An Introduction to the Euler-Lagrange Equations

An Introduction to the Euler-Lagrange Equations
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👋 Hey friends,

I am currently reading the first book of the "Theoretical Minimum" by the famous physicists Leonard Susskind and George Hrabovsky and I today want to share my experience with the Euler-Lagrange Equations which play a massive role in the concept of optimization and especially in the principle of least action. Today's episode focuses on the mathematical description of these differential equations. However, if you want to get a more historical and reader-friendly look at the usage of standard calculus take a look at the article I wrote about "The Art of Calculus".

But for all the math enthusiasts like me sit back and enjoy!  

(And for all E-Mail Readers: Reading the article on the website is easier since I couldn't get the equations to work in the E-Mail. I apologise for that!)

🏛️ Introduction and History

The Euler-Lagrange Equations were introduced by Euler and refined by Lagrange in the 18th century as a mathematical framework for describing the dynamics of mechanical systems based on the general principle of least action. To overcome misconception it is worth trying to understand what the principle of least action states before stepping into the description and further history of the Euler-Lagrange Equations.  

The principle of least action states that the general path taken by a physical system or simplified by a physical body like a ball for example, between two points in space and time is the one that minimizes the action. The rationalization of this would be to describe the principle of least action as the principle where the physical system, in our case the ball, is the most relaxed. So which path does the ball need to take to have the least amount of energy? The searched path optimizes for being the laziest path.  In more detail, action is defined as the integral of a function called the Lagrangian over time:

\( \mathcal{A} =\int_{t_0}^{t_1} (L)dt\)

We later learn how this actually makes sense when exploring the example of the shortest path between two points in a two-dimensional space.

Furthermore, we can historically say that the Euler-Lagrange Equations have become a fundamental tool in classical mechanics. Even in more fundamental theoretical theories such as string theory or quantum mechanics, the basic understanding relies upon the idea of the Euler-Legrange Equations because it still utilizes Legrangienges in order to solve complicated optimization problems.  

🧮 The shortest path between two points in the two-dimensional Cartesian plane

Let's say we live in a two-dimensional world with a deterministic property. This is not really relevant but it is a nice caveat to propose. Furthermore, we just leave out the fact that living in a two-dimensional world is absolutely impossible since no organism could ever be developed in such a world. Still, it is fun to show the example of a two-dimensional dog who would just be cut in half by its digestive tract since it needs to have input and output but it does not have any "protection".

The digestive tract of a 2D-Animal from Stephen Hawkins's book "A brief history of Time"

But let's get back to our two-dimensional space example. A particle moving between to points \( (x_0,t_0)\) and \( (x_1,t_1)\) can usually be specified by the the initial position \(x_0\) and its momentum \( \overrightarrow{p}\). Under the knowledge of all forces, we can determine the trajectory of the particle under the condition of Newton's equations.

However, the idea of action and the lagrangian results from thinking of the search for the trajectory of a particle as a result of the given start- and endpoint. The question of least action is now: Which trajectory from point \( (x_0,t_0)\) to \( (x_1,t_1)\) is the shortest? Seen in (Fig.1) are two possible trajectories of a particle where we now have to find the best one.

Fig.1: Two possible trajectories of a particle between x(t0) and x(t1)

The answer to this question is fairly easy and it is an everyday example. It is obviously the red line since it is a straight line and it is well-known that a straight line is the best possible way to come from one point to the other in a non-stretched two-dimensional space. Still, how would you go over proving this particular example? How do you find the optimal way between two points and how do you give a rigours definition for it?

Solving such an optimization problem will lead us to the Euler-Lagrange Equations but it is worth taking note that it again is only based on the ideas of Calculus which we have explored some time ago. Indeed to solve this particular problem we need something known as Variational Calculus or the Calculus of Variations which is extremely useful to solve optimization problems.

Ultimately the goal is to define some function \( x(t)\) which gives us an equation for the shortest path between \( (x_0,t_0)\) and \( (x_1,t_1)\) or in other words, the function which gives us the optimal way to go between these two points.

We start by defining infinitely many infinitely small intervals on both axes \( (dt,dx)\) in order to define an infinitely small resulting distance \( ds\). In this example, it is easier to think of \( ds\) as a literal distance instead of a part of the trajectory. Still, it is useful to keep that in mind.

In (Fig.2) exactly this is visualized. The searched function \( x(t)\) is colored red and the intervals \( dt,dx,ds\) are scaled up but you have to keep in mind that we make these infinitely small as the limit of \(dt\) and \( dx\) approaches zero: \( \lim_{dt,dx \to 0} \)

Fig.2: Visualisation of the intervals dx, dt and ds

How do we describe the small incremental change in the trajectory \( ds\)? \( ds\) is the modulus of the components \( dt\) and \( dx\). In order to compute ds we use the Pythagorean Theorem as follows:

\(ds = \sqrt{dt^{2} +dx^{2}}\)

By using the laws of algebra we modify this equation, as follows:

\( ds = \sqrt{1+(\frac{dx}{dt})^2}\cdot dx\)

As we now see this has a very practical application. Since \( ds\) only gives us knowledge about one infinitely incremental change to the function \( x(t)\) we need to add up or in calculus terms, we need to integrate over all possible \( ds's\) in order to get the full distance \( S\) between \( (x_0,t_0)\) and \( (x_1,t_1)\) . And don't forget, \( S\) is not a literal distance but a part of the general trajectory of the particle instead.

