Spin Time Questions

[Secular Sanity]

Using Earth as a convenient spinning thing..

Atomic clocks all run at the same speed at sea level whether you are at the poles or the equator. If you google this and find it isn't true we'll have to have a google fight to sort out the true sites from the false sites. Any site that disagrees with me is a false site (by definition) except when I am wrong.

Two atomic clocks. Bob at the North pole and Colin at the equator.

Thing is that the Earth is fatter in the middle than at the top

equatorial radius is 6378 km, but its polar radius is 6357 km

see https://imagine.gsfc.nasa.gov/features/c..._info.html

If we had a clock at the pole at the same radius as one at the equator it would be 21 miles (6378-6357) higher than sea level, at a higher gravitational potential* , and would run faster than the one at the equator.

If we took away the Earth's gravitational field entirely (still forcing the equator clock to stay on the same circular path) the clock at the pole at the same radius as the one at the equator would still be at a higher gravitational potential. With the Earth gone we can drop the pole clock down to the centre of rotation and Bob is your clock running faster at the centre of a rotating disc.

*Higher is more a 'less low' than actually high. Clocks entirely away from any gravitational field run fastest of all which is what the pole clock would be without the Earth being there. The equator clock is in a pseudo field caused by the  acceleration due to rotation and not being allowed to fly off in a straight line.

I’m a little spun around (no pun intended). Let me try to make sure I’m understanding you correctly. So, you’re saying that we’d measure a difference between clocks, simply due to the rotational velocity, right?

Recap→ Clocks in weaker gravity run faster. Due to the bulge, like you said you’re 21 km further out from the center of the earth at the equator. The strength of gravity weakens as you get farther away, but centrifugal force is also proportional to the tangential speed of the rotating reference frame. So, there’s even less gravity at the equator than the poles due to centrifugal force. Obviously, GR dominates SR between these two locations, but we’re dismissing Earth’s gravitational field, and just contending with Bob’s pole at the center of a flat earth disk.

So, yes, "theoretically" you’re right. You originally said the outside dot would age slower. With Bob being at the center his clock would run faster…but as where the gravity effect is measurable and has been confirmed through experiments with highly accurate atomic clocks, this would be so minuscule that we could not even measure it with our current technology.

Fun to think about, though.

Thanks, C2!

[confused2]

Hm. Not often (probably never) that anyone understands my 'explanations' so thanks for taking the time to work through it. I was kind of on a roll when I wrote it and didn't want to dive off into 'other things'. A major point is that clocks run at different rates because of the difference in gravitational potential not the field. You can always find the lowest potential point - it is where a billiard ball will roll to unless acted on by demonic [unusual] forces. So a clock at the centre of the Earth runs slower than one at the surface even though the gravitational field is zero at the centre. I knew clocks would run slower at the edge of a spinning disc because that is where a billiard ball rolls to when it isn't taking time out to do your demonic dancing billiard ball trick. ( https://www.youtube-nocookie.com/embed/3oM7hX3UUEU )

I could go on about gravitational potential and slow clocks but it would need some high school maths to make any sense - to do or not to do, that is the question.

[Secular Sanity]

Hm. Not often (probably never) that anyone understands my 'explanations' so thanks for taking the time to work through it. I was kind of on a roll when I wrote it and didn't want to dive off into 'other things'. A major point is that clocks run at different rates because of the difference in gravitational potential not the field. You can always find the lowest potential point - it is where a billiard ball will roll to unless acted on by demonic forces. So a clock at the centre of the Earth runs slower than one at the surface even though the gravitational field is zero at the centre. I knew clocks would run slower at the edge of a spinning disc because that is where a billiard ball rolls to when it isn't taking time out to do your demonic dancing billiard ball trick. ( https://www.youtube-nocookie.com/embed/3oM7hX3UUEU )

I could go on about gravitational potential and slow clocks but it would need some high school maths to make any sense - to do or not to do, that is the question.

Could we stick with the rotating disk and discuss the rigidity of the rotating disk? Staying within the realm of SR, with the disk closer to the speed of light, with a rigid disk, the two points should be synchronized. Hence, Bob should age at the same time as (what was the name?) oh, yeah, Colin, but instantaneous interactions are not allowed in SR.

IOW, I don’t really understand the resolutions to the Ehrenfest Paradox. Toss in length contractions on a rotating disk, and my intuitive understanding vanishes.

If the observer in an inertial frame, saw the length contraction increase from Bob to Colin, what would that look like?

[confused2]

Beyond my pay grade. I have spent literally years working on the little bits of SR and GR that I claim to have some small grasp of - the rest is 'There be dragons' and I don't attempt to go there. Anyone else (of course) is welcome to have a go.

Edit.. Loosely speaking Bob and Colin will be ageing at the same time ..but not at the same rate.

[Secular Sanity]

I knew clocks would run slower at the edge of a spinning disc because that is where a billiard ball rolls to when it isn't taking time out to do your demonic dancing billiard ball trick.

That part seems a little off.

[confused2]

I knew clocks would run slower at the edge of a spinning disc because that is where a billiard ball rolls to when it isn't taking time out to do your demonic dancing billiard ball trick.

That part seems a little off.

Off as in 'wrong' or off as in teasing?
edit .. being within edit time I have removed any references to demons.

[Secular Sanity]

Off as in 'wrong' or off as in teasing?
edit .. being within edit time I have removed any references to demons.

Wrong.

