Euler Lagrange equations in continuum

[JD_PM]

Homework Statement

Given

$$\frac{d}{dt} \frac{\partial L}{\partial \dot \phi_a^{(i j k)}} - \frac{\partial L}{\partial \phi_a^{(i j k)}} = 0 \tag{1}$$

Show that, in the limit $l \rightarrow 0$, we obtain

$$\partial_{\mu} \frac{\partial \mathscr{L}}{\partial(\partial_{\mu} \phi_a)} - \frac{\partial \mathscr{L}}{\partial \phi_a} = 0 \tag{2}$$

Relevant Equations

Please see main post

OK I've been stuck for a while in how to derive $(1)$, so I better solve a simplified problem first:

We work with

Where

$$\mathscr{L} = \mathscr{L}(\phi_a (\vec x, t), \partial_{\mu} \phi_a (\vec x, t)) \tag{3}$$

And $(3)$ implies that $\mathscr{L}(\vec x, t)$

We know that

$$L=l^3\sum_{(i j k)} \mathscr{L}^{(i j k)}(t) \tag{4}$$

Where

$$\lim_{l \rightarrow 0} L = \int d^3 \vec x \mathscr{L} \tag{5}$$

So, in analogy with $(3), \mathscr{L}^{(i j k)}$ depends on the fields $\phi_a^{(i j k)} (t)$, on the time derivative of the fields $\dot \phi_a^{(i j k)} (t)$ and on the partial derivative of the fields with respect to $x, y$ and $z$ i.e;

$$\frac{\phi^{(i+1, j, k)}(t)-\phi^{(i, j, k)}(t)}{l} \tag{6.1}$$

$$\frac{\phi^{(i, j+1, k)}(t)-\phi^{(i, j, k)}(t)}{l} \tag{6.2}$$

$$\frac{\phi^{(i, j, k+1)}(t)-\phi^{(i, j, k)}(t)}{l} \tag{6.3}$$Let's tackle the problem.

I would naively plug $(4)$ into $(1)$ and evaluate the terms to get

$$\frac{\partial}{\partial \dot \phi_a^{(i j k)}} \sum_{(i'j'k')} \Big[ l^3 \mathscr{L}^{(i' j' k')}(t) \Big] = l^3 \frac{\partial \mathscr{L}^{(i j k)}}{\partial \dot \phi_a^{(i j k)}} \tag{7.1}$$

$$\frac{\partial}{\partial \phi_a^{(i j k)}} \sum_{(i'j'k')} \Big[ l^3 \mathscr{L}^{(i' j' k')}(t) \Big] = l^3 \frac{\partial \mathscr{L}^{(i j k)}}{\partial \phi_a^{(i j k)}} \tag{7.2}$$

Where I've used the Kronecker delta.

Next I'd take the limit $l \rightarrow 0$$ of $\frac{\partial \mathscr{L}^{(i j k)}}{\partial \dot \phi_a^{(i j k)}}$ and $\frac{\partial \mathscr{L}^{(i j k)}}{\partial \phi_a^{(i j k)}}$ to get $\frac{\partial \mathscr{L}}{\partial \dot \phi_a}$ and $\frac{\partial \mathscr{L}}{\partial \phi_a}$ (respectively)

So I get

$$l^3 \Big( \frac{d}{dt} \frac{\partial \mathscr{L}}{\partial \dot \phi_a} - \frac{\partial \mathscr{L}}{\partial \phi_a} \Big) = 0 \tag{8}$$

Which of course does not yield $(2)$ when taking the limit $l \rightarrow 0$.

The issue is that I am missing the spatial components $(6.1), (6.2), (6.3)$...

I've been discussing this problem and related but we did not manage to really understand it.

Any help is really appreciated.

Thank you

 

[JD_PM]

Alright here's another attempt (after discussing with a colleague).

So it seems we are closer but we still do not see why certain terms in above's EQ. 5 should vanish when taking the limit $l \rightarrow 0$.

The issue we really have is that we do not fully understand how to deal with discretized terms...

@samalkhaiat , would you have time for this?