In his TASI notes Oliver DeWolfe starts with the KG equation on the Poincaré patch metric $$ ds^2=\frac{r^2}{L^2}(-dt^2+dx^2)+\frac{L^2}{r^2}dr^2. $$ When we use the ansatz$$ \phi(r\rightarrow\infty,x,t)=\frac{\alpha(x,t)L^{2\Delta_-}}{r^{\Delta_-}}+\dots+\frac{\beta(x,t)L^{2\Delta_+}}{r^{\Delta_+}}+\dots $$ on the KG equation $$ \left(-\frac{1}{\sqrt{-g}}\partial_\mu\sqrt{-g}g^{\mu\nu}\partial_\nu+m^2\right)\phi=0 $$ we end up with an equation which is an ODE in r since the $\partial_t^2$,$\partial_x^2$ terms have a factor ${1}/{r^2}$ which we can ignore.

The equation I end up with is $$ -\frac{3r}{L^2}\partial_r\phi-\frac{r^2}{L^2}\partial_r^2\phi+m^2\phi=0. $$

When I substitute the ansatz I don't confirm that the equation is satisfied since I have a lot of terms with different powers of $r$. How can I show that this is indeed satisfied?


To solve equations of the form $$ ar^2 \partial^2_{r}\phi + b r \partial_r\phi +c \phi=0 $$ with constants $a$, $b$, $c$ (which is what you have) one sets $$ \phi(r) =r^\lambda $$ so the equation becomes $$ (a\lambda(\lambda-1) +b\lambda +c)r^\lambda=0. $$ Thus one solves the quadradic equation $$ a\lambda(\lambda-1) +b\lambda +c=0 $$ to get the two values of $\lambda$.

  • $\begingroup$ I am only interested in the ansatz given. I understand that you can solve the equation as any ode but I am trying to understand why we use this ansatz. $\endgroup$
    – Penny
    May 9 '20 at 12:33
  • $\begingroup$ The ansatz given is identical to mine! The $\Delta_{\pm}$ in the ansatz are just my $-\lambda$.I don't see what your problem is. When one includes the $1/r^2$ terms it is more complicated, but it's still the usual Frobenius indicial equation at a regular singular point. $\endgroup$
    – mike stone
    May 9 '20 at 12:56

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