# Change of variables from FRW metric to Newtonian gauge

My question arises from a physics paper, where they state that if we take the FRW metric as follows, where $$t_c$$ and $$\vec{x}$$ are the FRW comoving coordinates:

$$ds^2=-dt_c^2+a^2(t_c)d\vec{x}_c^2$$

and perform the coordinate change given by:

$$\begin{cases}t_c=t-\dfrac{1}{2}H(t)\vec{x}^2\\ \vec{x}_c=\dfrac{\vec{x}}{a(t)}\bigg[1+\dfrac{1}{4}H^2(t)\vec{x}^2\bigg]\end{cases}$$

then the metric becomes, for $$|\vec{x}|\ll H^{-1}$$:

$$ds^2\simeq -[1-(\dot{H}+H^2)|\vec{x}|^2]dt^2+\bigg[1-\dfrac{1}{2}H^2|\vec{x}|^2\bigg]d\vec{x}^2$$

How should I proceed to reproduce this result? My attempt was to differentiate the expression of the change of variables, which leads to the following expression for $$dt_c$$:

$$dt_c=\dfrac{\partial t_c}{\partial t}dt+\dfrac{\partial t_c}{\partial\vec{x}}d\vec{x}=\bigg(1-\dfrac{1}{2}\vec{x}^2\dfrac{dH}{dt}\bigg)dt-H(t)\vec{x}d\vec{x}$$

I assume I would need to calculate the tensor product between $$dt_c$$ and $$dt_c$$ to get $$dt_c^2$$, which produces cross terms of the form $$dt_cd\vec{x}$$... I am unsure of what to do next.

Any advice on how to continue?

• I think your total differential method is a good start for the $dt_c$ part writing it as a differential of $t_c(d,\mathbf{x})$. Now do the same with the $\mathbf{x}_c = \mathbf{x}_c(t,\mathbf{x})$ differential. It looks like you will get a square root, and so use the expansion $\sqrt{1+ax}\simeq 1 + \frac{1}{2}ax + \mathcal{O}(x^2)$. I am guessing, based on some mental calculus, that your $d\mathbf{x}_c$ differential will be of the form $d\mathbf{x}\propto \sqrt{1-H^2|\mathbf{x}|^2}d\mathbf{x}_c$ or something like that. Hope that helps, I'll do a full answer if one isn't provided. Commented Jun 8 at 19:45
• $ds^2=-dt_c^2+a^2(t_c)d\vec{x}^2~$ should be $~ds^2=-dt_c^2+a^2(t_c)d\vec{x}_c^2~$ ?
– Eli
Commented Jun 9 at 15:12
• @Eli Yes, that's right, sorry about the typo! I'll correct it right away. Commented Jun 9 at 15:36

Firstly as you suggested,

$$d t_{c} = \frac{\partial t_{c}}{\partial t} d t+\frac{\partial t_{c}}{\partial \vec{x}} d \vec{x}$$

$$d t_{c} = d t-\frac{1}{2} \dot{H}(t) \vec{x}^{2} d t-H(t) \vec{x} \,d \vec{x}$$

and next,

$$d \vec{x}_{c} = \frac{\partial \vec{x}_{c}}{\partial t} d t+\frac{\partial \vec{x}_{c}}{\partial \vec{x}} d \vec{x}$$

$$d \vec{x}_{c} = - \vec{x}\frac{\dot{a}(t)}{a^{2}(t)} \left(1+\frac{1}{4} H^{2}(t) \vec{x}^{2}\right)d t + \frac{\vec{x}}{a(t)} \frac{1}{2} H(t) \dot{H}(t) \vec{x}^{2} dt + \frac{1}{a(t)}\left(1+\frac{1}{4} H^{2}(t) \vec{x}^{2}\right)d \vec{x} + \frac{\vec{x^2}}{2a} H^2 d\vec{x}$$

