[neilep (OP)]

Dearest Spinologists and those with degrees in Spinology !


As a sheepy I of course rotate, it's so that everyone gets their fair share of viewing me....


When we look out into the universe, motion seems to be everywhere. Planets spin on their axes, moons orbit planets, stars rotate, and entire galaxies swirl through space. Even tiny particles at the quantum level appear to exhibit forms of angular momentum. With motion so deeply woven into the fabric of the cosmos, it raises a fascinating question: Is there anything in the universe that isn?t spinning or rotating in some way?


whajafink ?

sheepy
xxxxxxxxx

[alancalverd]

The Higgs boson has a spin of zero, making it the only elementary particle with no spin. It is classified as a scalar boson, which means it has no intrinsic angular momentum.

What has a mass of 125 GeV, horns and a tail?

[neilep (OP)]

The Higgs boson has a spin of zero, making it the only elementary particle with no spin. It is classified as a scalar boson, which means it has no intrinsic angular momentum.

What has a mass of 125 GeV, horns and a tail?

Thank ewe Alan...........Higgs Bison?

[paul cotter]

The devil, obviously.

[Eternal Student]

Hi.

    As always, thanks for introducing a good topic and I guess our job is to say something interesting in reply and maybe find a few side-roads and tangents that come off from it.

    So, it's a problem isn't it?  A big problem in something that seemed like we really had it and understood it.

    If something isn't spinning then there aren't any centrifugal forces in its rest frame,  or to say it another way, its rest frame could be a genuine inertial reference frame.   So you'd think that you'd know if something was spinning.

     Let's assume for a moment that you're a person and not always a sheep, @neilep, then you know when you're spinning because there is some centrifugal force on your arms and they tend to lift out and away from your body.   Out in the field on a clear night, you can feel this in your arms even with your eyes closed.  You don't have to be looking up at the stars and seeing them whirr around you to know that you are spinning.
     However, let's suppose that we do a slightly different trick:   We'll have you remain as still as you can.    What we'll do now is just start the distant stars spinning and whirring around you.   Now when you open your eyes, it obviously looks as if you are spinning.   The question is, has a centrifugal force just appeared in your rest frame - are your arms naturally tending to be pulled outward and away from your body?   Remember, we haven't touched you at all, we've just made some distant stars spin around.

    I'll leave you (or any reader) to think about that, there are some hints under this spoiler:

Spoiler: show

   You may want to look up   "Mach's principle".
   You might also like to look up "frame dragging" in General Relativity - but that's optional.  Mach's principle doesn't have to be connected to GR,  it's just one very convenient explanation for it along with some evidence to support it (e.g. the experiments and data of Gravity Probe A and B that are in orbit around earth).

    It's relevant for your original question because you probably ought to be clear what it actually means when we think or say that something is, or is not, spinning.

   We could say more about elementary particle spins that @alancalverd mentioned and @neilep mentioned in the OP  -  but it's best just to introduce a thing and leave the other person to think about that first.

Best Wishes.

[paul cotter]

Here is a follow up question: in the total absence of external reference points would it be possible to tell if one was rotating or not? For example an imaginary planet with it's own source of light and a sky that is uniformly black with no stars at all. My guess, and it is just a guess, is that one could not tell and in the absence of an external reference can rotation even be defined?

[alancalverd]

Anyone on the surface of a completely spherical and uniform planet should be able to detect an apparent increase in gravitation as they approach the poles, due to a decrease of centrifugal force.

[paul cotter]

Very good, Alan, I did not think of that.

[Eternal Student]

Hi.

....in the total absence of external reference points would it be possible to tell if one was rotating or not?...

     Basically, no, it shouldn't be possible to know.   There shouldn't be any centrifugal forces in your rest frame frame and it should appear in all respects that you are not spinning.    Of course, it's never been possible to do the experiment, we can't delete the rest of the universe, so all we have is the theory.   The only relevant theory I know a little about is GR, so I'll be using that but as I stated before, Mach's principle doesn't HAVE to be connected to GR - there are other theories and methods of explanation proposed.

Anyone on the surface of a completely spherical and uniform planet should be able to detect an apparent increase in gravitation as they approach the poles, due to a decrease of centrifugal force.

    Yes and no.   That's ok for a typical planet in the sort of the universe that we do actually have.    However, let's now start doing a thought experiment.   We will steadily start deleting all the other matter in the universe until only our one planet is left.   As we start doing this, the centrifugal forces that would have been necessary to describe the motion of particles in the rest frame of the planet, should start reducing and eventually disappear.

