In Mo, Frank van den Bosch and Simon White's book, 'Galaxy Formation and evolution', for the Press Schrecter formalism, we introduce a filter/window function to smooth the density perturbation field. We assume the primordial perturbation is Gaussian, and thus is fully describable via its mean and variance:

$$\left\langle \delta(x,R)\right\rangle =0 $$

$$\sigma(R)^2:= \left\langle \delta(x,R)^2 \right\rangle = \frac{1}{2\pi^2}\int k^3 P(k) \tilde{W}(k,R)^2 d\ln k $$

We can apparently convert this using the mass $M(R)$ defined by $$M(R)=V(R)\int\rho(x')W(x-x',R)d^3x' $$

So that $\sigma^2(R)=\sigma^2(M)$. To prove then that, (6.40) in the text,

$$\sigma^2(M) = \left\langle \left(\frac{M(x,R)-\bar{M(R)}}{\bar{M}(R)}\right)^2\right\rangle=\sigma^2(R) $$

given $$\bar{M}(R)=\left\langle M(x,R) \right\rangle = \frac{1}{V}\int M(x,R) d^3 x $$

I'm a little unsure of how to reduce the above expression to $\sigma^2(M)$, and why it is even equivalent in the first place (given you start from a variation in the field as dependent on some length scale). They say in the text it follows from ergodicity, but it's not immediately clear to me why this is so.


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    – ACuriousMind
    Commented May 18, 2017 at 9:55

2 Answers 2


I think I got it

$$ M(x,R) = V(R)\int \rho(x')W(x-x',R)d^3 x' $$

So it's mean satisfies, taking it with respect to x, and using $\rho(x)=\bar{\rho}(1+\delta(x))$:

$$\bar{M}(R)=V(R)\int \left\langle \bar{\rho}(1+\delta(x')) \right\rangle_x W(x-x',R)d^3x' $$

But the mean of the overdensity is assumed gaussian so it vanishes. Hence

$$\bar{M}(R)=V(R)\int \bar{\rho}W(x-x',R)d^3 x' = \bar{\rho}V(R) $$

Thus $$\frac{M(x,R)-\bar{M}(R)}{\bar{M}(R)}=\int \frac{\bar{\rho}(1+\delta(x')-1)}{\bar{\rho}}W(x-x',R) d^3x'=\int\delta(x')W(x-x',R)d^3x'=:\delta_R(x,R) $$

Hence $\sigma^2(M)=\sigma^2(R)$.

Ergodicity is implicit because the density of states is actually equivalent to the integral over space


There's a connection between the size of the filter (which sets the length scale $R$) and the mass $M$. Imagine you have a uniform density field $\rho$, and a $W$-sharp filter of size $R$, then your second equation reduces to

$$ M(R) = \frac{4\pi}{3}R^3\rho $$

This establishes a connection between $R$ and $M$, and therefore

$$ \sigma(M) = \sigma(M(R)) = \sigma(R) $$


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