# How to derive this (Lagrangian of mattress) from Zee's book

#### Clara Chung

Problem Statement
attached
Relevant Equations
attached I am not sure where does the dy term, phi^2 and phi^4 terms come from.
I guess there are dx and dy because we have to account for the nearest neighbour pairs in the x and y axis?
I guess there is a phi^2 term because 2q_a*q_b=(q_a-q_b)^2-q_a^2-q_b^2, the term q_a^2-q_b^2?

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#### king vitamin

Gold Member
I guess there is a phi^2 term because 2q_a*q_b=(q_a-q_b)^2-q_a^2-q_b^2, the term q_a^2-q_b^2?
That is correct.
In his expression for $V$, he is only writing the first term in a power series, and the power series will really include all higher powers too. Although he is presumably assuming that the potential is symmetric under taking all the $q_a$ to $-q_a$, otherwise there would be a $\phi^3$ term too. (Actually, looking at my copy of Zee, it seems that the very next equation in the book does include $\phi^3$ so this might just be a typo.) So the next term will look like

$$V = \sum_{ab} \frac{1}{2} k_{ab} q_a q_b + \sum_{abcd} k_{abcd} q_a q_b q_c q_d + \cdots$$
Now when you take the continuum limit of this, you'll get a bunch of terms including a $\phi^4$ term (and also terms like $(\partial_x \phi)^2 \phi^2$ which are presumably included in the ellipses along with higher order terms). If I had included a cubic term it would also allow terms like $\phi^3$.

• Clara Chung

"How to derive this (Lagrangian of mattress) from Zee's book"

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