[katieHaylor (OP)]

Alex has a thought experiment...

If we have:
a) a sufficiently long rod (of negligible mass) spinning in a complete vacuum with a speed that makes its ends' linear velocity be close to the speed of light,
b) a laser emitter on one end of the rod firing packets of photons towards the other end
and
c) a laser detector on the other end of the rod facing the emitter...

Will the photons be always received on the detector as if the rod was stationary? If so, does it mean that to an external observer the space curves so the photons reach the moving detector? Or if not, what would it mean to an internal observer (the one positioned on the rod)?

To kind-of summarise, does the motion of the measurement system cause space-time deformation for that system (or for the system of an external observer)?

What do you think?

[Colin2B]

You can visualise this by imagining a rod on a pivot rotating over a pond. If the end of the rod drops a pebble into the pond, the ripples will move towards the pivot, but by the time they reach the pivot the rod (and also the detector) will be pointing in the wrong direction.
Note that this is not an exact model of your set up because the source and detector you are describing are very directional, but the pond rippples are not. However, it does illustrate what is happening.

[evan_au]

Relativity says that the rod will look completely normal, regardless of its velocity, if you are in an "Inertial Frame of Reference", which you can regard as being "in microgravity" (like an astronaut in the ISS). Under these conditions, flashes of light will pass between the ends of the rod just as if they were in the ISS (or, for most purposes, sitting in your living room).

An observer on your spinning rod would not be in microgravity - in fact they would be quite crushed by the centrifugal force.

In imagining this scenario, I expect that an observer on the end of this rod would not see a straight rod, but a curved rod. And flashes of a laser beam would not hit the other end of the rod, because the other end of the rod would no longer be there by the time the light pulse arrived.

[jeffreyH]

Once you introduce angular momentum (via the spinning of the rod) you can no longer expect a linear relationship between the path of light and the path of the rod.

[ALp]

 
I intuitively feel the point on angular momentum is right, but then does it mean that motion itself creates "wobbles" in space-time? Or because different parts of the rod move with different speeds, each infinitely small part would "experience" space-time differently?
I vaguely remember a point about light bouncing up and down, zigzaging between mirrors on the floor and the ceiling of a train moving close to the speed of light. In that case, as far as I remember, both to internal and external observer, light would bounce the same way from every mirror, i.e. not miss even one ...So what would be happening here?

Further on that point, what if we spin not a rod but a disc of material in which speed of light is slow enough and the speed of rotation is high enough to achieve the "Coriolis-style" effect? Would we see the laser beam miss the sensor completely?

[Colin2B]

I intuitively feel the point on angular momentum is right, but then does it mean that motion itself creates "wobbles" in space-time? Or because different parts of the rod move with different speeds, each infinitely small part would "experience" space-time differently?

Each part of the rod experiences spacetime in the same way, however, observers at different points on the rod (and outside of the rod) will measure the motion of other parts of the rod differently.

I vaguely remember a point about light bouncing up and down, zigzaging between mirrors on the floor and the ceiling of a train moving close to the speed of light. In that case, as far as I remember, both to internal and external observer, light would bounce the same way from every mirror, i.e. not miss even one ...So what would be happening here?

Imagine instead of the light clock someone on the train is tossing a ball up and down. An observer on the train sees the ball just goes up and down. If you are standing on the platform and the player throws the ball up as the carriage passes one end of the platform and catches it as the carriage passes the other end, you will see the ball follow a curved path, not straight up or down, you measure the distance travelled by the ball as being much greater than the person on the train does.
This doesn't cause a problem because we can add the speed of the train to that of the ball and understand what is happening, but with light the measured speed is the same for all observers so the calculations of what we measure are more complicated and very different from what we experience with everyday objects.

Further on that point, what if we spin not a rod but a disc of material in which speed of light is slow enough and the speed of rotation is high enough to achieve the "Coriolis-style" effect? Would we see the laser beam miss the sensor completely?

As has been pointed out in previous posts the beam does miss the sensor completely, and you don't need light speed to achieve this, nor do you need to assume spacetime is distorted. You just need the time taken by the laser pulse to be long enough for the rod to have rotated slightly. This is in fact a Coriolis effect, an outside observer says the light travels in a straight line, whereas someone travelling on the rod says the path of the light is curved.

[ALp]

In this case then, if time-of-flight makes the photons miss the detector because the system is rotating... why did Michelson–Morley experiment failed? Earth was and is moving quite fast and rotating too ...they were looking for such a shift? I always thought light is not susceptible to Coriolis ...

[yor_on]

That's a very tricky question. It reminds me of the question of spinning a disk in where its perimeter (outer side sort of) will be spinning faster than the center. Will the disk crack? The rim will contract. We have neutron stars that we presume to be spinning close to light speed if I remember correctly so it should possibly be testable.

