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Assuming the spatially flat FRW metric for simplicity: $$ds^2=c^2dt^2-a(t)^2(dx^2+dy^2+dz^2)$$ where $t$ is cosmological time, $a(t)$ is the scaling factor and $x,y,z$ are co-moving spatial Cartesian co-ordinates.

Light freely propagating along the $x$-direction follows the null geodesic with spacetime interval $ds=0$. Therefore for small finite co-ordinate intervals $\Delta t$ and $\Delta x$ along the null geodesic we have the relationship: $$\frac{c\ \Delta t}{a(t)}=\Delta x\tag{1}$$ The co-moving interval $\Delta x$ in this expression must be constant as it is the distance between co-moving spatial co-ordinates. This implies that the cosmological time interval $\Delta t$ must scale like $a(t)$.

Now let us imagine a model of a clock that consists of a rigid ruler of fixed proper length $\Delta l$ with an optical fiber attached along it.

Let us assume that the rigid ruler is co-moving with the expansion of the Universe.

Let us suppose that each tick of the clock consists of a light pulse sent down the optic fiber (assuming refractive index 1 for simplicity) which takes a constant time interval $\Delta\tau$ given by: $$\Delta \tau = \frac{\Delta l}{c}\tag{2}$$ Now the co-moving interval $\Delta x$ in the null geodesic of freely propagating light, equation $(1)$, and the proper length of the rigid ruler $\Delta l$ are both constant. Without loss of generality let us arrange for them both to be equal in magnitude so that we have: $$\Delta x=\Delta l\tag{3}$$ By substituting equations $(2)$ and $(3)$ into the null geodesic equation $(1)$ we obtain the relationship: $$\frac{\Delta t}{a(t)}=\Delta \tau\tag{4}$$ Thus each tick of our clock $\Delta \tau$ does not measure an interval of cosmological time $\Delta t$ as one might expect but instead it measures an interval $\Delta t/a(t)$ which is in fact an interval of conformal time.

Is this reasoning correct?

John Eastmond
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  • NB: use \tag{1} in place of \ \ \ \ \ \ \ \ \ \ \ \ (1) for better numbering (because it's automatically right-adjusted). – Kyle Kanos May 24 '15 at 18:33
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    I was always under the impression, that clocks measure proper time. – Sebastian Riese May 24 '15 at 18:41
  • Ok so what's wrong with my reasoning? – John Eastmond May 24 '15 at 18:48
  • I understand that for a co-moving observer in a FRW Universe the proper time elapsed is the same as the cosmological time $t$. But my model of a co-moving clock, which seems perfectly reasonable, measures conformal time $\tau=\int dt/a(t)$ instead. – John Eastmond May 24 '15 at 18:59
  • For a clock to count as a clock, the events that constitute ticks must have a timelike separation. Your clock has a lightlike separation between ticks. You would need to put a mirror at one end and count each back-and-forth journey as a tick for this to be a true clock. – user27118 May 24 '15 at 19:08
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    A "clock" in general relativity is rather by axiom (see this question) something that measures the proper time along its worldline. If what you have measures something else, it's not a clock. – ACuriousMind May 24 '15 at 19:18
  • @ACuriousMind Locally speaking the construction in the question used will reproduce proper time. – Prathyush May 24 '15 at 19:20
  • John Eastmond: "a model of a clock that consists of a rigid ruler of fixed proper length" -- Does this mean specifically two ends which were and remained bilaterally chronometrically rigid to each other?, i.e. such that the ping durations of either end wrt. the other remained (separately) constant (but not necessarily equal to each other). "[...] co-moving spatial Cartesian co-ordinates." -- Physics is co-ordinate independent. – user12262 May 31 '15 at 07:11

2 Answers2

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Firstly, notice that the Frequency of light used to measure time will not remain constant. I wont use this but do take note.

Secondly notice that If the object is rigid, then and One end is fixed at x = 0, then the other end in the co-ordinate systems used will be at

$\Delta x = \frac{\Delta L}{a(t)}$

So Now light using equation 1 from your question we get

$\Delta t = a(t)*\frac{\Delta x}{c} = \frac{\Delta L}{c}$

So both clocks reproduce cosmological time as expected.

I hope I understood your arguments correctly. I think the flaw in the argument is you cannot allow for both the rod to be rigid and its ends to be co-moving at the same time.

