Do clocks speed up in an expanding Universe?

[johne1618]

For simplicity I assume a flat radial FRW metric (with c=1):
<br /> ds^2 = -dt^2 + a^2(t)\ dr^2<br />
Now let us consider the path of a light ray, a null geodesic, with ds=0 so that we have:
<br /> dt = a(t)\ dr<br />
Now at the present time t_0 we define a(t_0)=1 so that we have:
<br /> dt_0 = dr<br />
The interval of radial co-ordinate dr does not depend on time so that it is the same in both equations.

We combine the two equations to eliminate dr giving:
<br /> dt = a(t)\ dt_0<br />
If dt is always a fixed interval of cosmological time t then from our perspective at the present time t_0 it is represented by the time interval dt_0 given by:
<br /> dt_0 = \frac{dt}{a(t)}<br />
Thus one second measured in the future at time t is equivalent to 1/a(t) seconds of our present time t_0.

Therefore in an expanding Universe clocks speed up from the perspective of our present time.

Is this effect real or apparent?

Are there any observations that could decide between the two?

The conventional view is that this effect is only apparent and is the cause of the cosmological redshift.

 

[phsopher]

I don't quite follow. Why are we considering a null-geodesic? A comoving observer will measure the proper time, that is, the "proper distance" in the time direction:

$$\mathrm{d}\tau \equiv \sqrt{-g_{00}}\mathrm{d}t \qquad (\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu})$$
Since for RW metric $g_{00} = -1$ the clocks always run at the same rate. If we had a metric for which $g_{00} = -b(t)$ is some function of t then clocks would run at different rates at different times. Compare for example to the Schwarchild metric where the time component of the metric depends on the distance from the center and so clocks run differently for different observers.

 

[johne1618]

I don't quite follow. Why are we considering a null-geodesic? A comoving observer will measure the proper time, that is, the "proper distance" in the time direction

I use a null-geodesic in order to relate time intervals measured by observers at different times.

As the observers are comoving then their proper times \tau are indeed the same as their cosmological times t. However I want to know how the rates of their proper/cosmological times compare with each other.

I think the situation is analogous to the case of special relativity. Imagine that two observers in different inertial frames observe the same light beam. Since the speed of the light remains the same in each frame, the observers' time and distance measures must change.

 

[phsopher]

OK, I guess you are thinking of an expanding clock. So a light ray travels a fixed comoving distance and the time it takes for it to do that is used as a time standard. Do I understand you correctly? In that case I would agree that by that clock the rates would be different. However, most clocks we use are not like that, they are held together by EM forces and so do not expand with the universe. I think by a conventional clock the rates would be the same.

 

[johne1618]

OK, I guess you are thinking of an expanding clock. So a light ray travels a fixed comoving distance and the time it takes for it to do that is used as a time standard. Do I understand you correctly? In that case I would agree that by that clock the rates would be different. However, most clocks we use are not like that, they are held together by EM forces and so do not expand with the universe. I think by a conventional clock the rates would be the same.

No - the time is the same time as measured by standard rigid clocks.

The analysis I use above is a short-hand version of the argument normally used to derive the cosmological redshift.

 

[phsopher]

To me your analysis seems very much to be saying that it will take longer for a light ray to travel a fixed comoving distance in the future than now. That is of course true, but only has relevance to the rate at which clocks run if one chooses to identify the traveling of a comoving distance by light as a clock.

The derivation of the cosmological redshift is indeed the consequence of the fact that it will take light more time to travel a fixed comoving distance at later times (e.g. for two consecutive wave crests). The time measured by the observer between two crests is longer than by the emitter is because the second crest had a longer way to travel because the universe has expanded. To try to bring this back to clocks, for a "rigid clock" both crests would have the same distance to travel, hence no time difference.

But if I'm wrong, perhaps someone else will have better insight.

 

[Mordred]

I'm heading out so don't have the time to explain however this site has the same essential calculation. You need to look at the difference between conformal time and proper time.

http://physics.stackexchange.com/questions/67898/conformal-time-or-cosmological-time

sorry on the run

 

[johne1618]

I'm heading out so don't have the time to explain however this site has the same essential calculation. You need to look at the difference between conformal time and proper time.

http://physics.stackexchange.com/questions/67898/conformal-time-or-cosmological-time

sorry on the run

I must admit this is a previous post of mine!

I think I describe my idea better in this thread.

 

[bapowell]

I think Lubos answered your question in that other thread. Why don't we start with why you think his answer is missing the mark?

