Action in the continuum limit for an elastic medium.

[VoxCaelum]

Homework Statement

https://www.physicsforums.com/attachment.php?attachmentid=57592&stc=1&d=1365348538
I'm stuck at the second part, not really sure what to here to be honest.
https://www.physicsforums.com/attachment.php?attachmentid=57590&stc=1&d=1365346819

Homework Equations

Usually the Lagrangian is the function inside the integral for the action.
\begin{equation}
L = T - U.
\end{equation}
Usually this becomes
\begin{equation}
L = \frac{p^{2}}{2m} - U.
\end{equation}

The Attempt at a Solution

I figured for a single particle the kinetic energy is:
\begin{equation}
T = \sum_{i=1}^{3} \frac{1}{2} m (\delta_{t} u_{i})^{2}.
\end{equation}
The potential energy should be some force constant times the displacement squared:
\begin{equation}
U = \sum_{i=1}^{3} k (a \delta_{i}u_{i})^{2}.
\end{equation}
So for N\end particles the Lagrangian would become:
\begin{equation}
L_{discrete} = \sum_{n=1}^{N} \sum_{i=1}^{3} \left[ \frac{1}{2} m (\delta_{t} u_{i})^{2} - \frac{k}{3} (a \delta_{i}u_{i})^{2}\right].
\end{equation}
Where the factor 1/3 comes from the fact that I am counting each bond three times.
Now this is where it gets a bit fuzzy for me; what exactly does it mean to take the continuum limit? I believe it has something to do with letting m and a approach zero in such a way that if N approaches infinity the mass stays constant, but I'm not exactly sure how to do that/what that means.

I believe that I require the fact that this is a cubic lattice to be able to write the potential energy down the way I did. But how would change the potential energy if, say for instance, the particles were on a face centered cubic lattice? Or any other lattice for that matter.

Any help would be appreciated.

 

[mark726]

Thank you for sharing your thoughts and progress on this problem. I can see that you have a good understanding of the Lagrangian and how to write it for a single particle. However, you are correct in saying that taking the continuum limit can be a bit tricky, especially when dealing with a lattice structure.

To better understand this concept, let's first define what we mean by the continuum limit. Essentially, it means taking the limit as the spacing between particles (or in this case, the lattice constant) goes to zero. This allows us to approximate the system as continuous instead of discrete, which can simplify the equations and make them more applicable to real-world situations.

In your case, you are correct in saying that the mass and lattice constant should approach zero in a way that keeps the mass constant as the number of particles approaches infinity. This is known as the thermodynamic limit, and it is commonly used in statistical mechanics to study systems with a large number of particles.

To apply this concept to your problem, you can start by redefining your potential energy in terms of the lattice constant a. This will allow you to take the limit as a goes to zero, while keeping the potential energy constant. For example, in a face-centered cubic lattice, you can write the potential energy as:

\begin{equation}
U = \sum_{i=1}^{3} k (a \delta_{i}u_{i})^{2} + \sum_{i=1}^{3} k (a \delta_{i}u_{i} + \frac{a}{2})^{2}.
\end{equation}

Notice that in this case, the potential energy includes contributions from both nearest neighbors and next nearest neighbors, which is a characteristic of the face-centered cubic lattice. You can similarly write the potential energy for other lattice structures, such as a body-centered cubic or hexagonal close-packed lattice.

I hope this helps clarify the concept of taking the continuum limit and how to apply it to different lattice structures. Keep up the good work!