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 (x1,y1,z1) (x_1,y_1,z_1) to (x2,y2,z2) (x_2,y_2,z_2) between time t1t_1 and time t2t_2 . Suppose also that space is potential-free (and thus force-free). The action is defined as (where EE is the kinetic energy and UU is the potential energy):

S=t1t2(EU)dt=t1t2EdtS = \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 NN constant-velocity periods. N N can be made arbitrarily large. For each period, the velocity is vkv_k . The action becomes:

S=m2Σk=1k=N(vxk2+vyk2+vzk2)ΔtΔt=t2t1N 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=m2Σk=1k=N(vxk2+vyk2+vzk2)Δt+λx((Σk=1k=NvxkΔt)(x2x1))+λy((Σk=1k=NvykΔt)(y2y1))+λz((Σk=1k=NvzkΔt)(z2z1)) 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:

Lvx1=mvx1Δt+λxΔt=0    vx1=λxmLvy1=mvy1Δt+λyΔt=0    vy1=λymLvz1=mvz1Δt+λzΔt=0    vz1=λzmLvx2=mvx2Δt+λxΔt=0    vx2=λxmLvy2=mvy2Δt+λyΔt=0    vy2=λymLvz2=mvz2Δt+λzΔt=0    vz2=λzmetc.    vx1=vx2=vxk    vy1=vy2=vyk    vz1=vz2=vzk \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
4 weeks, 1 day 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.

Karan Chatrath - 4 weeks, 1 day ago

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

Steven Chase - 4 weeks ago

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The problem is up now in the Calculus section

Steven Chase - 4 weeks ago

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

Karan Chatrath - 4 weeks ago

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