# Can the metric expansion of space be accounted purely by an expanding time?

I guess it's an either/or scenario, that is, either the metric expansion is purely along the spatial coordinates, or can be accounted only by the time component. Since $ds^2=a(t)(dx^2+dy^2+dz^2)-dt^2$, why are we implicitly assuming only the spatial components are affected by the metric expansion?

A friend suggested that the curvature cannot be absorbed by the time coordinate since it is impossible to come up with such a transform.

Since some people did not get my question; I might rephrase it: Is there an a priori reason why we choose co-moving coordinates in FLRW metric? Is it possible to construct a similar coordinate system where the metric expansion could be accounted by an expanding time?

Question was inspired from the answer of the third question in: (https://physics.stackexchange.com/q/24327)

Feel free to include as much math as you want. Thank you.

• I don't quite understand your question so this might not answer it. If in the line element that you wrote, the $g_{tt}$ component of the metric is some function of the time coordinate only, you can always re-scale the time coordinate and "absorb" that function in a new time coordinate... – PML Jul 21 '17 at 15:14
• Can you elaborate on your comment PML, I didn't quite get that. – Chetan Pandey Jul 21 '17 at 15:25
• In Bob Knighton's answer - and further comments - it is explained what I meant. If something is still troubling you or you don't understand something specific try to pin it down and I'm sure the PhySE community will be happy to help you. – PML Jul 21 '17 at 21:20

Just a little expansion of the good answer by @Bob Knighton and some of the comments. The assumption used comes from observations. That space is homogeneous and isotropic in the large. From that you have 6 Killing spacelike vector fields (translations and rotations in space, notice no Lorentz invariance), and you can mathematically and physically use the coordinates along 3 of the vectors as your spatial coordinate axis. That is what comoving was designed from. So FLRW is the most general spacetime in one covariantly defined (not observer at rest, observer in inertial coordinates, free floating) set of coordinates.

You have some freedom in the time coordinate as you saw in the answer to go to conformal time. And you can always use other spatial coordinates of value like spherical coordinates because the radius is then important and the physical value of it increases (but in comoving you keep than fixed, with a metric that depending on time that increases the radial physical distance by the scale factor (e.g., r is not physical distance radially).

But all of those coordinate systems are obtained from the symmetries and then some variations

I'm not really sure what you mean by "accounted for by time dilation," since time dilation is a result of a given spacetime metric, as opposed to a reason for it. However, I'll answer the second part of your question.

There is a transformation you can make that puts the metric in the form

$$\mathrm{d}s^2=a(\eta)^2\left(\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2-\mathrm{d}\eta^2\right),$$

where $\eta$ is a new time parameter given by

$$\eta(t)=\int^{t}\frac{\mathrm{d}t'}{a(t')}.$$

This new time parameter is known as conformal time. It is called this because, in this coordinate system, the metric takes the form $g_{\mu\nu}=\Omega(x)\eta_{\mu\nu}$, which is conformally related to the Minkowski metric. One useful fact about conformal time is that, in a spacetime diagram, light will travel in 45 degree lines, just like in flat space (since lightlike geodesics have $\mathrm{d}s^2=0$). This is great for making visualizations of say particle horizons or for motivating concepts like inflation.

I hope this helps!

• Sorry for a poor choice of words in the question, I will edit it. My question was that in FLRW metric, we assume that the metric expansion affects only the spatial coordinates, and not the time. Is there an a priori reason why we assume that, apart from the fact that it makes the equations easier to solve. – Chetan Pandey Jul 21 '17 at 15:23
• @ChetanPandey Do you mean why is the $dt^2$ coefficient $-1$ in $ds^2$? – J.G. Jul 21 '17 at 16:21
• No, not the metric signature. – Chetan Pandey Jul 21 '17 at 17:15
• The FLRW spacetime is the most general isotropic and homogeneous spacetime (assuming you add in a spatial curvature part) on a path connected spacetime. These coordinates are just a specific choice, and are convenient since $t$ becomes the time measured by a stationary observer. But as I pointed out, simply changing coordinates can give you a metric where the time component is hit (but now we have the caveat that $\eta$ isn't the time measured by stationary observers). – Bob Knighton Jul 21 '17 at 18:11