Cosmological fluctuations (Weinberg's cosmology, p. 284)

[jouvelot]

Hi everyone,

I'm unable to understand how to derive Formula (6.3.11) in Weinberg's cosmology book. It's a relation between time-related derivation (d/dt) and RW-scale-factor-related derivation (d/dy, where y = a(t)/a_EQ, a(t) is the RW scale factor in the metric and the EQ subscript denotes the matter-radiation equality condition). The given formula is

d/dt = (H_EQ/√2)(√(1+y)/y)d/dy ,

and I don't see how one gets this... Any suggestion?

Thanks in advance.

Bye,

Pierre

 

[kimbyd]

Hi everyone,

I'm unable to understand how to derive Formula (6.3.11) in Weinberg's cosmology book. It's a relation between time-related derivation (d/dt) and RW-scale-factor-related derivation (d/dy, where y = a(t)/a_EQ, a(t) is the RW scale factor in the metric and the EQ subscript denotes the matter-radiation equality condition). The given formula is

d/dt = (H_EQ/√2)(√(1+y)/y)d/dy ,

and I don't see how one gets this... Any suggestion?

Thanks in advance.

Bye,

Pierre

Can you please use LaTeX to make your equations more readable? A link to the LaTeX guide is right below the reply box.

 

[jouvelot]

Sure, no problem: $$ \frac{d}{dt} = \frac{H_{EQ}}{\sqrt{2}}\frac{\sqrt{1+y}}{y}\frac{d}{dy},$$ with $$y = \frac{a(t)}{a_{EQ}(t)}.$$ $H$ is the Hubble "constant", of course: $H = (da(t)/dt)/a(t)$.

Thanks for any help that might come :)

Bye,

Pierre

 

[kimbyd]

Sure, no problem: $$ \frac{d}{dt} = \frac{H_{EQ}}{\sqrt{2}}\frac{\sqrt{1+y}}{y}\frac{d}{dy},$$ with $$y = \frac{a(t)}{a_{EQ}(t)}.$$ $H$ is the Hubble "constant", of course: $H = (da(t)/dt)/a(t)$.

Thanks for any help that might come :)

Bye,

Pierre

Thanks.

First, I don't think $a_{EQ}$ is a function of time. I think it's just a value.

That said, $d/dy$ should be derivable using the chain rule. Consider:

$${d \over dt} f(y(t)) = {d \over dy} f(y(t)) {dy(t) \over dt}$$

Thus,
$${d \over dy} = \left({dy(t) \over dt}\right)^{-1} {d \over dt}$$

But this doesn't really make sense with respect to the question at hand. The square root is just confusing, for one. Do you think you could post some more context from the text? Because I'm honestly not sure what's going on here.

 

[jouvelot]

Hello Kimbyd,

I typed too fast my translation to LaTeX: indeed, $a_{EQ}$ is a constant. Sorry for the confusion.

I did try to use the chain rule already, but to no avail; in particular, the square root in the numerator eludes me. There are no additional explanations in the text, although Weinberg usually provides good intermediate steps when/if needed.

I'll try to post some more context information in the coming days.

Thanks for your help.

Bye,

Pierre

 

[George Jones]

Any suggestion?

Using the chain rule and the Friedmann equation, I get everything except the $1/\sqrt{2}$, so I am close. Have to do some work. After this, I will look for the missing $1/\sqrt{2}$.

 

[jouvelot]

Hi George,

I also managed to get it, up to this $1/\sqrt{2}$ factor :)

I'm not sure this is relevant, but this relation is obtained in the case where one tries to model the perturbations to density, speed and so on outside the horizon and also using the adiabatic mode, in which all perturbations to matter (baryonic and dark) and radiation (photon and neutrino) are assumed equal. The presence of these 4 components (instead of just 2, matter and radiation) may explain this factor, depending on how $H_{EQ}$ is defined.

Bye,

Pierre

 

[kimbyd]

Hi George,

I also managed to get it, up to this $1/\sqrt{2}$ factor :)

I'm not sure this is relevant, but this relation is obtained in the case where one tries to model the perturbations to density, speed and so on outside the horizon and also using the adiabatic mode, in which all perturbations to matter (baryonic and dark) and radiation (photon and neutrino) are assumed equal. The presence of these 4 components (instead of just 2, matter and radiation) may explain this factor, depending on how $H_{EQ}$ is defined.

Bye,

Pierre

Look for errata for the version of the textbook that you're using online. The $\sqrt{2}$ might be in there.

 

[jouvelot]

Hi kimbyd,

I already checked, and there are no errors mentioned on that page in the errata list.

I'll try to make sense of this factor when reading the following pages, since this notion is used again. Also, since $\rho_{EQ}$ is the common density of matter and radiation where they are equal, there is a factor of 2 somewhere when adding these two densities at radiation-matter equality (and thus a $\sqrt2$ for the related H).

Thanks a lot for your support.

Bye,

Pierre

 

[kimbyd]

Hi kimbyd,

I already checked, and there are no errors mentioned on that page in the errata list.

I'll try to make sense of this factor when reading the following pages, since this notion is used again. Also, since $\rho_{EQ}$ is the common density of matter and radiation where they are equal, there is a factor of 2 somewhere when adding these two densities at radiation-matter equality (and thus a $\sqrt2$ for the related H).

Thanks a lot for your support.

Bye,

Pierre

Yes, I think you've got it! $H_{EQ}^2 = 8 \pi G / 3 (2 \rho_{EQ})$ after all.

 

[jouvelot]

Yes, indeed :)

Thanks.

Bye,

Pierre