 

[Mordred]

I agree with Bapowell the solution he has given is correct. perhaps if you see the relation of the first equation on how to derive Hubble law will help.

http://en.wikipedia.org/wiki/Scale_factor_(cosmology)

note the the section after the Hubble's law

 

[Mordred]

Therefore in an expanding Universe clocks speed up from the perspective of our present time.

the descriptive above is conformal time.

d=ax where d is the physical proper distance x is the comoving distance. with convention c=1
\eta is the conformal time.

d\eta=\frac{dx}{a}=\frac{ctd}{a}

gives particle horizon

\eta = \int_{0}^{t} \frac{dt&#039;}{a(t&#039;)}

http://en.wikipedia.org/wiki/Particle_horizon

edit forgot for flat geometry (euclidean)

 

[Mordred]

Just a side not question, do we even use conformal distance and conformal time anymore. At one time the particle horizon used to be called the comoving horizon ( I have some pretty old textbooks and article:P) This at one time used to be thought of as the region of causal connection, (in older articles).? The reason I ask in my newer textbooks no longer include it.

 

[johne1618]

I think Lubos answered your question in that other thread. Why don't we start with why you think his answer is missing the mark?

I think my argument I gave in the "conformal time or cosmological time?" post was wrong. Therefore Lubos's criticism refers to a false argument.

Contrary to what I stated in that post observers do not measure intervals of conformal time dt/a(t), that vary with time, instead they simply measure fixed intervals of cosmological time dt. I now agree with Lubos that all observers will measure the speed of light to be c.

Let me summarise my new improved argument as follows:

The equation that I derive from the null geodesic in this thread:
<br /> dt_0 = \frac{dt}{a(t)},<br />
gives the time interval dt_0, at the present time t_0, that corresponds to a fixed time interval dt at some future time t.

Thus from our perspective today clocks in the future speed up. I think this effect is real rather than simply apparent. It implies that the energy scale of all physical processes is increasing with the scale factor a(t). I think this means that the Planck mass is proportional to a(t).

 

[phsopher]

Let me summarise my new improved argument as follows:

The equation that I derive from the null geodesic in this thread:
<br /> dt_0 = \frac{dt}{a(t)},<br />
gives the time interval dt_0, at the present time t_0, that corresponds to a fixed time interval dt at some future time t.

For the case where light travels a fixed comoving (coordinate) distance.

I think the question you need to answer is why a fixed comoving (i.e. coordinate) distance traveled by light would have something to do with the rate of a clock.

In contrast, consider a rigid clock comprised of two parallel mirrors with a photon bouncing between them. The time it takes for the photon to travel between the mirrors is taken as a time unit.

Since the clock is rigid, the proper distance between two mirrors is fixed, let's call it $d_M$. Suppose Alice, living today ($t_0$), observers one bounce of the photon. The photon travels on null-geodesics so it will tavel the proper distance $2d_M$ in cosmic time interval $\mathrm dt_{\mathrm{Alice}} = a(t_0)\mathrm dr_0 = 2d_M$, where $\mathrm dr_0$ is the comoving distance between the plates.

Suppose a million years pass and Bob finds this same clock in the future ($t_1$). Bob also observes one bounce of the photon which takes cosmic time $\mathrm dt_{\mathrm{Bob}} = a(t_1)\mathrm dr_1 = 2d_M$. The proper distance between the mirrors is fixed while the comoving (coordinate) distances at different times scale with the expansion of the universe, $$2d_M = a(t_1)\mathrm dr_1 = a(t_0)\mathrm dr_0 \quad \Leftrightarrow \quad \mathrm dr_1 = \frac{a(t_0)}{a(t_1)}\mathrm dr_0$$ Hence, the cosmic time corresponding to one bounce of the photon (one tick of the clock) is $\mathrm dt_{\mathrm{Bob}} = \mathrm dt_{\mathrm{Alice}}$ so the clocks run at the same rate.

 

[johne1618]

I think the question you need to answer is why a fixed comoving (i.e. coordinate) distance traveled by light would have something to do with the rate of a clock.

The light ray connects the time interval measured by Bob, dt_{\mathrm{Bob}}, with a corresponding time interval in Alice's present dt_0:
<br /> dt_0 = \frac{dt_{\mathrm{Bob}}}{a(t_{\mathrm{Bob}})}.<br />

 

[phsopher]

Let me put it this way: what is $dr$ for you and why do you think that it should be constant? Like I've been saying, all you have given is the different times it takes a light ray to travel a fixed coordinate distance (remember that coordinate distance is not a physical distance measured by observers). What exactly is the connection of this to clocks and how different observers measure time is entirely unclear (in my opinion there is none).

