Timeline for A Variation on Laplace's equation (context: Yang-Mills N-Instantons, Rajaraman's book)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2015 at 14:22 | vote | accept | Jonathan Rayner | ||
| May 29, 2015 at 14:22 | |||||
| Sep 5, 2014 at 15:41 | comment | added | adipy | In regards to (1), what I'm saying is that $\infty * 0$ is a meaningless quantity, which is what $\phi*0$ will be at the singular points of $\phi$. We're concerned with $\square\phi/\phi=0$. If you know that $\phi$ contains no infinities (singular points) then at every point the denominator is finite and so $\square\phi$ must be zero everywhere - so you can drop the $1/\phi$. If $\phi$ does contain infinities, then at those points $\square\phi$ can be finite - or even infinite so long as it's a "lesser infinity" like in $x/x^2$. | |
| Sep 5, 2014 at 15:08 | comment | added | Jonathan Rayner | 2) However, I'm not sure that 2) is correct. From the ansatz I've mentioned, the gauge fields corresponding to the linear solution would be $A_{\mu} = i \overline{\sigma}_{\mu\nu}\partial^{\nu} \ln (a_{\sigma} x^{\sigma} +b) = i \overline{\sigma}_{\mu\nu} a_{\nu}/(a_{\sigma} x^{\sigma} + b)$ which vanishes at infinity. | |
| Sep 5, 2014 at 13:33 | vote | accept | Jonathan Rayner | ||
| Sep 5, 2014 at 15:04 | |||||
| Sep 5, 2014 at 13:32 | comment | added | Jonathan Rayner | Nevermind on 3, have seen my mistake and corrected above. Thanks! | |
| Sep 5, 2014 at 13:16 | comment | added | Jonathan Rayner | For 1) This seems reasonable (I've never really formally learned this). Can you elaborate a little, maybe in a group theoretical sense why $0 * \phi$ is undefined for singular $\phi$? 2) Seems reasonable also. Not really clear from Rajaraman, so I wondered if there was another reason... 3) Sorry, I don't see this. Can you show explicitly? I have edited what I have in my question above. | |
| Sep 4, 2014 at 19:56 | history | answered | adipy | CC BY-SA 3.0 |