A major point is that clocks run at different rates because of the difference in gravitational potential not the field. You can always find the lowest potential point - it is where a billiard ball will roll to unless acted on by demonic forces. So a clock at the centre of the Earth runs slower than one at the surface even though the gravitational field is zero at the centre. I knew clocks would run slower at the edge of a spinning disc because that is where a billiard ball rolls to when it isn't taking time out to do your demonic dancing billiard ball trick.
I could go on about gravitational potential and slow clocks but it would need some high school maths to make any sense - to do or not to do, that is the question.

Yeah, I'd love to see some high school math showing the gravitational potential on a spinning disk.

[confused2]

Yeah, I'd love to see some high school math showing the gravitational potential on a spinning disk.

Straight lines are so much easier .. I wish I hadn't done what follows (circular motion) but you may as well see it anyway..

The gravitational potential (U) is (by definition) the work done moving a unit mass from infinity to the point.

From wiki
https://en.m.wikipedia.org/wiki/Gravitational_potential

In classical mechanics, the gravitational potential is a scalar field associating with each point in space the work (energy transferred) per unit mass that would be needed to move an object to that point from a fixed reference point.

Because we're dealing with unit mass we have m=1 so I'll ghost in the m as [m]

Since we know the velocity v at the circumference of a circle is
v=rw and the velocity at the centre is zero..

the work done bringing mass up to velocity v is (1/2)[m]v^2
so it looks like the equivalent of gravitational might be (1/2)[m](rw)^2
where m = 1 we get
(1/2)(rw)^2

I probably should be doing this in moments of inertia (circular) but with a point mass I think it all comes out the same..

radius of the Earth r is 6.371E6m
r^2=4.0589641e+13
Rate of rotation w=2*pi/86400=7.27E-6 radians/sec
w^2=5.2898E-9

So my prediction of potential at the circumference is..
U = (1/2)(rw)^2=107339 [this is a positve potential relative to zero - you don't see many of those in gravitational fields]

Since potential difference (close to the Earth) is gh
we get h = U/g = 107339/9.81 = 10941 m or 11km

'The book' says the difference is actually 21km. A 10km error in 6,370km but this is either right or wrong and it isn't right. I don't know why it isn't right but hopefully it is at least clear that the link between rotational and gravitational potential lies in something 'similar' to what I done above - preferably without losing either a factor of 2 or 10km.
Edit .. maybe a textbook would help *- I don't have one and this seems to be off piste on the Internet after a quick search.
* or an actual physicist!

Edit2 .. see if you can track down rpenner - he's one of the few you can get any sense out of. Say Confused2 says "Hi" and all best wishes.

[Secular Sanity]

All I’m saying, C2, is that when discussing a flat rotating disk, such as an LP, gravity potential doesn’t come into play. You were only right due to the SR effects, but even if we think of a flat disk as large as our planet, at the current speed (remember neutron stars aren’t even close enough to make much of a difference), it would be something like a millionth of a second over course of a year. It’s immeasurable.

The Ehrenfest Paradox might be above your paid grade, but you’re retired now, right? And it’s fun to think about. It’s not associated with GR but the discussions contributed to the expansion of Einstein’s thinking about the nature of spacetime.

"Two questions arose in the first few decades after the birth of relativity: (a) What would a rotating disk look like, particularly a rigid disk?  (b) What is a good set of spacetime coordinates for life on a rotating platform?  In other words, we can enquire into the physics of a rotating physical disk, and we can do likewise for systems of rotating coordinates."

https://math.ucr.edu/home/baez/physics/R..._disk.html

Or you could just stick with demons. Dragons might be out of your field.  

[confused2]

All I’m saying, C2, is that when discussing a flat rotating disk, such as an LP, gravity potential doesn’t come into play. You were only right due to the SR effects, but even if we think of a flat disk as large as our planet, at the current speed (remember neutron stars aren’t even close enough to make much of a difference), it would be something like a millionth of a second over course of a year. It’s immeasurable.

The Ehrenfest Paradox might be above your paid grade, but you’re retired now, right? And it’s fun to think about. It’s not associated with GR but the discussions contributed to the expansion of Einstein’s thinking about the nature of spacetime.

"Two questions arose in the first few decades after the birth of relativity: (a) What would a rotating disk look like, particularly a rigid disk?  (b) What is a good set of spacetime coordinates for life on a rotating platform?  In other words, we can enquire into the physics of a rotating physical disk, and we can do likewise for systems of rotating coordinates."

https://math.ucr.edu/home/baez/physics/R..._disk.html

Or you could just stick with demons. Dragons might be out of your field.  

I asked my favourite AI how to deal with gravitational potential and rotating discs .. it said find a good textbook. So I asked it to comment on my solution..

Yes, I must admit that your explanation of the problem is quite simple and straightforward. Bringing the mass from 0 to rw does indeed involve work, assuming we ignore the complexities of relativistic velocities..

The social dynamics are complex but in machine speak "I must admit.." means "I [should have/could have] thought of that given a bit more cpu time.". It recognised that what I suggested was a possible solution to the problem. As a comment on its level of understanding I'd like to be able to say "At a rudimentary level it.." but there's nothing rudimentary about this machine. In due course (for fun) I'll probably ask it to do the calculation but it takes so long to get it in the mood to calculate that it's quicker to do it myself.

Edit.. this is a very simple 'spot on a disc' problem. I have the wrong answer - I really don't think it is wrong because of or in spite of Born rigidity. If it were me I would look for the right answer. I've been down the Born rigidity rabbit hole .. I'd suggest correcting my calculation will get you a lot further forward than spending time in some (other) rabbit hole.