 

[Mordred]

Let me ask you a simple question, if conformal time was real and not simply a consequence of the coordinate system. Then how do you explain that when we measure a standard candle at some faraway location. Let's say in a galaxy near the CMB, we measure a type 1a supernova.
Type Ia supernovae have a characteristic light curve, their graph of luminosity as a function of time after the explosion. see the graph on this page

http://en.wikipedia.org/wiki/Type_Ia_supernova. regardless of how far away we measure the luminosity to time relation. The ratio of luminosity to time remains the same. Keep in mind the further we look the further back in time we look.

If time was running slower as your trying to prove, then interactions we observe would be running slower. (our rate of time would be faster) Keep in mind scientists do monitor reaction rates all the time in the past.

Why do they not see a time dilation?

Simple reason is there is no time dilation involved, conformal time is a mathematical construct of the coordinate system being used, nothing more. Conformal time and conformal distance has been known to scientists since the FLRW metric was developed. Do you honestly believe your the first one to misconstrue conformal time as a time dilation? Trust me your not its a well argued and well proven fact that conformal time does not = time dilation.

This is easily proven as any form of known reaction rate can serve as our rigid clock

 

[johne1618]

Let me ask you a simple question, if conformal time was real and not simply a consequence of the coordinate system. Then how do you explain that when we measure a standard candle at some faraway location. Let's say in a galaxy near the CMB, we measure a type 1a supernova.
Type Ia supernovae have a characteristic light curve, their graph of luminosity as a function of time after the explosion. see the graph on this page

http://en.wikipedia.org/wiki/Type_Ia_supernova. regardless of how far away we measure the luminosity to time relation. The ratio of luminosity to time remains the same. Keep in mind the further we look the further back in time we look.

If time was running slower as your trying to prove, then interactions we observe would be running slower. (our rate of time would be faster) Keep in mind scientists do monitor reaction rates all the time in the past.

Why do they not see a time dilation?

Simple reason is there is no time dilation involved, conformal time is a mathematical construct of the coordinate system being used, nothing more. Conformal time and conformal distance has been known to scientists since the FLRW metric was developed. Do you honestly believe your the first one to misconstrue conformal time as a time dilation? Trust me your not its a well argued and well proven fact that conformal time does not = time dilation.

This is easily proven as any form of known reaction rate can serve as our rigid clock

Ned Wright reports that time dilation in supernova light curves has been observed:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#TD

He thinks it's an "apparent" effect - I think it's a real effect.

 

[johne1618]

Let me put it this way: what is $dr$ for you and why do you think that it should be constant? Like I've been saying, all you have given is the different times it takes a light ray to travel a fixed coordinate distance (remember that coordinate distance is not a physical distance measured by observers). What exactly is the connection of this to clocks and how different observers measure time is entirely unclear (in my opinion there is none).

Ok - Here is my argument in terms of light clocks.

I start from the null geodesic equation relating an interval of cosmological time $dt$ to an interval of radial co-ordinate $dr$:
<br /> c\ dt = a(t)\ dr<br />
Now a rigid clock always has a proper length of $dr$ as it does not expand with the Universe.
The velocity of light is always $c$.
Therefore the time interval $d\tau$ measured by this rigid lightclock is:
<br /> d\tau = \frac{dr}{c}<br />
I substitute this expression into the null geodesic equation above to obtain:
<br /> d\tau = \frac{dt}{a(t)}<br />
Therefore lightclocks (in fact all types of physical clock) measure conformal time.

 

[Mordred]

Ned Wright reports that time dilation in supernova light curves has been observed:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#TD

He thinks it's an "apparent" effect - I think it's a real effect.

all those papers he referenced are 14+ years old lol. This is one of the reasons why you don't see conformal distances used anymore. that and those papers don't account for the effects of dark energy

 

[Mordred]

Ned Wright reports that time dilation in supernova light curves has been observed:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#TD

He thinks it's an "apparent" effect - I think it's a real effect.

if you won't listen to the professionial cosmologists when they tell you its Apparent. Then we certainly cannot convince you otherwise

 

[phsopher]

Ok - Here is my argument in terms of light clocks.

Now a rigid clock always has a proper length of $dr$ as it does not expand with the Universe.
The velocity of light is always $c$.

