The Work-Energy Theorem reformulates Newton’s Second Law in a way that makes certain classes of problems easier. You may wonder if there can be reformulations of the Work-Energy Theorem. Indeed, there are a few different variations that each have their own benefits.
Recall two important facts about energy:
Proposition 1.1 (Lagrangian Proposition ) If we know the energy of the system, we can tell how the system will evolve given the two principles above.
So the question is: how? Suppose you throw a ball, giving it some initial velocity. What path does it take?
You’re probably thinking about forces and uniform acceleration motion, but we want something in terms of energy. Remember conservation of energy?
\[ U_0 + K_0 = U + K \]
The sum of kinetic and potential energy doesn’t change. So maybe we should reformulate this idea in a new way. Let’s fix the endpoints of our path.
Clearly it is the same path that is taken every time. One would think there’s something unique about this path then.
Well, what can energy tell us about the path taken? We know that nature doesn’t “like” high energy states (forces are directed to minimize energy). So you might say that nature just wants to minimize energy. But which kind of energy?
Let’s summarize what we know in an equation.
\[ \mathcal{S}(t) = \defint{t_0}{t}{\mathcal{L}(U, K)}{t} \]
\[ x(t) = \arg\min_{x(t)} \mathcal{S}(t) \]
We can make our lives a little easier if we write energy in terms of the path. If we assume that mass is constant, then we just have to know velocity to calculate kinetic energy. And potential energy functions by definition only require position. So given position and velocity we can recover the energies later given the constant mass. Now we just have to solve an extrema-finding problem.
One slight problem: in regular single-variable calculus, a function takes in a real number and spits out a real number. Thus, to find an extrema, you look for a place in the function where a small first-order change (i.e. \(f{x}\)) results in no output change (i.e. \(\diff{y}\) is zero).
But here, we don’t have a function that takes in a real number. We have a function that takes in a path…which is a function. A function that takes in a function and outputs a real number is called a functional.5 The branch of mathematics that deals with optimizing functionals is known as the calculus of variations So we now have to look in the path-space for a place where a first-order change in the path results in no output change (i.e. the best path is one where no simple modifications could be made to improve the performance given the reward or cost of the functional).
To make a long story short, the optimal path follows this condition, known as the Euler-Lagrange equation (it actually works for the problem of finding maxima/minima of any arbitrary \(\mathcal{L}\)):
\[ \pderiv{\mathcal L}{x} = \deriv{}{t}\pderiv{\mathcal L}{\dot x} \]
Proof (Lagrangian Proof by Analogy).
Energy is a result (or a restatement) of Newton’s Second Law, which predicts the path of objects
We have a mystery quantity based on energy that gives us the path if we know the start and endpoints
We should set Newton’s Second Law as the currently working theory and then show that our new theory gives the same results
Which means we should show that for a specific \(\mathcal L\), Euler-Lagrange equation is the same as Newton’s Second Law
That right hand term looks like it would be a force (if \(\mathcal L\) has a potential energy term, differentiation w.r.t position yields force) and the right hand side…
\[ \deriv{}{t}\deriv{K}{v} = \deriv{}{t}\left[\frac{m}{2}\pderiv{v^2}{v}\right] = \deriv{}{t}[mv] = ma \]
So we recovered \(F = ma\) if \(\mathcal L = K - U\).