So the overall distance of, the function \( x(t)\) becomes the indefinite integral of \( ds\):

\( S = \int_{}^ {}\sqrt{1+(\frac{dx}{dt})^2}dx\)

Furthermore, the distance between \( (x_0,t_0)\) and \( (x_1,t_1)\) is than given by the definite integral between \(t_0\) and \( t_1\):

\( S = \int_{t_0}^ {t_1}\sqrt{1+(\frac{dx}{dt})^2}dx\)

So far we have used only regular calculus since we haven't really used any optimization analysis. In order for the above distance to be the shortest and best trajectory possible it needs to have zero change. This implies the fact that the change of the found-out distance has to be zero in order to ensure that it is the best option.

It is similar to finding the stationary point of any function. With a given function \( f(x)\) we try to find it's stationary points by searching for when the rate of change, so the derivative of the function is zero:

\( \frac{d f(x)}{dx}=0\)

We ask the same question for the function under the integral. When the change of the integrated function is equal to zero it takes the shortest path possible since it avoids a change in the general trajectory. Implying this back to our particle we can say that if the change of the integral function equals zero the particle "uses" (It cannot think, of which way is better. So uses sounds kind of misleading) the path with the least action. Mathematically we say:

\( \partial\int_{t_0}^ {t_1}\sqrt{1+(\frac{dx}{dt})^2}dx=0 \)

So for our two-dimensional particle, we say that the best possible way to get from \( (x_0,t_0)\) to \( (x_1,t_1)\) is via a straight line since the equation \( \partial\int_{t_0}^ {t_1}\sqrt{1+(\frac{dx}{dt})^2}dx=0 \) holds true and says that under the condition the trajectory between \( (x_0,t_0)\) and \( (x_1,t_1)\) is linear the change of the integral is zero and the particle follows the way of least action.

🚀 The Lagrangian, Action and the Euler-Lagrange Equation

We have seen how we can use variational calculus to determine the best possible trajectory for a particle between two given points, but how do we define the best possible way a particle can take where it really uses the least amount of action? This involves finding an integrated function where the energy state of a particle follows the path with the least action if we want to describe exactly that we need to consider two very important parameters, the Action and the Legrangian.

Formulating the specific action principle involves knowing the mass and the velocity of a particle, as in Newtonian mechanics, since we want to have knowledge about the two fundamental types of energy, kinetic energy and potential energy.

As we will see the action will be the integral from the start of a given particle trajectory \( t_0\) to the end of a particle trajectory \( t_1\), similar to the integral we saw in the example earlier.

We illustrate the example of action and the lagrangian on a particle moving on a straight line so we only focus on the \( x(t)\) and \( v_{x}(t)\) since we need to have knowledge about the two fundamental energies, potential and kinetic energy:

\( E_{pot}= V(x)\)

\( E_{kin} = T = \frac{1}{2}m(v_x)^{2}\)

The \( V(x)\) stands for a potential energy function depending on the position which we do not need to know at that point.

The action of a one-dimensional moving particle is now written:

\( \mathcal{A} =\int_{t_0}^{t_1} (T-V)dt\)

\( \mathcal{A} =\int_{t_0}^{t_1} (\frac{1}{2}m(v_x)^{2}-V(x))dt\)

The integrated property is the so-called Lagrangian and it is denoted as an L. And similar to Newtonian mechanics we also need to have knowledge about the position and the velocity of a particle since the Lagrangian is a function depending on the potential and kinetic energy, making it dependent on the initial position and velocity of the particle:

\( L(x,v_x)\)

The principle of least action seems very remarkable and unnatural. Does the particle have supernatural powers to feel out all the possible trajectories and pick the one that makes the action stationary? Stationary describes action A when it has no measurable change, so \(\partial \mathcal{A}=0\) holds true and the action function does not change which makes the particle on the way of least action.

But how do we really describe the process of minimizing the action? To do so we use a set of equations known as the Euler-Lagrange Equation. These equations come in a differential form and if you ever heard about differential calculus, you know that differential equations describe changing behaviour, so they tell how the particle is moving over time. This means that the particle does not have supernatural powers and we can rely on the fact that the Newtonian description of reality fits into the picture of least action.  

Since interactivity is a little harder when writing an article I want to give all the enthusiastic readers a little homework till next Sunday. For a one-dimensional particle the Euler-Lagrange Equation is the following:

\( \frac{d }{dt}\frac{\partial L}{\partial v_{x}}-\frac{\partial L}{\partial x}=0\)

From the idea of the Lagrangian, the action principle and the newly defined Euler-Lagrange Equation can you find a way to derive Newton's Equations of motion, in particular, \( F=ma\) considering the fact, that the force of acting on a particle at position x is equal to the negative derivative of the potential function, as follows:

\( F(x)=-\frac{d V(x)}{dx}\)

In order to answer this question think of the action integral shown before not as a smoothly integrated property but as an incremental change to the time and position of a particle instead. This will lead you to a different way of thinking about the Euler-Lagrange Equation which helps derive the Newtonian formulation of reality.

If you find an answer or any kind of solution I will be happy to receive your idea under the email:

Send your results to me!:)

On 23.04.23 we will go over the solution, derive the Euler-Lagrange Equation and explore its beauty even more!

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Thanks again and I'll see you soon.


Victor 👋 :) (@observethecosmos)

🎥 Youtube Video

Beyond Polynomials, a deep dive into NEWTON's Method

✍️ Quote of the week

“I would guess that there are limits to what we can understand. But old people always think there are limits to what we can understand. It's the young people who push past those limits.”
-Leonard Susskind

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