No, the proper (physical) distance is $\mathrm ds = a(t)dr$. This is what is constant for a rigid clock. $\mathrm dr$ is the coordinate distance.

 

[johne1618]

No, the proper (physical) distance is $\mathrm ds = a(t)dr$. This is what is constant for a rigid clock. $\mathrm dr$ is the coordinate distance.

At the present time, $t_0$, with $a(t_0)=1$, imagine a rigid rod adjacent to an equal length of space each with the same proper length $ds_0=a(t_0)dr=dr$.

Now at a later time, $t$, the space will have expanded to a proper length $ds=a(t)dr$ but the rigid rod will continue to have its original proper length $ds_0=dr$.

 

[Mordred]

For simplicity I assume a flat radial FRW metric (with c=1):
<br /> ds^2 = -dt^2 + a^2(t)\ dr^2<br />
Now let us consider the path of a light ray, a null geodesic, with ds=0 so that we have:
<br /> dt = a(t)\ dr<br />
Now at the present time t_0 we define a(t_0)=1 so that we have:
<br /> dt_0 = dr<br />
The interval of radial co-ordinate dr does not depend on time so that it is the same in both equations.

We combine the two equations to eliminate dr giving:
<br /> dt = a(t)\ dt_0<br />
If dt is always a fixed interval of cosmological time t then from our perspective at the present time t_0 it is represented by the time interval dt_0 given by:
<br /> dt_0 = \frac{dt}{a(t)}<br />
Thus one second measured in the future at time t is equivalent to 1/a(t) seconds of our present time t_0.

Therefore in an expanding Universe clocks speed up from the perspective of our present time.

Is this effect real or apparent?

Are there any observations that could decide between the two?

The conventional view is that this effect is only apparent and is the cause of the cosmological redshift.

Ok I am done hinting...you obviously aren't catching the hints. This above does not work as it does not work beyond the particle horizon. Nor does it consider the cosmological constant

the correct formula you should be using is

d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]

in the notation from Barbera Ryden's "Introductory to Cosmology"

the particle horizon used to be the comoving horizon or the region of causality. Now our region of causality is the cosmic event horizon. Thanks to the cosmological constant

http://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

edit noticed the previous article was copyrighted so I removed it, note none of the calculations this far, either posted or referenced include \Lambda. In particular conformal distance or time

 

[phsopher]

At the present time, $t_0$, with $a(t_0)=1$, imagine a rigid rod adjacent to an equal length of space each with the same proper length $ds_0=a(t_0)dr=dr$.

Now at a later time, $t$, the space will have expanded to a proper length $ds=a(t)dr$ but the rigid rod will continue to have its original proper length $ds_0=dr$.

The point is that the coordinate distance corresponding to the distance between the two ends of the rod will change. If one end of the rod is fixed at the origin and the other end is at coordinate distance $\mathrm dr_0$ from it today then the proper length of the rod is $\mathrm ds_0 = a(t_0)\mathrm dr_0 = \mathrm dr_0$ (for $a(t_0) = 1$). It will take light the time $\mathrm dt_0 = a(t_0)\mathrm dr_0 = \mathrm dr_0$ to travel from one end of the rod to the other.

At some later time $t_1$, the other end of the rod will be at coordinate distance $\mathrm dr_1$ from the origin (where the other end is fixed), and the proper length of the rod is $ds_1 = a(t_1)\mathrm dr_1$. It will take light a time interval $\mathrm dt_1 = a(t_1)\mathrm dr_1$ to travel from one end to the other.

Now we require that the rod is rigid, i.e., that its proper length remains constant ($\mathrm ds_0 = \mathrm ds_1$), from which it follows that

$$a(t_1)\mathrm dr_1 = a(t_0)\mathrm dr_0 = dr_0 \quad \Leftrightarrow \quad \mathrm dr_1 = a^{-1}(t_1) \mathrm dr_0.$$

Thus at time $t_1$ it will take light to travel from one end of the rod to the other a time interval

$$\mathrm dt_1 = a(t_1)\mathrm dr_1 = a(t_1) a^{-1}(t_1) \mathrm dr_0 = \mathrm dr_0 = \mathrm dt_0,$$

the same as at time $t_0$. Therefore if you build a clock which measures time by counting how many times a light ray travels the length of the rigid rod, for example, this clock will be always running at the same rate.

 

[johne1618]

The point is that the coordinate distance corresponding to the distance between the two ends of the rod will change.

