Question about rotating reference frames

[Amaterasu21]

Hello,

I have a few questions about rotation and relative motion.

1. Suppose we transport the proverbial spinning ice skater used to demonstrate conservation of angular momentum to beginning physics students to a universe with only her and two planets. She is now spinning in deep space. Presumably she can measure her rotation and angular momentum relative to the two planets. However, now suppose we can somehow remove these planets, leaving the ice skater spinning by herself... spinning relative to what? Does the concept of "rotation" mean anything when there's nothing around you to rotate relative to? However, let's suppose she was to draw in her arms and decrease her radius. What happens? What is her angular momentum?
2. Let's suppose a star is 10 light years away, and when I look at it, I spin round with a period of one second. From the star's reference frame, I'm spinning round once a second. No problem there. But from my reference frame, doesn't the star complete a 31.4...etc. light-year long circuit around me in one second? But wouldn't that mean the star is traveling almost a billion times faster than the speed of light? How can both my reference frame and the star's then be equally valid?

 

[Quantum Defect]

Hello,

I have a few questions about rotation and relative motion.

1. Suppose we transport the proverbial spinning ice skater used to demonstrate conservation of angular momentum to beginning physics students to a universe with only her and two planets. She is now spinning in deep space. Presumably she can measure her rotation and angular momentum relative to the two planets. However, now suppose we can somehow remove these planets, leaving the ice skater spinning by herself... spinning relative to what? Does the concept of "rotation" mean anything when there's nothing around you to rotate relative to? However, let's suppose she was to draw in her arms and decrease her radius. What happens? What is her angular momentum?
2. Let's suppose a star is 10 light years away, and when I look at it, I spin round with a period of one second. From the star's reference frame, I'm spinning round once a second. No problem there. But from my reference frame, doesn't the star complete a 31.4...etc. light-year long circuit around me in one second? But wouldn't that mean the star is traveling almost a billion times faster than the speed of light? How can both my reference frame and the star's then be equally valid?

I think you are confusing things. The issues with rotating reference frames come about when the observer does not realize that they are rotating. I.e. when we think that our position on the Earth is stationary. If we treat the rotating frame as stationary, we will see some new, unexpected forces -- coriolis, etc. We can tweak our models and insert these imaginary forces to get the physics right in our fictional non-rotating frames.

1. The skater is rotating in space -- a perfectly fine inertial frame. If she pulls in her arms, she will twirl faster, just as she does on the ice. Her angular momentum remains the same. She is very cold, however, since she is not dressed for deep space...

 

[Amaterasu21]

I think you are confusing things. The issues with rotating reference frames come about when the observer does not realize that they are rotating. I.e. when we think that our position on the Earth is stationary. If we treat the rotating frame as stationary, we will see some new, unexpected forces -- coriolis, etc. We can tweak our models and insert these imaginary forces to get the physics right in our fictional non-rotating frames.

1. The skater is rotating in space -- a perfectly fine inertial frame. If she pulls in her arms, she will twirl faster, just as she does on the ice. Her angular momentum remains the same. She is very cold, however, since she is not dressed for deep space...

I'm aware of concepts like centrifugal force and the Coriolis force, but surely no force - fictitious or not - could ever accelerate an object to a speed greater than c? If I treat my reference frame as stationary when I'm rotating, then how could any force be pushing this star 10 light years away to circle round me at 990 million times c?

Also, can you treat space as an inertial frame that you can move and rotate relative to? Doesn't that contradict the idea that there's no preferred/absolute reference frame in relativity?

 

[PeterDonis]

Does the concept of "rotation" mean anything when there's nothing around you to rotate relative to?

This is really a philosophy question, not a physics question. We can never actually run this experiment because we can't empty the universe of everything except the skater, so there's no way to investigate the question scientifically. You can assume a particular answer for purposes of modeling a scenario. See further comments below.

let's suppose she was to draw in her arms and decrease her radius. What happens? What is her angular momentum?

