Timeline for Interpretation of self-interacting terms in the expansion of a pure YM Lagrangian?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jun 3 at 0:31 | history | became hot network question | |||
| Jun 2 at 19:04 | vote | accept | Keith | ||
| Jun 2 at 18:00 | answer | added | Qmechanic♦ | timeline score: 6 | |
| Jun 2 at 16:39 | comment | added | J.G. | Sorry: you're right, an interacting $F$ is analogous to $p$, whereas it would just be $\dot{A}$ in the case of Maxwell's equations. | |
| Jun 2 at 16:20 | comment | added | Keith | @J.G. The analogy is not perfect, I agree. Still, what I roughly understand is that $F$ is analogous not to $\dot{A}$ but to $p$. So, you are right after all that the canonical momentum (=$p$ or $F$) is different from the kinetic momentum (= $\dot{A}$ or spacetime derivatives of $A$ in general). | |
| Jun 2 at 15:47 | comment | added | Prahar | @Keith - Precisely! The same is true for YM. It doesn't have a cubic term by itself. There's also the quartic term, which makes everything nice and stable. | |
| Jun 2 at 15:46 | comment | added | Prahar | @J.G. - No, I meant what I wrote. | |
| Jun 2 at 15:46 | comment | added | J.G. | I derived $p$ from $p=\frac{\partial L}{\partial A}$ with $L$ as I defined in my first comment mentioning it. Note $F$ is actually analogous to $\dot{A}$. It's not a perfect analogy to field theory because an antisymmetric tensor requires multiple dimensions, i.e. field theory rather than discrete mechanics. But the main point here is that the resulting Hamiltonian is lower-bounded because the momentum isn't just a derivative. In jargon, the canonical momentum isn't the same as the kinetic momentum. | |
| Jun 2 at 15:37 | comment | added | Keith | @J.G. Oh, thank you for your clarification. I see that such a choice of $p$ corresponds to the field strength tensor $F$ in my post. Did you choose such $p$ for some sort of gauge invariance? | |
| Jun 2 at 15:05 | comment | added | J.G. | @Keith since $L=\frac12p^2$ with $p=\dot{A}+CA^2$, $H+L=p\dot{A}=p^2-CA^2p$ so $H=\frac12p^2-CA^2p$. Now complete the square. | |
| Jun 2 at 15:03 | comment | added | J.G. | @Prahar Did you mean $L_\text{int}=(\dot{\phi}+\phi^2)^2$? | |
| Jun 2 at 14:52 | history | edited | Qmechanic♦ |
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| Jun 2 at 14:45 | comment | added | Keith | @Prahar Yes, now I understand. What I originally thought is a theory whose interaction is given by a "single" cubic term. | |
| Jun 2 at 14:21 | comment | added | Prahar | @Keith - There is no such statement that theories with cubic interactions are unstable. This is something you have to analyze case-by-case. For example, the theory with interaction $L_{int} = ( \phi + \phi^2)^2$ is completely stable even though it has a cubic interaction term. | |
| Jun 2 at 13:58 | comment | added | Keith | @J.G. How does the Hamiltonian have $C^2 A^4$ term as an extra? How exactly did you perform the Legendre transform? | |
| Jun 2 at 13:32 | comment | added | J.G. | Let's try a simpler example where you might worry about a cubic. Consider the Lagrangian $L=\frac12(\dot{A}+CA^2)^2$ with momentum $\dot{A}+CA^2$ and Hamiltonian $H=\frac12(p-CA^2)^2+\frac12C^2A^4$. There's nothing unstable about that. | |
| Jun 2 at 13:17 | comment | added | Keith | @Prahar Sorry for being rather vague... My understanding is that a "cubic" interaction is not physically plausible because there is no stable vacuum. I wonder if this issue applies to the above interaction term. If not, exactly what difference does the presence of a derivative make here? | |
| Jun 2 at 13:00 | comment | added | Prahar | I don't know what sort of answer you are looking for. Yes, it's a cubic interaction with a derivative. What specifically do you want to know about this? Also, why does the presence of a derivative complicate things for you? That is a very standard type of interaction term. | |
| Jun 2 at 12:47 | history | asked | Keith | CC BY-SA 4.0 |