Lagrangian reformulations

Question

I am reading a book on Group Theory and got stuck at the very first equation (even before the formal text begins):

$$I = \int L(q,q',t)dt $$

How did he come up with this formulation and why it is correct? I have been searching relevant materials online but most of them are QM, which is out of my reach.

Could you please help the math newbie to find a material that is easier to understand the basic terms such as "action"?

Why not define "action" as $\int L(q,q',q'',t)dt$ ? Or even

$ \int L(|q\rangle,t)dt $ where $ |q\rangle = (q,q^{(1)},...q^{(n)},...) $

Thanks

Answer 1

Mathematically, there is nothing preventing you from defining "action" as $I = \int{L(q,q',q'',t)}$. However in physics, the action is usually defined as $I=\int{L(q,q',t)}$, because the phase space of a particle is $(q,q')$. In other words, in a physical dynamics system, the particle's position and velocity are sufficient to determine its future trajectory. It is just how the physical world works. It's possible that in another universe, knowing the particle's position and velocity is not enough. You also need to know its acceleration in order to determine its future trajectory. Then the action definition would be like $I = \int{L(q,q',q'',t)}$.

Answer 2

We could also query, why have Newton wrote down the second principle of dynamic as $F=m \ddot x$, instead of F=m x? It is not a differential equation, it would be simpler to solve! We could even wondering why have not Newton wrote is second law as a third order ode?

The reason is that the nature is how it is. Newton postulate his second law in the form of a second order ordinary differential equation, because it is in accordance with the experiments.

Similarly Hamilton formulate the principle of least action as it is, because the stationary points of such a functional must satisfy the Euler-Lagrange equation. Which may also be derived from the D'alembert principle (totally equivalent to the second Newton's law).

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