Time Measurement in Friedman Metric: Physically Possible?

In summary: It has an expanding component too. So the equation for proper time would be different in a Friedmann spacetime.What "both events" do you mean?The first event is the moment of the clock's placement, and the second event is the moment the clock is actually measuring.
  • #1
exmarine
241
11
TL;DR Summary
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?
 
Physics news on Phys.org
  • #2
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##).
 
  • Like
Likes exmarine
  • #3
exmarine said:
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.
 
  • Like
Likes vanhees71 and exmarine
  • #4
PeterDonis said:
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).
 
  • Like
Likes exmarine
  • #5
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.
 
  • #6
exmarine said:
It was about how a physical clock could occupy the spatial locations for both events as space expands.

What "both events" do you mean?

exmarine said:
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.
 

1. What is the Friedman metric and how does it relate to time measurement?

The Friedman metric is a mathematical model used to describe the expansion of the universe. It is based on the theory of general relativity and is used to measure the rate of change of distance between objects in space. Time measurement in the Friedman metric is closely related to the concept of cosmic time, which is the time measured by an observer at rest in the expanding universe.

2. How does the Friedman metric account for the effects of time dilation?

The Friedman metric takes into account the effects of time dilation, which is the difference in the passage of time between two objects moving at different speeds. This is due to the fact that the metric is based on the theory of general relativity, which describes how gravity affects the fabric of space-time. Time dilation is a result of the curvature of space-time caused by the presence of massive objects.

3. Is time measurement in the Friedman metric affected by the expansion of the universe?

Yes, time measurement in the Friedman metric is affected by the expansion of the universe. This is because the metric is specifically designed to measure the changing distances between objects in an expanding universe. As the universe expands, the distance between objects increases, and this is reflected in the time measurement of an observer at rest in the universe.

4. Can time be measured in the Friedman metric without reference to a specific point in space?

Yes, time can be measured in the Friedman metric without reference to a specific point in space. This is because the metric is based on the concept of cosmic time, which is the time measured by an observer at rest in the expanding universe. This means that time can be measured without any fixed reference point, as it is relative to the observer's position in the universe.

5. How does the Friedman metric account for the effects of gravitational time dilation?

The Friedman metric takes into account the effects of gravitational time dilation, which is the difference in the passage of time between two objects located at different gravitational potentials. This is due to the fact that the metric is based on the theory of general relativity, which describes how gravity affects the fabric of space-time. Objects in stronger gravitational fields experience time at a slower rate compared to objects in weaker gravitational fields, and this is accounted for in the Friedman metric.

Similar threads

  • Special and General Relativity
3
Replies
95
Views
4K
  • Special and General Relativity
Replies
17
Views
2K
  • Special and General Relativity
Replies
6
Views
178
  • Special and General Relativity
Replies
14
Views
583
  • Special and General Relativity
2
Replies
48
Views
3K
  • Special and General Relativity
Replies
10
Views
1K
  • Special and General Relativity
3
Replies
88
Views
3K
  • Special and General Relativity
Replies
9
Views
870
  • Special and General Relativity
Replies
11
Views
1K
  • Special and General Relativity
Replies
5
Views
845
Back
Top