In 't Hooft's original paper: http://igitur-archive.library.uu.nl/phys/2005-0622-152933/14055.pdf

he takes $N \rightarrow \infty $ while $ g^2 N$ is held fixed. Is this just a toy model? Or is there some reason to believe that $g^2$ goes like $\frac{1}{N}$ for large coupling? Thanks a lot.


1 Answer 1


The parameters $g^2$ and $N$ are independent of each other so it is meaningless to ask whether $g^2$ goes like $1/N$ for large or small coupling "in general". In general, it may go but it doesn't have to go.

But if it does go, i.e. if $g^2 N$ is kept fixed, then one may say new interesting things. If one introduces a new symbol $\lambda = g^2 N$ for the 't Hooft coupling, the condition simply says that $\lambda$ is kept fixed – which is natural, especially in the dual stringy interpretation of the same physics.

I need to dedicate a special paragraph to an error in your question which is probably not just a typo, due to the complicated work you had to go through to write the fractions. What is kept fixed in 't Hooft's limit isn't $g^2/N$; it is $g^2 N$. It's the product, not the ratio! So $g^2$ indeed goes like $1/N$ and not $N$ if $N$ is sent to infinity.


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