Changing variables between two different metric ansatzes in the calculation of the Klein-Gordon equation

Question

My question concerns changing variables in the calculation of the Klein-Gordon equation for a scalar field given two different "guesses" for the metric.

I consider the following Einstein tensor, which describes an action with a scalar field $\phi(r)$ with potential $V(\phi)$:

$$ R_{\mu \nu}-\kappa \left(\partial_\mu \phi\partial_\nu \phi+g_{\mu \nu}V(\phi)\right)=0 $$

Now, we'll propose a metric of the form

$$ ds^2=-f(r)dt^2+f^{-1}(r)dr^2+a^2(r)d\sigma^2 $$

This gives us a Klein-Gordon equation of the form (Eqn. 1)

$$ \square \phi=g^{11}\phi''-(g^{00}\Gamma_{00}^1+g^{11}\Gamma_{11}^1+g^{22}\Gamma_{22}^1+g^{33}\Gamma_{33}^1)\phi'\notag\\ =f(r)\phi''(r)+\left(f'(r)+2\frac{a'(r)}{a(r)}f(r)\right)\phi'(r)=\frac{dV}{d\phi} $$

Now, we can also define the metric in the form

$$ ds^2=p(r')\left\{ -b(r')dt^2+\frac{1}{b(r')}dr'^2+r'^2d\sigma^2 \right\} $$

The Klein-Gordon equation now yields (Eqn. 2)

$$\frac{b(r')}{p(r')}\phi''(r')+\phi'(r')\left\{\frac{b(r')p'(r )}{p^2(r')}+\frac{2b(r')}{r'p(r')}+\frac{b'(r')}{p(r')}\right\}=\frac{dV}{d\phi} $$

where now the derivative is taken with respect to r'.

My goal is to go from Eqn 1. from Eqn 2. with the following change of variables:

$$f(r')=p(r')b(r'),\quad a(r')=r'\sqrt{p(r')},\quad \frac{dr'}{dr}=\frac{1}{p(r')}$$

For example, for the second derivative term,

$$f\phi''=pb \frac{d^2\phi}{dr^2}\notag\\ =pb\frac{d}{dr}\frac{dr'}{dr}\frac{d\phi}{dr'}\notag\\ =pb \frac{d}{dr'}\left(\frac{1}{p}\frac{d\phi}{dr'}\right)\frac{1}{p}\notag\\ =-\frac{p'b}{p^2}\phi'+\frac{b}{p}\phi'' $$

where the variables are implied and primed-derivative notation is reserved for the $r'$ coordinate. The term proportional to $\phi''$ matches the form in Eqn. 2, but the term proportional to $\phi'$ seems a bit off. Doing the exact same thing for the other terms in $\phi'$ and adding them altogether, I find that

$$f'+2\frac{a'}{a}f=\frac{2b}{r'}+\frac{2bp'}{p}+b' $$

Putting this altogether, I find

$$\frac{b(r')}{p(r')}\phi''(r')+\phi'(r')\left\{ -\frac{b(r')p'(r')}{p^2(r')}+\frac{2b(r')}{r'}+\frac{2b(r') p'(r')}{p(r')}+b'(r') \right\} $$

Although I am close, I am off by a few signs and factors. However, I have checked it several times and all seems to be fine. Is my mapping incorrect? When I change variables in the derivative, am I doing something incorrect, or am I missing something more subtle?

Answer 1

I have discovered the issue: the derivative of $\phi'$ was not transformed properly. The correct calculation is given below:

$$\left(f'+2\frac{a'}{a}f\right)\phi'\notag\\ =\left(b\frac{p'}{p}+b'+\frac{2b}{r'}+\frac{bp'}{p}\right)\frac{1}{p}\phi'\notag\\ =\frac{bp'}{p^2}+\frac{b'}{p}+\frac{2b}{r'p}+\frac{bp'}{p^2} $$

The first term cancels the $\frac{d\phi}{dr'}$ term that pops out of $\frac{d^2\phi}{dr^2}$, yielding the correct Eqn. 2 given in the question.