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I'm trying to follow a derivation of the Euler-Lagrange equation at the continuum limit, and find some details hard to understand. The 1D lattice has a mono-atomic basis with atomic spacing $\mathfrak{a}$ and atom mass being $m$. The relative displacement of the nth atom is $\delta \eta_{n}$. The Lagrangian at the discrete limit is written as

$$\begin{aligned} \mathfrak{L} &=\sum_{n=1}^{N} \mathfrak{a} \frac{1}{2}\left[\frac{m}{\mathfrak{a}}\left(\delta \dot{\eta}_{n}\right)^{2}-\kappa \mathfrak{a}\left(\frac{\delta \eta_{n+1}-\delta \eta_{n}}{\mathfrak{a}}\right)^{2}-\frac{m}{\mathfrak{a}} \Omega^{2}\left(\delta \eta_{n}\right)^{2}\right] \\ &=: \sum_{n=1}^{N} \mathfrak{a} L_{n} \end{aligned}$$

where $\kappa$ and $\Omega^2$ are the strength of the atomic bonding and the external field, respectively. At the limit of $N\rightarrow\infty$, the time derivative of the relative displacement $\delta \eta_{n}$ at time $t$ has been replaced by the value of the time derivative $\left(\partial_{t} \varphi\right)$ at the space-time coordinate $(x, t)$. And we have periodic boundary condition of

$$\varphi(x+L, t)=\varphi(x, t), \quad x \in] 0, L], \quad \forall t \in \mathbb{R}.$$

If we let

$$\mu:=\frac{m}{\mathfrak{a}}, \quad Y:=\kappa \mathfrak{a}$$

and replace the discrete sum $\sum_{n}$ by the integral $\int \mathrm{d} x /\mathfrak{a}$ over the semi-open interval $] 0, L]$, then we write the Lagrangian as

$$\begin{aligned} \mathfrak{L}=& \int_{0}^{L} \mathrm{d} x \frac{1}{2}\left[\mu\left(\frac{\partial \varphi}{\partial t}\right)^{2}-Y\left(\frac{\partial \varphi}{\partial x}\right)^{2}-\mu \Omega^{2} \varphi^{2}\right] \\ =&: \int_{0}^{L} \mathrm{d} x \mathcal{L} \end{aligned}.$$

where $\mathcal{L}$ is the "Lagrangian density". From Christopher Mudry's book one obtains the continuum limit of Euler-Lagrange equations as

$$\partial_{t} \frac{\delta \mathcal{L}(x, t)}{\delta\left(\partial_{t} \varphi\right)(y, t)}+\partial_{x} \frac{\delta \mathcal{L}(x, t)}{\delta\left(\partial_{x} \varphi\right)(y, t)}=\frac{\delta \mathcal{L}(x, t)}{\delta \varphi(y, t)}\tag{1}.$$

To get (1) I first expand $\mathcal{L}$ in terms of $\varphi,\left(\partial_{x} \varphi\right),$ and $\left(\partial_{t} \varphi\right)$ to give

$$\begin{aligned} \delta \mathcal{L} &=\mathcal{L}\left[\varphi+\delta \varphi,\left(\partial_{x} \varphi\right)+\delta\left(\partial_{x} \varphi\right),\left(\partial_{t} \varphi\right)+\delta\left(\partial_{t} \varphi\right)\right]-\mathcal{L}\left[\varphi,\left(\partial_{x} \varphi\right),\left(\partial_{t} \varphi\right)\right] \\ &=\frac{\partial \mathcal{L}}{\partial \varphi} \delta \varphi+\frac{\partial \mathcal{L}}{\partial\left(\partial_{x} \varphi\right)} \delta\left(\partial_{x} \varphi\right)+\frac{\partial \mathcal{L}}{\partial\left(\partial_{t} \varphi\right)} \delta\left(\partial_{t} \varphi\right)+\cdots \end{aligned}\tag{2}.$$

