JD_PM
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- TL;DR
- I want to understand discretization of 3D space (lattice) and the summation-by-parts method. To do so I am deriving the discrete Euler-Lagrange equations. I am basically stuck in how to proceed with summation by parts.
I want to derive the discrete EL equations
$$\frac{d}{dt} \frac{\partial L}{\partial \dot \phi_a^{(i j k)}} - \frac{\partial L}{\partial \phi_a^{(i j k)}} = 0$$
We deal with a Lagrange density which only depends on the fields themselves and their first order derivatives.
We discretize space, so that any spatial vector can be written as
$$\vec x = il \hat e_1 + jl \hat e_2 + kl \hat e_3$$
Where $l$ is the distance between consecutive lattice points.
The Lagrangian is no longer ##L= \int d^3 \vec x \mathscr{L}## but ##L= \sum_{(i j k)} l^3 \mathscr{L}^{(i j k)}##. When we take the limit ##l \rightarrow 0## we recover ##L= \int d^3 \vec x \mathscr{L}##.
The fields are no longer ##\phi_a (\vec x, t)## but ##\phi_a^{(i j k)} (t)##. When we take the limit ##l \rightarrow 0## we recover ##\phi_a (\vec x, t)##.
The idea I have is that we will not have to integrate by parts but sum by parts.
As we know, the action is defined as follows
$$S= \int dt L; \ \text{where} \ L=l^3\sum_{(i j k)} \mathscr{L}^{(i j k)}$$
We extremize the action (i.e. ##\delta S =0##)
$$\delta S = \int dt \ l^3 \sum_{(i j k)} \delta \mathscr{L}^{(i j k)}=0$$
Let's work out the term
$$\sum_{(i j k)} \delta \mathscr{L}^{(i j k)} \tag{*}$$
I know that, for the fields ##\phi_a## with spacetime coordinate dependence, we have
$$\delta \phi_a = \frac{\partial \phi_a}{\partial x^{\mu}} \delta x^{\mu}$$
So I would naively proceed as follows
$$\delta \mathscr{L}^{(i j k)} = \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \phi_a^{(i j k)}} \delta \phi_a^{(i j k)}+ \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \dot \phi_a^{(i j k)}} \delta \dot \phi_a^{(i j k)} + \sum_b^3 \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_b \phi_a^{(i j k)}} \delta_b \phi_a^{(i j k)} \tag{**}$$
Where ##b=x,y,z##
I am stuck in the following.
The idea is to perform summation by parts; i.e.
$$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_n g_{n+1} - f_m g_m) - \sum_{k=m+1}^n g_k (f_k -f_{k-1})$$
To the terms ##\frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \dot \phi_a^{(i j k)}} \delta \dot \phi_a^{(i j k)}## and ##\sum_b^3 \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_b \phi_a^{(i j k)}} \delta_b \phi_a^{(i j k)}##
I am a bit lost here. As an example
$$\sum_{(i j k)} \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_x \phi_a^{(i j k)}} \delta_x \phi_a^{(i j k)}=\sum_{(i j k)} \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_x \phi_a^{(i j k)}} \Big(\frac{\phi_a^{(i+1, j, k)}-\phi_a^{(i, j, k)}}{l} \Big) \tag{***}$$
Where I have used the definition of derivative on the infinitesimal term. I am confused, as I've got two derivatives before performing summation by parts, instead of 1; could you please shed some light on how to perform the summation by parts ##(***)##? Once that is understood, I should be able to derive the (discrete EL equations).
Thank you
$$\frac{d}{dt} \frac{\partial L}{\partial \dot \phi_a^{(i j k)}} - \frac{\partial L}{\partial \phi_a^{(i j k)}} = 0$$
We deal with a Lagrange density which only depends on the fields themselves and their first order derivatives.
We discretize space, so that any spatial vector can be written as
$$\vec x = il \hat e_1 + jl \hat e_2 + kl \hat e_3$$
Where $l$ is the distance between consecutive lattice points.
The Lagrangian is no longer ##L= \int d^3 \vec x \mathscr{L}## but ##L= \sum_{(i j k)} l^3 \mathscr{L}^{(i j k)}##. When we take the limit ##l \rightarrow 0## we recover ##L= \int d^3 \vec x \mathscr{L}##.
The fields are no longer ##\phi_a (\vec x, t)## but ##\phi_a^{(i j k)} (t)##. When we take the limit ##l \rightarrow 0## we recover ##\phi_a (\vec x, t)##.
The idea I have is that we will not have to integrate by parts but sum by parts.
As we know, the action is defined as follows
$$S= \int dt L; \ \text{where} \ L=l^3\sum_{(i j k)} \mathscr{L}^{(i j k)}$$
We extremize the action (i.e. ##\delta S =0##)
$$\delta S = \int dt \ l^3 \sum_{(i j k)} \delta \mathscr{L}^{(i j k)}=0$$
Let's work out the term
$$\sum_{(i j k)} \delta \mathscr{L}^{(i j k)} \tag{*}$$
I know that, for the fields ##\phi_a## with spacetime coordinate dependence, we have
$$\delta \phi_a = \frac{\partial \phi_a}{\partial x^{\mu}} \delta x^{\mu}$$
So I would naively proceed as follows
$$\delta \mathscr{L}^{(i j k)} = \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \phi_a^{(i j k)}} \delta \phi_a^{(i j k)}+ \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \dot \phi_a^{(i j k)}} \delta \dot \phi_a^{(i j k)} + \sum_b^3 \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_b \phi_a^{(i j k)}} \delta_b \phi_a^{(i j k)} \tag{**}$$
Where ##b=x,y,z##
I am stuck in the following.
The idea is to perform summation by parts; i.e.
$$\sum_{k=m}^n f_k (g_{k+1}-g_k) = (f_n g_{n+1} - f_m g_m) - \sum_{k=m+1}^n g_k (f_k -f_{k-1})$$
To the terms ##\frac{\partial \mathscr{L}^{(i' j' k')}}{\partial \dot \phi_a^{(i j k)}} \delta \dot \phi_a^{(i j k)}## and ##\sum_b^3 \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_b \phi_a^{(i j k)}} \delta_b \phi_a^{(i j k)}##
I am a bit lost here. As an example
$$\sum_{(i j k)} \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_x \phi_a^{(i j k)}} \delta_x \phi_a^{(i j k)}=\sum_{(i j k)} \frac{\partial \mathscr{L}^{(i' j' k')}}{\partial_x \phi_a^{(i j k)}} \Big(\frac{\phi_a^{(i+1, j, k)}-\phi_a^{(i, j, k)}}{l} \Big) \tag{***}$$
Where I have used the definition of derivative on the infinitesimal term. I am confused, as I've got two derivatives before performing summation by parts, instead of 1; could you please shed some light on how to perform the summation by parts ##(***)##? Once that is understood, I should be able to derive the (discrete EL equations).
Thank you