Confusing with the equation $(2.4)$ and $(2.5)$ of Landau and Lifshitz, Mechanics, Chapter 1, The principle of Least Action

Question

I'm a 12th Grader and I'm interested in Lagrangian Mechanics and having a bit of knowledge about the Newtonian Mechanics. So, I found a book of Landau and Lifshitz's Mechanics and started reading from the very first chapter, but I've encountered some serious doubts here!

I'm just writing the symbol/variable meaning and conventions here first!

For instance, let's imagine a particle. $q$ is it's radius vector magnitude (it's scalar), $\dot q$ is the derivative of position vector or velocity (scalar), $t$ as time duration, $S$ as action.

So, $$S = \int \limits_{t_1}^{t_2} L(q, \dot q, t) dt.\tag{2.1}$$

Let variation of function be $\delta q(t)$, so now

$$\delta q(t_1) = \delta q (t_2) =0 \tag{2.3}$$

$$\Rightarrow \qquad\delta S = \int \limits_{t_1}^{t_2} L(q+ \delta q, \dot q + \delta \dot q, t)dt - \int \limits_{t_1}^{t_2} L (q, \dot q, t)dt = 0.\tag{2.3b}$$

So, after the next few lines, it changes to:

$$\delta S = \delta \int \limits_{t_1}^{t_2} L(q, \dot q, t)dt =0.\tag{2.4}$$

This is doubtful as $\delta$ isn't a number which can be multiplied both sides of equal sign both ways,

$$\int \left ( \frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot q} \delta \dot q \right)dt = 0.\tag{2.4b}$$

From which multiverse the above thing in concluded even I can't understand after hours of thinking, please help me with these concepts.

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Ref: https://archive.org/details/Mechanics_541/page/n11

Answer 1

What I have understood is that you want to know why the $\delta$ can be applied on both sides of the equation. Inform me if I have misinterpreted.

$\delta$ expresses the change of a function by a small amount. What this means is that for every input the the function gives a different output, except at the end points from which one gets the relation $$ \delta q(t_1)=\delta q(t_2)= 0 $$ where $t_1$ and $t_2$ are the end points. If I define a new function $$ \bar{q} = q + \epsilon \eta $$ Where $\eta $ is some arbitrary function of $t$ where $\eta(t_1)=\eta(t_2)=0$. The variation in $q$ is $$ \delta q = \eta \epsilon$$ Taking the derivative $$ \delta \dot{q} = \eta \epsilon$$ Now the change in Lagrangian is $$\delta L = \big(\partial_{q} L \big)\delta q+ \big(\partial_{\dot{q}} L \big)\delta \dot{q}$$ The same can also be achieved by taking derivative of Lagrangian w.r.t. $\epsilon$. $$\frac{d}{d\epsilon}L = \big(\partial_{\bar{q}} L \big)\frac{d\bar{q}}{d\epsilon} + \big(\partial_{\dot{\bar{q}}} L \big)\frac{d \dot{\bar{q}}}{d\epsilon}$$ The end result in both cases will be the same. In this method the derivative of $S$ w.r.t. $\epsilon$ will be zero. Taking the derivative or using variational method is effectively the same. Both sides of an equation can be varied just like you take the derivative of the equation on both sides. (just an analogy, do not misinterpret $\delta$ as derivative).

If you want further proof of Euler Lagrange equation you can get it on Wikipedia or Physics SE.

Hint (if you want to do it on your own): add and subtract

$$ \delta q \frac{d}{dt} \big( \partial_{\dot{q}} L \big) $$

in the fifth equation given in the question.

Hope this helps.