Unable to derive exact Coleman-de Luccia instanton differential equation 3.2

Question

As an Amazon Associate, I earn from qualifying purchases. For more practice problems like this, see Schaum’s Outline of Calculus, 7th Edition.

1. What the problem asks

We are given the “master” equation (2.10) of the paper (with the spacetime dimension set to (d=4))

[ \kappa\Bigl(\frac{d\phi}{dz}\Bigr)^{2} = \frac{2}{a^{2}}\Bigl(\frac{da}{dz}\Bigr)^{2} -\frac{1}{a}\,\frac{d^{2}a}{dz^{2}}-1 .\tag{2.10\;(d=4)} ]

The scale factor is taken to be

[ a(z)=\ell\,\frac{f(\lambda)}{\cosh z}, \qquad \lambda\equiv\tanh z . ]

The goal is to rewrite (2.10) only in terms of the new variable (\lambda) and the function (f(\lambda)) and to show that it reduces to

[ \boxed{\; \frac{\kappa^{2}}{4}\Bigl(\frac{d\phi}{d\lambda}\Bigr)^{2} = \Bigl(\frac{1}{f}\frac{df}{d\lambda}\Bigr)^{2} -\frac{1}{2f}\frac{d^{2}f}{d\lambda^{2}} \;}\tag{3.2} ]

All we need is ordinary calculus (chain rule, product rule) – no cosmology.

2. Step‑by‑step derivation

2.1 Useful identities for the change of variable

[ \lambda=\tanh z \quad\Longrightarrow\quad \frac{d\lambda}{dz}= \operatorname{sech}^{2}z = 1-\lambda^{2}. \tag{1} ]

Because

[ \operatorname{sech}z=\frac{1}{\cosh z}= \sqrt{1-\tanh^{2}z} =\sqrt{1-\lambda^{2}}, ]

we can write

[ \frac{1}{\cosh z}= \sqrt{1-\lambda^{2}} . \tag{2} ]

Thus the scale factor becomes a pure function of (\lambda):

[ a(z)=\ell\,\frac{f(\lambda)}{\cosh z} =\ell\,f(\lambda)\sqrt{1-\lambda^{2}} \equiv \ell\,\mathcal{A}(\lambda),\qquad \mathcal{A}(\lambda)=f(\lambda)\sqrt{1-\lambda^{2}} . \tag{3} ]

All derivatives with respect to (z) are obtained from derivatives with respect to (\lambda) by the chain rule (1).

2.2 First derivative (da/dz)

[ \frac{da}{dz}= \frac{da}{d\lambda}\frac{d\lambda}{dz} = (1-\lambda^{2})\,\frac{da}{d\lambda}. \tag{4} ]

From (3),

[ \frac{da}{d\lambda}= \ell\Bigl[ f’(\lambda)\sqrt{1-\lambda^{2}} + f(\lambda)\,\frac{1}{2}(1-\lambda^{2})^{-1/2}(-2\lambda) \Bigr] = \ell\Bigl( \sqrt{1-\lambda^{2}}\,f’ - \frac{\lambda f}{\sqrt{1-\lambda^{2}}}\Bigr). \tag{5} ]

Insert (5) into (4) :

[ \boxed{ \frac{da}{dz}= \ell(1-\lambda^{2}) \Bigl( \sqrt{1-\lambda^{2}}\,f’ - \frac{\lambda f}{\sqrt{1-\lambda^{2}}}\Bigr) = \ell\Bigl[(1-\lambda^{2})^{3/2}f’ -\lambda(1-\lambda^{2})^{1/2}f\Bigr]. } \tag{6} ]

2.3 Second derivative (d^{2}a/dz^{2})

Differentiate (4) once more:

[ \frac{d^{2}a}{dz^{2}} = \frac{d}{dz}!\Bigl[(1-\lambda^{2})\,\frac{da}{d\lambda}\Bigr] = \frac{d}{d\lambda}!\Bigl[(1-\lambda^{2})\frac{da}{d\lambda}\Bigr] \frac{d\lambda}{dz}. ]

Using (\frac{d\lambda}{dz}=1-\lambda^{2}) again,

[ \frac{d^{2}a}{dz^{2}} =(1-\lambda^{2})\Bigl[-2\lambda\frac{da}{d\lambda} +(1-\lambda^{2})\frac{d^{2}a}{d\lambda^{2}}\Bigr]. \tag{7} ]

Now we need (d^{2}a/d\lambda^{2}).
From (5),

[ \frac{d^{2}a}{d\lambda^{2}} =\ell\Bigl[ -\frac{\lambda}{\sqrt{1-\lambda^{2}}}f’ +\sqrt{1-\lambda^{2}}\,f’’ -\frac{f}{\sqrt{1-\lambda^{2}}} +\frac{\lambda^{2}f}{(1-\lambda^{2})^{3/2}} \Bigr]. \tag{8} ]

Insert (5) and (8) into (7) and simplify (all factors of (\ell) will cancel later).
A straightforward but lengthy algebraic manipulation gives

[ \frac{d^{2}a}{dz^{2}} =\ell\Bigl[ (1-\lambda^{2})^{2}f’’ -2\lambda(1-\lambda^{2})f’ -(1-\lambda^{2})f \Bigr]. \tag{9} ]

2.4 The combinations that appear in (2.10)

Recall that

[ a(z)=\ell f\sqrt{1-\lambda^{2}},\qquad a^{2}= \ell^{2}f^{2}(1-\lambda^{2}). \tag{10} ]

First term

[ \frac{2}{a^{2}}\Bigl(\frac{da}{dz}\Bigr)^{2} =\frac{2}{\ell^{2}f^{2}(1-\lambda^{2})} \bigl[\ell^{2}\bigl((1-\lambda^{2})^{3/2}f’ -\lambda(1-\lambda^{2})^{1/2}f\bigr)^{2}\bigr] ]

[ \boxed{ = \frac{2(1-\lambda^{2})}{f^{2}} \Bigl[(1-\lambda^{2})f’ -\lambda f\Bigr]^{2}} . \tag{11} ]

Second term

[ \frac{1}{a}\,\frac{d^{2}a}{dz^{2}} =\frac{1}{\ell f\sqrt{1-\lambda^{2}}} \,\ell\Bigl[(1-\lambda^{2})^{2}f’’ -2\lambda(1-\lambda^{2})f’-(1-\lambda^{2})f\Bigr] ]

[ \boxed{ = \frac{(1-\lambda^{2})^{3/2}}{f} \Bigl[f’’ -\frac{2\lambda}{1-\lambda^{2}}f’ -\frac{f}{1-\lambda^{2}}\Bigr]} . \tag{12} ]

2.5 Insert everything into the master equation

Equation (2.10) with (d=4) reads

[ \kappa\Bigl(\frac{d\phi}{dz}\Bigr)^{2} = \frac{2}{a^{2}}\Bigl(\frac{da}{dz}\B

Original question: Unable to derive exact Coleman-de Luccia instanton differential equation 3.2 on Mathematics Stack Exchange, licensed CC BY-SA.