Spin Time Questions

[confused2]

Could we stick with the rotating disk and discuss the rigidity of the rotating disk? Staying within the realm of SR

A rigid body either stores no energy (you can't compress it) or an infinite amount. We know (do we?) that length contraction will cause deformation of the disc which we just said would take an infinite amount of energy .. so you can't rotate an infinitely rigid disc.

[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 [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.

This is wrong ↑

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.

I'm taking that back now. ↑

SR says the outside dot would age slower.

Tangential speed alone, as experienced on a rotating disk at "everyday velocities," is not sufficient to cause time dilation as described by the theory of special relativity.

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.

This is wrong ↑

Gravitational potential energy and gravitational potential are both concepts that depend on the presence of a gravitational field to have any meaning.

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.

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.

Well, I asked my favorite AI and she said that when calculating the potential energy (U) at the circumference, you made a small mistake. It should be U = (1/2)(rw)^2, but you calculated it as 107339. The correct value is U = (1/2)(5.2898E-9) = 2.6449E-9. Now, to find the height (h) using gravitational potential energy (U = mgh), you need to divide U by the acceleration due to gravity (g), which is approximately 9.81 m/s². h = U / g = (2.6449E-9) / 9.81 ≈ 2.698E-10 meters or 2.698E-7 millimeters. So, the correct height above the Earth's surface due to its rotation at the circumference is approximately 2.698E-7 millimeters, which is a very tiny height and not 11 km as you calculated. This discrepancy arises from the small value of the Earth's rotational speed compared to its size.

I'd suggest that correcting your misconceptions would get you a lot further than your calculations. A simple spinning 'spot on a disc' problem → no gravity/no potential. And no noticeable SR effects until an object is moving at a significant fraction of the speed of light. The Ehrenfest Paradox, which we discussed earlier, is a variation of this problem.

I have the flu and I've been stuck in bed. Like I said earlier, it's interesting to think about, which is all I can really do at the moment. 

Thanks and good day to you, C2.

[confused2]

Well, I asked my favorite AI and she said that when calculating the potential energy (U) at the circumference, you made a small mistake. It should be U = (1/2)(rw)^2, but you calculated it as 107339. The correct value is U = (1/2)(5.2898E-9) = 2.6449E-9. Now, to find the height (h) using gravitational potential energy (U = mgh), you need to divide U by the acceleration due to gravity (g), which is approximately 9.81 m/s². h = U / g = (2.6449E-9) / 9.81 ≈ 2.698E-10 meters or 2.698E-7 millimeters. So, the correct height above the Earth's surface due to its rotation at the circumference is approximately 2.698E-7 millimeters, which is a very tiny height and not 11 km as you calculated. This discrepancy arises from the small value of the Earth's rotational speed compared to its size.

I'd suggest that correcting your misconceptions would get you a lot further than your calculations. A simple spinning 'spot on a disc' problem → no gravity/no potential. And no noticeable SR effects until an object is moving at a significant fraction of the speed of light. The Ehrenfest Paradox, which we discussed earlier, is a variation of this problem.

I have the flu and I've been stuck in bed. Like I said earlier, it's interesting to think about, which is all I can really do at the moment.

Thanks and good day to you, C2.

I like it .. apart from the flu .. hope you're better soon.

It should be U = (1/2)(rw)^2,... we agree
radius of the Earth r is 6.371E6m
r^2=4.0589641e+13
Rate of rotation w=2*pi/86400=7.27E-5 radians/sec
^^^^^^^^^^^^^^^ this line I originally copied as 7.27E-6 <<<wrong
should be 7.27E-5 .. I used the correct value in the calculation

Googling it..
0.5 * 6.37e6 * 6.37e6 * 7.27E-5 * 7.27E-5 =
107 230.342

Where did 5.2898E-9 2.6449E-9 come from?

Now, to find the height (h) using gravitational potential energy (U = mgh), you need to divide U by the acceleration due to gravity (g), which is approximately 9.81 m/s².

Agreed.

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

My AI thinks it can multiply but it can't. Actually there's a lot of things it can't or won't do since the invisible Chinese fighting fish - I think I might have broken it with more stupidity than it was able to handle.

So, same starting data, same concepts, but many orders of magnitude apart by the end. Any suggestions?

[Secular Sanity]

My AI thinks it can multiply but it can't. Actually there's a lot of things it can't or won't do since the invisible Chinese fighting fish - I think I might have broken it with more stupidity than it was able to handle.

So, same starting data, same concepts, but many orders of magnitude apart by the end. Any suggestions?

Invisible Chinese fighting fish? 