Nota bene, you could see a disk as a 'infinite assembly of rods' so it's not that farfetched to think of it that way.

Well, or a 'finite assembly' if you like :)

One more point, something spinning is defined as a acceleration,  by Newton as well as Einstein.

[yor_on]

And no, all uniform motion is equivalent being still. What you call motion needs something to define it from, and that is what we do driving a car for example, we define it relative the earth (ground). There are two forms of motion, uniform and accelerations. It's only in the later case you can prove it locally, you do 'feel' it. A uniform motion is not provable locally, only by comparing to some other 'object' that you then define to be 'still'.

[Colin2B]

In this case then, if time-of-flight makes the photons miss the detector because the system is rotating... why did Michelson–Morley experiment failed? Earth was and is moving quite fast and rotating too ...they were looking for such a shift?

No, that's not what they were looking for.
Even if they were, the lengths of the arms in their experiment are very short and light is very, very fast compared to rotation of earth.

I always thought light is not susceptible to Coriolis ...

It's not a question of light being susceptible to the effect, it is the apparent paths as observed by two different observers, one on the rotating system and one outside of it. What you observe is frame dependant just like centrifugal force.

[jeffreyH]

If you had a large, vertically oriented disk, like a Ferris wheel, you could do the following. Have sensors mounted around the spindle and as the disk rotates have photons fired from the edges. They should all hit the centre, even though the emitters are rotating. That is, if you get the aim right. Can this measure the one way speed of light? That is the question.

[jeffreyH]

BTW Colin the middle of the coaxial cable was a brilliant idea.

[ALp]

Hmm, I kind-of get the point... In Jeffrey's Ferris Wheel example the photons should all hit the center detectors because in the frame of reference of the Ferris Wheel, for both internal and external observer, the beams fire at a static point, which is why in my thought experiment I have placed the detectors on the other side of the wheel and not in the middle.
Now if I am thinking of that, to the internal observer in my experiment (frame of reference is the rod) the light would appear to bend slightly in one direction, then bend back and pass through the middle point and then continue bending and miss the sensor on the other end?

Also, Earth is moving around the sun in an elliptical orbit with apogee and perigee and with a mean speed of around 30km/s if I remember correctly. The point on the surface of the Earth at the equator is moving about 40000km in roughly 24 hours, so ±11 km/s. C is roughly 300000km/s, so our speed in the Sun's frame of reference would roughly be of C or 0.01% of C... If today's CCDs have pixel sizes of of 4x4 μM, what would be the length of the beam needed to see the 4μM shift on the detector?
And also ... hmm ... being in Earth frame of reference, how would we know light has shifted in the first place? ... Tough That would mean that in my original query someone who spends all of his life and his reality in the frame of reference of the spinning rod, would just point the beam so it hits the detector on the other end without realizing the beam is bent in the first place... HMMMM...

[David Cooper]

Hmm, I kind-of get the point... In Jeffrey's Ferris Wheel example the photons should all hit the center detectors because in the frame of reference of the Ferris Wheel, for both internal and external observer, the beams fire at a static point, which is why in my thought experiment I have placed the detectors on the other side of the wheel and not in the middle.

If you use lasers and point them at the centre from the edge, the light will miss the centre due to aberration, so you'd need to point all the lasers away from the centre in order to hit the centre, and you'd have to point them further away from the centre the faster you spin the wheel. To hit the a sensor directly opposite a laser (meaning directly opposite when the wheel isn't rotating, but trying to hit it once it is rotating), you'd have to aim the laser perhaps three times further away from the centre.

That would mean that in my original query someone who spends all of his life and his reality in the frame of reference of the spinning rod, would just point the beam so it hits the detector on the other end without realizing the beam is bent in the first place... HMMMM...

If you are not rotating with the rod but are co-moving with its centre of mass, it can appear to be straight (if you're looking at it from a long way off and perpendicular to the plane in which it's rotating). If you are moving with one end of the rod though, it will look warped and you will know that you are having to aim your laser away from the centre to hit it. Not only would you have to aim the laser away from where the centre is, you would see the centre to the opposite side of where the centre actually is, so even if you could cancel out the perceived acceleration force you would still be able to work out that you're in a rotating frame which behaves differently from a normal frame of reference.

[yor_on]

You can ignore all 'motion' except the postulate you make about the rod spinning, to keep it simple.

Although :)
In a empty universe, how would you prove it spinning?

[Colin2B]

Although
In a empty universe, how would you prove it spinning?

Centrifugal force. Accelerating frame, not inertial.

[jeffreyH]

Note my point about the Ferris wheel "That is, if you get the aim right." Maybe I should have been more precise. Anyway David has provided the required adjustment.