I think that is true for a fixed proper distance of expanding space but not for a rigid rod.

Expanding space obeys the relation:
<br /> ds = a(t)\ dr<br />
whereas a rigid rod obeys the relation:
<br /> ds = dr.<br />
Well that's what I believe at the moment - I might be wrong!

 

[Mordred]

You don't seem to want to include the cosmological constant or realize that the metrics your using is outdated or rather usable only in specific circumstances. Anyways here is a solid article showing the correct metrics with the cosmological constant and it also mentions the problem of measurements beyond the Hubble sphere.

see section 11

Distance measures in cosmology
David W. Hogg
http://arxiv.org/pdf/astro-ph/9905116.pdf?origin=publication_detail
see equation 15 for the comoving distance
equation 28 for the comoving volume

comoving volume is proper volume*three factors of the relative scale factor

(1+z)^3

edit by the way AFIAK even with the metrics above, there is still an "apparent" time dilation, however I could be wrong on that.

 

[johne1618]

You don't seem to want to include the cosmological constant or realize that the metrics your using is outdated or rather usable only in specific circumstances.

As far as I'm aware the only assumption about the metric that I'm making is that space is flat ($k=0$) which is born out by current observations.

The cosmological constant does not enter into the metric and I'm only using the metric in my argument. I must admit that I am a bit skeptical about the cosmological constant as I tend towards believing in a linearly expanding universe. I know that's very heretical.

See Fulvio Melia's preprints on arxiv.org for a defense of a linear cosmology (he calls it $R_h = c\ t$ where $R_h$ is the Hubble radius):

http://arxiv.org/find/grp_physics/1/au:+melia_fulvio/0/1/0/all/0/1

 

[Mordred]

fair enough I can understand that, however we do observe far beyond Hubble radius. Measurements do agree on the accelerating expansion of that there is no doubt. For that reason this relation does not describe expansion with k=0 outside of the metrics your using. The Hubble radius defined by the equations your using only covers a portion of the observable universe.
<br /> ds = a(t)\ dr<br />
The comoving volume of the observable universe is roughly 1.3*10^3 Gpc^3. The Observable universe is 46 Gly in radius roughly, R_h=ct would only give you 13.789 not sure I follow how Fulvio Melia accounts for larger volumes. I'll stick to LCDM

 

[phsopher]

I think that is true for a fixed proper distance of expanding space but not for a rigid rod.

Expanding space obeys the relation:
<br /> ds = a(t)\ dr<br />
whereas a rigid rod obeys the relation:
<br /> ds = dr.<br />
Well that's what I believe at the moment - I might be wrong!

A rigid rod measures a fixed proper distance in space so they are one and the same. You can't have $\mathrm ds = \mathrm dr$ continuously, something must change. If you have two points in space that are comoving, that is, their coordinate separation $\mathrm dr$ stays constant, then the physical distance between them will change as $\mathrm ds = a(t) \mathrm dr$. This is the case for example for vastly separated galaxies (though for large distances you would actually need to integrate over the scale factor to get the propagation of light instead of just considering $\mathrm dt$ like we have).

If on the other hand the physical distance is constant between two points in space, as would be the case for the two ends of a rigid rod, then the comoving (coordinate) separation between these two points can't possibly be the same as what it was in the beginning. I have already shown this in the calculation above.

 

[johne1618]

If on the other hand the physical distance is constant between two points in space, as would be the case for the two ends of a rigid rod, then the comoving (coordinate) separation between these two points can't possibly be the same as what it was in the beginning.

Ok - I accept your point that the co-ordinate separation between the ends of a rod with a fixed proper length must change. But I'm not using the rod's changing co-ordinate separation in my argument - I only use the rod's constant proper length.

At time $t_0$ I start with a rigid rod with proper length $l_0$ adjacent to an equal proper length of space $ds_0$.

As $a(t_0)=1$ we have:
<br /> ds_0 = a(t_0)\ dr_0 \\<br /> ds_0 = dr_0<br />
where $dr_0$ is the co-ordinate length of the space. Therefore at time $t_0$ we have:
<br /> ds_0 = dr_0 = l_0.<br />
Now at time $t$ the proper length of the space is given by:
<br /> ds = a(t)\ dr_0<br />
The rod is rigid so that its proper length is still $l_0$ (I don't care that its co-ordinate length has shrunk accordingly.)

The space is co-moving so that its co-ordinate length $dr_0$ has not changed.

