✍️Derivation of the Euler-Lagrange Equations

✍️Derivation of the Euler-Lagrange Equations

👋 Hey friends,

I gave you a little introduction to the Euler-Lagrange Equations and to the principle of least action in last week's newsletter. Since I have sent you home with a bit of homework trying to derive Newton's equations of motion with the Euler-Lagrange Equation and some more detailed information, we shall today explore how we derive these absurd-looking equations and how we can define the equations of motion through the eyes of the principle of least action.

If you haven't read the last episode yet I highly recommend reading it first since it provides you with all the necessary knowledge in order to understand today's article. It also provides you with a brief historical overview which can be very motivational for some students among you:

But now let's start where we have left off with the principle of least action and let's try to derive the Euler-Lagrange Equations for a single degree of freedom!

(And again 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!)

🧮 The Euler-Lagrange Equations for a single degree of Freedom

Last week I have given you the Euler-Lagrange equation, so you were able to work on the task I gave you, but now let us derive this absurd-looking equation:

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

To review the concept of least action it is helpful to remind us what we did last week. To find the trajectory of a particle where it follows the path of least action we need a branch of mathematics known as variational calculus. I am not going over the explicit explanation again (click here if you would like to read that), but we have left off with the result that the overall action can be described as a time-dependent integral of something known as the lagrangian:

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

The Lagrangian is a function which is dependent on the initial position and velocity of a particle:

\( L(x,v_x)\)

What we now want to know is when the variation in the action is zero. In order words, we want to know when the action is stationary which leads us to the fact that the particle follows the trajectory of least action:

\(\partial \mathcal{A} =0\)

The \(\partial \) stands for the variation in the function \(\mathcal{A} \).

Given these parts of information, we shall now try to derive the Euler-Lagrange Equations. The first idea is replacing the continuous time, followed by the integral, with stroboscopic time. We set time intervals of the stroboscopic time equal to a very small instant of time, call it \( \Delta t \).

We have replaced the infinite sum of the integral with a finite sum of identified intervals and considered this interval to be very small. Namely, we achieve that the action integral becomes a successive sum:

\(\int_{}^{} L dt =\sum_{}^{}L \Delta t\)

\(\overset{.}{x}=v_x=\frac{x_{n+1}-x_n}{\Delta t}\)

We have discussed the first replacement but what about the new definition of velocity? The velocity is now the change over time of two neighbouring points, so the difference between two points is divided by this very small instant of time.

In order two find an explicit sum over the small intervals between the neighbouring points but not the points themselves, we need an expression that describes the position halfway between the instants:


We have now replaced the expressions for the two inputs of the Lagrangian and newly define our Lagrangian as:

\(L(\frac{x_{n}+x_{n+1}}{2}, \frac{x_{n+1}-x_n}{\Delta t})\)

We can newly define the action as the sum of all the incremental contributions:

\(\mathcal{A}=\sum_{n}^{}L(\frac{x_{n}+x_{n+1}}{2}, \frac{x_{n+1}-x_n}{\Delta t})\Delta t\)

Now suppose we want to minimize the action by varying any one of the \(x_n\) and setting the result equal to zero. Let's pick any of them, say \(x_8\). This appears to be very complicated but notice that \(x_8\) only appears in two terms in the above equation:

\(\mathcal{A}=\sum_{n}^{}L(\frac{x_{8}+x_{9}}{2}, \frac{x_{9}-x_8}{\Delta t})\Delta t + L(\frac{x_{7}+x_{8}}{2}, \frac{x_{8}-x_7}{\Delta t})\Delta t\)

Now all we have to do is minimize \(x_8\) by differentiating with respect to \(x_8\) which luckily only appears in two ways within each term. It firstly appears through the velocity dependence and through the \(x\) dependence, making the derivative the following:

\(\frac{\partial A}{\partial x_8}= \frac{1}{\Delta t} (-\frac{\partial L}{\partial \overset{.}{x}}+ \frac{\partial L}{\partial \overset{.}{x}})+\frac{1}{2}(\frac{\partial L}{\partial \overset{}{x}}+ \frac{\partial L}{\partial \overset{}{x}})\)

To minimize the action with respect to variations of \(x_8\) we set \(dA/dx\) equal to zero. But before we can do that let's see what happens to \(dA/dx\) when \(\Delta t\) tends to zero.

The first part of the equation has the form of the difference between a quantity evaluated at two neighbouring times, divided by a small separation between them. More formally this tends to the following derivative:

\(\frac{1}{\Delta t} (-\frac{\partial L}{\partial \overset{.}{x}}+ \frac{\partial L}{\partial \overset{.}{x}})\longrightarrow -\frac{d}{dt}\frac{\partial L}{\partial \overset{.}{x}}\)

The second term of the equation also has a very simple limit. It is half the sum of \(\frac{\partial L}{\partial x}\) evaluated at the neighbouring times. As the separation between these points shrinks to zero, we just get \(\frac{\partial L}{\partial x}\). The condition that \(\frac{\partial A}{\partial x_8} = 0 \) becomes the Euler-Lagrange equation for a single degree of freedom:

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

This derivation shows that there is no magic involved in the ability of the particle to feel out the entire path before deciding which way to go. At each stage along the trajectory, the particle has only to minimize the action between a point in time and a neighbouring point in time. The principle of least action has now become a differential equation which can be solved to understand the motion of a given particle.

📚Interactivity, homework and next week's episode  

Last week, I promised to also go over the derivation of Newtons Equations given the Euler-Lagrange Equation but since this article has already been way too long again, I will do a final part of my little series about the Euler-Lagrange Equation next week where we learn and explore how these equations can be served as the equations of motion of a particle.

Still, it is worth taking note of the fact that maybe working out the equations of motion yourself helps understand the topic. As always I will have an open E-Mail block where you can ask me questions, send in results or just chat with me after all:

Send your results to me!:)

On 30.04.23 we will go over the solution, derive Newton's Equations of Motion 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

"Life is fragile: it thrives only in a narrow range of temperatures between freezing and boiling. How lucky that our planet is just the right distance from the sun: a little farther, and the death of the perpetual Antarctic winter - or worse - would prevail; a little closer, and the surface would truly fry anything that touched it."
-Leonard Susskind

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