Principle of Stationary Action with Standard Lagrangian implies Newton's Laws of Motion

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Theorem

Let $\MM$ be an $n$-dimensional Euclidean manifold.

Let $P$ be a physical system composed of a countable number of classical particles with inertial masses $m_i$ with $i \in \N$.

Let $\mathbf x = \map {\mathbf x} t$ be twice-differentiable vector-valued function embedded in $\MM$.

Suppose ${\mathbf x}_i$ represents the position of the $i$th particle of $P$.

Suppose the action of $P$ is of the following form:

$\ds S = \int_{t_0}^{t_1} L \rd t$

where $L$ is the standard Lagrangian.

Suppose, all (internal or external) forces ${\mathbf F}_i$ acting upon $P$ are of the form:

${\mathbf F}_i = -\dfrac {\partial U} {\partial \mathbf x_i}$

where:

$U = \map U {t, \set {\mathbf x_i} }$ is a differentiable real function
$\set {\mathbf x_i}$ denotes dependence on the positions of all the particles.


Then the stationary point of $S$ implies Newton's Second Law of Motion.


Proof

Standard Lagrangian is of the following form:

$\ds L = \sum_{i \mathop = 1}^n \frac {m_i} 2 \dot {\mathbf x}_i^2 - U$

By the Principle of Stationary Action, equations of motion of $P$ are:

$\forall i \in \N: m_i {\ddot {\mathbf x} }_i + \dfrac {\partial U} {\partial \mathbf x_i} = 0$

By definition of velocity:

${\dot {\mathbf x} }_i = {\mathbf v}_i$

By assumption:

${\mathbf F}_i = -\dfrac {\partial U} {\partial {\mathbf x}_i}$

Then it follows that the equations of motion of $P$ can be rewritten as:

$\map {\dfrac \d {\d t} } {m_i {\mathbf v}_i } = {\mathbf F}_i$

By definition, these equations together represent Newton's Second Law of Motion.

$\blacksquare$


Sources

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