Ignore the higher-order terms in (2) and notice that $\delta\partial_x\varphi=\partial_x\delta\varphi$, we can integrate (2) by parts to give

$$ \begin{aligned} \delta\mathfrak{L}&=\int^L_0dx\delta\mathcal{L}\\ &=\int^L_0dx\{\frac{\partial\mathcal{L}}{\partial\varphi}\delta\varphi+\partial_x\left(\frac{\partial\mathcal{L}}{\partial(\partial_x\varphi)}\delta\varphi\right)+\partial_t\left(\frac{\partial\mathcal{L}}{\partial(\partial_t\varphi)}\delta\varphi\right)-\delta\varphi\partial_x(\frac{\partial\mathcal{L}}{\partial(\partial_x\varphi)})-\delta\varphi\partial_t(\frac{\partial\mathcal{L}}{\partial(\partial_t\varphi)})\} \end{aligned}\tag{3}. $$

In order to get (1) from (3), the following integral must be satisfied:

$$\int^L_0dx\partial_x\left(\frac{\partial \mathcal{L}}{\partial\left(\partial_{x} \varphi\right)} \delta \varphi\right)=\left[\frac{\partial \mathcal{L}}{\partial\left(\partial_{x} \varphi\right)} \delta \varphi\right]^L_0=0\tag{5},$$

$$\int^L_0dx\partial_t\left(\frac{\partial \mathcal{L}}{\partial\left(\partial_{t} \varphi\right)} \delta \varphi\right)=\frac{d}{dt}\int^L_0dx\left(\frac{\partial \mathcal{L}}{\partial\left(\partial_{t} \varphi\right)} \delta \varphi\right)=0\tag{6}.$$

While I have no problem with (5) as we don't change the endpoints, I don't know why Eqn. (6) is valid. Is it just because we treat time $t$ as a fixed parameter in (6)?

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  • $\begingroup$ Hi Lonitch, I am studying the continuous limit of the Euler-Lagrange equations as well and I was wondering what is the bibliography you used (setting aside Christopher Mudry's book). $\endgroup$
    – JD_PM
    Oct 12, 2020 at 16:28
  • $\begingroup$ @JD_PM, I used Mudry's book to get familiar with some QFT problems. If your goal is to learn the Euler-Lagrange equation and its application in analytical mechanics, I would recommend "The variational principles of mechanics" by Cornelius Lanczos as a starter. If this is not enough for you, you can try "Analytical Mechanics: An Introduction" by Antonio Fasano and Stefano Marni. Their book has a very good coverage on the integrability. In case you're interested in the relationship between analytical and quantum mechanics, Oliver Johns' OGT book has a good intro. $\endgroup$
    – Lonitch
    Oct 12, 2020 at 17:51
  • $\begingroup$ Lonitch thanks for the recommendation. Actually, I am particularly interested in how to derive the discrete Euler Lagrange equations (more details physicsforums.com/threads/…). I recently learned about 3D lattice discretization and I am still getting the hang of it. In my undergrad courses on Mechanics (we focused on Gregory's Classical Mechanics book) we did not cover such a technique, that is why I am searching for bibliography. Does any of your recommended sources deal with 3D lattice discretization? $\endgroup$
    – JD_PM
    Oct 12, 2020 at 18:11
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    $\begingroup$ @JD_PM It sounds to me that you're trying to learn the variational integrator on a 3D grid. None of the books gives such. You might want to check some papers for that. For example, the second chapter of Dr. Mathew West's thesis(thesis.library.caltech.edu/2492/1/west_thesis.pdf) $\endgroup$
    – Lonitch
    Oct 12, 2020 at 18:35

2 Answers 2

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OP is right. In general, to make the action principle work, we only need to impose boundary conditions (BCs) at the boundary of spacetime. Hence we shouldn't impose eq. (6) in the interior of spacetime.

In other words, OP should preferably redo their above analysis using the action rather than the Lagrangian.

