 Let us consider a particle in the gravitational field which is thrown vertically upwards from the height h1 equals 0 at the time t1 equals 0. It moves upwards in the straight line and arrives at the ground at the same location h2 equals 0 at time t2. In the position time diagram, the particle has started at height h1 at time t1. Let us denote this starting point with a. And the particle has arrived at the end point b at height h2 at time t2. The connection between a and b must be a parabola in this problem. But why is this path h of t a parabola and not any other path? Why does nature choose exactly this path and not for example this or this one? To be able to answer this question, we need a quantity called action, which is abbreviated with the letter s. We can assign an action value to each of these conceivable paths. The action takes a whole function h in the argument and spits out a number, namely the value of the action for the corresponding function h. Such a mathematical object that eats a function h and spits out a number is called a functional. Since the functional spits out the action, s of h is also called an action functional. To distinguish the action functional from usual functions, let's take just the number in the argument we use square brackets. The action has the unit joule seconds. For example, this path could then have the value 3.5 joule seconds, this path the value of 5.6 joule seconds, and the parabolic path 2 joule seconds. So now back to the question. Why is the path a parabola? Experience shows that nature is extremal. That means if we calculate the action for all possible paths h1, h2, h3 and so on between a and b, then nature takes the action value, which is either maximum, minimum or a saddle point. Nature chooses one of these paths. Which of these extremal paths nature takes concretely depends on the considered problem. In the case of a particle thrown upwards in the gravitational field, the path has the smallest action, so a minimum. Since the parabolic path has the smallest action, the particle chooses this path in the position versus time diagram. But how do we calculate these values of the action? For this, we need the Lagrange function l. It depends on the time t, on the function value h of t, and on the velocity h dot of t. The Lagrange function has the unit of energy, that is, joules. If we integrate the Lagrange function over the time t between t1 and t2, we get a quantity that has the unit joules seconds. This is exactly the action we need. Thus, we can calculate concretely the value of the action for each possible path h, if we specify the Lagrange function. Mostly, you use the letter q instead of h, and q dot instead of h dot, and call q as generalized coordinate, and the derivative q dot as generalized velocity. What is meant by generalized, you will learn in another video. For us, it is only important to know that q can be, for example, a height above the ground, or an angle, or any other quantity which can depend on the time t. Of course, it is totally cumbersome to calculate the integral for all possible paths to take the path that results in the smallest value of the integral. To avoid such a huge task, the so-called Euler-Lagrange equation comes into play. It can be derived from the definition of the action and the principle that nature is extremal. In the video description, you will find the link to the derivation. In this video, we just want to know how to determine the unknown path q of t using the Euler-Lagrange equation. First, let's take a close look at how the Euler-Lagrange equation is composed. It contains the partial derivative of the Lagrange function with respect to the generalized velocity q dot. This derivative of l with respect to q dot is also called the generalized momentum and is abbreviated with p. The generalized momentum is then differentiated with respect to t. The other term is the derivative of the Lagrange function with respect to the generalized coordinate q. If we bring the time derivative of the momentum to the other side, we can read from the Euler-Lagrange equation whether the momentum is conserved. For this, its time derivative must be zero. So we only have to calculate whether the derivative of l with respect to q is equal zero. The Lagrange function l is a scalar function that cannot be derived but can only be guessed or motivated. If you think you have discovered a suitable Lagrangian for a problem, be it from quantum mechanics, classical mechanics or relativity, you can easily check whether the Lagrangian you found describes your problem correctly or not by using the Euler-Lagrange equation. In most cases of classical mechanics, the Lagrange function is the difference between the kinetic energy and the potential energy of a particle. Let's look at our example and how we can calculate the parabola from the Lagrange function and the Euler-Lagrange equation. For this, we have to do five steps. Step number one, set generalized coordinates q and q dot. In our example, q is equal to the height h and q dot is equal to velocity v. Velocity here is nothing else than the time derivative of the path, so h dot. Step number two, set up the Lagrange function l. The kinetic energy is one-half mv squared or one-half mh dot squared. The potential energy in the gravitational field is mgh. Thus the Lagrange function is one-half mh dot squared minus mgh. Step number three, insert the determined Lagrange function into the Euler-Lagrange equation and calculate the derivatives. The partial derivative of l with respect to h is minus m times g. The partial derivative of l with respect to h dot is mh dot. The time derivative of mh dot is mh dot dot. Let's combine our results. Let's cancel the mass m and bring h dot dot to the other side. We get a differential equation for the function h we are looking for. Here you see the meaning of the Euler-Lagrange equation. It tells us which differential equation we have to solve to find out the unknown function h of t. Note that our example is a one-dimensional problem, so we only get one differential equation out of it. In more complex multidimensional problems, we get several differential equations. Step number four, solve the differential equation for the unknown function. The solution of this differential equation is easy to determine. For this, we integrate both sides over time t. Then we get h dot is equal to minus g times t plus c1, where c1 is an integration constant. Then we integrate again over time. Then we get h is equal to minus gt squared plus c1t plus c2, where c2 is another integration constant. Step number five, insert the boundary conditions of the considered problem and determine the unknown constants c1 and c2. In our problem, we have thrown the particle at the time t1 equals zero from the height h1 equals zero. So let's set t and h equal to zero. Thus c2 must be equal to zero. To find out the constant c1, we need a second boundary condition. We know that the path of h ends at the point b. The point b corresponds to the time t2, at which the particle has landed on the ground at h2 equals zero. So insert t2 for t and zero for h. Rearrange for the constant c1 and you get c1 is equal to one-half times g times t2. Insert the two determined constants to get the function h of t. If you plot the result in the position versus time diagram, you get a parabola. Thus the Euler Lagrange equation yields exactly the curve we expected. So that's it. The Euler Lagrange equation together with the Lagrange function helps us to set up differential equations for a concrete problem. The solution of these differential equations yields the shape of the function that nature allows. With this in mind, bye and see you next time.