Deriving Newton's 1st Law for Three-Dimensional Motion

Here is a derivation of Newton's 1st law of motion in three-dimensions. The derivation is based on the classical least-action principle. It should be evident by the end that the same process works for any number of dimensions.

Suppose a particle goes from $(x_1,y_1,z_1)$ to $(x_2,y_2,z_2)$ between time $t_1$ and time $t_2$. Suppose also that space is potential-free (and thus force-free). The action is defined as (where $E$ is the kinetic energy and $U$ is the potential energy):

$S = \int_{t_1}^{t_2} ( E - U) \, dt = \int_{t_1}^{t_2} E \, dt$

The particle will follow the path between the two points which minimizes the action. Suppose we discretize the path into $N$ constant-velocity periods. $N$ can be made arbitrarily large. For each period, the velocity is $v_k$. The action becomes:

$S = \frac{m}{2} \, \Sigma_{k=1}^{k=N} (v_{xk} ^2 + v_{yk} ^2 + v_{zk} ^2) \, \Delta t \\ \Delta t = \frac{t_2 - t_1}{N}$

The actual path must minimize the action while satisfying the constraints corresponding to the changes in position. We can use the method of Lagrange multipliers with the following Lagrangian:

$L = \frac{m}{2} \, \Sigma_{k=1}^{k=N} (v_{xk} ^2 + v_{yk} ^2 + v_{zk} ^2) \, \Delta t \\ + \lambda_x \Big ((\Sigma_{k=1}^{k=N} v_{xk} \, \Delta t) - (x_2 - x_1) \Big) \\ + \lambda_y \Big ((\Sigma_{k=1}^{k=N} v_{yk} \, \Delta t) - (y_2 - y_1) \Big) \\ + \lambda_z \Big ((\Sigma_{k=1}^{k=N} v_{zk} \, \Delta t) - (z_2 - z_1) \Big)$

Taking partial derivatives with respect to the period-specific velocities and setting to zero (per standard practice) results in:

$\frac{\partial{L}}{\partial{v_{x1}}} = m \, v_{x1} \, \Delta t + \lambda_x \, \Delta t = 0 \implies v_{x1} = -\frac{\lambda_x}{m} \\ \frac{\partial{L}}{\partial{v_{y1}}} = m \, v_{y1} \, \Delta t + \lambda_y \, \Delta t = 0 \implies v_{y1} = -\frac{\lambda_y}{m} \\ \frac{\partial{L}}{\partial{v_{z1}}} = m \, v_{z1} \, \Delta t + \lambda_z \, \Delta t = 0 \implies v_{z1} = -\frac{\lambda_z}{m} \\ \frac{\partial{L}}{\partial{v_{x2}}} = m \, v_{x2} \, \Delta t + \lambda_x \, \Delta t = 0 \implies v_{x2} = -\frac{\lambda_x}{m} \\ \frac{\partial{L}}{\partial{v_{y2}}} = m \, v_{y2} \, \Delta t + \lambda_y \, \Delta t = 0 \implies v_{y2} = -\frac{\lambda_y}{m} \\ \frac{\partial{L}}{\partial{v_{z2}}} = m \, v_{z2} \, \Delta t + \lambda_z \, \Delta t = 0 \implies v_{z2} = -\frac{\lambda_z}{m} \\ etc. \\ \implies v_{x1} = v_{x2} = v_{xk} \\ \implies v_{y1} = v_{y2} = v_{yk} \\ \implies v_{z1} = v_{z2} = v_{zk}$

Thus, for motion in potential-free (force-free) three-dimensional space, the least-action principle dictates that the vector velocity must be constant.