabla\times\vec{A}) = 0.\]
Rearrange:
\[m\ddot{\vec{x}}
= q\left(-\nabla\phi – \frac{\partial\vec{A}}{\partial t}\right)
+ q\dot{\vec{x}}\times(\nabla\times\vec{A}).\]
Using the standard definitions
\[\vec{E} = -\nabla\phi – \frac{\partial\vec{A}}{\partial t},
\qquad
\vec{B} = \nabla\times\vec{A},\]
we obtain the Lorentz force law:
\[m\ddot{\vec{x}} = q\left(\vec{E}+\dot{\vec{x}}\times\vec{B}\right).\]
Equivalently, since $\vec{F}=m\ddot{\vec{x}}$ and $\vec{v}=\dot{\vec{x}}$,
\[\vec{F}=q(\vec{E}+\vec{v}\times\vec{B}).\]
The total derivative of a vector field $\vec{A}(t,\vec{x}(t))$ is defined by
\[\frac{d\vec{A}}{dt}
= \lim_{\Delta t\to 0}
\frac{\vec{A}(t+\Delta t,\vec{x}(t+\Delta t))-\vec{A}(t,\vec{x}(t))}{\Delta t}.\]
Use the first-order Taylor expansion around $(t,\vec{x}(t))$:
\[\vec{A}(t+\Delta t,\vec{x}(t+\Delta t))
\approx
\vec{A}(t,\vec{x}(t))
+ \Delta t\frac{\partial\vec{A}}{\partial t}
+ \Delta\vec{x}\cdot\nabla\vec{A},\]
where
\[\Delta\vec{x}=\vec{x}(t+\Delta t)-\vec{x}(t).\]
Substituting into the limit,
\[\frac{d\vec{A}}{dt}
= \lim_{\Delta t\to 0}
\left(
\frac{\partial\vec{A}}{\partial t}
+ \frac{\Delta\vec{x}}{\Delta t}\cdot\nabla\vec{A}
\right).\]
Since
\[\lim_{\Delta t\to 0}\frac{\Delta\vec{x}}{\Delta t}=\dot{\vec{x}},\]
we get
\[\frac{d\vec{A}}{dt}
= \frac{\partial\vec{A}}{\partial t} + (\dot{\vec{x}}\cdot\nabla)\vec{A}.\]
Let
\[\vec{U}=U_x\hat{i}+U_y\hat{j}+U_z\hat{k},
\qquad
\vec{V}=V_x\hat{i}+V_y\hat{j}+V_z\hat{k},\]
where each component may depend on $(x,y,z)$. The dot product is
\[\vec{U}\cdot\vec{V}=U_xV_x+U_yV_y+U_zV_z.\]
The $x$-component of $\nabla(\vec{U}\cdot\vec{V})$ is
\[\partial_x(U_xV_x+U_yV_y+U_zV_z)
= V_x\partial_xU_x+V_y\partial_xU_y+V_z\partial_xU_z
+ U_x\partial_xV_x+U_y\partial_xV_y+U_z\partial_xV_z.\]
Now compute the $x$-component of the right-hand side:
\[[(\vec{U}\cdot\nabla)\vec{V}]_x
= U_x\partial_xV_x+U_y\partial_yV_x+U_z\partial_zV_x,\]
\[[(\vec{V}\cdot\nabla)\vec{U}]_x
= V_x\partial_xU_x+V_y\partial_yU_x+V_z\partial_zU_x,\]
\[[\vec{U}\times(\nabla\times\vec{V})]_x
= U_y(\partial_xV_y-\partial_yV_x)-U_z(\partial_zV_x-\partial_xV_z),\]
and
\[[\vec{V}\times(\nabla\times\vec{U})]_x
= V_y(\partial_xU_y-\partial_yU_x)-V_z(\partial_zU_x-\partial_xU_z).\]
Adding these four expressions cancels the terms containing $\partial_yV_x$, $\partial_zV_x$, $\partial_yU_x$, and $\partial_zU_x$, leaving exactly
\[\partial_x(U_xV_x+U_yV_y+U_zV_z).\]
The $y$- and $z$-components follow in the same way, so the full vector identity is verified.
Starting from the electromagnetic Lagrangian
\[L = \frac{1}{2}m\dot{\vec{x}}^2 + q\dot{\vec{x}}\cdot\vec{A} – q\phi,\]
the Euler-Lagrange equation gives
\[m\ddot{\vec{x}} = q(\vec{E}+\dot{\vec{x}}\times\vec{B}).\]
The electric force comes from the scalar potential and the explicit time dependence of the vector potential, while the magnetic force comes from the curl of the vector potential.
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