# Length of ruler in expanding Universe

1. Jun 28, 2013

### johne1618

Imagine a standard ruler (made of atoms) at the present epoch.

Assume its comoving length is $dx=x_1 - x_2$ where $x_1$ and $x_2$ are the comoving coordinates of its ends at the present time.

As the scale factor $a=1$ then its proper length $ds=a \ dx$ is equal to its comoving length $dx$.

Now imagine that ruler persists to a later epoch with $a=2$.

As the ruler doesn't expand with the Universe is it correct to say that its proper length $ds$ is still equal to its comoving length $dx$ even though the space around it has expanded by a factor of two?

Last edited: Jun 28, 2013
2. Jun 28, 2013

### Naty1

An expanding universe depends on homogeneous and isotropic characteristics assumed in our cosmological model. The model doesn't apply at even at galactic size distances.

Some research papers discussed in these forums argue that the change in distance is infinitesimally small, much too small to be detected; but those are simplified models with different assumption and don't seem to match those of the cosmological model.

A practicing cosmology Wallace had this to say in a post:

Some may dissent from this view...we'll have to wait and see.

3. Jun 28, 2013

### Mordred

In regards to Naty's reply. The cosmological constant has an energy density of 6.0-10 joules per m3.
The strong nuclear force of the ruler easily has a higher energy density so can easily overpower expansion energy, in the same manner that gravity does with gravitationally bound objects.

This is often a confusing point, DE exists everywhere in a homogeneous and isotropic distribution, however the amount of expansion in a given region is a sum of energy density or pressure.

4. Jun 28, 2013

### johne1618

Does the FRW model apply for an astronaut floating in space far away from any planets, stars or galaxies?

If it does is the proper length of his ruler always equal to its comoving length?

5. Jun 28, 2013

### Naty1

That's the only place it really does apply....read Wallace's post which I quoted above.

6. Jun 28, 2013

### johne1618

Then is it correct to say that the proper length of the astronaut's ruler is always equal to its comoving length?

7. Jun 29, 2013

### Chalnoth

It's more correct to say that applying a comoving length to something that isn't expanding doesn't make any sense. It's just plain "length". Comoving length is for things that expand with the universe, such as the typical separation between galaxies that is used for the Baryon Acoustic Oscillation measurement (also a standard ruler measurement, though this one is a comoving ruler).

8. Jun 29, 2013

### johne1618

The reason for my question is that I want to investigate what happens if one uses a rigid ruler to define an interval of time (one's "second").

I want to define a unit of time $d\tau$ as the time it takes light to travel along a rigid ruler of proper length $ds = dx$. If we assume $c=1$ we have
$$d\tau = ds = dx$$
In an expanding metric light travels on a null geodesic given by
$$\frac{a dx}{dt} = 1$$
If I substitute $d\tau$ for $dx$ in the above equation I get
$$d\tau = \frac{dt}{a}$$
Thus if the astronaut uses a rigid ruler to define his time scale, which seems a reasonable thing to do, then his time will speed up relative to cosmological time as the Universe expands. His "second" will get shorter and shorter compared to ours now. This seems an interesting and surprising result to me.

Last edited: Jun 29, 2013
9. Jun 30, 2013

### timmdeeg

I don't think that the astronaut defines his time, he rather collects information regarding the expansion of the universe.

10. Jun 30, 2013

### johne1618

Well I think measuring time by timing light along a fixed length is a reasonable model of a clock.

The concept of a light clock has been extensively used in explanations of time dilation due to special relativity.

This is just another use of the concept but this time in a cosmological context.

11. Jun 30, 2013

### Naty1

John, you have progressed 'above my paygrade'.....!!

There are too many subtleties for me to be reasonably certain about a reply....

[My novice reaction is that ds =dx likely isn't the FLRW metric required in cosmology for
OUR universe.....]

This discussion might enable you to draw some conclusions about your approach:

Last edited: Jun 30, 2013
12. Jun 30, 2013

### Naty1

By chance I stumbled across some prior posts which may apply...they reflect my concern about 'ds'....

Hope this helps..it's funny that some questions elicit pages of expert discussion, others so little.

13. Jun 30, 2013

### Chalnoth

Sorry for the lack of response, been busy.

The first thing I'd point out is that this can't be correct, because it would imply that the speed of light changes over time in proportion to the scale factor. And we know that isn't the case.

14. Jun 30, 2013

### johne1618

But I would say that I am assuming that the speed of light is constant.

