Normal Lagrangian for the Brachistochrone problem When solving the Brachistochrone problem most textbooks approach it as minimization problem (variational calculus) of the integral between points A and B:
$$ Time = \int_{A}^{B} \ dt = \int_{A}^{B} \frac{ds}{v}  $$
Expanding $ds$ as arc length and obtaining $v$ using conservation of energy, an integral to minimized is obtained. From there, the Euler-Lagrange equation is applied, problem solved.
What I don't understand is why isn't the standard approach of writing the Lagrangian $$ L = T - V $$ done and then applying the Euler-Lagrange equation. I've tried to do it that and seem to get nowhere by writing the following Lagrangian:
$$ L = T - V = \frac{1}{2} m \big(\dot{x}^2 + \dot{y}^2\big) - ygm$$
What am I missing?
 A: *

*On one hand, in usual point mechanics, the background geometry is fixed, and we use equations of motion to find the particle trajectories.
In the brachistochrone problem (without friction), for fixed particle path, the point mechanical problem is trivial: It is in principle trivial to find the position as a function of time (or vice-versa) from energy conservation alone. One does not need Newton's 2nd law/Lagrange equations.


*On the other hand, in the brachistochrone problem, the interesting part is not the point mechanical problem per se, but instead to change the background geometry in order to minimize the travelling time. That's a different problem, which requires a different functional to minimize.
A: So, here's a clue to understanding. If we take your Lagrangian $L=T-V,$ and solve the EL equations, we'll end up with the following:
\begin{align*}
\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}&=0\\
\frac{d}{dt}\frac{\partial L}{\partial\dot{y}}-\frac{\partial L}{\partial y}&=0,\\
m\ddot{x}&=0\\
m\ddot{y}+mg&=0,\\
\ddot{x}&=0\\
\ddot{y}&=-g,\\
x&=x_0+v_{0x}t\\
y&=y_0+v_{0y}t-\frac12 gt^2.
\end{align*}
These are the equations for free-fall! Evidently, then, using $L=T-V,$ at least for the $V$ you specified, is incorrect. The underlying problem is that the wire on which the bead is supposed to travel is going to exert force on the bead, which in turn will change $V$ according to the usual formula
$$V=-\int_{\gamma}\mathbf{F}\cdot \mathrm{d}\mathbf{x}. $$
But this isn't reflected in your $V.$ If you were able, somehow, to come up with the right formula for $\mathbf{F}$ and include its negative integral in your potential energy $V,$ then theoretically it should work. But I'm thinking that would be horrendously complicated. Hence the usual approach you outlined at first.