Prathyush
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  • But in my model of a light clock the light pulse is sent down an optic fiber of fixed proper length $\Delta L$. The interval of time that it takes $\Delta \tau=\Delta L / c$ is not itself an interval of proper time $\Delta t$ because it does not propagate through space so it does not obey the null geodesic equation $c\Delta t/a(t)=\Delta x$. – John Eastmond May 24 '15 at 21:11
  • Ok so what is the co-ordinate length? – Prathyush May 24 '15 at 21:12
  • The co-ordinate length $\Delta x$ is an arbitrary constant co-moving co-ordinate length used in the null geodesic equation of freely propagating light that by definition has the same size as the fixed proper length $\Delta L$ of the clock. I use the equality of $\Delta x$ and $\Delta L$ to write down a relationship between an interval of proper time $\Delta t$ and an interval of time measured by the clock $\Delta \tau$. In my argument I do not need to consider the changing co-ordinate length of the fixed proper length clock itself. – John Eastmond May 24 '15 at 21:22
  • This is exactly where you are going wrong,Constant Comoving length is different from Constant Proper length. Proper length = Comoving length*a(t) – Prathyush May 24 '15 at 21:26
  • i know proper length = comoving length * a(t) – John Eastmond May 24 '15 at 21:28
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    You said "In my argument I do not need to consider the changing co-ordinate length of the fixed proper length clock itself." This is where you are going wrong you have to. – Prathyush May 24 '15 at 21:37
  • I assume that the light pulse used in the clock travels through an optic fiber so I do not have to use a FRW null geodesic equation for that light pulse. Therefore I don't need to consider the co-moving length of the clock itself. – John Eastmond May 24 '15 at 21:45
  • Oh I see, This is an unexpected twist. So exactly how does it travel ? I guess it is reasonable to surrender the expectation that it will reproduce cosmological time. – Prathyush May 24 '15 at 21:48
  • Hi @Prathyush I think I accept your answer now. A fixed proper length light clock will measure cosmological time $t$. – John Eastmond May 28 '15 at 12:33
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Do clocks measure conformal time?

No. I would venture to say that clocks don't literally measure time at all. A clock isn't some cosmic gas meter with time flowing through it. What clocks do is "clock up" some kind of regular cyclical local motion and show you a cumulative result that you call the time. A pendulum clock clocks up the swings of a pendulum. A quartz wristwatch clocks up the vibrations of a crystal. Your light clock clocks up how many times light moves back and forth, just like the parallel-mirror light clock. And in all cases, the clock rate is affected by macroscopic motion. Keep parallel-mirror light-clock A on Earth and send B on a fast out-and back trip, and the clock B reading is less than the clock A reading. Because the local motion is reduced by the macrosopic motion, because the maximum motion is c:

enter image description here Public domain image by Mdd4696

Imagine you're panning your gedanken telescope and watching that local motion in clock B: it's slower. The spacetime interval is the same for both clocks because the light-path lengths are the same. The ds=0 only tells you that the motion of light less the motion of light is zero. We talk about time dilation, but there is no actual thing call time being stretched, just rates of motion being reduced.

That's what happens with gravitational time dilation. See Einstein talking about the speed of light being spatially variable here. Nowadays we say the coordinate speed of light varies with gravitational potential, or spatial energy density. And when it comes to conformal time we're talking about 46.9 billion years, wherein "the particle horizon is equal to the conformal time η that has passed since the Big Bang, times the speed of light c". Only we liken the early universe to a black hole, and swap space and time such that the expanding universe is like pulling away from a black hole. Only at the event horizon the coordinate speed of light is zero. So we can reason that the speed of light isn't constant over space and time, so multiplying by c doesn't work, and at the time of the big bang your light-clock wasn't ticking. As for it being co-moving, let's simplify matters by making it stationary with respect to the universe. As the universe expands the clock starts ticking, slowly at first, then faster. It's motionless with respect to the CMB rest frame, and it now shows a clock reading of 13.8 billion years. Not 46.9 billion years. So it isn't showing conformal time.

What's it showing? The universe is flat now and always has been, which is why you use the flat FRW metric. See what Mortimer said on Physics Forums: proper time is the time as measured by a clock in motion or in a gravity field, while coordinate time is the time as measured by a stationary observer at infinity where there is no gravity. And there is no overall gravity in the universe. So your clock is showing coordinate time. Even though for all you know, it's been around for ever.

John Duffield
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