Now it is given that $$|\vec{x}| \ll H^{-1}$$, which implies that $$|\vec{x}|$$ can be treated as being small w.r.t. the characteristic length set by the Hubble parameter $$H$$. This allows us to make certain approximations. In particular, we may ignore higher-order terms in $$\vec{x}$$. This gives

$$d \vec{x}_{c} = - \vec{x}\frac{\dot{a}(t)}{a^{2}(t)} d t + \frac{d \vec{x}}{a(t)}$$

As OP suggested, Taylor expanding $$a(t_c)$$ along the lines $$f(x \pm h)=f(x) \pm h f^{\prime}(x)$$ we get $$a^2\left(t_{c}\right)= \left[ a(t-\frac{1}{2} H \vec{x}^{2}) \right]^2 \simeq \left[a(t)-\frac{1}{2} H \vec{x}^{2} \dot{a}(t) \right]^2 \simeq a^2(t) - H \vec{x}^{2} \dot{a}(t) a(t) \\ = a^2(t)[ 1 - H^2 \vec{x}^{2} ]$$

where in the last expression I have used $$H = \frac{\dot a}{a}$$. Thus

$$a^{2}(t_{c}) d \vec{x}_c^{2} = a^2(t)[ 1 - H^2 \vec{x}^{2} ] \left( - \vec{x}\frac{\dot{a}(t)}{a^{2}(t)} d t + \frac{d \vec{x}}{a(t)} \right)^2 = [ 1 - H^2 \vec{x}^{2} ] \left( - \vec{x} H d t + d \vec{x} \right)^2$$

where again we can use $$H = \frac{\dot a}{a}$$ to get

$$a^{2}(t_{c}) d \vec{x}_c^{2} \simeq [ 1 - H^2 \vec{x}^{2} ] d \vec{x}^2 -2 H \vec{x} dt d\vec{x} + \vec{x}^2H^2dt^2$$

$$dt_c^2 = \left( [d t - H(t) \vec{x} \,d \vec{x}] + \frac{1}{2} \dot{H}(t) \vec{x}^{2} d t \right)^2 \\ \simeq dt^2 + H^2 \vec{x}^2 d \vec{x}^2- 2 H \vec{x} dt d\vec{x} + \dot{H}\vec{x}^2 dt^2$$

Substituting these in the metric, we find that

$$d s^{2} \simeq-\left[1 + \left(\dot{H} - H^{2}\right)\vec{x}^{2}\right] d t^{2}+\left[1-2H^{2} \vec{x}^{2}\right] d \vec{x}^{2}$$

Although this looks similar to what you needed, I am right now unable to match the factor of "1/2" in front of the $$H^2$$ term in $$dx^2$$, and I also don't know why the negative signs do not match in the $$dt^2$$ part.

• Thank you! But when you compute the expression for $a^2(t_c)d\vec{x}_c^2$, you are cancelling $a^2(t_c)$ with $1/a^2(t)$, and I think this is not correct since they are evaluated at different times. Commented Jun 9 at 17:18
• I think we need to use that: $$a(t_c)=a\bigg(t-\frac{1}{2}H(t)\vec{x}^2\bigg)\simeq a(t)-\frac{1}{2}H(t)\vec{x}^2\dot{a}(t)$$ Commented Jun 9 at 17:48
• @WildFeather Thank you for catching this. I have modified my previous response. Also have left a note for you at the end of the answer.
– S.G
Commented Jun 9 at 22:10
• Thank you! This is a great starting point for me to keep working on getting the correct expression. You deserve the bounty! Have a great day! :) Commented Jun 14 at 22:05

I'll show how you can do this with Mathematica very quickly. If you don't have access to Mathematica through your institution I believe WolframCloud is free.

