Lagrangian mechanics: deriving Newton's 1st law with just math

By Ryan Ng

Today, we’ll derive Newton’s first law of motion using nothing but mathematics. We’ll do this with Lagrangian mechanics, which is a formulation of classical mechanics that is nearly purely mathematical. This is opposed to Newtonian mechanics, as Newton’s laws are results of experimentation.

The first thing we need to start with is coordinates. These define the positions of particles. We’ll use the Cartesian coordinates here—x, y, z.

For our model, we’ll look at a singular particle. Because we want to know where the particle is located at every instant in time, we’ll define the coordinates as a function of time, that outputs a position vector:

This allows us to then define the velocity, which is the change in coordinates with respect to time. This is simply v(t) = dr/dt. Here comes the magic. 

Let’s suppose we have a function L(r, v, t). We don’t know what this function is, but it is a function of the position and the velocity of our singular particle, and the time. We’ll also give the function a crucial constraint.

Say our particle is at the coordinate r1 at time t1, and at coordinate r2 at time t2. We want to know how the particle will move between these coordinates in this time interval, to go from r1 to r2. This movement would be defined by the function r(t).

The constraint, then, is as follows: the particle’s movement between the time interval t1 and t2 will be such that the following integral is minimal.

This may sound confusing. What we mean here is this: we don’t know what the function L is. However, we know that according to the laws of physics (that we still “don’t know” at this point), a particle placed in the same situation will move between two coordinates the same way—the laws of physics don’t randomly change. This means that there is one specific coordinates function r(t) that will correctly model the particle’s movement from r1 to r2 from time t1 to t2. When we plug this coordinates function, and its time derivative v(t) into the mythical function L, the integral S will be minimized. This means that if the coordinates function was changed to anything else, the value of S would be greater.

Notice that we haven’t included any assumptions thus far. All we’ve suggested is that a particle placed in the same exact situation will move in the same way every time. But we now have an integral to mess around with!

This minimality condition will actually tell us a lot. Let’s suppose that the function r(t) is the specific coordinates function, moving from coordinates r1 to r2 during time t1 to t2 that minimizes the integral. This means that if we replaced r(t) with anything else, the value of S would increase. So let’s replace r(t) by r(t) + δr(t) where δr(t) is an extremely small function. We’ve modified the particle’s trajectory just a tiny bit.
Since we still have our constraint that the particle starts at r1 and ends at r2, this new modified function must follow the constraint too. Then, we see that

and the same goes for δr(t2). 

When we replace the function, our new integral is

and let’s define this difference as

To further evaluate this, we’d like to expand the first term as a Taylor series in (r, v). Using the two-variable Taylor series expansion we get 

We can apply integration by parts on the second term, integrating δv and differentiating ∂L/∂v:

Since δr(t1) = δr(t2) = 0, the first term on the right is zero. So substituting back, we have

Now, because our added variation δr(t) is extremely small, and S is a continuous function, and we had a minimum with r(t), then δS should be zero too. This is best visualized by thinking of a parabola. At its minimum, its derivative is zero/its tangent line is horizontal, because it is decreasing from the left and increasing to the right. Because S is continuous (and has a smooth derivative), its tangent line at the minimum is also horizontal. 

Therefore, we have δS = 0. This can only happen if the first term in the product is zero, because if not, we could replace the small function δr(t) with something slightly different and violate the condition. Therefore:

This is known as the Euler-Lagrange equation.

How does this relate to Newton’s first law? Newton’s first law states that “A body remains at rest, or in motion at a constant speed in a straight line, unless it is acted upon by a force.” This means that there is no force field located anywhere in the system (or anywhere at all). We know that the laws of physics don’t vary from place to place: if I had a system here, the particle would move in the same way as if I moved the system 10 kilometers away, everything else remaining equal. 
Since there is no force field located anywhere, the specific coordinates of the particle should not have an impact on its motion, because we don’t have to consider the particle’s position relative to any force field. This means our function L should also not depend on the positions, so ∂L/∂r = 0. This simplifies the equation to

This means that ∂L/∂v doesn’t change over with time, and since L is a function of only the velocity (we saw earlier that L does not depend on position), ∂L/∂v is also a function of velocity, that is constant over time. This implies that velocity is constant over time, and hence, Newton’s first law has been proven with almost nothing but math.