# Is Cosmological Time Dilation real?

1. Aug 2, 2013

### johne1618

The cosmological redshift can be understood in terms of time dilation.

In an expanding Universe light travels on a null-geodesic (ds=0) so that:
$$dr = \frac{c\ dt}{a(t)},$$
where $dr$ is an element of co-moving distance along its path, $dt$ is an element of time and $a(t)$ is the Universal scaling factor.

Thus if a photon is emitted from a co-moving galaxy it starts out with an element of co-moving distance $dr$ given by
$$dr = \frac{c \ \delta t_{em}}{a(t_{em})}$$

By the time we observe the photon an element of co-moving distance $dr$ is given by
$$dr = \frac{c \ \delta t_{ob}}{a(t_{ob})}$$
If we equate the two expressions for $dr$ we find an expression for an element of time now when we observe the photon, $\delta t_{ob}$, in terms of an element of time when the photon was emitted, $\delta t_{em}$ :
$$\delta t_{ob} = \frac{\delta t_{em}}{a(t_{em})}$$
where I take the current scale factor $a(t_{ob})=1$.

Thus if $a(t_{em})=1/2$ when the photon was emitted in the past then one second at time $t_{em}$ is equivalent to 2 seconds now at time $t_{ob}$.

Therefore my clock now is running twice as fast as the same clock at time $t_{em}$.

This interpretation seems at least as valid as the redshift interpretation. Instead of photons being somehow stretched by expanding space as they travel it seems that the passage of time itself is speeding up. I personally could only imagine photon wavelengths being stretched if one had standing waves in an expanding box.

Furthermore if all atomic frequencies are twice as high now as they were at time $t_{em}$ then surely all energies are twice as high now as at time $t_{em}$?

Thus the redshift that we observe when we observe photons emitted at time $t_{em}$ is due to our energy scale at time $t_{ob}$ being higher than the energy scale at time $t_{em}$.

Last edited: Aug 2, 2013
2. Aug 2, 2013

### PAllen

Generally speaking, there is nothing wrong with your line of reasoning. Redshift (which is invariant) can always be related to time dilation and energy (which are frame variant). Any time red shift is observed, a corresponding clock will be observed to be slow.

Where I would challenge you is that the direct observable here is the red shift. That is as real as anything gets in physics. Frame variant interpretations (time dilation, energy scale, stretching waves) have, at best, pedagogical value for certain purposes.

3. Aug 2, 2013

### Staff: Mentor

[Edit: corrected some things below, the original statements I made weren't strong enough.]

I'm not sure I agree.

This seems to me to be a key [STRIKE]missing piece[/STRIKE] error: [STRIKE]what justifies[/STRIKE] equating the two expressions for $dr$ is not justified. Since you are using the standard FRW chart, in which the time coordinate $t$ is the same as proper time for all "comoving" observers, and you are modeling both observers as "comoving" (since they are both at rest in the chart), [STRIKE]surely the obvious assumption[/STRIKE] the correct statement is that $\delta t_{ob} = \delta t_{em}$, and [STRIKE]that[/STRIKE] any change in the photon's observed wavelength is due to a change in $a(t)$ as it multiplies $dr$ -- in other words, it's a change in how much actual proper length corresponds to a given increment of the $r$ coordinate. The FRW metric also bears this out, since the scale factor $a(t)$ multiplies the spatial part of the metric; the time part of the metric is unchanged.

In other words, your statement that $dr$ is the "comoving distance" is not correct; the comoving distance is $a(t) dr$.

I also don't see any justification for this interpretation. First, there is the issue I stated above. Second, there is no global notion of "energy scale" in a non-stationary spacetime, which this is.

Last edited: Aug 2, 2013
4. Aug 2, 2013

### Staff: Mentor

As you can see from what I just posted, I would challenge even more than that; but I agree with what you say here about the redshift being a direct observable vs. frame-variant quantities.

5. Aug 2, 2013

### johne1618

The calculation I use is a short-hand for the standard derivation of cosmological redshift using two integrals to express the same co-moving distance between an emitting atom and an observer at times that are one period apart. i.e.
$$\large r = \int dr = \int^{t_{ob}}_{t_{em}} \frac{cdt}{a(t)} = \int^{t_{ob}+\delta t_{ob}}_{t_{em} + \delta t_{em}} \frac{cdt}{a(t)}$$
This leads to the same expression relating time intervals:
$$\frac{c \delta t_{em}}{a(t_{em})} = \frac{c \delta t_{ob}}{a(t_{ob})}$$
There is no stretching of wavelength due to space expansion in this derivation. The observed redshifted wavelength is determined by the dilated oscillation period.

6. Aug 2, 2013

### johne1618

My interest in the idea that cosmological time dilation implies an increase in energy scale stems from the fact that this might affect the mass/energy density of matter.

Normally one argues that the mass/energy density of matter goes like $1/a^3$ but if energy is increasing with the scale factor then the mass/energy of matter should go like $a/a^3=1/a^2$ instead. A matter dominated Universe would then scale linearly with time rather than the standard $t^{2/3}$ dependence.

