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Is the speed of light constant for a comoving observer? 
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#1
Oct3012, 06:53 AM

P: 373

Using a simplified (radial coord only, spatially flat) FRW metric with the usual coordinates of cosmological time [itex]t[/itex] and comoving radial distance [itex]r[/itex]:
[itex] ds^2 = c^2 dt^2 + a(t) dr^2 [/itex] we find the path a lightbeam takes by setting [itex]ds=0[/itex] to obtain [itex] \frac{dr}{dt} = \frac{c}{a(t)} [/itex] Therefore if a comoving observer measures the speed of light by measuring the time [itex]dt[/itex] light takes to travel a fixed length [itex]dr[/itex] 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 comoving observer experiences conformal time [itex]\tau[/itex] such that [itex] d\tau = \frac{dt}{a(t)} [/itex] so that when he measures the speed of light over a fixed length he gets [itex] \frac{dr}{d\tau} = c [/itex] so that he does not experience the speed of light decaying as the Universe gets older. 


#2
Nov212, 04:16 AM

P: 939

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
Nov212, 05:40 AM

P: 373

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. 


#4
Nov512, 08:08 AM

P: 669

Is the speed of light constant for a comoving observer?



#5
Nov512, 12:09 PM

P: 373




#6
Nov512, 01:19 PM

P: 669




#7
Nov612, 05:26 AM

P: 939




#8
Nov612, 10:46 AM

P: 669




#9
Nov712, 02:04 AM

P: 939

Suppose the stick is one lightyear 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
Nov712, 06:42 AM

P: 669




#11
Nov712, 08:40 AM

P: 373




#12
Nov712, 11:25 AM

P: 669




#13
Nov712, 01:01 PM

PF Gold
P: 184

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
Nov712, 01:29 PM

P: 669

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. 


#15
Nov712, 02:18 PM

P: 669

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.



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