Lagrange's equations of motion

That Lagrange's equations of motion are equivalent to Newton's second law in Cartesian coordinates with conservative forces can be shown directly by defining the Lagrangian (kind of an arbitrary quantity but turns out to be extremely useful):

L = T - V = \frac{1}{2} m \vec{v} \cdot \vec{v} - V (\vec{r})

and writing out:

\begin{aligned} \frac{\partial L}{\partial x} & = - \frac{\partial V}{\partial x} = F_{x} \\ \frac{\partial L}{\partial \dot{x}} & = m v_{x} = p_{x} \\ so, {\dot{p}}_{x} = F_{x} & ⟹ \frac{d}{d t} \frac{\partial L}{\partial \dot{x}} = \frac{\partial L}{\partial x} \end{aligned}

We can derive the equivalence in more general terms (generalized coordinates, generalized forces, multiple particles, under holonomic constraints) by starting from D'Alembert's virtual displacement framework.

Derivation from D'Alembert's principle

In classical mechanics, it is often more productive to work with generalized coordinates $q_{j}$ instead of natural Cartesian coordinates ${{\vec{r}}_{1}, {\vec{r}}_{2}, \dots {\vec{r}}_{n}}$ of many particles in general constrained systems. The Lagrange equations can actually be derived from D'Alembert's principle based on virtual displacements (infinitesimal changes consistent with constraints). Newton's equations come out to:

\sum_{i} ({\vec{F}}_{i}^{(a)} - m_{i} {\vec{a}}_{i}) \cdot δ {\vec{r}}_{i} = 0

The generalized coordinates are expressed by the system of equations ${\vec{r}}_{i} = {\vec{r}}_{i} (\vec{q}, t)$ so that the virtual displacement is

δ {\vec{r}}_{i} = \sum_{j} \frac{\partial {\vec{r}}_{i}}{\partial q_{j}} δ q_{j}

and

{\vec{v}}_{i} = \frac{d {\vec{r}}_{i}}{d t} = \sum_{j} \frac{\partial {\vec{r}}_{i}}{\partial q_{j}} {\dot{q}}_{j} + \frac{\partial {\vec{r}}_{i}}{\partial t} ⟹ \frac{\partial {\vec{v}}_{i}}{\partial {\dot{q}}_{j}} = \frac{\partial {\vec{r}}_{i}}{\partial q_{j}}

and a generalized force can be defined as

Q_{j} = \sum_{i} {\vec{F}}_{i} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{j}} $ $ T h i s y i e l d s $ $ \sum_{i} {\vec{F}}_{i}^{(a)} \cdot δ {\vec{r}}_{i} = \sum_{j} Q_{j} δ q_{j} $ $ a n d $ $ \begin{aligned} \sum_{i} m_{i} {\vec{a}}_{i} \cdot δ {\vec{r}}_{i} & = \sum_{i, j} m_{i} {\vec{a}}_{i} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{j}} δ q_{j} \\ = \sum_{i, j} [\frac{d}{d t} (m_{i} {\vec{v}}_{i} \cdot \frac{\partial {\vec{r}}_{i}}{\partial q_{j}}) - m_{i} {\vec{v}}_{i} \cdot \frac{d}{d t} (\frac{\partial {\vec{r}}_{i}}{\partial q_{j}})] δ q_{j} \\ = \sum_{i, j} [\frac{d}{d t} (m_{i} {\vec{v}}_{i} \cdot \frac{\partial {\vec{v}}_{i}}{\partial {\dot{q}}_{j}}) - m_{i} {\vec{v}}_{i} \cdot \frac{\partial {\vec{v}}_{i}}{\partial {\dot{q}}_{j}}] δ q_{j} \\ = \sum_{j} [\frac{d}{d t} {\frac{\partial}{\partial {\dot{q}}_{j}} (\frac{1}{2} \sum_{i} m_{i} {\dot{v}}_{i} \cdot {\dot{v}}_{i})} - \frac{\partial}{\partial {\dot{q}}_{j}} (\frac{1}{2} \sum_{i} m_{i} {\dot{v}}_{i} \cdot {\dot{v}}_{i})] δ q_{j} \end{aligned}