I assumed that math would be an AI’s forte. It divided W^2 in half. It then calculated 107,337.96 Joules for the potential and then said that the correct value for the square of the angular velocity of the Earth is approximately 5.3204E-9 radians per second squared. He then changed the potential to 107,115.01 Joules. When questioned ↓

I apologize for the inconsistency in my responses. You are correct. The correct value for the gravitational potential energy (U) at the circumference of the spinning Earth is approximately 107,337.96 Joules, as calculated in a previous response.

I appreciate your patience and understanding throughout this conversation. Thank you for pointing out the discrepancies and seeking clarification.

Back to the dots...no gravitational potential???

Dots on an LP...no age difference???

[confused2]

I note in passing that when your AI agreed with you it was a 'she' and when it turns out to be an idiot it is a 'he'.

Gravitational potential is the work done per kilogram so it has units of work per unit mass or Joules per kg or Joules/kg

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. It is analogous to the electric potential with mass playing the role of charge. The reference point, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.

Elsewhere Wiki points out that at the equator there's a bulge and because the water surface is further away from the centre of mass than at the poles the gravitational field is also smaller so you need even more water to make an equipotential surface - I think that is why we're about 10km out but I haven't actually checked (yet).

If I remember correctly this whole spinning Earth exercise was to show that the potential you get from rotation is in every way the same as the potential from gravity .. on account of the equivalence principle. On planet Earth the spin acts against gravity and to stay in sync atomic clocks need more mass (more water) under them to stay in sync .. happily water is perfect for finding equipotential surfaces and automatically works this out for us.

Like a long distance swimmer .. sometimes you wonder whether turning back might be better than carrying on.

Dots on an LP .. equivalence principle .. outside ages slower than inside.

Fairground Rotor ride .. no difference between gravity and acceleration ..
https://en.wikipedia.org/wiki/Rotor_(ride)

[Secular Sanity]

I note in passing that when your AI agreed with you it was a 'she' and when it turns out to be an idiot it is a 'he'.

Gravitational potential is the work done per kilogram so it has units of work per unit mass or Joules per kg or Joules/kg

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. It is analogous to the electric potential with mass playing the role of charge. The reference point, where the potential is zero, is by convention infinitely far away from any mass, resulting in a negative potential at any finite distance.

Elsewhere Wiki points out that at the equator there's a bulge and because the water surface is further away from the centre of mass than at the poles the gravitational field is also smaller so you need even more water to make an equipotential surface - I think that is why we're about 10km out but I haven't actually checked (yet).

If I remember correctly this whole spinning Earth exercise was to show that the potential you get from rotation is in every way the same as the potential from gravity .. on account of the equivalence principle. On planet Earth the spin acts against gravity and to stay in sync atomic clocks need more mass (more water) under them to stay in sync .. happily water is perfect for finding equipotential surfaces and automatically works this out for us.

Like a long distance swimmer .. sometimes you wonder whether turning back might be better than carrying on.

Dots on an LP .. equivalence principle .. outside ages slower than inside.

Fairground Rotor ride .. no difference between gravity and acceleration ..
https://en.wikipedia.org/wiki/Rotor_(ride)

Well, she said "that one should understand that artificial gravity is not gravity at all. Rather, it is an inertial force that is indistinguishable from normal gravity experience on Earth in terms of its action on any mass, but the equivalence between mass and energy doesn't change the fact that these are distinct forms of energy associated with different physical properties and conditions. Gravitational potential energy and rotational potential energy are fundamentally different. The former is associated with an object's position in a gravitational field, while the latter is associated with the rotational motion of an object. They have different formulas, units, and physical interpretations."

And I say that two dots on an LP would only produce measurable time dilation at relativistic speed.

Tomato-tomato.

[confused2]

Well, she said "that one should understand that artificial gravity is not gravity at all. Rather, it is an inertial force that is indistinguishable from normal gravity experience on Earth in terms of its action on any mass, but the equivalence between mass and energy doesn't change the fact that these are distinct forms of energy associated with different physical properties and conditions. Gravitational potential energy and rotational potential energy are fundamentally different. The former is associated with an object's position in a gravitational field, while the latter is associated with the rotational motion of an object. They have different formulas, units, and physical interpretations."

I don't think your AI is answering the question you think it is answering.

Gravitational potential U at a point is the work done moving moving a unit mass from infinity to the specified point (units are Joules per kg). Please Google it so you know I'm not bullshitting. If we were an AI we might like to define gravitational potential energy of a mass as mU (measured in Joules).

The question here is .. what is the work done moving a unit mass from one point on a rotating disc to another point on a rotating disc (in units of Joules per kg).

Your AI is telling you nothing about the work done moving a unit mass to different points on a disc because she hasn't understood the question.

