The Principle of Least Action

An interactive lesson by Ethan Uppal

$$ \Delta \int_{t_0}^{t_1} { T - U } \, dt = 0 $$

Contents (Loading...)

1. Introduction
2. Part 1
3. Part 2
4. Part 3
5. Part 4
6. Part 5
7. Part 6
8. Notes

Introduction

Consider a particle moving from point $A$ to point $B$. What path does it take?

Typically, you'd approach this kind of problem by writing down all the forces on the object and solving Newton's second law for $\vec{r}$.

$$ \vec{F}_{net} = m \frac{d^2 \vec{r}}{dt^2} $$

We're going to explore a different way of looking at the same problem: the principle of least action. Where Newton analyzed forces, we're going to investigate energies and the interplay between them. But first, it's story time!

Feynman's Story

The great Richard Feynman recounts a story of his old high school physics teacher, Mr. Bader, who summoned him on a boring day with the mysterious promise of “something interesting”.

That “something interesting” turned out to be the principle of least action, and it continued to fascinate him long after. So fascinating, in fact, that he interrupts the second volume of his lectures to share it. The Maxwell Equations, Chapter 18. Solutions of the Maxwell Equations, Chapter 20. And, of course, The Principle of Least Action, Chapter 19. This lesson draws heavily on Feynman's incredible lecture, so be sure to check it out!

What exactly is this principle though? And why did it intrigue Feynman so much?

The Principle of Least Action!

A particle in motion has kinetic energy and often experiences potential energy. The interplay between these two forms of mechanical energy together reflect the ebb and flow of the particle.

Typically when encountering energies, we add them together; by the law of conservation of energy, this sum should remain constant. We're not going to do that, though. Instead, we'll subtract them.

$$L = T - U$$

Here, $L$ is the Lagrangian, defined as the difference between kinetic energy $T$ and potential energy $U$.

Your first reaction should be something along the lines of “What?”, “How?”, or “Why?”, with varying degrees of punctuation marks and vocal inflections. Unfortunately, you're going to have to stick with it and see where it could possibly lead.

Let's define a new quantity, $S$, called the action, to be the integral of $L$ over time. Since $L$ has units of joules, this quantity has units of joule-seconds, which is the same as the Planck constant!

$$ S = \int_{t_0}^{t_1} { L } \, dt \qquad [S] = J \cdot s $$

What's an integral?

An integral tells you the area under a curve. In this case, we can plot the Lagangian $L$ against time $t$, and the integral from $t_0$ to $t_1$ is the area under $L$ between those instances.

Area under the x-axis gets counted as negative.

It turns out that the path the particle takes is that for which the action is minimized or otherwise stationary.

Intuition

Let's get a feel for why this is true because, at first, it seems almost magical. Consider a particle in a vacuum with initial velocity $\vec{v_0}$. We know that after $t$ seconds, the particle will take the path $$ \vec{r}(t) = \vec{r_0} + \vec{v_0}t $$ i.e., a straight line. However, there are many other equally valid trajectories for the particle.

Here, we have a red particle moving between two points. Since it's in a vacuum, there is no potential energy at play. Try out a bunch of different paths to see how the action changes for each one by dragging the blue control point.

As you can see, it's straight line motion that minimizes the action. The graph of the Lagrangian provides a visual explanation. When making a deviation to the path from the straight line, the Lagrangian does indeed show curvature, but the increased curvature is higher than the decreased curvature, so they don't cancel each other out.

A particle moving in a vacuum is a fairly straightfoward (pun unintended but appreciated) scenario. Let's add potential energy into the mix by considering a ball thrown in the air. Neglecting air resistance, the only force acting upon the ball is gravity. That means our Lagrangian is

$$ L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} m \dot{y}^2 - mgy $$

where $\dot{x}$ is the horizontal velocity, $\dot{y}$ is the vertical velocity, and $y$ is the height of the ball. Essentially, we're just adding up the kinetic energy in both directions and subtracting off gravitional potential energy. Now what path is it going to take? You know the drill: play around with the blue dot and try to minimize the action.

You'll find that the result is a parabola, or at least a portion of it. To understand what's going on here, we return to the Lagrangian $L$ and explore the tug-of-war between kinetic energy $T$ and potential energy $U$.

$$ L = T - U $$

Now, our equation factors in gravitational potential energy, which increases as the ball goes higher. When the ball is thrown, it already has a large amount of kinetic energy, and since we want the least difference between kinetic and potential energy, it helps to quickly gain a lot of potential energy to balance that out. In other words, the ball should go up.

However, by the law of conservation of energy, there's only a fixed amount of kinetic and potential energy. As potential energy increases, kinetic energy has to decrease, making $L$ even more negative. This starts to shift the action in the wrong direction, so we need to reallocate more energy to the kinetic side of things. In other words, the ball should start to go down again.

So, now that you've got a handle on the concept, let's see if we can rigorously prove that this way of thinking about physics actually checks out.

Proof

In ordinary calculus, when you're asked to find the minimum of a function, what do you do? You take the derivative and set it equal to zero to solve for the input value that minimizes it.

In this case, though, we are trying to find the function that minimizes something. This is a problem for the calculus of variations, a branch of math that applies the ideas of calculus to sets of functions.

But what does it even mean for a function to be a minimum? Looking at the image above, we see that if we construct the tangent line to the curve at a minimum, moving left or right on the line does not change the value at all. In other words, to first order, the function does not change.

To say this more mathematically, if we Taylor-expand $f(x)$ around a local minimum at $x = c$

$$ f(x) = f(c) + f'(c)(x - c) + \frac{1}{2} f''(c)(x - c)^2 + \dots $$

the leading order term vanishes, and we are left with

$$ f(x) = f(c) + \frac{1}{2} f''(c)(x - c)^2 + \dots $$

If we plug in a value extremely close to $c$, such as $c + h$, then the $c$'s cancel out and we get

$$ f(x) = f(c) + \frac{1}{2} f''(c)h^2 + \dots $$

If $h$ is very small, higher powers of $h$ become negligible, and the right side reduces to $f(c)$, the local minimum.

