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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.

Thanks.

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  • $\begingroup$ Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual question over those just asking for a specific computation. Also please define the notation you use within your post so that people will know what exactly you're talking about. $\endgroup$
    – ACuriousMind
    Commented May 18, 2017 at 9:55

2 Answers 2

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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

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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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