Timeline for Deriving Yang-Mills Equations
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2023 at 9:45 | comment | added | G. Blaickner | @PunkZebra See my answer there. | |
| Sep 29, 2023 at 9:03 | comment | added | PunkZebra | Is there a more explicit way to show why the $d_A\alpha+[A\wedge\alpha]$ corresponds to $d_A \alpha_M$? This would answer the question I posed here physics.stackexchange.com/questions/782375/… | |
| Oct 4, 2021 at 12:07 | history | edited | G. Blaickner | CC BY-SA 4.0 |
Typo fixed
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| Oct 4, 2021 at 10:57 | vote | accept | AaronMaroja | ||
| Sep 19, 2021 at 22:00 | comment | added | G. Blaickner | Oh sorry. In the definition of the curvature form, I wrote $F^{A}=\mathrm{d}A+\frac{1}{2}[A\wedge A]$, where $A\in\Omega^{1}(P,\mathfrak{g})$. In this context, $\mathrm{d}$ is just the usual exterior derivative of vector-valued forms. This is defined coordinate-wise. Take a basis $\{T_{a}\}_{a}$ of $\mathfrak{g}$ and write $A=\sum_{a}A_{a}T_{a}$ where $A_{a}\in\Omega^{1}(P)$. Then $\mathrm{d}A:=\sum_{a}(\mathrm{d}A_{a})T_{a}$. | |
| Sep 19, 2021 at 21:57 | comment | added | DanielC | Just a comment. When you define the curvature form, you use "d" as a notation for the differential, and one understands it as the exterior differential in the De Rham bundle over $\mathcal M$, but it is actually an operator in the principle bundle ("space of gauge fields" in physics). You need to link/project this to the exterior differential somehow. How? | |
| Sep 15, 2021 at 21:17 | history | edited | G. Blaickner | CC BY-SA 4.0 |
added 766 characters in body
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| Sep 15, 2021 at 21:01 | history | answered | G. Blaickner | CC BY-SA 4.0 |