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As usual I'm asking a question about boundary conditions for AdS${}_3$, based on the thesis by Porfyriadis.

He is solving equations $\mathcal{L}_\xi g_{\mu\nu}$ for AdS${}_3$ metric, with a given boundary conditions that are basically a fall off conditions in $r$, since we're interested in the asymptotic behavior (as $r\to\infty$). I've finally gotten to the part where he, using the ansatz for diffeomorphism $\xi^\mu=\sum\limits_n \xi^\mu_n(t,\phi)r^n$ ($\mu=t,r,\phi$), gets a set of 6 equations for coefficients. I'll write the component of metric that was used to obtain the equation next to the equation, for clarification:

$$(tt)\qquad \xi^r_{n-1}+l^2\xi^t_{n,t}+\xi^t_{n-2,t}=0,\ n\ge 2$$ $$(tr)\qquad l^4(n+1)\xi^t_{n+1}-l^4\xi^r_{n,t}+3l^2(n-1)\xi^t_{n-1}+2(n-3)\xi^t_{n-3}=0,\ n\ge 3$$ $$(t\phi)\qquad l^2\xi^t_{n,\phi}+\xi^t_{n-2,\phi}-l^2\xi^\phi_{n-2,t}=0,\ n\ge 1$$ $$(rr)\qquad l^2(n+1)\xi^r_{n+1}+(n-2)\xi^r_{n-1}=0,\ n\ge 2$$ $$(r\phi)\qquad l^2(n+1)\xi^r_{n+1}+(n-3)\xi^\phi_{n-3}+l^2\xi^r_{n,\phi}=0,\ n\ge 3$$ $$(\phi\phi)\qquad \xi^r_{n-1}+\xi^\phi_{n-2,\phi}=0,\ n\ge 2$$

I got that, and I understand how the author got that. But how did he find the $\xi^r,\ \xi^t$, and $\xi^\phi$ I don't get :\

He said: for ($rr$) equation, using backwards induction, since for large $n$ the series for $\xi^r$ must truncate, we get that the components $\xi^r_{2m}=0,\ m\ge 1$, and $\xi^r_{2m+1}=0,\ m\ge 1$, so the most general form of $\xi^r$ is

$$\xi^r=\xi^r_1(t,\phi)r+\xi^r_0(t,\phi)+\mathcal{O}(r^{-1})\quad (\star)$$

How did he get that? I mean, I tried putting from n=10 towards n=2, and for n=2 I get

$$3l^2\xi^r_3+0\cdot\xi^r_1=0$$

And that means that for n=even my odd terms are 0, if $n\ge 2$?

I could kinda relate this to the fact that the $\xi^r=\sum\limits_n \xi^r_n r^n$, and for n=2 I'll get 0, so only n that are less than that will contribute, since I am making a expansion around infintiy. But I don't know if I'm right about this. And how did he make other equations? I tried using the same 'reasoning' but cannot get what he gets. In equations ($tr$) and ($r\phi$) he just drops off the $\xi^r_n$ terms. Why? :\ And then all of a sudden he gets

$$\xi^t=\xi^t_0(t,\phi)+\xi^t_{-1}(t,\phi)\frac{1}{r}+\mathcal{O}(r^{-2})$$ $$\xi^\phi=\xi^\phi_0(t,\phi)+\xi^\phi_{-1}(t,\phi)\frac{1}{r}+\mathcal{O}(r^{-2})$$

How?! :( I'm desperate :(

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Alright so for this part you actually just need to look at three equations: $$(rr)\qquad l^2(n+1)\xi^r_{n+1}+(n-2)\xi^r_{n-1}=0,\ n\ge 2$$ $$(r\phi)\qquad l^2(n+1)\xi^r_{n+1}+(n-3)\xi^\phi_{n-3}+l^2\xi^r_{n,\phi}=0,\ n\ge 3$$ $$(tr)\qquad l^4(n+1)\xi^t_{n+1}-l^4\xi^r_{n,t}+3l^2(n-1)\xi^t_{n-1}+2(n-3)\xi^t_{n-3}=0,\ n\ge 3$$

Let's start with the first one: $$(rr)\qquad l^2(n+1)\xi^r_{n+1}+(n-2)\xi^r_{n-1}=0,\ n\ge 2.$$ The guy assumes that you don't keep having non-zero $\xi^\mu_n$ for arbitrarily large $n$. In other words, there must be some $N$ so that $\xi^\mu_m = 0$ for $m>N$. Now let's assume that this upper limit $N$ is just 10.

