# Is the speed of light constant for a co-moving observer?

1. Oct 30, 2012

### johne1618

Using a simplified (radial co-ord only, spatially flat) FRW metric with the usual co-ordinates of cosmological time $t$ and co-moving radial distance $r$:

$ds^2 = -c^2 dt^2 + a(t) dr^2$

we find the path a lightbeam takes by setting $ds=0$ to obtain

$\frac{dr}{dt} = \frac{c}{a(t)}$

Therefore if a co-moving observer measures the speed of light by measuring the time $dt$ light takes to travel a fixed length $dr$ then he will derive a light speed that changes as the Universe grows older.

Is that right?

I think the light speed measured by such an observer should not change with cosmological time.

I think a co-moving observer experiences conformal time $\tau$ such that

$d\tau = \frac{dt}{a(t)}$

so that when he measures the speed of light over a fixed length he gets

$\frac{dr}{d\tau} = c$

so that he does not experience the speed of light decaying as the Universe gets older.

Last edited: Oct 30, 2012
2. Nov 2, 2012

### clamtrox

So, what you are describing is the following: suppose you have two galaxies which are comoving with the expansion (this corresponds to a fixed Δr). You measure the time light takes to travel between the two galaxies. It takes longer than before. You conclude that the speed of light has decreased. However, you could equally well say that the universe is expanding and the distance between the two galaxies has increased.

Indeed, if the light propagation time between two points was the only thing you were interested in, then the two effects are indistinguishable. However, there are numerous different things happening (for example, objects further away look dimmer and smaller) which allow you to find out what's happening.

3. Nov 2, 2012

### johne1618

But what happens if I measure the time Δτ it takes light to travel along a yardstick with proper length Δs? This seems like a reasonable definition of a time interval in a comoving frame.

Thus

Δτ = Δs / c

The yardstick is not expanding with the Universe therefore its proper length Δs is always the same as its comoving length Δr.

Therefore we have

Δτ = Δr / c

According to the null geodesic from the FRW metric a cosmic time interval Δt is related to a comoving interval Δr by

Δt = a Δr / c

Therefore combining these two equations we find that the time interval Δτ for a lightbeam to travel a fixed yardstick is given by

Δτ = Δt / a

where Δt is an interval of cosmic time.

Thus it seems to me that time is speeding up for a comoving observer though he can't detect that fact from any distance/time measurement directly.

Last edited: Nov 2, 2012
4. Nov 5, 2012

### RUTA

No, dr is the coordinate difference, the distance is a(t)dr. See American Journal of Physics paper

5. Nov 5, 2012

### johne1618

But what happens if you are measuring the speed of light over a distance say on earth which doesn't expand with the Universal scale factor?

6. Nov 5, 2012

### RUTA

Since infinitesimal distance is a(t)dr, you plug in t = now and use that. If you do the measurement over a time span for which a(t) is not essentially a constant (dr is not infinitesimal, but very large), then you don't have a local measurement and you won't obtain c.

7. Nov 6, 2012

### clamtrox

If the length of the stick is fixed, then the coordinate distance between its endpoints decreases in the usual FRW coordinates. If I understand your argument correctly, that's the thing throwing you off here.

8. Nov 6, 2012

### RUTA

Careful, if the space expands during the measurement process, you will measure a speed less than c, i.e., the expanding space will drag the photon away. I think that's the point the OP was making. To get c, you have to assume no expansion involved in measurement process, i.e., small enough (flat) spacetime region. See Figs 1 & 2 of http:\\users.etown.edu\s\stuckeym\AJP1992a.pdf

Last edited: Nov 6, 2012
9. Nov 7, 2012

### clamtrox

But that can't be right...

Suppose the stick is one light-year long in its inertial frame, which is same in all points of the stick as it is a solid object. An observer standing on the stick will always see the photon moving passing by with the speed c wrt. the stick. Since any such observers share the same inertial frame, they all agree that the photon always moves with the speed of light wrt the stick. Therefore the time it takes is one year, according to all observers. How does expansion of the universe affect that?

10. Nov 7, 2012

### RUTA

If the observers are to maintain the same spatial distance between themselves, they have velocities wrt the comoving observers and do not occupy the same inertial frame. They are in fact accelerating, i.e., their worldlines are not geodesics. Consider the observer trying to maintain a constant spatial distance wrt the origin (you) at a distance R such that c = HR, i.e., that observer must be moving at the speed of light relative to comoving observers there. There was a good Am J Phys paper on this coordinate system some years ago, but I don't have the title or author, so I don't know that I can find it. I'll see.

11. Nov 7, 2012

### johne1618

But what happens if I measure the speed of light along a meter rule that I'm holding in my hand as I float in space as a co-moving observer? Surely each part of the meter rule occupies my co-moving frame? The speed of light that I measure should always be c regardless of the expansion of the Universe.

Last edited: Nov 7, 2012
12. Nov 7, 2012

### RUTA

Then v = a(t)dr/dt for some t = now is the speed you measure for any object moving radially (wrt coordinate system) and you will measure v = c for a photon. And all observers at your location will measure v = c for that photon, even if they are moving wrt you. However, that photon is not moving at c relative to observers elsewhere, due to cosmological expansion. The photon need only move at c locally per special relativity, as special relativity obtains locally in general relativity.

13. Nov 7, 2012

### BillSaltLake

If you're talking about a "global" comoving frame, then the 1 lyr ruler, which is rigid, will be shrinking by just over 1 mile/day. The photon will always take a year to travel the length of the ruler, even though the ruler will have "shrunk" ~430 miles WRT the global comoving frame during that year.

14. Nov 7, 2012

### RUTA

If a photon is emitted radially towards the origin at a distance d = Xcy (X light-years) from the origin, the time it takes for the photon to reach the origin will not be Xy (X years) because the expansion of the universe will "slow it down" as it tries to "move against the expanding space." See the derivation of equation 22 in this AJP paper. If you were somehow able to construct a giant ruler from the origin to r = d that did not participate in the expansion of the universe, then its length would not change (by supposition). The comoving coordinate separation r would change, but so would a(t) such that a(t)r would remain equal to the fixed length of the rigid ruler. That's the idea behind "Schwarzschild coordinates" in FRW cosmology. I'm still trying to find that paper. I want to say it was written by Ron Gautreaua, but I could be confusing that with a paper he did on making a coordinate system from free-falling observers in the Schwarzschild spacetime. Sorry, I'll let you know if I find it.

Anyway, the problem of trying to "measure" a speed for the photon as it moved from one end of this giant ruler to the other would then reduce to establishing a proper time. I assumed you were using comoving proper time in which case you would "measure" an average speed of light over the ruler that is less than c. The speed of the photon as it passed your immediate location would be c.

Last edited: Nov 7, 2012
15. Nov 7, 2012

### RUTA

Try this paper: American Journal of Physics -- December 1996 -- Volume 64, Issue 12, pp. 1457 "Cosmological Schwarzschild radii and Newtonian gravitational theory," Ronald Gautreau, Physics Department, New Jersey Institute of Technology, Newark, New Jersey 07102.