EulerLagrange Equation and How to Use it in 5 Easy Steps
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Let us consider a particle in the gravitational field which is thrown vertically upwards from the height \(h_1 = 0\) at the time \(t_1 = 0\). It moves upwards in a straight line and arrives at the ground at the same location \(h_2 = 0\) at time \(t_2\).
In a positiontime diagram, the particle has started at height \(h_1\) at time \(t_1\). Let us denote this starting point with \(\text{A}\). And the particle has arrived at the end point \(\text{B}\) at height \(h_2\) at time \(t_2\).
The connection between \(\text{A}\) and \(\text{B}\) must be a parabola in this problem. But why is this path \(h(t)\) a parabola and not any other path? Why does nature choose exactly this path? And not for any other path as shown in illustration 2?
Action Functional
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 \(S\) takes a whole function \(h\) in the argument and spits out a number \(S[h]\), namely the value of the action for the corresponding function.
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, \(S[h]\) is also called an action functional. To distinguish the action functional from usual functions that take just a number in the argument, we use square brackets.
The action has the unit Joulesecond (\(\mathrm{Js}\)). For example, one path might have the value \( 3.5 \, \mathrm{Js}\), another path might have the value \( 5.6 \, \mathrm{Js}\), and the parabolic path might have \( 2 \, \mathrm{Js}\).
So now back to the question: Why is the path a parabola? Experience shows that:
That means, if we calculate the action for all possible paths \(h_1(t)\), \(h_2(t)\), \(h_3(t)\) 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 vs. 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 \(h(t)\) and on the velocity \( \dot{h}(t) \). The Lagrange function has the unit of energy, that is, joule (\(\mathrm{J}\)). 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.
If we integrate the Lagrange function over the time \(t\) between \(t_1\) and \(t_2\), we get a quantity that has the unit joulesecond:
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 \( \dot{q} \) instead of \( \dot{h} \) and call \(q\) as generalized coordinate and the derivate \( \dot{q} \) as generalized velocity. What is meant by "generalized" you will learn in another lessen. 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 \(t\).
EulerLagrange equation to determine the path
Of course, it is totally cumbersome to calculate the integral for all possible paths and to take the path that results in the smallest value of the integral. To avoid such a huge task, the socalled EulerLagrange equation comes into play:
The derivation of the EulerLagrange equation is based on the definition of action 2 and the stationaryaction principle. In this lesson, we do not want to know how to obtain the EulerLagrange equation, but how to use it to determine the path \(q(t)\) we are looking for.
The EulerLagrange equation contains the partial derivative of the Lagrangian function \(L\) with respect to the genaralized velocity \(\dot{q}\): \(\frac{\partial L}{\partial \dot{q}}\). This derivative \(\frac{\partial L}{\partial \dot{q}}\) is also called generalized momentum and abbreviated with \(p\). The generalized momentum is then differentiated with respect to \(t\).
The second term in the EulerLagrange equation is the derivative of the Lagrangian function \(L\) with respect to the generalized coordinate \(q\): \(\frac{\partial L}{\partial q}\).
If we bring the time derivative of the momentum to the other side, we can read from the EulerLagrange equation whether the momentum is conserved.
For this, the time derivative of the momentum must be zero. So we only have to calculate if \(\frac{\partial L}{\partial q}\) is equal zero.
Lagrange function as ingredient
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 EulerLagrange equation.
In most cases of classical mechanics, the Lagrange function is the difference between the kinetic energy \(W_{\text{kin}} \)and the potential energy \(W_{\text{pot}} \) of a particle:
Thus, if we know the kinetic and potential energies of a particle, we can determine the Lagrange function \(L\) and then use \(L\) in the EulerLagrange equation 3
.
Example: How to use EulerLagrange equation
Let's look at our example and how we can calculate the parabola from the Lagrange function and the EulerLagrange equation. For this we have to do 5 steps.
Step #1: Choose generalized coordinates
First, we need to know what \(q\) and \(\dot{q}\) represent. In our example, \(q = h \) and \(\dot{q} = v\), where \(v(t)\) is the velocity of the thrown particle. Velocity here is nothing else than the time derivative of the path: \(\dot{q} = \dot{h}\).