Therefore at time $t$ we still have:
<br /> l_0 = dr_0<br />
and so at time $t$ we can assert:
<br /> ds = a(t)\ l_0.<br />
The cosmological time interval is given by $dt=a(t)dr_0/c=ds/c$ whereas the time interval measured by a rigid light clock, of length $l_0$, is given by $d\tau=l_0/c$ so that we have finally:
<br /> d\tau = \frac{dt}{a(t)}.<br />

 

[phsopher]

The cosmological time interval is given by $dt=a(t)dr_0/c=ds/c$ whereas the time interval measured by a rigid light clock, of length $l_0$, is given by $d\tau=l_0/c$ so that we have finally:
<br /> d\tau = \frac{dt}{a(t)}.<br />

The cosmological time interval of what?. What we want to consider is the time it takes for a light ray to travel the length of a rigid rod (that's the basis of our clock remember). You need the coordinate separation of the two ends of the rod at time $t$ in order to use the null-geodesic condition $\mathrm dt = a(t)\mathrm dr$. That thing you say you don't care about is precisely what you need in order to calculate the path of a light ray.

 

[johne1618]

The cosmological time interval of what?. What we want to consider is the time it takes for a light ray to travel the length of a rigid rod (that's the basis of our clock remember). You need the coordinate separation of the two ends of the rod at time $t$ in order to use the null-geodesic condition $\mathrm dt = a(t)\mathrm dr$. That thing you say you don't care about is precisely what you need in order to calculate the path of a light ray.

I am comparing a time interval, $dt$, measured by an expanding light clock with a time interval, $d\tau$, measured by a rigid light clock assuming that initially at time $t_0$ both light clocks have the same length.

Therefore we have:
<br /> dt = a(t)\ d\tau.<br />
As expected the time interval $dt$ measured by the expanding light clock scales with the scale factor $a(t)$.

I think the expanding light clock is the "natural" time keeper in a co-moving reference frame and measures intervals of cosmic time $t$.

One second measured by the expanding clock is equivalent to $1/a(t)$ seconds measured by the rigid clock.

Therefore from the perspective of rigid clocks and atomic systems using time $\tau$, that is our perspective, cosmic time $t$ is speeding up.

 

[phsopher]

Now I'm confused; when I asked you in the beginning of the thread if you were considering an expanding clock you answered no and said that the time would the same as that measured by rigid clocks.

Putting that aside, yes, the rate of an expanding clock will change; however, since the actual clocks we are using in reality do not expand with the universe, those will always tick at the same rate. The rigid clocks measure the cosmic time $t$ and continue to do so at the same rate.

 

[johne1618]

Now I'm confused; when I asked you in the beginning of the thread if you were considering an expanding clock you answered no and said that the time would the same as that measured by rigid clocks.

Putting that aside, yes, the rate of an expanding clock will change; however, since the actual clocks we are using in reality do not expand with the universe, those will always tick at the same rate. The rigid clocks measure the cosmic time $t$ and continue to do so at the same rate.

I admit that I've changed my mind.

I now think that a "natural" clock in co-moving co-ordinates is an expanding light clock just as a natural length measure is an expanding ruler. If those were the only measuring instruments we have then we would not notice that times/distances are increasing as the Universe expands.

However we have fixed rulers so that we can detect that co-moving distances are increasing. I would also say that we have fixed light clocks so that we can detect that time intervals are increasing. I think this implies that one second of cosmic time is equivalent to $1/a(t)$ seconds of fixed light clock time. As our time is fixed clock time then from our point of view one second of cosmic time is getting shorter and shorter.

 

[johne1618]

However, since the actual clocks we are using in reality do not expand with the universe, those will always tick at the same rate. The rigid clocks measure the cosmic time $t$ and continue to do so at the same rate.

Thinking about it again maybe you're right.

I've been barking up the wrong tree!

 

[johne1618]

Actually I now think there might be some truth in my argument: See my post about energy conservation in an expanding Universe:

https://www.physicsforums.com/showthread.php?p=4753126#post4753126

 

[Mordred]

conservation of entropy and the arrow of time, has been and still is a common argument for an evolving time. Or time reversal in some extreme cases. You didn't quite state that in the other post, however the similarity is there. The arrow of time argument in regards to the second law of thermodynamics has been around for as long as the concept of entropy was first introduced. To my knowledge though its never been universally accepted, nor proven. For one thing disorder to order does not mean time reverses, like the mathematics the model would have one believe. Entropy used as the measure of order.