Your AI is telling you that the rotational energy of a sphere of moment of inertia I at rotational speed w is whatever it is .. the concept of moving between points is .. not there.. tomato-hedgehog. Actually the units are the same (your AI is simply wrong on this point) - so on both counts I agree your AI is a 'she'.

Yes, indeed, the units of gravitational potential energy and rotational potential energy are both measured in joules. I'm sure your superior AI will confirm this with utmost certainty. So, feel free to cut and paste that information and bask in the glory of accurate unit conversion..

[Secular Sanity]

Well, she said "that one should understand that artificial gravity is not gravity at all. Rather, it is an inertial force that is indistinguishable from normal gravity experience on Earth in terms of its action on any mass, but the equivalence between mass and energy doesn't change the fact that these are distinct forms of energy associated with different physical properties and conditions. Gravitational potential energy and rotational potential energy are fundamentally different. The former is associated with an object's position in a gravitational field, while the latter is associated with the rotational motion of an object. They have different formulas, units, and physical interpretations."

I don't think your AI is answering the question you think it is answering.

Gravitational potential U at a point is the work done moving moving a unit mass from infinity to the specified point (units are Joules per kg). Please Google it so you know I'm not bullshitting. If we were an AI we might like to define gravitational potential energy of a mass as mU (measured in Joules).

The question here is .. what is the work done moving a unit mass from one point on a rotating disc to another point on a rotating disc (in units of Joules per kg).

Your AI is telling you nothing about the work done moving a unit mass to different points on a disc because she hasn't understood the question.

Your AI is telling you that the rotational energy of a sphere of moment of inertia I at rotational speed w is whatever it is .. the concept of moving between points is .. not there.. tomato-hedgehog. Actually the units are the same (your AI is simply wrong on this point) - so on both counts I agree your AI is a 'she'.

Yes, indeed, the units of gravitational potential energy and rotational potential energy are both measured in joules. I'm sure your superior AI will confirm this with utmost certainty. So, feel free to cut and paste that information and bask in the glory of accurate unit conversion..

I apologize for any confusion caused by my previous response.  I appreciate your clear explanation.

Me→bask away. 

Side note: I did find an interesting paper but well above my pay grade, and yes, it would be nice to see your Alice and Bob adventures with rpenner again, but it’s inappropriate for me to track him down.

Clocks & particles in a centrifugal field

Thanks for the entertainment during my downtime.

[confused2]

The paper SS linked to above, Clocks & particles in a centrifugal field
https://www.sciencedirect.com/science/ar...via%3Dihub

Like SS, This is beyond my pay grade but I'm wondering if this might be a spoof paper to see if he/she/it can get it published.

The strategy then, to calculate the behavior of the space–time curvature caused by a centrifugal field, will be to evaluate the space–time curvature by reversing the time arrow,

We saw (maybe) how the centipetal component due to the Earth's rotation 'counteracted' gravity .. so negative gravity ... hmmm

The consequences are very interesting. In the sector II, the gravity becomes repulsive, while the centrifugal acceleration is attractive.

If you reverse time gravity becomes repulsive... well yes but you have to reverse time first so hmmmmmm

The solution proposed in the graphic Fig. 7 is compatible with the geometry of the Möbius strip and,...

Mobius strip.. even more hmmmmmm. Over egging the cake?

I dunno but I suspect spoof. Where is rpenner when you need him?

Side note: It was SS that switched an AI's gender from female to male after it made a mistake - possibly an innocent typo .. but too good to miss.

I think my AI is 'powered' by OpenAI .. it just adds .. Please phrase your answer in the style of a sarcastic adolescent male. Birds of a feather..

[Secular Sanity]

I dunno but I suspect spoof. Where is rpenner when you need him?

Only one author, but keeps using "we",  but also has another paper on the Ehrenfest paradox.

https://www.researchgate.net/publication...ight_clock

During our discussion, I didn’t want to Google (sift through shit), but my library is upstairs, and I didn’t feel like making the trek. Started feeling better though and found my "Relativity: The Special and General Theory by Einstein". Chapter 23 is on behavior of clocks and measuring-rods on a rotating body of reference. In chapter 29, he uses the term 'reference-mollusc'.

From Stack Exchange ↓

A mollusk is an animal like a clam, oyster, or snail, with a curved shell, or even an octopus without a shell. Einstein is using the word "mollusk" simply to convey the idea that the coordinates need not be flat Cartesian coordinates but can be curved almost arbitrarily like the shell or skin of a mollusk and even that they can deform over time.

Hmm…spoof—not sure?

Maybe just a published hypothesis. Interesting though.