# Explanation for term in formula, QFT

1. Jul 19, 2006

### beta3

Hi

I've got a question about a term in a formula I've found in Mandl&Shaw's QFT book

It's about equation 2.18 on page 31

$$L(t) = {\sum_i \delta \bf{x}_i {\cal L}_i \ ...$$

Why is there a delta x_i when summing over all lagrangians for getting the lagrange-function for the whole system?
And what operator is that delta in this particular equation?
The difference between two different points? (wouldn't that rather be $$\Delta \bf{x}$$ ?)
$$\delta$$ serves only as functional derivative AFAIK

2. Jul 19, 2006

### Rach3

I don't have Mandl & Shaw, this looks like a derivation of a field Lagrangian from a system of finite oscillators? The continuum limit $$\delta x_i \rightarrow 0$$ means you're spacing the components closer and closer together, which means their masses (and thus Lagrangians) must be scaled down as $$\delta x_i$$ if the mass density is to remain unchanged. There's a nice discussion in the final chapter of Goldstein, I assume a QFT book would not go into much detail about this.

As for the notation, it looks like he's summing over infinitesimals $$\delta x_i$$, which is a physicist's way of doing calculus. It essentially means $$\int dx \, {\cal L} (x)$$.

Last edited by a moderator: Jul 19, 2006
3. Jul 19, 2006

### beta3

yep, exactly, that's what is described in the book

oh, IC now the reason of my wrong understanding. He uses $$\delta x_i$$ because he wants to show explicitly that he's talking about seperate cells (with the index i as one of the arguments of the field operator). On the page before that he declares explicitly his -in my opinion, awkward - notation.

Thanks ;)