We can do a similar process with functions. Instead of a single deviation $h$, we have a whole function deviation $\eta(t)$ ("eta") that we add to our original function. If the new function does not change to first order, the original function must be a local minimum!

We require $\eta(t_0) = 0$ and $\eta(t_1) = 0$ because we don't want the deviation to change the endpoints. Similarly to the small nudge $h$, we have $\eta^2 = 0$ for all $t$ that we care about.

Since we're trying to minimize the action, a good first step would be to actually write out the action. If we call our trajectory $x(t)$ and we assume that potential energy depends only on the position of the particle, the action is

$$ S = \int_{t_0}^{t_1} { \underbrace{\frac{1}{2}m\dot{x}^2 - U(x)}_{L} } \, dt $$

Note that $x$ and $\dot{x}$ are implicit functions of time. Now, we want to show that adding on a deviation $\eta$ (also an implicit function of time) does not change the action to first order.

Our new path is $x(t) + \eta(t)$, and its derivative is $\dot{x}(t) + \dot{\eta}(t)$. Plugging these new expressions in in, we get

$$ S_{new} = \int_{t_0}^{t_1} { \frac{1}{2}m \left(\dot{x} + \dot{\eta} \right)^2 - U(x + \eta) } \, dt $$

We can expand the first term to $\frac{1}{2}m \left(\dot{x}^2 + 2\dot{x}\dot{\eta} + \dot{\eta}^2\right)$. The $\eta^2$ dies, leaving us with

$$ S_{new} = \int_{t_0}^{t_1} { \frac{1}{2}m \dot{x}^2 + m \dot{x} \dot{\eta} - U(x + \eta) } \, dt $$

On to the second term. If we Taylor-expand $U$ around $x$ to first order, we can rewrite it as $U(x) + U'(x)\eta$. (If we expanded further, we'd get terms like $\frac{1}{2} U''(x) \eta^2$, which would vanish)

$$ S_{new} = \int_{t_0}^{t_1} { \underbrace { \frac{1}{2}m \dot{x}^2 }_{T} + m \dot{x} \dot{\eta} - \underbrace{ U(x) }_{U} - U'(x)\eta } \, dt $$

You may be getting a bit of déja-vu. Indeed, part of what we have recovered is $T - U$, the original Lagrangian! Subtracting off $S$, which is the integral of that, yields

$$ S_{new} - S = \Delta S = \int_{t_0}^{t_1} { m \dot{x} \dot{\eta} - U'(x)\eta } \, dt $$

It would be nice if we could factor out $\eta$, but the derivative is getting in our way. If only there was a calculus tool that let us trade integrals and derivatives in a product...

Exactly, we'll be using integration by parts here.

Since we want to integrate $\eta$, we'll choose $dv = \dot{\eta}$ and $u = \dot{x}$. $v$ is then $\eta$ and $du$ is $\ddot{x}$. The $uv$ hops out of the integral, leaving us with

$$ \Delta S = m x \eta \Big|_{t_0}^{t_1} + \int_{t_0}^{t_1} { - m \ddot{x} \eta - U'(x)\eta } \, dt $$

We can actually kill the term outside the integral because it involves evaluating $\eta$ at $t_0$ and $t_1$, where it vanishes.

$$ \Delta S = \int_{t_0}^{t_1} { - m \ddot{x} \eta - U'(x)\eta } \, dt $$

Finally, as promised, we can factor out a common $\eta$.

$$ \Delta S = \int_{t_0}^{t_1} { \eta [ \underbrace{ - m \ddot{x} - U'(x) }_{\text{must be zero}} ] } \, dt = 0$$

Keep in mind that this is the change to the action, and we want it to vanish for any $\eta$. That's only true if the stuff in the middle is zero.

$$ - m \ddot{x} - U'(x) = 0 $$

Your senses may be tingling now; as it turns out, the negative derivative of potential energy is force. Substituing that in, we get

$$ - m \ddot{x} + F = 0 $$

$$ F = m \ddot{x} $$

Behold! We have Newton's second law, straight from the principle of least action. Hopefully, this proof gives some validation to the idea.

The avid reader among you might notice that we haven't actually established we found a minimum. All we've shown is that the derivative of the action vanishes. This could mean it's a saddle point or even a maximum, so the principle is more aptly called the principle of stationary action!

Quantum Mechanics

Wow, that was a lot of math. Let's take a breather and look at some weird $\langle$quantum$|$mechanics$\rangle$.

So far, we've been assuming that the particle only takes one path. Of course, quantum weirdness messes that up: in Feynman's path integral formulation, a particle takes not one path but every possible path!

You might think that this breaks the principle of least action, but it actually helps it make a lot more sense. In the classical least action principle, the endpoints are fixed because that's the only way the proof makes sense. If you think about it, however, that requires knowing where the particle will end up, which is exactly what we are trying to figure out.

Feynman's description solves this by assigning each possible path a strength or amplitude. As the particle goes along each path, the actual path it takes is the sum of all of them, weighted by their amplitude, akin to wave interference. The one that wins out is the classical action-minimizing path.

Conclusion

The principle of least action provides a fascinating and powerful way to think about physics in terms of energies instead of forces. It may seem magical at first, but the idea of energy flux gives it an intuitive meaning. And it is equivalent to Newton's world.

"I cannot go to school today," said little Peggy Ann McKay. Well, she now has a reason grounded in high-level physics!

This is my submission for the Summer of Math Exposition, 2023 :)

Notes