Classical Mechanics - 1: Derivation of Euler Lagrange Equation

In this blog post we will discuss the Lagrangian approach to Classical Mechanics, this forms a major chapter of almost every Classical Mechanics textbook.

This formulation is very important and plays an important role in Classical physics.

I am solely following John Taylor’s book on Classical Mechanics.

Refer image below,
Figure 1



In the above image, the curve $y(x)$ is the correct path and the curve $Y(x)$ is the incorrect path (basically $y(x)+\text{error}(x)$).

We will for now will only concentrate on curve $y(x)$,

The length of a small segment on $y(x)$ is given by,
$\mathrm{d}s = \sqrt{\mathrm{d}x^2+\mathrm{d}y^2}\tag*{(1)}$
And since, $y=y(x)$,
$\mathrm{d}y = \frac{\mathrm{d}y}{\mathrm{d}x}\mathrm{d}x$
Which is,
$\mathrm{d}y = y'(x)\mathrm{d}x \tag*{(2)}$
Use $(2)$ in $(1)$,
$\mathrm{d}s = \sqrt{1+(y'(x))^2}\mathrm{d}x \tag*{(3)}$

Now, to find the total length of the curve $y(x)$, say the total length is $\mathrm{L}$, so,
$\mathrm{L}=\int_1^2 \mathrm{d}s = \int_{x_1}^{x_2}\sqrt{1+(y'(x))^2}\mathrm{d}x \tag*{(4)}$
To go further, we will introduce the concept of least time given by Fermat. It’s called Fermat’s principle.
In 1650, Fermat discovered a way to explain reflection and refraction as the consequence of one single principle. It is called the principle of least time or Fermat’s principle.
Fermat’s principle says that the correct path from point 1 to 2 (refer figure) is the path for which the time of travel is the least. So, the correct $y(x)$ will the one for which,
$\text{(time of travel)}=\int_{1}^{2}dt = \int_{1}^{2}\frac{\mathrm{d}s}{v}=\frac{1}{c}\int_{1}^{2}n\mathrm{d}s \tag*{(5)}$
is minimum.

For the sake of generalization, we assume that $n$ is variable and we, from now on, denote it as $n(x,y)$.
Now, using $(3)$ in $(5)$, we have (neglecting $c$ because it is a constant - the speed of light) ($n$ assumed as a variable),
$\int_{1}^{2}n(x,y)\mathrm{d}s=\int_{x_1}^{x_2}n(x,y)\sqrt{1+(y'(x))^2}\mathrm{d}x \tag*{(6)}$
So, our task is to find a $y(x)$ such that $(6)$ is minimum.
Eqn $(6)$ can be denoted as integral $\mathbb{S}$ for the sake of simplicity of reference,
$\boxed{\mathbb{S}=\int_{x_1}^{x_2}n(x,y)\sqrt{1+(y'(x))^2}\mathrm{d}x}$
For the sake of analysis, we write $\mathbb{S}$ as
$\mathbb{S}=\int_{x_1}^{x_2}\mathbb{f}[x, y(x), y'(x)]\mathrm{d}x \tag*{(7)}$
We must appreciate now that, $y(x)$ is still unknown.
From the figure, we assume $Y(x)$ is the wrong path with error value $\eta(x)$, thus,
$Y(x) = y(x)+\eta(x)$
Understand that $\eta(x)$ is the error function and therefore,
$\eta(x_1)=\eta(x_2)=0 \tag*{(8)}$
The $\eta(x)$ can be of any form and type.
The integral $\mathbb{S}$ taken along the wrong curve $Y(x)$ must be larger than the right curve $y(x)$, to express we introduce a factor $\alpha$, thus, $Y(s)$ becomes,
$Y(s) = y(s) + \alpha \eta(x) \tag*{(9)}$
Therefore, the integral $\mathbb{S}$ will now be written as $\mathbb{S(\alpha)}$,
the requirement is $\mathbb{S(\alpha)}$ is minimum for right curve $y(x)$, therefore, at $\alpha=0$, $\mathbb{S(\alpha)}$ is minimum.

Get into the mathematics now,

$\mathbb{S}(\alpha)=\int_{x_1}^{x_2}f(Y,Y', x)\mathrm{d}x$
Which is,
$\mathbb{S}=\int_{x_1}^{x_2}f(y+\alpha\eta, y'+\alpha\eta',x)\mathrm{d}x \tag*{(10)}$

Now, we need to find the differential of $\mathbb{S}$ with respect to $\alpha$,
$\frac{\mathrm{d}\mathbb{S}}{\mathrm{d}\mathbb{\alpha}}=\int_{x_1}^{x_2}\frac{\partial f}{\partial \alpha}\mathrm{d}x \tag*{(11)}$
Now,
$\frac{\partial f}{\partial \alpha}=\frac{\partial f(y+\alpha\eta, y'+\alpha\eta',x )}{\partial \alpha}=\eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'}\tag*{(12)}$
Use $(12)$ in $(11)$,
$\frac{\mathrm{d}\mathbb{S}}{\mathrm{d}\mathbb{\alpha}}=\int_{x_1}^{x_2}\frac{\partial f}{\partial \alpha}\mathrm{d}x=\int_{x_1}^{x_2}(\eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'})\mathrm{d}x \tag*{(13)}$
For finding minimum, we equate $\frac{\mathrm{d}\mathbb{S}}{\mathrm{d}\mathbb{\alpha}}$ to $0$.
Thus,
$\frac{\mathrm{d}\mathbb{S}}{\mathrm{d}\mathbb{\alpha}}=\int_{x_1}^{x_2}\frac{\partial f}{\partial \alpha}\mathrm{d}x=\int_{x_1}^{x_2}(\eta\frac{\partial f}{\partial y}+\eta'\frac{\partial f}{\partial y'})\mathrm{d}x =0\tag*{(14)}$
We try to simplify the second part of $(14)$,
$\int_{x_1}^{x_2}\frac{\partial f}{\partial y'}\eta'(x)\mathrm{d}x=[\frac{\partial f}{\partial y'}\int(\eta'(x))\mathrm{d}x-\int \frac{\mathrm{d}}{\mathrm{d}x}(\frac{\partial f}{\partial y'})\mathrm{d}x\int\eta'(x)\mathrm{d}x]_{x_1}^{x_2}$

$=[\eta(x)\frac{\partial f}{\partial y'}-\int \eta(x)\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\partial f}{\partial y'})\mathrm{d}x]_{x_1}^{x_2} \tag*{(15)}$
Using condition $(8)$, we get,
$\int_{x_1}^{x_2}\frac{\partial f}{\partial y'}\eta'(x)\mathrm{d}x=-\int_{x_1}^{x_2}\eta(x)\frac{\mathrm{d}}{\mathrm{d}x}(\frac{\partial f}{\partial y'})\mathrm{d}x\tag*{(16)}$
Using $(16)$ in $(14)$,
$\frac{\mathrm{d}\mathbb{S}}{\mathrm{d}\mathbb{\alpha}}=\int_{x_1}^{x_2}\frac{\partial f}{\partial \alpha}\mathrm{d}x=\int_{x_1}^{x_2}(\eta\frac{\partial f}{\partial y}-\eta\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'})\mathrm{d}x =0$
Thus, we have,
$\int_{x_1}^{x_2}\eta(x)(\frac{\partial f}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'})\mathrm{d}x =0 \tag*{(17)}$
The above equation should satisfy for any value of $\eta(x)$, thus,
$\boxed{\frac{\partial f}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial f}{\partial y'}=0} \tag*{(18)}$
Equation $(18)$ is called the Euler Lagrange Equation.

Cheers.

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