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?

  • $\begingroup$ What are $\vec X$ and $\vec q$? $\endgroup$
    – Py-ser
    Apr 28, 2015 at 15:43
  • $\begingroup$ $\vec{X}$ 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. $\endgroup$
    – Spencer
    Apr 28, 2015 at 21:06
  • $\begingroup$ 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? $\endgroup$
    – Py-ser
    May 3, 2015 at 9:39
  • $\begingroup$ I've asked a duplicate of this question on math.se to see if it can be resolved there. math.stackexchange.com/questions/1271789/… $\endgroup$
    – Spencer
    May 7, 2015 at 18:16

1 Answer 1


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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