In Weinberg's Cosmology, on page 412 eq. 8.1.42 he provides the power spectrum as;

$$P(k)=\frac{4(2\pi)^3 N^2 C^2(\Omega_\Lambda / \Omega_M)}{25 \Omega_M^2H_0^4 k_*^{n_s-1}}k^{n_s}\mathcal{T}^2(\sqrt{2}k/k_\text{eq}).$$

Assuming $n_s =1$ we can simplify as,

$$P(k)=\frac{4(2\pi)^3 N^2 C^2(\Omega_\Lambda / \Omega_M)}{25 \Omega_M^2H_0^4 }k\mathcal{T}^2(\sqrt{2}k/k_\text{eq}).$$

He then goes on to calculate the maximum of the powerspectrum in equation 8.1.49 as

$$P_{max} = 7.2 \times 10^{13} (\Omega_M h^2)^{-1} |N|^2 \,\text{Mpc}^3.$$

However, as he notes, $C(\Omega_\Lambda / \Omega_M) = 0.767$ and the quantity $k\mathcal{T}^2(\sqrt{2}k/k_\text{eq})$ at maximum should be something like $0.74k_\text{eq}/\sqrt{2}=.005$.

Plugging in all these numbers I find something many orders of magnitude less. Where did all the powers of 10 come from in Weinberg's expression? I have tried tracing my units without any luck...

I assume Weinberg is right, since it provides a value of the power spectrum amplitude consistent with the $N$ values found by Plank and others.


1 Answer 1


The reason comes from the annoying habit of setting $c=1$ in equations. I don't know why so many physicists consider this convenient, I find it pointless and an endless source of confusion and mistakes. From the equation of the maximum value, it is clear that $P(k)$ should have dimensions of $[L]^{3}$. However, since $k$ has dimensions of $[L]^{-1}$ and $H_0$ has dimensions of $[T]^{-1}$, the equation of $P(k)$ appears to be of dimension $[L]^{-1}[T]^{4}$. So something is missing: a factor $c^4$. In other words, the "actual" equation is $$P(k)=c^4\frac{4(2\pi)^3 N^2 C^2(\Omega_\Lambda / \Omega_M)}{25 \Omega_M^2H_0^4 }k\mathcal{T}^2(\sqrt{2}k/k_{\text{eq}}).$$ On page 414, it is mentioned that $$ k\mathcal{T}^2(\sqrt{2}k/k_{\text{eq}}) = \frac{\Omega_Mh^2}{19.3}\kappa\mathcal{T}^2(\kappa)\;\text{Mpc}^{-1}, $$ and that $\kappa|\mathcal{T}(\kappa)|^2$ has a maximum value of 0.74. Also, $$ \begin{align} C(\Omega_\Lambda / \Omega_M) &= 0.767,\\ H_0 &= 100\,h\;\text{km}\,\text{s}^{-1}\,\text{Mpc}^{-1},\\ c &= 3\times 10^5\; \text{km}\,\text{s}^{-1}. \end{align} $$ Putting it all together, we get $$ P_\text{max} = \left(8.1\times 10^{21}\right)\left(\frac{4(2\pi)^3 (0.767)^2}{25\times 10^8\,\Omega_M^2h^4}\right)\left(\frac{0.74\,\Omega_Mh^2}{19.3}\right)N^2\;\text{Mpc}^3, $$ and you'll find the correct answer.


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