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  • $\begingroup$ Thanks for your advice. The way I think about this is that at any given time instance the disturbance introduced to the path $\delta\varphi$ should only be a function of $x$. And the Euler-Lagrange equation will be Eqn. (1) without the $\partial_t$ term. My attempt to correct this is shown in my answer. $\endgroup$
    – Lonitch
    Aug 9, 2020 at 14:45
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    $\begingroup$ Note that the infinitesimal variation $\delta \varphi(x,t)$ can also depend on $t$. $\endgroup$
    – Qmechanic
    Aug 9, 2020 at 14:48
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My attempt to derive the Euler-Lagrange equation above starts from the action,

$$S=\int_{t_1}^{t_2}\int_0^L\mathcal{L}(\varphi,\partial_x\varphi,\partial_t\varphi)dxdt\tag{1}$$

and I introduce the disturbance to the path, $\delta\varphi(x,t)$ to write

$$S_{\lambda}=\int_{t_1}^{t_2}\int^L_0dxdt\mathcal{L}(\varphi+\lambda\delta\varphi,\partial_x\varphi+\lambda\partial_x\delta\varphi,\partial_t\varphi+\lambda\partial_t\delta\varphi).$$

Then

$$ \begin{aligned}\frac{\partial S_{\lambda}}{\partial\lambda}&=\int^{t_2}_{t_1}\int_0^Ldxdt\left\{\delta\varphi\frac{\partial \mathcal{L}}{\partial(\varphi+\lambda\delta\varphi)}+\partial_x(\delta\varphi)\frac{\partial\mathcal{L}}{\partial(\partial_x\varphi+\lambda\partial_x(\delta\varphi))}\right.\\ &\left.+\partial_t(\delta\varphi)\frac{\partial \mathcal{L}}{\partial(\partial_t\varphi+\lambda\partial_t(\delta\varphi))}\right\}\end{aligned}\tag{2}$$

and

$$\begin{aligned}\frac{\partial S_{\lambda}}{\partial \lambda}|_{\lambda=0}&=\int^{t_2}_{t_1}\int_0^Ldxdt\left\{\delta\varphi\frac{\partial \mathcal{L}}{\partial\varphi}+\partial_x(\delta\varphi)\frac{\partial \mathcal{L}}{\partial(\partial_x\varphi)}+\partial_t(\delta\varphi)\frac{\partial \mathcal{L}}{\partial(\partial_t\varphi)}\right\}\\ &=\int_0^L\left[\delta\varphi\frac{\partial \mathcal{L}}{\partial(\partial_t\varphi)}\right]^{t_2}_{t_1}dx+\int_{t_1}^{t2}\int^L_0dxdt\left\{\delta\varphi\frac{\partial \mathcal{L}}{\partial\varphi}-\delta\varphi\partial_t\left[\frac{\partial \mathcal{L}}{\partial(\partial_t\varphi)}\right]+\partial_x(\delta\varphi)\frac{\partial \mathcal{L}}{\partial(\partial_x\varphi)}\right\}\\ &=\int_{t_1}^{t_2}\int_0^Ldxdt\left\{\delta\varphi\left[\frac{\partial L}{\partial\varphi}-\partial_t\left(\frac{\partial L}{\partial(\partial_t\varphi)}\right)-\partial_x\left(\frac{\partial L}{\partial(\partial_x\varphi)}\right)\right]\right\}+\int_{t1}^{t2}\left[\delta\varphi\frac{\partial \mathcal{L}}{\partial(\partial_x\varphi)}\right]^{L}_{0}dt\\ &=\int_{t_1}^{t_2}\int_0^Ldxdt\left\{\delta\varphi\left[\frac{\partial L}{\partial\varphi}-\partial_t\left(\frac{\partial L}{\partial(\partial_t\varphi)}\right)-\partial_x\left(\frac{\partial L}{\partial(\partial_x\varphi)}\right)\right]\right\} \end{aligned}\tag{3}.$$

So

$$ \frac{\partial S_{\lambda}}{\partial \lambda}|_{\lambda=0}=\int^{t_2}_{t_1}\int_0^Ldxdt\left\{\delta\varphi\left[\frac{\partial L}{\partial\varphi}-\partial_t\left(\frac{\partial L}{\partial(\partial_t\varphi)}\right)-\partial_x\left(\frac{\partial L}{\partial(\partial_x\varphi)}\right)\right]\right\}=0\tag{4}. $$

From (4) we have

$$ \frac{\partial\mathcal{L}}{\partial\varphi}=\partial_t\left(\frac{\partial L}{\partial(\partial_t\varphi)}\right)+\partial_x\left(\frac{\partial L}{\partial(\partial_x\varphi)}\right)\tag{5}. $$

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