In the QFT book by Itzykson and Zuber, the variation of the action functional $I=\int_{t_1}^{t_2}dtL$ is written as:

$$\delta I=\int_{t_1}^{t_2}dt\frac{\delta I}{\delta q(t)}\delta q(t)$$

How is this justified? In particular, why is there an integration sign?


The integration sign in $\delta I$ is there because the integration sign was there in the original $I$ to start with. The term "variation" means the addition of $\delta$ in front of the object. It means we study an infinitesimal differential of the object; the rules that obey the variation are identical to those for other derivatives including, for example, the Leibniz rule for the variation of a product.

The only possible way how the integral sign could disappear would be if we were taking the derivative of the function $I$ with respect to $t_2$ or $t_1$ (the upper or lower limit; the lower limit would pick a natural minus sign). But the variation isn't a derivative with respect to a particular variable such as $t_2$, the upper limit. It is the object that knows about the derivatives of $I$ with respect to everything that can vary. The main thing we want to vary are the values of $\delta q(t)$ for any value of $t$, not just a single upper limit $t_2$.

  • $\begingroup$ Would it be correct to say this is the functional continuous equivalent of $df = \Sigma \frac{\delta f}{\delta x_i}\delta x_i$ ? $\endgroup$ – Whelp Jan 7 '12 at 12:59
  • $\begingroup$ Dear Whelp, not quite. The left hand side of your latest equation is "infinitesimal", you know, its size is 0.0000000001, so to say, while the right hand side is finite, comparable to 1, so they're clearly not equal. The symbol $d$ or $\delta$ in front of a variable only represents the "numerator". The valid equation is $\delta I = \sum_i (\delta I / \delta x_i) \delta x_i$. In this case, there are continuously many variables $x$, so $\sum_i$ must become $\int dt$ and $x_i$ becomes $x(t)$. I am confused: I am explaining you the very equation you started with. Why are you trying to damage it? $\endgroup$ – Luboš Motl Jan 7 '12 at 13:04
  • $\begingroup$ Now I see you fixed you eqn in the subcomment before the deadline. So yes, it's valid for finitely or countably many variables $x_i$, but in the case you started with, functional analysis, there are continuously infinitely many variables $x_i(t)$ that depend on a continuous $t$ and not just discrete $i$, so there has to be the integral over $t$ as well. The integral $\int dt$ has nothing to do with the variation itself - it's a continuous generalization of the summation $\sum_i$. $\endgroup$ – Luboš Motl Jan 7 '12 at 13:05
  • $\begingroup$ Yeah I had a bit of trouble with the Latex code while I was writing the comment. I think it's clear in my mind now. $\endgroup$ – Whelp Jan 7 '12 at 13:49

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