<p>The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question. </p>

<p>The $a$ is the occurring in the Fourier transform is the scale factor <em>which has a time dependence</em>. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,</p>

<p>$$
\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q} 
$$</p>

<p>$$
 = \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$</p>

<p>$$
 = \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$</p>

<p>$$
 = \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$</p>

<p>The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel. </p>

CC BY-SA 3.0
2015-05-08T19:36:33.667Z
2015-05-08T19:36:33.667Z
Spencer
<p>I was reading <strong>Appendix F</strong> of Steven Weingberg's book "Cosmology". In this Appendix he works out the perturbations to a cosmological fluid described by non-relativistic hydrodynamics and Newtonian gravity. </p>

<p>It turns out that the first order perturbations satisfy,</p>

<p>$$
\frac{\partial \delta \rho }{\partial t } + 3 H \delta \rho + H \vec{X} \cdot \nabla \delta \rho + \bar{\rho} \nabla \cdot \vec{v} = 0, \qquad \tag{1}
$$</p>

<p>$$
\frac{\partial \delta \vec{v}}{\partial t } +  H \vec{X} \cdot \nabla \delta \vec{v} + H \delta \vec{v} = - \nabla \delta \phi, \qquad \tag{2}
$$</p>

<p>$$
\nabla^2 \delta \phi = 4\pi G \delta \rho. \qquad \tag{3} 
$$</p>

<p>Weinberg applies the following Fourier transform to these equations,</p>

<p>$$ f(\vec{X},t)  = \int \exp \left(  \frac{i \vec{q} \cdot \vec{X}}{a} \right)  f_{\vec{q}}(t) \ \mathrm{d}^3\vec{q} $$,</p>

<p>where $f(\vec{X},t)$ is a place holder for $\delta \vec{v}, \delta \rho, $ and $\delta \phi$. </p>

<p>The resulting equations he gets are,</p>

<p>$$
\frac{\mathrm d \delta \rho_{\vec{q}}}{\mathrm d t } + 3 H \delta \rho_{\vec{q}} + \frac{i\bar{\rho}}{a}\ \vec{q} \cdot \delta \vec{v}_{\vec{q}} = 0 \qquad \tag{1'}$$</p>

<p>$$
\frac{\mathrm d \delta \vec{v}_{\vec{q}}}{\mathrm d t } +  H \delta \vec{v}_{\vec{q}} = -\frac{i}{a}\ \vec{q}  \delta \phi_{\vec{q}} \qquad \tag{2'}$$</p>

<p>$$
\vec{q}^2 \delta \phi_{\vec{q}} = -4\pi G a^2 \delta \rho_{\vec{q}} \qquad \tag{3'}$$.</p>

<p>For the most part these new equations can be obtained by making the substitution $\nabla \rightarrow i \vec{q}/a$. </p>

<hr>

<p><strong>My question</strong> : There doesn't seem to be any terms in the transformed equations which correspond to the terms $ H \vec{X} \cdot \nabla \delta \rho$ and $H \vec{X} \cdot \nabla \delta \vec{v}$. Weinberg makes no comment about their absence. Is anyone aware of a legitimate mathematical reason for these terms to disappear in the transformed equations?</p>

CC BY-SA 3.0
2015-04-28T15:43:59.947Z
What are $\vec X$ and  $\vec q$?
Py-ser
CC BY-SA 3.0
2015-04-28T21:06:50.187Z
$\vec{X}$ is just the ordinary position vector, for instance the components of $\nabla$  are derivatives with respect to the components of $\vec{X}$. Weinberg used a capital $X$ to distinguish from the comoving coordinates $\vec{x}= \vec{X}/a$. The vector  $\vec{q}$ is the frequency vector associated with the Fourier transform, i.e., the wave vector, which is more commonly denoted by $\vec{k}$. I see that I made a typo in the definition of the Fourier transform which has been fixed.
Spencer
CC BY-SA 3.0
2015-05-03T09:39:19.113Z
Uhm, I really can not see the reason... The only thing that come in my mind, is that the product of any operator nullifies. Perhaps, did you check that?
Py-ser
CC BY-SA 3.0
2015-05-07T18:16:03.340Z
I've asked a duplicate of this question on math.se to see if it can be resolved there. http://math.stackexchange.com/questions/1271789/missing-terms-in-weinbergs-treatment-of-perturbations-on-newtonian-cosmology
Spencer
<cosmology>
Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology
Spencer
# Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology

(Asked by: username_0 on 2015-04-26)

I was reading **Appendix F** of Steven Weingberg's book "Cosmology". In this Appendix he works out the perturbations to a cosmological fluid described by non-relativistic hydrodynamics and Newtonian gravity.

