Is the Euler-Lagrange equation a special case of the principle of least action?

Question

Is the Euler-Lagrange equation a special case of the principle of least action? I have some confusion after reading a few dozen stackexchange articles of the "principle of least action".

I follow the derivation of the Euler-Lagrange equation , it appears to treat the functional in a generalized geometric and algebraic fashion. But now here I get lost after the derivation.

Is the principle of least action an application of the Euler-Lagrange equation with the functional being the Lagrangian ( i.e. the difference between the kinetic and potential energy ) as applied to some specific case?

Answer 1

If we have functional defined as $$S({\boldsymbol {q}})=\int _{t_0}^{t_1}L(t,{\boldsymbol {q}}(t),{\boldsymbol {\dot {q}}}(t))\,\mathrm {d} t$$ Then we have a theorem that says that a function $\mathbf{q}$ for which this functional is stationary must satisfy the Euler-Lagrange equations.

Now the principle of least action simply states that physical system must evolve in such a way between times $t_0$ and $t_1$ such that the action (which is functional with the same form as above where $L$ is the Lagrangian and $\mathbf q$ are the generalized coordinates) must be stationary. Now we can use the above theorem to say that $\mathbf q$ will satisfy the Euler-Lagrange equation which gives a more practical way to find such function by solving this equation.

Answer 2

1. The Euler-Lagrange (EL) expression $\frac{\delta F[\phi]}{\delta \phi^{\alpha}(x)}$ is the functional/variational derivative of a functional $F[\phi]$.

2. The usual$^1$ principle of least action (which more precisely should be called the principle of stationary action) asks for $\phi$-configurations with vanishing functional derivatives $\frac{\delta S[\phi]}{\delta \phi^{\alpha}(x)}\approx 0$ of the action functional $S[\phi]$, i.e. $\phi$-solutions to the EL equations.

   So to answer OP's title question (v2): The EL equations follow from the principle of stationary action.

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$^1$ Be aware that some authors (e.g. Goldstein) attach a different meaning to the principle of least action, cf. my Phys.SE answer here.

Answer 3

The Euler-Lagrange equations solve a particular variational problem where you want to extremize a functional is of the form

$$ F = \int_a^b f(x(t), \dot{x}(t), t) dt $$

One such example for $f$ is the Lagrangian $L$, in which case $F$ would be the action $S$. From my point of view, this is the principle of least action and it is solved by the Euler-Lagrange equations for a particular choice of the function under the integral.

Something that seems to vary from person to person is what exactly the Euler-Lagrange equations are. Certainly the solution $\delta F = 0$ for the above functional is the Euler-Lagrange equation, but some people define the Euler-Lagrange equations to be the solution to any function under the integral.

One case of particular interest for physics is the case that the Lagrangian may be defined in terms of a Lagrangian density $\mathcal{L}$ as $L = \int \mathcal{L} d^3x$. When this is the case and you vary the action, you get what are essentially equations of motion for a field. You can also take the function under the integral to have multiple space coordinates $x^1, x^2, ...$ to obtain equations of motion for a system of particles. You can also choose to include higher-order derivatives in the function, but I'm not aware of any use of that in physics. All three of these are fairly straightforward once you understand varying a functional and are a decent exercise to try out.