How are Lagrangian mechanics equivalent to Newtonian mechanics?

Question

I didn't study Lagrangian mechanics yet but I did study Newtonian mechanics, and someone said to me that later we would study analytic mechanics (which contain Lagrangian mechanics) and that it contain some equations that are equivalent to Newton's laws but are more fundamental. I want just one single example of this without complicated maths if possible, but I will study the core and pure thing later on.

Answer 1

The Euler-Lagrange equation: $$\frac{\mathrm d\mathcal{L}}{\mathrm dx}=\frac{\mathrm d}{\mathrm dt}\frac{\mathrm d\mathcal{L}}{\mathrm d\dot{x}}$$ is equivalent to Newton's second law of motion which states: $$\mathbf{ F}=m\vec a.$$

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I will not go into the pure maths way since I will only present the procedures to show that our first equation is equivalent to Newton's second law.
The Lagrangian (denoted $\mathcal L$) is simply the kinetic energy of a body minus its potential energy which can be written as follows, $\mathcal L=1/2mv^2-\mathrm V(x)$. Now if you take the derivative of this Lagrangian w.r.t $x$ you will simply get the derivative of the potential energy $\mathrm{V}(x)$ w.r.t $x$, which is a force. So the LHS of the first equation is equivalent to the LHS of the second.
$\dot{x}$ just means the derivative of $x$ w.r.t time, so if you differentiate the Lagrangian with respect to $\dot{x}$, which is velocity, you will get $m\dot{x}$ since $$\frac{\mathrm d\mathcal{L}}{\mathrm d\dot{x}}=\frac{\mathrm d}{\mathrm d\dot{x}}\frac12 mv^2-\mathrm{V}(x)=\frac{\mathrm d}{\mathrm d\dot{x}}\frac12 m\dot{x}^2=m\dot{x}.$$ And the derivative of the latter w.r.t time will be $m\ddot{x}$ which is of course $m\vec{a}$. So the RHS of the first equation is equivalent to the RHS of the second equation. Therefore:

$$\frac{\mathrm d\mathcal{L}}{\mathrm dx}=\frac{\mathrm d}{\mathrm dt}\frac{\mathrm d\mathcal{L}}{\mathrm d\dot{x}}\iff \mathbf{ F}=m\vec a.$$