Due to some other threads going on I’m interested in improving my understanding of relativity as it works with rotating reference frames. Now I’m prepared to be told even my initial setup for discussion is flawed, but I’m hoping it isn’t.
Let’s say I build myself a large cylindrical spaceship/spacestation of length l and radius r and call it, just at random, Rama. I set up a sort of light clock along the central axis with lasers and reflectors. The time between each pulse is equal to the time it takes a pulse to traverse the cylinder and they’re synchronised, from a position on the cylinder wall, halfway from the circular ends, I see two flashes, the firing laser and the reflection of the incoming previous pulse, from the two ends arriving at the same time. This links the clock to the length of the cylinder and light speed. Change the length or change light speed and the synchronisation breaks.
Next I spin up the cylinder so I’m moving at some speed while I stand on the cylinder wall. For convenience I could make it so I experience 1G. Now one knee jerk, and I believe false, application of SR would be to say that I’m now moving at some speed, so time is slowed, and I observe the still synchronised pulses arriving at my location closer together, so light is either moving faster than light, or the cylinder is shorter.
But SR is about relative motion, and me and the cylinder center are not moving relative to each other, so I think using SR is wrong.
But I am accelerated which, I think, means GR says time slows down compared to inertial reference frames, so I am observing the “faster than light” change in my giant light clock. My understanding though is that there are two sensible ways to look at this situation. Either I ignore all the signs that I’m in a rotating cylinder and say that I’m in a gravity well and that the light lock is obviously not as deep in that well and should be ticking faster, or I say that I am in a rotating cylinder and this is just the gravity-like behaviour I should expect.
Am I horribly off on any points in this description?
I’m confused by the parts in bold, since they seem contradictory. Where is the laser, on the axis or on the surface of the cylinder? Are you standing (or floating) next to the laser?
Since there’s no gravity, you can use SR as long as you’re in an inertial frame.
Yeah, that should have been divided into a couple of sentences, here’s a new and improved version:
Let’s say I build myself a large cylindrical spaceship/spacestation and call it, just at random, Rama. I set up a sort of light clock along the central axis with lasers and reflectors. The time between each pulse is equal to the time it takes a pulse to traverse the cylinder and they’re synchronised so that, from a position on the cylinder wall, halfway from the circular ends, I’ll see two flashes from each end, scatter from the firing laser and the reflection of the incoming previous pulse. With light from both ends arriving at my position at the same time. This links the clock to the length of the cylinder and light speed. Change the length or change light speed and the synchronisation breaks.
I’m not in an inertial frame, I’m on the inner wall of the rotating cylinder.
You can still have an observer in an inertial frame. From the point of view of someone stationary WRT Rama, you’re moving, and you will experience time dilation. So that person knows that you will observe the pulses a little closer together than before the station was spinning.
If you want to use your frame of reference, on the inner wall of the rotating cylinder, then yes, you are lower in a gravitational well than the laser/clock, so you will expect the clock to be running a little fast. Note that it’s how far down the well you are that matters, not what force it is that you are feeling.
I’m not sure what you mean by ‘the “faster than light” change in my giant light clock’, though.
I’m also not sure why you need the laser pulses exactly timed to the length of the cylinder, but if that is important, you might need to take into account that the light from the laser to the cylinder end and then to you is travelling a distance sqrt(L^2 + r^2), not just L. For the “reflection from the previous pulse”, I’m not sure what that path is.
Think of it this way: whilst the person orbiting the central axis of the cylinder is not an inertial observer, at any given instant of their time there is always an inertial observer who is at rest relative to them and their instantaneous observations must match those of these inertial observers who we will call ICIOs (instantaneously co-moving inertial observers). Similarly instantaneous observations of the spinning observer must match what would be observed of their ICIO (though we must be careful how we handle any delay in signal propagation).
From the point of view of an inertial observer at rest at the centre of the central axis of the cylinder, this family of ICIOs have two universal properties:
[ol]
[li]whilst they have different velocities, the absolute value of their velocities is the same.[/li]
[li]they coincide with the observer on the cylinder at their point of closest approach to the central observer[/li][/ol]
Now imagine then the observer spinning around with the cylinder carries a light source around with them. From the point of view of the observer in the centre, the light emitted will always be emitted when the ICIO is at the closest point of approach and the absolute value of the velocity will be constant.
In special relativity, from the pov of the receiver, when light is emitted at the closest point of approach it will always be red-shifted by an amount that is purely a function of the absolute value of the velocity of the emitter at the time of emission. This is called the transverse Doppler effect. As the absolute value of the velocity is constant, the amount of red-shift seen by the central observer will be constant; and as frequency is simply the inverse of time, that the observer at the centre will see any clock measuring the time of the observer spinning around with the cylinder to be slowed by the same constant rate compared to their own clock.
At this point though it would be tempting to think that the the spinning observer also observes the central observer’s clock to be slowed. However this would clearly result in a paradox as both observers can agree on what is one revolution and they cannot both measure less than the other for the spinning observer to complete one revolution. In fact there is no paradox, the spinning observer actually sees the clock of the observer to run faster than they observe their own clock to run and by the same amount as which the observer in the centre sees the spinning observer’s clock slowed.
The reason for this can be seen if this time we have the observer at the centre with a light source. From the point of view of the ICIO when the light is received it will be received at the point of closest point of approach and the relative speed will be constant for all ICIOs. The transverse Doppler effect for light received at the closest point of approach says such light will always be blue-shifted by the same amount at which light emitted at the closest point of approach is red-shifted. Using the same arguments as before, the spinning observer must see the central observers clock to run at a rate faster than their own.
To tie your example closer to mine, f you like the frequency at of the light source of the central observer could be governed by your light clock.
I think the observation of the non-symmetry of only the rotating reference frame being time dilated is what’s throwing off a certain other poster, even if the setup is confusing and possibly more complicated. If I set up this experiment, not knowing about time dilation, wouldn’t one possible interpretation be that light had sped up?
I don’t know if I need it either, but it seems to me that makes the clock more tamper proof. As long as the two lasers are seen to be synchronized you can’t have a change in frequency without a change in either path length from end to end in the cylinder or speed of light.
The reflection from the previous pulse would also travel sqrt(L^2 + r^2), I’d draw a picture, but it doesn’t actually matter.