Euler-Lagrange Equation in Physics
Overview
This article assumes that the reader has already gone through the category of classical mechanics, specifically Lagrangian Mechanics and Hamilton’s Principle of Least Action. While this article will attempt to re-explain notations and content, even if they have been covered before, it is recommended to refer to the linked document for any unexplained notation.
The integral of the Lagrangian over a path of motion is called the action and is denoted by $J$. If position is designated as $y$, then
$$ J =\int_{t_{1}} ^{t_2} Ldt=\int_{t_{1}}^{t_2}L\big( y^{\prime}(t),\ y(t),\ t \big)dt $$
In this case, the principle of Hamilton states that the action has a minimum value for the Lagrangian corresponding to the actual path of motion.
Euler-Lagrange Equation
The Lagrangian $L$ for the actual path of motion, which makes the action take an extremum (maximum or minimum) value, satisfies the following equation.
$$ \dfrac{ \partial L}{\partial y}-\dfrac{d}{dt} \left( \dfrac{\partial L}{\partial y^{\prime}} \right)=0 $$
This equation is called the Euler-Lagrange equation.
Explanation
It is important to note that the converse is not true. Differentiating at a maximum or minimum yields $0$, but having a derivative that equals $0$ does not necessarily mean that it is at a maximum or a minimum. A derivative equal to $0$ could also indicate a point of inflection. Similarly, although the actual path of motion satisfies the Euler-Lagrange equation, not all paths that satisfy the Euler-Lagrange equation represent actual paths of motion.
Proof
Let’s consider an arbitrary path of motion as $y(\alpha, t)=y(0,t)+\alpha\eta (t)$, where $y(0,t)$ represents the actual path of motion. Since the starting and ending points of all paths are the same, $\eta (t_{1})=\eta (t_2)=0$ holds true. Let’s also state the Lagrangian and the action for an arbitrary path of motion as follows.
$$ L = L \big( y(\alpha, t)^{\prime},\ y(\alpha, t),\ t \big) $$
$$ J(\alpha) = \int_{t_{1}}^{t_2} L \big( y(\alpha, t)^{\prime},\ y(\alpha, t),\ t\big) dt $$
$y(\alpha, t)$ becomes the actual path of motion when $\alpha=0$, and since the action takes an extremum value for the actual path of motion, differentiating with respect to $\alpha=0$ yields $0$.
$$ \dfrac{\partial J}{\partial \alpha} \bigg|_{\alpha=0}=0 $$
The above formula is often simply represented as $\delta J=0$ in mechanics textbooks or other sources.$(\delta=\frac{\partial}{\partial \alpha}$) Upon calculating the derivative above, the following formula is obtained.
$$ \begin{align*} \dfrac{\partial J}{\partial \alpha} &= \dfrac{\partial}{\partial \alpha} \int_{t_{1}}^{t_2} L \big( y(\alpha, t)^{\prime},\ y(\alpha, t),\ t \big) dt \\ &= \int_{t_{1}}^{t_2} \dfrac{\partial}{\partial \alpha} L \big( y(\alpha, t)^{\prime},\ y(\alpha, t),\ t \big) dt \\ &= \int_{t_{1}}^{t_2} \left( \dfrac{\partial L}{\partial y^{\prime}}\dfrac{\partial y^{\prime}}{\partial \alpha}+\dfrac{\partial L}{\partial y}\dfrac{\partial y}{\partial \alpha} +\dfrac{\partial L}{\partial t}\dfrac{\partial t}{\partial \alpha} \right) dt \end{align*} $$
Note that $y^{\prime}$ in ${}^{\prime}$ refers to the differentiation with respect to $t$. Since $y(\alpha, t)=y(0,t)+\alpha\eta (t)$, $y^{\prime}(\alpha, t)=y^{\prime}(0,t)+\alpha\eta^{\prime}(t)$ holds. Therefore, the following is true.
$$ \dfrac{\partial y^{\prime}}{\partial \alpha}=\eta^{\prime}(t), \quad \dfrac{\partial y}{\partial \alpha}=\eta (t),\quad \dfrac{\partial t}{\partial \alpha}=0 $$
Since $t$ and $\alpha$ are independent variables, differentiating yields $0$. Hence, we have:
$$ \dfrac{\partial J}{\partial \alpha} = \int_{t_{1}}^{t_2} \left( \dfrac{\partial L}{\partial y^{\prime}}\eta^{\prime}(t)+\dfrac{\partial L}{\partial y}\eta (t) \right) dt $$
If only the first term is calculated using partial integration, the result is as follows.
$$ \begin{align*} \int_{t_{1}}^{t_2} \dfrac{\partial L}{\partial y^{\prime}}\eta^{\prime}(t) dt &= \left. \dfrac{\partial L}{\partial y^{\prime}}\eta (t) \right]_{t_{1}}^{t_2}-\int_{t_{1}}^{t_2} \dfrac{d}{dt}\left( \dfrac{\partial L}{\partial y^{\prime}}\right) \eta (t) dt \\ &= -\int_{t_{1}}^{t_2} \dfrac{d}{dt}\left( \dfrac{\partial L}{\partial y^{\prime}}\right) \eta (t) dt \end{align*} $$
The reason for the disappearance of the definite integral term is because of $\eta (t_{1})=\eta (t_2)=0$. Substituting this into the original equation and arranging it with respect to $\eta (t)$ gives the following formula.
$$ \dfrac{\partial J}{\partial \alpha} = \int_{t_{1}}^{t_2} \left(- \dfrac{d}{dt}\dfrac{\partial L}{\partial y^{\prime}}+\dfrac{\partial L}{\partial y} \right) \eta (t)dt $$
However, the value of $\dfrac{\partial J}{\partial \alpha}$ when the path is the actual motion path, at $(\alpha=0$, yields $)$ $0$. Therefore, if we set $\alpha=0$, we obtain the following formula.
$$ \dfrac{\partial J}{\partial \alpha} = \int_{t_{1}}^{t_2} \left(- \dfrac{d}{dt}\dfrac{\partial L}{\partial y^{\prime}}+\dfrac{\partial L}{\partial y} \right) \eta (t)dt=0 $$
This formula represents the action for the Lagrangian of the actual path and must always hold for any $\eta (t)$. In other words, for any $\eta (t)$, the integral value must come out as $0$. For this to be possible, the value inside the brackets multiplied must be $0$. Thus, the Lagrangian for the actual path satisfies the following equation.
$$ \dfrac{ \partial L}{\partial y}-\dfrac{d}{dt} \left( \dfrac{\partial L}{\partial y^{\prime}} \right)=0 $$
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