Now, kinetic energy $T = \frac{1}{2} \sum_{i} m_{i} v_{i}^{2}$ so that

\begin{aligned} \sum_{i} ({\vec{F}}_{i}^{(a)} - m_{i} {\vec{a}}_{i}) \cdot δ {\vec{r}}_{i} & = 0 \\ ⟹ \sum_{j} [\frac{d}{d t} (\frac{\partial T}{\partial {\dot{q}}_{j}}) - \frac{\partial T}{\partial {\dot{q}}_{j}} - Q_{j}] δ q_{j} & = 0 \end{aligned}

Under holonomic constraints, the coordinates $q_{j}$ can be chosen to be mutually independent and therefore we get that each coefficient above should vanish.

\frac{d}{d t} (\frac{\partial T}{\partial {\dot{q}}_{j}}) - \frac{\partial T}{\partial {\dot{q}}_{j}} - Q_{j} = 0

If the external forces are conservative,

Q_{j} = - \frac{\partial V}{\partial q_{j}}

and the potential $V = V (\vec{q})$ is not explicitly dependent on time (conservative) so that $\frac{\partial V}{\partial {\dot{q}}_{j}} = 0$ , then

\frac{d}{d t} (\frac{\partial T}{\partial {\dot{q}}_{j}}) - \frac{\partial (T - V)}{\partial {\dot{q}}_{j}} = 0

\frac{d}{d t} (\frac{\partial L}{\partial {\dot{q}}_{j}}) - \frac{\partial L}{\partial {\dot{q}}_{j}} = 0

where $L = T - V$ is the Lagrangian. This can alternately be derived starting from the least action principle.

Atypical forces and potentials

The above holds even for certain velocity-dependent potentials like with the electromagnetic forces on moving charges with generalized potentials $U (\vec{q}, \dot{\vec{q}})$ instead of the typical potentials $V (\vec{q})$ . This can be done when one is able to express the generalized potential as $Q_{j} = - \frac{\partial U}{\partial q_{j}} + \frac{d}{d t} (\frac{\partial U}{\partial {\dot{q}}_{j}})$ .

Not all (generalized) forces satisfy this requirement and hence can't be derived from a potential. In this case, Lagrange equations are just written as:

\frac{d}{d t} (\frac{\partial T}{\partial {\dot{q}}_{j}}) - \frac{\partial T}{\partial {\dot{q}}_{j}} = Q_{j}

Dissipative effects like friction are useful to treat specifically, where the forces are given by

\begin{aligned} F & = \frac{1}{2} \sum_{i} {\dot{q}}_{i}^{2} \\ {\vec{F}}_{f} & = - \nabla_{v} F \end{aligned}

In this case, the Lagrange equations become:

\frac{d}{d t} (\frac{\partial L}{\partial {\dot{q}}_{j}}) - \frac{\partial L}{\partial {\dot{q}}_{j}} + \frac{\partial F}{\partial {\dot{q}}_{j}} = 0

Three useful categories of generalized forces:

Type	Form	Energy	Lagrangian-compatible?
Conservative	$Q = - \partial V (q) / \partial q$	Conserved	Yes, $L = T - V$
Generalized potential	$Q = - \partial U / \partial q + \frac{d}{d t} \partial U / \partial \dot{q}$	Conserved (if $\partial U / \partial t = 0$ )	Yes, $L = T - U$
Dissipative	$Q = - \partial F / \partial \dot{q}$	Decreasing	No — added as separate term

Links

Sources

Classical Mechanics by Herbert Goldstein, Charles Poole and John Safko
Classical Mechanics (John R. Taylor)