You can't answer this question without adopting a model, and a model implies that you've assumed an answer to the question of what the "rotation" of the skater means. In this particular case, you've assumed that there is a background spacetime with respect to which the skater is rotating. Under that assumption, you can easily show that the skater's angular velocity, relative to that same background spacetime, will increase when the skater pulls in her arms. But you have to make the assumption to do that.

wouldn't that mean the star is traveling almost a billion times faster than the speed of light?

Yes, but that's ok because in non-inertial frames the speed of light "speed limit" doesn't work the same way as it does in inertial frames; you have to reformulate it. If you consider a light ray emitted by the star in the direction of its motion relative to you, that light ray will travel even faster than the star does, so the star is still traveling slower than light; it's just that the speed of light itself, in a non-inertial frame, can be "faster than light", i.e., faster than the value $c$ that we obtain for the speed of light in an inertial frame.

 

[Amaterasu21]

Yes, but that's ok because in non-inertial frames the speed of light "speed limit" doesn't work the same way as it does in inertial frames; you have to reformulate it. If you consider a light ray emitted by the star in the direction of its motion relative to you, that light ray will travel even faster than the star does, so the star is still traveling slower than light; it's just that the speed of light itself, in a non-inertial frame, can be "faster than light", i.e., faster than the value $c$ that we obtain for the speed of light in an inertial frame.

Ahh, that's a little confusing. Isn't the speed of light measured at the same value by all observers?

If that doesn't apply to non-inertial reference frames, how can that ever be tested, since just about everything in the Universe is accelerating? My understanding is the most famous and well-publicised test of invariant light speed was the Michelson-Morley experiment and that was about Earth's motion around the Sun, which isn't at a constant velocity!

 

[A.T.]

If that doesn't apply to non-inertial reference frames, how can that ever be tested

Just like anything else in physics, which is just an idealization, that applies approximately in certain situations.

 

[Amaterasu21]

So just to clarify, the speed of light is the same for all observers in inertial reference frames, but doesn't have to be for accelerating observers?

I thought General Relativity extended the Principle of Relativity to accelerating observers, though?

 

[PeterDonis]

Isn't the speed of light measured at the same value by all observers?

Locally, yes--if you measure the speed of a light ray as it is passing you, you will always get $c$, regardless of where you are in spacetime and what your state of motion is.

But as soon as you start talking about the speed of light that is spatially separated from you, that no longer holds; you can make the "speed" of light distant from you assume any value you like by choosing appropriate coordinates.

I thought General Relativity extended the Principle of Relativity to accelerating observers, though?

It does. But the extended version of the principle of relativity only says what I said above--that local measurements of the speed of light will always give $c$. It doesn't say that the speed of light is always $c$ for non-local measurements.

 

[Amaterasu21]

Ahh, got it! Thank you everyone, I think my questions have been resolved. (I'm still not entirely comfortable with the skater rotating "relative to space/space-time", since it seems to imply an absolute reference frame. But PeterDonis seemed to imply that "there is an absolute space-time which the skater rotates relative to" is a modelling assumption I'd have to make for this scenario to mean anything?)

 

[PeterDonis]

"there is an absolute space-time which the skater rotates relative to" is a modelling assumption I'd have to make for this scenario to mean anything?

Exactly. Without that assumption (or something equivalent to it), you can't make any predictions at all about what will happen, which basically means the scenario is meaningless.

 

[stevendaryl]

Ahh, got it! Thank you everyone, I think my questions have been resolved. (I'm still not entirely comfortable with the skater rotating "relative to space/space-time", since it seems to imply an absolute reference frame. But PeterDonis seemed to imply that "there is an absolute space-time which the skater rotates relative to" is a modelling assumption I'd have to make for this scenario to mean anything?)

Spacetime doesn't give an absolute criterion for rest, but it does give an absolute criterion for inertial motion. For an analogy, take a plain white piece of paper. If you draw a curve on the piece of paper, you can define the "slope" of the curve at any point to be the ratio of the changes in vertical and horizontal directions as you move along the curve. There is no absolute standard for what is "vertical" and what is "horizontal", and so there is no absolute notion of "slope", only relative to a coordinate system. However, there is an absolute notion of "constant slope" (namely, a straight line).