It is our definition of time that must change inversely with the scale factor in order to maintain the constancy of the speed of light.

15. Jun 30, 2013

### johne1618

I guess I'm saying that if we take a co-ordinate system such that the spatial interval is $dx$ then we should take the time interval as $d\tau=dt/a(t)$ to ensure that all future observers, using that co-ordinate system, measure the speed of light to be constant.
$$\frac{dx}{d\tau}=\frac{dx}{dt/a(t)}=\frac{a(t)\ dx}{dt}=c$$

Actually, contrary to what I have stated in earlier posts, I now recognize that what I'm describing is a cosmological time dilation effect analogous to the time dilation effects in special and general relativity. From our perspective, at the present time, clocks in the future will tick slower and slower as the Universe expands.

Last edited: Jul 1, 2013
16. Jul 1, 2013

### timmdeeg

Well, and the expansion of the universe is the cause of cosmological time dilation. This simply what the astronaut measures, nothing else. And this is the difference to flat spacetime, there you are right.

17. Jul 1, 2013

### Naty1

John, your a[dx]/t = 1 seems to set distance equal to time...the units seem wrong...

Check out how Leonard Susskind uses a[t] in this youtube lecture.....

He says the 'x' designations don't change, a[t] changes...and is a function of time....

This is because the 'x's designate comoving points.....points that expand according to the scale factor. They are not the usual axis coordinates.

Try the first 15 or 20 minutes....

Post any conclusions you reach....

Last edited by a moderator: Sep 25, 2014
18. Jul 1, 2013

### Naty1

What do you mean by this?

The purpose of co moving observers is to set some standard of time that does NOT change. these are the preferred observers with dipole symmetry in the CMB….at rest with respect to the CMBR. Such observers agree with one another on the amount of clock time since the Big Bang.

19. Jul 1, 2013

### Chalnoth

I figured it out. The issue is a misunderstanding of the coordinates. Take the FRW metric, ignoring two spatial dimensions and setting $c=1$:

$$ds^2 = dt^2 - a^2 dx^2$$

Consider, for a moment, what this means: if we have two objects that are at constant $x$ coordinate over time, when the universe expands from $a=1/2$ to $a=1$, the proper distance will have doubled.

So the distance across a rigid rod is not a constant distance in coordinate $x$, but a constant distance in $ax$: for a rigid rod as the universe expands, $a$ increases while the coordinate distance $\Delta x$ decreases, leaving the product $a\Delta x$ unchanged. Use this as your distance measure for a rigid rod, and you'll find that the light time travel distance across the rod does not change with expansion.

20. Jul 2, 2013

### timmdeeg

Perhaps I miss, what you want to say.

The cosmological time dilation, johne talked about, depends on a(t). The time the light travels along the rigid ruler yields the relative increase of a during this time.

There is no disagreement with the proper time of comoving observers.

21. Jul 2, 2013

### johne1618

Hmm - sounds plausible.

That might well be the answer.

Thanks.

Last edited: Jul 2, 2013
22. Jul 2, 2013

### Naty1

John..if you watch the Susskind video you'll see he explains the complementary expansion distance D around minute 8 of the video: D = a[t] (delta x)

I was not so clever as Chronos as to be able to make the simple translation from Susskind's description to yours....

a fun exercise...good luck....

23. Jul 2, 2013

### tiny-tim

hi johne1618!
i think it does

using the familiar balloon analogy:

the individual galaxies stars or atoms are not fixed to a particular point on the fabric of the expanding balloon … although they are forced to move with the surface, they are free to move along the surface, and their relative positions will depend on their mutual interactions, and the way those interactions decrease with distance

imagine a circle ABCDEF with six springs on it, three very weak ones initially of length 119° (AB CD and EF), and three very strong ones (in between) of length 1° (BC DE and FA) (total 360°)

expand the circle ten times: obviously, the lengths AC CE and EA will always be 120°

but the very strong springs will now be a lot less than 1° …

if we use them as our "rulers", then we measure the long springs as being (almost) ten times as long as before!

so yes, our one-metre rulers are expanding as the universe expands, but much more slowly (because gravity is a "spring" whose strength decreases with distance), and so we do measure an expansion of distances between galaxies

24. Jul 2, 2013

### Naty1

But that does not happen even at scales as large as galactic scales....

There are several extensive discussions on this issue in the forums.

25. Jul 2, 2013

### tiny-tim

ok, expand the circle 1.000000001 times …

the weak springs will still expand more, proportionately, than the strong springs