I am currently studying Newtonian Perturbation Theory in cosmology.

We have introduced the relation between the physical coordinates r and the comoving coordinates x in an expanding universe:

$\bf r$$=a(τ)$$\bf x$

The space and time derivatives of r and x are no longer independent:

$\nabla_\bf r$ $= a^{-1}$$\nabla_\bf x$


$\frac{\partial}{\partial t}|_{\bf r} = \frac{\partial }{\partial t}|_{\bf x} + (\frac{\partial} {\partial t}|_{\bf r}\bf x$)$\cdot \nabla_{\bf x}$

I don't seem to understand how we get the last relation (time derivatives). Why do we need to add the term $(\frac{\partial} {\partial t}|_{\bf r}\bf x$)$\cdot \nabla_{\bf x}$ ? I don't know where this comes from. I tried reading in textbooks (e.g Mo, van den Bosch & White) but I don't seem to find a more detailed calculation.

  • $\begingroup$ That looks like the chain rule $\endgroup$
    – jacob1729
    Commented May 30, 2019 at 14:27

1 Answer 1


Formally, this just comes from the vector transformation law when changing coordinate system $\mu \rightarrow \mu'$:

$$\frac{\partial}{\partial x^{\mu'}} = \frac{\partial x^{\mu}}{\partial x^{\mu'}} \frac{\partial}{\partial x^{\mu}}$$

In this case, we are switching from proper to comoving coordinates, so $t=t'$ and $\mathbb{x} = a(t)\mathbb{x}'$. From now on I will change the notation, which initially might be a bit confusing, and write '$\mathbb{x}$' as '$\mathbb{r}$' and '$\mathbb{x'}$' as '$\mathbb{x}$'. I apologize for what is admittedly a confusing switch in notation, but above I wanted to highlight the relation between our coordinate systems (primed and unprimed, or 'r' and 'x' coordinates) and draw connections to the general vector transformation law. In any case, the above law gives us:

$$ \frac{\partial}{\partial t_x} = \frac{\partial t_r}{\partial t_x} \frac{\partial}{\partial t_r} + \frac{\partial r}{\partial t_x} \frac{\partial}{\partial r} $$

where the subscripts on the t are just a reminder of what coordinate system the operator lives in, or alternatively, for the operator, which variable is being held fixed. From our change of variables, we can calculate $\partial t_r/\partial t_x = 1$ and $\partial r/\partial t_x = \dot{a}\mathbb{x}$. So the above expression becomes:

$$\frac{\partial}{\partial t_x} = \frac{\partial}{\partial t_r} + \dot{a}\mathbb{x} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial t_x} =\frac{\partial}{\partial t_r} + \frac{\dot{a}\mathbb{x}}{a} \frac{\partial}{\partial x} $$

I briefly note that the $\frac{\partial}{\partial r} =\frac{1}{a} \frac{\partial}{\partial x}$ relationship can be found in the same way as above. Now it is a simple rearranging of terms that yields: $$\frac{\partial}{\partial t_r} = \frac{\partial}{\partial t_x} - \frac{\dot{a}\mathbb{x}}{a} \frac{\partial}{\partial x} $$

and in more than one spatial dimension

$$\frac{\partial}{\partial t_r} = \frac{\partial}{\partial t_x} - \frac{\dot{a}\mathbb{x}}{a} \nabla_x $$

This can perhaps be shown a bit more directly by carrying out the same process, starting in the $\mathbb{x}$ space and going to the $\mathbb{r}$ space, but the result is the same either way. I initially misread your question which is why I chose to start as I did.

This is all a bit formal, so I can give a bit of an intuitive picture too. When we change coordinates, we must be careful to change operators as well. That is because:

$$ \left (\frac{\partial}{\partial t}\right)_\mathbb{r} \rho(\mathbb{r}, t) \neq \left (\frac{\partial}{\partial t}\right)_\mathbb{r} \rho(\mathbb{x}, t)$$

where the subscript on the operator indicates the coordinate being held fixed. This can be seen a bit more explicitly by writing $\mathbb{x} = \mathbb{r} / a(t)$. We can see now that there are two sources of time dependence that we must account for if we would like to find the total change of $\rho(\mathbb{x}, t)$ with time as $\mathbb{r}$ is held fixed. That is:

$$ \left (\frac{\partial}{\partial t}\right)_\mathbb{r} \rho(\mathbb{r}/a(t), t) = \text{change in } \rho(\mathbb{x}, t) \text{ due only to the second argument, first held fixed} + \text{change in } \rho(\mathbb{x}, t) \text{ due only to the first argument} $$

Or in symbols

$$ \left (\frac{\partial}{\partial t}\right)_\mathbb{r} \rho(\mathbb{r}/a(t), t) = \color{#DA5526}{ \left( \frac{\partial \rho(\mathbb{x},t)}{\partial t}\right)_{\mathbb{x}=\mathbb{r}/a(t)}} + \color{#FEBC38}{ \left( \frac{\partial x}{\partial t} \right)_\mathbb{r}} \color{#697F98}{ \frac{\partial \rho(\mathbb{x},t)}{\partial{x}}} $$

Forgive my use of color, but I want to highlight what each piece is saying here to drive home the point further. The $\color{#DA5526}{\text{copper}}$ bit collects the contribution from the second argument alone, by holding the first argument $\mathbb{x}$ constant. This is the change due to the explicit time dependence. The $\color{#FEBC38}{\text{gold}}$ and $\color{#697F98}{\text{silver}}$ portions together collect the change due to the fact that $\mathbb{x}$ has a time dependence and that when holding only $\mathbb{r}$ fixed we must still account for that dependence. The $\color{#697F98}{\text{silver}}$ term is how $\rho(\mathbb{x},t)$ changes with $\mathbb{x}$, and it is weighted by the $\color{#FEBC38}{\text{gold}}$ term, which is how $\mathbb{x}$ changes with time when $\mathbb{r}$ is held fixed.

  • $\begingroup$ Thank you very much for your answer! $\endgroup$
    – relaxon
    Commented Jun 1, 2019 at 8:07

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