Time and Gravity if rotating faster than a critical frequency

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If the earth rotates faster, distant stars traveling across the sky would eventually look like they are traveling faster than the speed of light. From the distant star's perspective, we are just rotating like crazy. How would time and gravity change for the two reference points of view if the speed of light can't be exceeded?
If a person was rotating on a verticle axis from head to toe like the Earth or quasar. If nothing can go faster than light, from the person's perspective looking at the stars traveling across the night sky, if you increase the rotation of the earth, stars further than a certain critical distance will have to appear to be traveling faster than the speed of light in the sky. The person on Earth wouldn't know that he is rotating. They would think they are standing still and everything is rotating around them. From the distance star, they would think they are standing still, and we were rotating super fast like quasars do. How does gravity and time work in this scenario?
 

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  • #2
Ibix
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The rule that nothing can exceed ##c## applies to inertial frames only. You don't need excessive rotation to cause distant objects to exceed ##c##. Just turn in place. If you can turn a full turn in one second, the nearest star traveled nearly 25ly in one second in your frame. But it will never overtake a light ray, which is the true physical restriction.

So there's nothing particularly special about something spinning in terms of time. You will, of course, be able to detect your rotation with sensitive enough experiments - meteorologists and artillery gunners routinely have to account for Earth's rotation.
 
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Orodruin
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What will be larger than c is the coordinate speed of light. This is an artefact of a coordinate choice. What cannot exceed the speed of light is the speed relative to other nearby objects.
 
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  • #4
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If the Earth rotates faster, distant stars traveling across the sky would eventually look like they are traveling faster than the speed of light.
That's already true for our actual Earth for stars that are far enough away. But the speed you are thinking of here, as others have already commented, is a coordinate speed in a non-inertial frame, and coordinate speeds in non-inertial frames are not limited to the speed of light.
 
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Ibix
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That's already true for our actual Earth for stars that are far enough away.
In fact, it's true of anything more than ##1/2\pi## light days away, which includes Neptune and Pluto if my arithmetic is reliable.
 
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Orodruin
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In fact, it's true of anything more than ##1/2\pi## light days away, which includes Neptune and Pluto, if my arithmetic is reliable.
So ... all stars except one. ;)
 
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  • #7
DrGreg
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In fact, it's true of anything more than ##1/2\pi## light days away, which includes Neptune and Pluto if my arithmetic is reliable.
So ... all stars except one. ;)
To be pedantic, only for bodies on the celestial equator. The limit is really$$ \frac {1} {2\pi \,\cos \theta} $$light days, where ##\theta## is angle above the celestial equator.
 
  • #8
jbriggs444
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To be pedantic, only for bodies on the celestial equator. The limit is really$$ \frac {1} {2\pi \,\cos \theta} $$light days, where ##\theta## is angle above the celestial equator.
How many stars would we expect to find in the observable universe that fall within the cylinder so defined?

I get in the neighborhood of 80 stars. A cylinder 90 billion light years in length, a cross sectional area of a circle ##\frac{1}{365\ 2\pi}## light years in radius and a stellar density of 0.004 stars per cubic light year.
 
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  • #9
anuttarasammyak
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Your argument does not rely on magnitude of the rotation. Everything far from the ##\rho## distance of ##\frac{c}{\omega}## in spinning cylindrical coordinate ##\rho, \phi, z##, cannot stop in that rotation system. For an example, 1 light year ##\rho## distance star has speed of ##2\pi/24## light year / hour > c in the Earth spinning frame of reference as we observe in the sky.

[Edit]
I would add to say time in rotational frame of reference share time with IFR clocks. Residents in rotational FR do not own their clocks but use time of "rotating" clocks which are at rest in the IFR where the center of rotation is at rest. This is the reason why 24 hours on the Earth is denominator on the above calculation.
The clocks at rest in rotational FR which cannot be synchronized are inconvenient to use.
The IFR clocks move faster than light speed as well as everything where
[tex]\rho > c/\omega [/tex] .
 
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