It turns out that the first order perturbations satisfy,

$$
\frac{\partial \delta \rho }{\partial t } + 3 H \delta \rho + H \vec{X} \cdot \nabla \delta \rho + \bar{\rho} \nabla \cdot \vec{v} = 0, \qquad \tag{1}
$$

$$
\frac{\partial \delta \vec{v}}{\partial t } + H \vec{X} \cdot \nabla \delta \vec{v} + H \delta \vec{v} = - \nabla \delta \phi, \qquad \tag{2}
$$

$$
\nabla^2 \delta \phi = 4\pi G \delta \rho. \qquad \tag{3}
$$

Weinberg applies the following Fourier transform to these equations,

$$ f(\vec{X},t) = \int \exp \left( \frac{i \vec{q} \cdot \vec{X}}{a} \right) f\_{\vec{q}}(t) \ \mathrm{d}^3\vec{q} $$,

where $f(\vec{X},t)$ is a place holder for $\delta \vec{v}, \delta \rho, $ and $\delta \phi$.

The resulting equations he gets are,

$$
\frac{\mathrm d \delta \rho\_{\vec{q}}}{\mathrm d t } + 3 H \delta \rho\_{\vec{q}} + \frac{i\bar{\rho}}{a}\ \vec{q} \cdot \delta \vec{v}\_{\vec{q}} = 0 \qquad \tag{1'}$$

$$
\frac{\mathrm d \delta \vec{v}\_{\vec{q}}}{\mathrm d t } + H \delta \vec{v}\_{\vec{q}} = -\frac{i}{a}\ \vec{q} \delta \phi\_{\vec{q}} \qquad \tag{2'}$$

$$
\vec{q}^2 \delta \phi\_{\vec{q}} = -4\pi G a^2 \delta \rho\_{\vec{q}} \qquad \tag{3'}$$.

For the most part these new equations can be obtained by making the substitution $\nabla \rightarrow i \vec{q}/a$.

---

**My question** : There doesn't seem to be any terms in the transformed equations which correspond to the terms $ H \vec{X} \cdot \nabla \delta \rho$ and $H \vec{X} \cdot \nabla \delta \vec{v}$. Weinberg makes no comment about their absence. Is anyone aware of a legitimate mathematical reason for these terms to disappear in the transformed equations?

**Tags:** cosmology

> **Comments:**
> **username_1**: What are $\vec X$ and  $\vec q$?  
> **username_0**: $\vec{X}$ is just the ordinary position vector, for instance the components of $\nabla$  are derivatives with respect to the components of $\vec{X}$. Weinberg used a capital $X$ to distinguish from the comoving coordinates $\vec{x}= \vec{X}/a$. The vector  $\vec{q}$ is the frequency vector associated with the Fourier transform, i.e., the wave vector, which is more commonly denoted by $\vec{k}$. I see that I made a typo in the definition of the Fourier transform which has been fixed.  
> **username_1**: Uhm, I really can not see the reason... The only thing that come in my mind, is that the product of any operator nullifies. Perhaps, did you check that?  
> **username_0**: I've asked a duplicate of this question on math.se to see if it can be resolved there. http://math.stackexchange.com/questions/1271789/missing-terms-in-weinbergs-treatment-of-perturbations-on-newtonian-cosmology  

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## 1 answers

### Answer by username_0 on 2015-05-08. Score: 0 | Accepted answer

The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question.

The $a$ is the occurring in the Fourier transform is the scale factor *which has a time dependence*. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,

$$
\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho\_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q}
$$

$$
= \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho\_{\vec{q}}(t) + \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho\_{\vec{q}}}{\mathrm{d} t } \ \mathrm{d}^3 \vec{q}
$$

$$
= \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho\_{\vec{q}}(t) + \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho\_{\vec{q}}}{\mathrm{d} t } \ \mathrm{d}^3 \vec{q}
$$

$$
= \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho\_{\vec{q}}(t) + \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho\_{\vec{q}}}{\mathrm{d} t } \ \mathrm{d}^3 \vec{q}
$$

The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel.


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