# 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.

Note by Steven Chase
11 months, 1 week ago

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Insightful analysis.

On a slightly unrelated yet interesting note, I suggest you look up Fermat's principle of least time. The principle goes on to say that the actual path traversed by a 'ray' of light is that which can be traversed in the shortest duration of time. Using this method, the laws of reflection and refraction of light can be derived. This is another variant of the Least Action Principle.

- 11 months, 1 week ago

I am going to post a problem on the "least time" principle soon. Should be fun

- 11 months, 1 week ago

The problem is up now in the Calculus section

- 11 months, 1 week ago

Just solved the problem. Thanks for posting it. It was a nice one.

- 11 months, 1 week ago