Is My Proper Time the Same as Cosmological or Conformal Time?

[johne1618]

As an observer who is simply traveling with the Hubble flow is my proper time the same as the cosmological time or is it equal to the conformal time?

 

[mfb]

If you are at rest relative to the CMB, otherwise you will get a different value. For typical velocities relative to the CMB (just the motion of galaxies, not relativistic spacecraft s), the difference is very small compared with the current timing uncertainties.

 

[johne1618]

If you are at rest relative to the CMB, otherwise you will get a different value. For typical velocities relative to the CMB (just the motion of galaxies, not relativistic spacecraft s), the difference is very small compared with the current timing uncertainties.

I think I am a free-falling co-moving observer rather than simply a co-moving observer.

My local spacetime should therefore be flat - in other words my local frame is inertial.

Starting with the FRW metric with cosmological time [itex]t[/itex] and co-moving spatial co-ordinates:

[itex] \large ds^2 = -dt^2 + a(t)^2 [ \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2)] [/itex]

I rewrite the FRW metric with conformal time [itex]\tau[/itex] and co-moving spatial co-ordinates:

[itex] \large ds^2 = a(t)^2(-d\tau^2 + \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2)) [/itex]

where an element of conformal time [itex]d\tau[/itex] is given by

[itex] \large d\tau = \frac{dt}{a(t)} [/itex]

The worldline of a radial lightbeam according to the re-written metric is given by

[itex] \large d\tau = \frac{dr}{\sqrt{1-kr^2}}[/itex]

For small [itex]r[/itex] this metric describes a locally flat spacetime in which light travels on diagonals on a spacetime diagram. This is consistent with the co-ordinate system of a free-falling observer with a local inertial frame.

Therefore I think my proper time is conformal time.