Timeline for How to solve a recursion relation on tensors including derivatives and traces?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2017 at 7:14 | comment | added | Jules Lamers | @IvanV. In that section I don't think there's anything to do with AdS, that comes later in the notes | |
| Oct 26, 2017 at 12:49 | comment | added | lel | Thanks I'll check those lectures, might be useful, although I'm not doing anything in AdS (I mean Vasiliev's theory), at least not now. | |
| Oct 26, 2017 at 11:30 | comment | added | Jules Lamers | @IvanV. Thanks, I indeed suspected it might be the Fronsdal tensor from the case $n=1$: up to conventions about symmetrization, $\mathcal{F}_1$ coincides with (4.8) in Vasiliev's lecture notes available at math.chalmers.se/~julesl/higherspin.html. Although I don't recognize your expression for $n=2$, Section 4 of those notes might be useful more generally. | |
| Oct 25, 2017 at 21:25 | answer | added | Willie Wong | timeline score: 2 | |
| Oct 25, 2017 at 20:58 | comment | added | Willie Wong | If you define $\mathcal{G}_n:= \Box^{n-1} \mathcal{F}_n$, then by induction every term in the expression for $\mathcal{G}_n$ have the schematic form of $\partial^{2n} \phi$ with $n$ contractions/traces. By your commutation relationships you can always move the traces as far "right" as possible, so you have a total of $n(n+1)/2$ different types of terms. From this you get a recursive rule on their coefficients. I guess how this simplify is a matter of combinatorics. | |
| Oct 25, 2017 at 18:59 | history | edited | Willie Wong |
edited tags
|
|
| Oct 25, 2017 at 16:33 | comment | added | lel | P.S. Perhaps it's worth mentioning that, physically, $\mathcal{F}_1$ is the Maxwell tensor if $s=1$ and it's the linearized Riemann tensor if $s=2$. Or it's simply the Fronsdal tensor if you happen to be familiar with higher-spin theory. | |
| Oct 25, 2017 at 16:31 | comment | added | lel | Your $\mathcal{F}_1$ is not quite right, up to multiplicative constants on the second and third term. I wrote out both the $\mathcal{F}_1$ and $\mathcal{F}_2$ in the edit at the end of my question, take a look. That ought to drive the point home. | |
| Oct 25, 2017 at 16:28 | history | edited | lel | CC BY-SA 3.0 |
Added some further clarification.
|
| Oct 25, 2017 at 12:20 | comment | added | Jules Lamers | And could you perhaps give $\mathcal{F}_2$ as well? That should give enough pointers to ensure I understand what's going on precisely. | |
| Oct 25, 2017 at 12:07 | comment | added | Jules Lamers | As another check: I find $\mathcal{F}_1= \Box \phi-\partial(\partial\cdot\phi)/2+\partial^2\phi'/6$. Is that right? | |
| Oct 25, 2017 at 11:47 | comment | added | lel | Yes, that is correct. | |
| Oct 25, 2017 at 10:08 | comment | added | Jules Lamers | Just to check if I understand what you mean, is it true that $\phi^{[2]} = \phi''$? | |
| Oct 24, 2017 at 7:30 | review | First posts | |||
| Oct 24, 2017 at 7:45 | |||||
| Oct 24, 2017 at 7:27 | history | asked | lel | CC BY-SA 3.0 |