7. Aug 2, 2013

### Staff: Mentor

Yes, there is, because $r$ is not a proper distance; it's a coordinate distance. Moving the $a(t)$ factor to the RHS of the integral doesn't make $r$ a proper distance. And the fact that $a(t)$ changes during the light's travel means that space does expand; the proper distance between the emission and observation points is larger when the light is observed than when it is emitted.

8. Aug 2, 2013

### Staff: Mentor

You appear to think that there is some wiggle room in this, but there isn't. The fact that the mass/energy density of matter goes like $1/a^3$ is not "argued" based on an interpretation that could be accepted or rejected; it's read directly off the stress-energy tensor. The SET component $T_{00}$, which is the mass/energy density of matter, goes like $1/a^3$, not $1/a^2$. There's no room for interpretation there.

9. Aug 2, 2013

### Staff: Mentor

Strictly speaking, it goes like $1/a^3$ for non-relativistic matter; it goes like $1/a^4$ for highly relativistic matter and radiation; and it is constant for dark energy.

10. Aug 3, 2013

### Staff: Mentor

On re-reading, I realized that I was wrong here; if this condition were true there would be no redshift!

There is a good discussion of cosmological redshift in this thread; in particular, this post by George Jones goes through the same derivation, but using hyperspherical coordinates ($\chi$ instead of $r$), which makes it clear that the coordinate interval $\chi$ is not a proper distance.

11. Aug 5, 2013

### johne1618

Apparently cosmological time dilation has been observed.

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

This is more than expansion of photon wavelength.

All my physical processes are running twice as fast as those same processes back at the time when the Universe was 1/2 the present size.

Last edited: Aug 5, 2013
12. Aug 5, 2013

### jartsa

A supernova that takes 20 days to decay will appear to take 40 days to decay when observed at redshift z=1.

A supernova that takes 20 days to decay will appear to take 40 days to decay when observed at relativistic Doppler redshift z=1, from a window of a spaceship that is escaping at approximate speed 0.5 c, relative to the exploding star.

Last edited: Aug 5, 2013
13. Aug 5, 2013

### Staff: Mentor

By this same logic, your physical processes are running twice as fast as those of an person that is receding from you at 0.866c (the relative velocity at which the time dilation factor is 2).

14. Aug 5, 2013

### Staff: Mentor

It's not 0.5c, it's 0.866c; that's the relative speed for which $\gamma = 1 / \sqrt{1 - v^2 / c^2}$ is 2.

15. Aug 5, 2013

### Staff: Mentor

Also, if this were true, there should be evidence that physical processes on Earth now are running twice as fast as physical processes on Earth a billion or so years ago (however long ago light from supernovas at z = 1 was emitted). I'm not aware of any evidence supporting that, and there's quite a bit of evidence against it.

16. Aug 5, 2013

### johne1618

In your example the two observers are indeed equivalent.

But in the expanding Universe example the emitter and the observer are not equivalent. The observer is in the emitter's causal future and not the other way round.

17. Aug 5, 2013

### Staff: Mentor

That is also true for you when you receive Doppler shifted light signals from someone moving away from you; the event of you receiving the signal is in the causal future of the event of the other person sending it. But you are not in the causal future of the other person "now".

Similarly, we on Earth, receiving light from a supernova with z = 1, are in the causal future of the event when the supernova emitted that light. But we are not in the causal future of the supernova "now".

(The point being that the relationship of being in the causal past or causal future is a relationship between events, not objects or observers.)

Also, we here on Earth now are in the causal future of the Earth a billion years ago (or whenever the light from supernovas at z = 1 was emitted), so my other point, that there should be evidence of physical processes on Earth now running twice as fast as a billion years ago, still stands even if we agree that the causal past/causal future relationship is relevant.

18. Aug 5, 2013

### johne1618

I think the only physical evidence would be the total number of oscillations of a physical system.

Last edited: Aug 5, 2013
19. Aug 5, 2013

### jartsa

Yes, but the z is 1 when the v is about 0.5 c. There is a 100 % frequency change, although there are almost no relativistic effects.

http://arxiv.org/abs/0808.1081v2
http://arxiv.org/abs/0707.0380v1

Now if we interpret cosmological redshift to be just kinematical redshifts, which is possible according to the two papers above, then no relativistic effects are involved.

20. Aug 5, 2013

### Staff: Mentor

The redshift formula is

$$1 + z = \sqrt{\frac{c + v}{c - v}}$$

where $v$ is the recession velocity. So $1 + z = 2$ when $c + v = 4 ( c - v )$ or $v = 3/5 c$. So it's somewhat larger than 0.5c. But you're correct that the key parameter is the observed redshift, not the time dilation factor; in my previous post I incorrectly used a time dilation factor of 2 to derive $v$, I should have used a redshift $1 + z$ of 2, as I did here.