I am struggling with deriving the equations of motion for the non-linear sigma model that becomes the WZW model later on, in the CFT book by Francesco et. al. The relevant snippet is below. The action is given by $$ S = \frac{1}{4a^2}\int d^2 x \text{ Tr}(\partial_\mu g^{-1}\partial^\mu g)$$ where apparently we get by varying $g \rightarrow g+ \delta g$ $$ \delta S = \frac{1}{2a^2} \int d^2x \text{ Tr}(g^{-1} \delta g \partial^\mu(g^{-1}\partial_\mu g)). $$ I can't seem to arrive at this result.
I tried to compute it as follows
$$ \begin{align} \delta \text{ Tr}(\partial_\mu g^{-1}\partial^\mu g) & = \text{Tr}(\partial_\mu \delta g^{-1}\partial^\mu g + \partial_\mu g^{-1}\partial^\mu \delta g) \\ & = \text{Tr}(-\partial_\mu(g^{-1}\delta g g^{-1})\partial^\mu g + \partial_\mu g^{-1} \partial^\mu \delta g) \\ & = \text{Tr}(-(\partial_\mu g^{-1})\delta g g^{-1}\partial^\mu g -g^{-1} \delta g (\partial_\mu g^{-1}) \partial^\mu g) + 2 \partial_\mu g^{-1}\partial^\mu \delta g) \\ & = \dots \end{align}$$ using $ \partial^\mu g^{-1} = - g^{-1}\partial^\mu g g^{-1}$ and the cyclicity on line three. Even though I tried to partially integrate it in different ways, I cannot arrive at the same result.
I am guessing there must be a better way to derive this. In particular I do not understand the comment in the book (below) in which they derive some formula for $A$, $B$ independent of $g$, as it is exactly $g$ that appears in the above variation. Maybe someone could point me in the right direction, I would gladly complete the solution here.
Answer
You could do worse than study the Gürsey 1960-1 papers where he discovers these chiral models (in 4D). Without the telltale topological term, what you write is not the WZW model--which makes your title somewhat bizarre. It is the plain chiral model.
In any case, you started right, but did not pursue your calculation to its conclusion. Integrating by parts inside the integral, cycling inside the trace, and using the identity for the variation and derivative of the inverse (and its egregious consequence $\partial g^{-1} \partial g=g^{-1}\partial g \partial g^{-1} g$ in the final step), you obtain $$ \int \delta \text{ Tr}(\partial_\mu g^{-1}\partial^\mu g) = \text{Tr}[\partial_\mu \delta g^{-1}\partial^\mu g + \partial_\mu g^{-1}\partial^\mu \delta g ~] \\ = \int \text{Tr}[-\partial_\mu(g^{-1}\delta g g^{-1})\partial^\mu g -(\partial^\mu \partial_\mu g^{-1}) ~ \delta g] \\ = \int\text{Tr}[-\partial_\mu (g^{-1}\delta g)~ g^{-1}\partial^\mu g -g^{-1} \delta g ~ \partial_\mu g^{-1} \partial^\mu g + \partial_\mu (g^{-1}\partial^\mu g~ g^{-})~ \delta g] \\ = \int\text{Tr}[ g^{-1}\delta g~ \partial_\mu (g^{-1}\partial^\mu g) -g^{-1} \delta g ~ \partial_\mu g^{-1} \partial^\mu g + \partial_\mu (g^{-1}\partial^\mu g)~ g^{-} \delta g+ (g^{-1}\partial^\mu g) \partial_\mu g^{-} ~ g g^{-1}\delta g ] \\ = 2 \int \text{Tr}[g^{-1}\delta g ~~\partial_\mu(g^{-1}\partial^\mu g )] ~. $$ In the final step, the second term cancels the fourth.
These are standard maneuvers for chiral models, and in a very limited range: there aren't as many. They are not unrelated to (15.9), of course, but if you follow it and it does not help you, simply slug through the above formal cascade, here.
And, yes, there is a better way: Pros normally utilize currents, $j^\mu=g^{-1} \partial^\mu g$, so that the action takes the transparent Sugawara form, $$4a^2~S=-\int \text{Tr}[j^\mu j_\mu] \qquad \Longrightarrow 4a^2~\delta S=-2\int \text{Tr}[j^\mu \delta j_\mu] .$$
Now, since $\delta j_\mu= g^{-1}\partial_\mu \delta g -g^{-1}\delta g ~j_\mu$, one simply has $$ 4a^2~\delta S= 2\int \text{Tr}[ g^{-1}\delta g ~j_\mu j^\mu +\partial_\mu(j^\mu g^{-1})\delta g ]\\ = 2\int \text{Tr}[ g^{-1}\delta g ~j_\mu j^\mu -j_\mu j^\mu ~g^{-1}\delta g +\partial_\mu(j^\mu ) ~ g^{-1}\delta g ]= 2\int \text{Tr}[g^{-1}\delta g ~\partial_\mu(j^\mu )]. $$
One's sanity could well be spared by this language upon the introduction of, e.g., a 7D-embedded WZW term.
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