# Deriving Newton's first law from the principle of least action

Newton's first law states that if the net force on an object is zero, then this object moves with constant velocity.

I'm interested in the derivation of this law from the principle of least action. What ought to be proved is pretty straight forward: given an object that is in an isolated system with the absence of any force fields, the Lagrangian of that object is simply its kinetic energy.

Let the position function of that object be given by $x(t)$. This is an arbitrary non-linear function. its kinetic energy is given by $\dfrac{1}{2}m[x'(t)]^2$. given a time interval $[t_1, t_2]$, the action of this object is the time integral of its Lagrangian (its kinetic energy in this case) from $t_1$ to $t_2$:

$$\text{action}=S=\int_{t_1}^{t_2}\dfrac{1}{2}m[x'(t)]^2 dt \, .$$

If the object moved in a uniform motion along $x(t_2)-x(t_1)$ in the interval $[t_1,t_2]$ then its velocity is simply given by the average velocity: total distance by time :

$$v_{\text{average}} = \dfrac{x(t_2) - x(t_1)}{t_2 - t_1} \, .$$

Using the principle of least action, proving Newton's first law is equivalent to proving that the action given by

$$\int_{t_1}^{t_2}\dfrac{1}{2} m \left[ \dfrac{x(t_2) - x(t_1)}{t_2 - t_1} \right]^2 dt$$

is always less than that of

$$\int_{t_1}^{t_2}\dfrac{1}{2}m[x'(t)]^2 dt.$$

This is my understanding of the problem. Feynman in his lectures gives an argument to convince us why the above result I stated must be true(the result is that the time integral of kinetic energy of an object in uniform motion is always less than that of an object in non-uniform motion):

“As an example, say your job is to start from home and get to school in a given length of time with the car. You can do it several ways: You can accelerate like mad at the beginning and slow down with the brakes near the end, or you can go at a uniform speed, or you can go backwards for a while and then go forward, and so on. The thing is that the average speed has got to be, of course, the total distance that you have gone over the time. But if you do anything but go at a uniform speed, then sometimes you are going too fast and sometimes you are going too slow. Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; so the kinetic energy integral would always be higher if you wobbled your velocity than if you went at a uniform velocity. So we see that the integral is a minimum if the velocity is a constant (when there are no forces)."

The part that I understand: its true that the action of an object that is moving uniformly is given by the integral of "the square of the mean" of velocity multiplied by $\dfrac{1}{2}m$. I think by square of the mean he means: $v_{\text{average}}^2 = \left[ \dfrac{x(t_2) - x(t_1)}{t_2 - t_1} \right]^2$.

I do not understand how the Lagrangian of an object that is moving non-uniformly is equal to "mean square of something that deviates around an average" (again, multiplied by $m/2$).

I presume he means by the mean square: the mean(or average) of kinetic energy of that body: that is :

$$\dfrac{\dfrac{1}{2}m[x'(t_2)]^2-\dfrac{1}{2}m[x'(t_2)]^2}{t_2-t_1}.$$

Is my understanding correct? and if so, How it is the case that this "mean square" expression is equal to the lagrangian of an object with non-uniform motion?

• You are not deriving Newton's first law. You are deriving a special case. Newton's law also applies to situations where there is a scalar potential that happens to be flat where the particle is. And in situations where there are multiple forces but zero net force. Also, you aren't going to derive the first law since the Lagrangian formulation derives the second law, which doesn't imply the first law. – Timaeus Jul 26 '15 at 19:34

First a couple general results: given a function $f(t)$ and an interval $T = [t_1,t_2]$, the square of the mean of $f$ on $T$ is $$\langle f\rangle_T^2 = \biggl(\frac{1}{t_2 - t_1}\int_{t_1}^{t_2}f(t)\,\mathrm{d}t\biggr)^2$$ and the mean of the square of $f$ on $T$ is $$\langle f^2\rangle_T = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2}\bigl[f(t)\bigr]^2\,\mathrm{d}t$$ In this case, the relevant function is the velocity, so you have (or can show) that \begin{align} \langle v\rangle_T^2 &= \biggl(\frac{x(t_2) - x(t_1)}{t_2 - t_1}\biggr)^2 & \langle v^2\rangle_T &= \frac{1}{t_2 - t_1} \int_{t_1}^{t_2}\bigl[v(t)\bigr]^2\,\mathrm{d}t \end{align} Your mistake was thinking that $$\langle v^2\rangle_T = \frac{\bigl[v(t_2)\bigr]^2 - \bigl[v(t_1)\bigr]^2}{t_2 - t_1}\tag{not correct}$$ It's not always the case that the mean is some final value minus some initial value divided by the interval. That only works when the thing you're calculating the mean of can be integrated.

With that in mind, it should be clear how the mean of the square is relevant: it shows up directly in the Lagrangian.

Hints:

1. By mean/average Feynman means temporal mean/average defined as $$\tag{1} \langle f \rangle ~:=~\frac{ \int_{t_i}^{t_f}\! dt~ f(t)}{t_f-t_i} .$$

2. Inequality: The mean square is always greater than the square of the mean $$\tag{2}\langle f^2 \rangle ~\geq ~ \langle f \rangle^2.$$ There are several proofs of ineq. (2), e.g. the variance is always non-negative.

3. Finally put $f(t)=v(t)$.

• Your hints were very helpful, I appreciate it. – Omar Nagib Jul 26 '15 at 19:07