We're starting with the FRW metric $$ds^2 = - dt_c^2 + a^2 (t_c) d\vec x_c^2$$ And we have the following coordinate transform: $$t_c = t - \frac 1 2 H(t) \vec x^2$$ $$\vec x_c = \frac{\vec x}{a(t)}\left[1 + \frac 1 4 H^2(t) \vec x^2 \right]$$ Lets first implement these definitions in Mathematica with

tc = t - 1/2  H[t]  x^2;
xc = x/a[t] (1 + 1/4  H[t]^2  x^2);


From here we can very quickly compute $$dt_c$$ and $$dx_c$$ in terms of $$t,x,dt,dx$$. Specificlaly we want to compute $$dt_c = \frac{\partial t_c}{\partial t}dt + \frac{\partial t_c}{\partial \vec x}d\vec x$$ $$d\vec x_c = \frac{\partial \vec x_c}{\partial t}dt + \frac{\partial \vec x_c}{\partial \vec x}d\vec x$$ Which in code you'd implement as

dtc = D[tc, t] dt + D[tc, x] dx;
dxc = D[xc, t] dt + D[xc, x] dx;


From here we can immediately compute $$ds^2$$ in these new coordinates with

ds2=-dtc^2 + a[tc]^2 dxc^2;


However this will be very messy until we apply the approximation.

Now to implement the approximation $$|\vec x| \ll H^{-1}$$ we rearrange and define a expansion parameter $$\alpha$$ as $$\alpha \equiv |\vec x| H \ll 1.$$ We can rewrite our ds2 in terms of this expansion parameter by making the repalcement $$x\rightarrow \alpha / H(t)$$. In code this is done by (see here for explanation of syntax)

ds2 = ds2 /. {x -> \[Alpha]/H[t]};


Now we can use the Series function to Taylor expand painlessly

ds2Approxed = Series[ds2, {\[Alpha], 0, 2}];


And if you print out ds2Approxed you will find the result reported in the paper[1]: $$\boxed{ds^2= -[1-(H^2 + \dot H)|\vec x|^2]dt^2 + \left[1 - \frac 1 2 |\vec x|^2 H^2 \right]d\vec x^2+O(\alpha^3)}$$ The notebook without all this extra commentary can be found here. Apologies if you're looking for a derivation by hand but I hope this is useful.

[1]:If you just look at the raw output and are willing to do some trivial algebra (relative to what you'd have to if you did everything by hand) you get the result. However, you can make use a few more functions to make the result obvious:

Normal[ds2Approxed] /. {\[Alpha] -> x   H[t]};
Collect[%, {dt, dx}]

• Thank you very much! I was indeed looking for a derivation by hand, but I also have to learn to use Mathematica for my work (and have access to it through my university) so this is very useful. Thanks again! Commented Jun 14 at 22:07

$$\def \b {\mathbf}$$

from $$ds^2=-dt_c^2+a(t)\,dx_c^2$$

$$ds^2=\underbrace{\begin{bmatrix} dt_c & dx_c \\ \end{bmatrix}}_{\mathbf q_p^T} \underbrace{\begin{bmatrix} -1 & 0 \\ 0 & a(t)^2 \end{bmatrix}}_{\mathbf{G}} \underbrace{\begin{bmatrix} dt_c \\ dx_c \\ \end{bmatrix}}_{\mathbf q_p}$$ with the new coordinates
$$\mathbf{T}=\begin{bmatrix} t_c \\ x_c \\ \end{bmatrix}\quad, t_c=t-1/2\,H \left( t \right) {x}^{2}\quad, x_c={\frac {x \left( 1+1/4\, \left( H \left( t \right) \right) ^{2}{x}^{2 } \right) }{a \left( t \right) }} \tag 1$$

from here with

$$\b q_p=\b J\,\b q_t\quad,\b q_t= \begin{bmatrix} dt \\ dx \\ \end{bmatrix}\\ \b J=\frac{\partial \b T}{\partial \b q}~,\b q= \begin{bmatrix} t \\ x \\ \end{bmatrix}~,\text{~\b J~ is the Jacobian matrix}$$

thus the line element is now

$$ds_t^2=\b q_t^T\,\b J^T\,\b G\,\b J\,\b q_t$$

substitute $$~\dot a=H(t)\,a(t)~$$ and linearized you obtain

$$ds_t^2=\left(-1+\left(H^2+\dot H\right)\,x^2\right)dt^2+\left(1+\frac 12 H^2\,x^2\right)dx^2+O \left( {x}^{3} \right)$$

Notice

to obtain the result $$~a(t_c)\mapsto a(t)~$$ and I use MAPLE symbolic program