Basic question on Zee-QFT in a nutshell

[LAHLH]

Hi,

In the very first chapter of Zee, he talks about the mattress analogy and gives the Lagrangian:

$$L=\frac{1}{2} \{ \sum_{a} m \dot{q}^2_{a}-\sum_{a,b} k_{ab}q_{a}q_{b}+... \}$$

This obviously leads to the equation of motion:

$$m \ddot{q_a}=- \sum_{b} k_{ab}q_{b}$$


I don't understand why this is the correct EOM for an oscillator in this 2D set, vibrating purely vertically, surely the restoring force on the a'th oscillator should on depend on it's distance from equilibrium $$q_{a}$$. Or perhaps it should depend on things like $$q_{a}-q_{b}$$, i.e. the total distance between it and the neighbouring particles it is directly connected to.

 

[Physics Monkey]

Hi LAHLH,

I don't understand why this is the correct EOM for an oscillator in this 2D set, vibrating purely vertically, surely the restoring force on the a'th oscillator should on depend on it's distance from equilibrium $$q_{a}$$. Or perhaps it should depend on things like $$q_{a}-q_{b}$$, i.e. the total distance between it and the neighbouring particles it is directly connected to.

Do you think the equation of motion you wrote is somehow inconsistent with this expectation? For example, suppose you want the potential energy between neighboring oscillators in a one dimensional chain to be proportional to $$(q_{n+1} - q_n)^2$$. Unless I misunderstood you, this is what you were expecting to see. But the EOM you wrote includes this possibility. The entries of the K matrix in this case should be something like $$K_{n \,n+1} = K_{n+1 \,n} = -1$$ and $$K_{n \, n} = 2$$ for all $$n$$ to reproduce the energy I wrote above.

You can check this with a three particle chain with periodic boundary conditions labeled by $$n = 1, 2, 3$$. The potential energy is proportional to $$(q_3 - q_2)^2 + (q_2 - q_1)^2 +(q_1 - q_3)^2$$ which when expanded out is simply $$2 q^2_3 + 2 q^2_2 + 2 q_1^2 - 2 q_1 q_2 - 2 q_2 q_3 - 2 q_3 q_1$$ giving exactly the K matrix entries I wrote down above.

Hope this helps.

 

[LAHLH]

Ahh yes of course, thanks a lot