Timeline for Infrared Divergence for Yang-Mills Theory
Current License: CC BY-SA 3.0
6 events
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| Oct 9, 2017 at 18:07 | comment | added | john mangual | He is saying Taylor expand $e^{-8\pi^2/g^2}$ around $g=0$, and then interchange the integral and summation $\int_0^\infty \sum_0^\infty \approx\sum_0^\infty \int_0^\infty \approx \sum_0^\infty \int_0^\Lambda $ and finally approximate $ \Lambda \approx \infty$ which can only end in disaster. He's saying the coefficients are infinite so you can't do resurgence at all. And the cutoff prescription gives the wrong answer. | |
| Oct 9, 2017 at 17:01 | comment | added | john mangual | Thanks @WillieWong if $b_1 \geq 4$ there's an issue. I guess the "cutoff prescription" is to approximate $\int_0^\infty$ with $\int_0^\Lambda$ and truncate the infrared range $\rho \gg \Lambda$. Strong coupling and Weak coupling usually have to do with the size of $g$. He says "Of course, Yang-Mills theory on $\mathbb{R}^4$ is strongly-coupled in the IR and hence the perturbative expansion breaks down." so I'd like to understand what is going wrong with perturbation theory here. | |
| Oct 9, 2017 at 16:56 | comment | added | Willie Wong | In terms of the infrared divergence: $b_1$ is usually a pretty big integer. In 't Hooft's article which your article cites for the vacuum energy, the leading term has $b_1 = 3 N^f$ where $N^f$ is generally some integer at least $2$ (it is the number of doublets). (I don't know what any of this means, but the divergence of the integral is pretty clear as soon as you factor in $b_1$.) | |
| Oct 9, 2017 at 16:55 | history | edited | john mangual | CC BY-SA 3.0 |
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| Oct 9, 2017 at 16:48 | history | edited | john mangual | CC BY-SA 3.0 |
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| Oct 9, 2017 at 16:41 | history | asked | john mangual | CC BY-SA 3.0 |