# Confusion about what the Euler-Lagrange equation says

I roughly understand the concept of the Lagrangian $L = T - V$ as well as the idea of stationary action $\delta \mathcal{S} =0$. However, I am confused what the Euler-Lagrange equation actually says.

Consider the Euler-Lagrange equation: \begin{equation} \frac{\partial L}{\partial q} - \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}}\right) = 0 \end{equation}

Here's my confusion:

To me, this looks like an empty mathematical exercise. If I know the partial derivatives $\frac{\partial L}{\partial q}$ and $\frac{\partial L}{\partial \dot{q}}$ and can take the derivative with respect to $t$ of the latter, what is the use of plugging all that into this setup? That's like saying after I show $2 + 3 = 5$, then show $x + y = z$, where $x=2,y=3,z=5$. In short, an empty exercise since it would be the same proof showing $x + y = z$ as showing $2 +3 = 5$ in this case.

Can someone explain what I am misunderstanding?

• I'm not sure I understand what you're asking. Given a Lagrangian, the E-L equations tell you what the equations of motion are.
– Jold
Jan 22 '15 at 5:11
• Yeah, I think that's what I'm not grasping. For instance, consider dropping an object of mass $m$ from a height $h$ off the surface of the earth. $mgh$ is PE $\frac{1}{2}mv^2$ is KE. What does E-L tell me about the path that I don't already know given a vector field model of gravity. I know the path is towards the earths surface already. What can E-L tell me? Jan 22 '15 at 5:14
• Here is a good introduction. damtp.cam.ac.uk/user/tong/dynamics.html Jan 22 '15 at 5:28

The reason algebra ($x + y = z$) is more useful than arithmetic ($2 + 3 = 5$) is that it allows you to treat large classes of problems in the same way, and to obtain very general results about them without having to do a separate calculation for every individual case. Same goes for the stationary action principle (a.k.a. action minimization). It allows you to treat a large class of problems - in this case, all of physics - in the same way, by solving $\delta S = 0$. Different choices of $S$ correspond to different areas of physics, but once you have $S$, the procedure for solving is always the same, namely the Euler-Lagrange equations.

Perhaps you've only been exposed to one formula for the action, $S = \int (T - V)\,\mathrm{d}t$ (or equivalently $L = T - V$), which is the action for a Newtonian system, and in that case it's understandable that the stationary action principle seems kind of useless. Similarly, if you had only been exposed to the one example of $x + y = z$ where $x = 2$, $y = 3$, and $z = 5$, it would seem pretty pointless to use letters. But it makes more sense once you see how general the formalism is.

• Oh, that is probably exactly the problem. I learned about applications in my geodesic exercises in general relativity today and for the first time I started to see how it could be more powerful than just the Newtonian case, as you say. Jan 22 '15 at 5:57
• What are some fundamental example uses for the Euler-Lagrange equation that go beyond the Newtonian cases so I can start to expand my understanding? I assume any elementary text will cover them, but will help if I know which examples to pay particular attention to. Jan 22 '15 at 5:59
• Any area of physics you want to explore, there's probably an action for it. For example general relativity uses the Einstein-Hilbert action. In this case the Euler-Lagrange equations give you the Einstein field equations. There are also actions for electromagnetism, relativistic particles, quantum field theory, etc. Jan 22 '15 at 7:36

Euler-Lagrange equations are a necessary condition if the action integral reaches an extremal through the trajectory described by $q(t)$ and you should obtain an expression for $q(t)$ for a given Lagrangian $L$ by solving Euler-Lagrange equations. Take for example the simple Lagrangian for a free particle with constant mass: $L=m\dot{q}^2/2$ then E-L equations are

\begin{equation} \frac{d}{dt}\left(m\dot{q}\right)=m\ddot{q}=0 \end{equation}

which has for solution

\begin{equation} q(t)=at+b \end{equation}

a straight line, something you already knew from Newton first law of motion.

Try solving E-L equations with the Lagrangian $L=m\dot{z}^2/2-mgz$ for the particle in free fall (taking partial derivatives with respect to $z$ and $\dot{z}$ instead of $q$ and $\dot{q}$) to obtain the well known trajectory $z(t)=gt^2/2+\dot{z}_0t+z_0$, where $z_0$ and $\dot{z}_0$ are the initial position and velocity.

• Ohh, wait. Okay, maybe I am starting to understand. So are they merely a necessary condition or are they a sufficient condition to tell us if the action integral reaches an extremal through the trajectory described by $q(t)$? Jan 22 '15 at 5:37
• They are not a sufficient condition, but i can't remember of a counterexample right now. To derive the E-L equations you assume that there exist a trajectory which minimizes the action and then you find out that this trajectory must satisfy the E-L equations. In fact there are several ways to derive these equations, look at the book "Emmy Noether's Wonderful Theorem" the author is Dwight E. Neuenschwander, there you'll find a good discussion about the equivalence of the principle of minimum action and Newton's second law. Jan 22 '15 at 7:40