Time Measurement in Friedman Metric: Physically Possible?

[exmarine]

TL;DR

If a proper time measuring clock goes along for the ride between events, then is such a clock physically possible as the scale factor changes / increases in the Friedman metric?

If a proper time measuring clock goes along for the ride between events, then is such a clock physically possible as the scale factor changes / increases in the Friedman metric? How could any clock have zero spatial changes for that situation?

 

[vanhees71]

Proper time is shown by any clock fixed as a (quasi) point like object that can be considered a "test particle", i.e., whose effect on the gravitational field (metric) can be neglected, i.e., a "point particle" moving through a given "background spacetime". Then the proper time shown by the clock simply is
$$\tau=\int_{\lambda_0}^{\lambda} \mathrm{d} \lambda' \sqrt{g_{\mu \nu}(x(\lambda')) \dot{x}^{\mu}(\lambda') \dot{x}^{\nu}(\lambda')},$$
where $\lambda$ is an arbitrary parameter of the world line of the particle. Note that proper time is only well-defined for time-like curves (and indeed any real-world object can move only along time-like trajectories) and that the expression is general, i.e., you can use any $\lambda$, and $\tau$ is independent of the choice of this arbitrary world-line parameter, because the integrand is a homogeneous function of degree 1 in $\dot{x}^{\mu}$ (where the dot stands for the derivative with respect to $\lambda$).

 

[PeterDonis]

If a proper time measuring clock goes along for the ride between events, then is such a clock physically possible as the scale factor changes / increases in the Friedman metric? How could any clock have zero spatial changes for that situation?

It seems like you are asking whether real clocks are affected by tidal gravity (that's what I think you mean by "zero spatial changes"), i.e., by geodesic deviation--the fact that geodesics converge or diverge, so any object held together by internal forces, which would have to include any real clock, will have to have its internal forces change along its worldline to compensate for geodesic deviation.

If this is what you are asking, the answer is that, while in principle any real object is affected by geodesic deviation, in practice it is perfectly possible to make real objects, including real clocks, that are small enough for the effects of geodesic deviation on them to be negligible in any spacetime geometry accessible to us.

 

[Orodruin]

It seems like you are asking whether real clocks are affected by tidal gravity

Let me put extra emphasis on ”seems like”. From the OP, it is not at all clear to me what OP is asking. Clarification would be desirable (or just a confirmation that Peter’s interpretation was the intended one).

 

[exmarine]

Thanks for the responses. My question was not about the tidal effects on a physical clock.

It was about how a physical clock could occupy the spatial locations for both events as space expands.

And how do you get from this equation (which I think is the definition of proper time?) (and I can't seem to get the equation to display very well - sorry)

$${(cd\tau)^2}-{a^2(dx'=0)^2}-{a^2(dy'=0)^2}-{a^2(dz'=0)^2}=
{(cdt)^2}-{(adx)^2}-{(ady)^2}-{(adz)^2}$$

to the integral in the first response above? Or please point to some reference for me. I have a lot of textbooks by now, MTW, Zee, and others.

Thanks.

 

[PeterDonis]

It was about how a physical clock could occupy the spatial locations for both events as space expands.

What "both events" do you mean?

this equation (which I think is the definition of proper time?)

It's the equation for the spacetime interval in flat spacetime. But the Friedmann spacetime is not flat.