Step #2: Set up Lagrange function
Next, we need to specify the Lagrange function 6
as a function of \(h\) and \(\dot{h}\). The kinetic energy \(W_{\text{kin}}(t, h, \dot{h})\) of the thrown particle is given by:
~&=~ \frac{1}{2} \, m \, {\dot{h}}^2 \end{align} $$
The potential energy \(W_{\text{pot}}(t, h, \dot{h})\) of the particle in the gravitational field is given by:
Thus, the Lagrange function is \(L(t, h, \dot{h}) \) for our problem:
Step #3: Calculate derivatives in the EulerLagrange equation
Now we can use the specified Lagrangian function 9
to calculate the derivatives occurring in the EulerLagrange equation 3
. We first write the EulerLagrange equation using \(h\) and \(\dot{h}\):

Differentiate the Lagrange function
9
with respect to \(h\):Formula anchor $$ \begin{align} \frac{\partial L}{\partial h} ~&=~ \frac{\partial}{\partial h} \left( \frac{1}{2} \, m \, {\dot{h}}^2 ~~ m \, g \, h \right) \\\\
~&=~  m \, g \end{align} $$ 
Differentiate the Lagrangian function
9
with respect to \(\dot{h}\):Formula anchor $$ \begin{align} \frac{\partial L}{\partial \dot{h}} ~&=~ \frac{\partial}{\partial \dot{h}} \, \left( \frac{1}{2} \, m \, {\dot{h}}^2 ~~ m \, g \, h \right) \\\\
~&=~ \frac{1}{2} \, m \, 2 \, \dot{h} \\\\
~&=~ m \, \dot{h} \end{align} $$ 
Differentiate the momentum \( \frac{\partial L}{\partial \dot{q}} = m \, \dot{h} \) with respect to time \(t\):
Formula anchor $$ \begin{align} \frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{h}} ~&=~ \frac{\text{d}}{\text{d}t} \, \left( m \, \dot{h} \right) \\\\
~&=~ m \, \ddot{h} \end{align} $$
Let us substitute the calculated derivatives into the EulerLagrange equation:
Let's cancel the mass \(m\) and bring \(\ddot{h}\) to the other side:
What we get is a differential equation for the function we are looking for \(h\). Here you can see the meaning of the EulerLagrange equation! It tells us which differential equation we have to solve to find out the time behavior of \(h\).
Note that our example is a onedimensional problem, so we only got one differential equation out of it. For more complex multidimensional problems, we get several differential equations.
Step #4: Solve the differential equation
Now we have to solve the differential equation 15
set up with the help of the EulerLagrange equation. To do this, we integrate both sides over time \(t\):
Then we get:
This results in two integration constants. We combine these to one integration constant \(C_1\). Then we integrate both sides again over time to eliminate the time derivative of \(h\):
~\leftrightarrow~ \int \text{d}t \, \dot{h} ~&=~ \int \text{d}t \, g \, t ~+~ \int \text{d}t \, C_1 \end{align} $$
Then we get:
Here \(C_2\) is a combined integration constant. Now we have solved the differential equation for \(h\)!
Step #5: Use boundary conditions
As a last step, we have to insert the boundary conditions of the considered problem into the solution of the differential equation and determine the unknown constants \(C_1\) and \(C_2\).
In our problem, we threw the particle from the height \(h_1 = 0\) at the time \(t_1 = 0\). So the first constraint is: \(h(0) = 0\). Let's insert it into the result 19
:
~\leftrightarrow~ C_2 ~&=~ 0 \end{align} $$
Thus we have figured out the integration constant \(C_2 = 0\). Let's insert it into the Eq. 19
:
To find out the constant \(C_1\), we need a second boundary condition. We know that the path of \( h(t) \) ends at the point \(\text{B}\). The point \(\text{B}\) corresponds to the time \(t_2 \) at which the particle landed on the ground at \(h_2 = 0 \). The second boundary condition is therefore: \(h(t_2) = 0 \). Insert it into the differential equation:
~\leftrightarrow~ C_1 ~&=~ \frac{1}{2} \,g \, t_2 \end{align} $$
Insert \(C_1\) into the Eq. 21
:
If you plot the result in the positiontime diagram, you get a parabola. The EulerLagrange equation delivers exactly the result we expected!
The EulerLagrange 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.