Then looking at the $(rr)$ equation for $n = 11$, we find that the left term on the LHS must be zero, because of the upper limit. Thus the right hand term, and thus $\xi^r_{10}$ must be zero. Similarly, the case $n=10$ tells us $\xi^r_9$ is zero.

Now we can consider $n=9$, we find again that the first term is zero, because we already established that $\xi^r_{10}$ is zero. We keep doing this all the way down to and including $n=3$. We find that $\xi^r_m$ is zero for $m\ge 2$, and so $\xi^r = \xi^r_1 r + \xi^r_0 + ...$ . We would have arrived at this conclusion regardless of how big $N$ was.

Ok good. What about the $(r\phi )$ equation? It is $$(r\phi)\qquad l^2(n+1)\xi^r_{n+1}+(n-3)\xi^\phi_{n-3}+l^2\xi^r_{n,\phi}=0,\ n\ge 3$$ We have already shown that $\xi^r_{n+1}$ and $\xi^r_{n}$ are zero for $n \ge 2$, and since this equation only applies to $n \ge 3$, we can drop these terms. The equation then becomes $$(r\phi)\qquad (n-3)\xi^\phi_{n-3}=0,\ n\ge 3$$ Now by considering this equation for $n \ge 4$, we find that $\xi^\phi_m=0$ for $m \ge 1$, and so $\xi^\phi = \xi^\phi_0 + \xi^\phi_{-1}/r + ...$ .

Finally let's consider the $(tr)$ equation:$$(tr)\qquad l^4(n+1)\xi^t_{n+1}-l^4\xi^r_{n,t}+3l^2(n-1)\xi^t_{n-1}+2(n-3)\xi^t_{n-3}=0,\ n\ge 3$$ Since the equation only applies for $n\ge 3$, and we know $\xi^r_{n}=0$ for $n\ge 3$, we can just ignore this term. The equation becomes $$(tr)\qquad l^4(n+1)\xi^t_{n+1}+3l^2(n-1)\xi^t_{n-1}+2(n-3)\xi^t_{n-3}=0,\ n\ge 3.$$ Let's again assume $N=10$, and consider the equation for $n=13$. Then the two leftmost terms disappear and we are left with $\xi^t_{10}=0$. Proceeding with $n=12$, $n=11$, all the way down to $n=4$ tells us that $\xi^t_m = 0$ for $m\ge 1$, and so $\xi^t = \xi^t_0 + \xi^t_{-1}/r + ...$ .

I think these were the parts you didn't get. Notice we didn't use three of the equations. The next thing the guy does is to milk some additional information out of those equations, but that wasn't your question I don't think. If you have questions about my answer be sure to ask.

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  • $\begingroup$ so I pick a number (say n=5), and input them in equations until I get an equation which will give me $0\cdot\xi^\mu_n$ where n is the number I got to from 5. Then I look at, what the coefficient is next to that $0\cdot\xi^\mu_n$ (in case of ($rr$) equation it was $\xi^r_1$, and in the case of ($r\phi$) equation it was $\xi^\phi_0$), and say that only nonzero contribution is from that coefficient lower in expansion? Am I understanding this right? $\endgroup$ – dingo_d Sep 18 '13 at 12:15
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    $\begingroup$ ya that is right. $\endgroup$ – Brian Moths Sep 18 '13 at 14:38
  • $\begingroup$ Thanks, you saved me again xD I need to understand how they get this so that I can use this in 4D case with near horizon extreme Kerr metric which is more complicated, and I have system of 10 equations like this. I hope I'll be able to reproduce their result. Thanks again :) $\endgroup$ – dingo_d Sep 18 '13 at 20:46

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