    Let's try and do this the easy way first, just with words and an appeal to a notion of what "frame dragging" is about.   

    If a massive spherical shell of material starts to spin around a point in space, then it tends to "drag" or twist the local inertial frame centred around that point with it.   To paraphrase this, the spinning massive shell of material drags the fabric of space around with it.    This local inertial frame centred at our point in space is genuinely "inertial" in the usual sense, there will be no centrifugal forces in this frame.   However, if we were able to overlay the old local inertial frame over the top of this (before the hollow shell of material was made to spin), we would see that the new inertial frame is seemingly dragged around and turns with the spinning massive hollow shell of material.
    Now, I need to make it clear that we can't just overlay the old and new local inertial frames and actually "see" that the new frame is being dragged around:   Space has fundamentally changed once the massive hollow shell started to spin around and any frame you tried to lay down there will naturally tend to twist and follow in sympathy with the rotating massive shell of material.   None-the-less it's useful as a concept, if we could make the old frame immune to the way space has now been warped,  then we would see the new inertial frame is being dragged around with the spinning hollow shell.

    So, as we start deleting more of the matter in the universe there are less things dragging space around.   The motion of our one planet becomes a more important contribution and an inertial frame centred at the planet will be frame dragged more by the planet than by anything else.   Eventually the only thing left is our planet, it's the only thing dragging space around.   When there is only that one (previously spinning) planet left, then the local inertial frame centred at that planet will be exactly that frame which is completely dragged around in sympathy or in keeping with the planet.   That is an inertial frame in the usual sense of "inertial frame", there just won't be any centrifugal forces in the rest frame of the planet.   

    Now it's your choice:   You can either say that the planet is no longer spinning   OR ELSE   we can say that spacetime has been altered / warped appropriately so that the hallmarks of rotation / spinning have been lost.   Either way you want to view it, the practicality is the same:   In the local inertial frame centred at this planet, we don't need to include any centrifugal force terms into the mechanics that describes how test particles would move about on the surface of that planet, it's not there, it will be exactly as if the planet is not spinning.

Best Wishes.

LATE EDITING:    The gist is OK but we might actually need the planet to be just a hollow shell.  A solid ball can mess it up a little because there will be bits of space with momentum and stress that differ in numerical value, for example the bit of space at the centre of ball is very different to the edge of the ball.

[alancalverd]

My scenario derives from theoretical physics, not planetary engineering: my planet was a uniform solid sphere spinning in an infinite vacuum. My test particle has mass m << M of the sphere.

There is one unique point in a stationary sphere - the center. If it spins, there are two other unique points at the poles.

Gravitational force = GMm/r_p² everywhere on the surface, but the test particle wants to fly off tangentially (Newton) so experiences a centrifugal force mr_rω² opposing gravity.

Annoyingly, this invokes Hamdani's concept of "rotational radius"  that I spent so much time decrying in the discussion of torque! It is clear that the radius of rotation at the equator is the radius of the planet r_r = r_p, but r_r = 0 at the poles.

There are, of course, plenty of engineering examples.  Surface tension acts very similarly to gravitation, and oldfashioned window glass was made by spinning a spherical blob of viscous material into a thin disc. All the forces involved are much greater than any due to the environment, so this approximates to the ideal planet with a test mass on the surface.

It gets more complicated if the spin axis is also precessing, which provides two more unique points in space where the axis of precession exits the sphere, but these are not unique points on the surface of the sphere. Calculation of the resultant net force on m at an arbitrary point on the surface is left as an exercise to the reader but if I recall correctly (it's in "takeoff characteristics of taildraggers" but that exam was a long time ago!) it involves something sinusoidal,

[paul cotter]

A two axis rotation would eliminate the poles.

[alancalverd]

You beat me to the draw by 2 minutes - see the last paragraph edit of reply #9!

[paul cotter]

Guess i'm guilty of carelessness. I should have read your post completely before commenting.

[Eternal Student]

Hi,

My scenario derives from theoretical physics, not planetary engineering:

    No it doesn't,  it's almost exactly the opposite of that.  It derives from an engineers bias that the universe should always be modelled with the equations and formulae they learnt at school and college.

   Half the problem is that when there is only that bit of mass in the universe that is (was) the planet,  then space is very different.   As you know, mass bends space.   So, it's not straightforward to know what sort of geometry will apply or exist in this space.  In particular, it's not clear that the planet would still look spherical or have properties like a sphere in Euclidean space.  Moreover, it's not clear that conventional rotational dynamics would apply (e.g. using formula like mω²r  for the centrifugal force at the surface of the planet).

     I've tried to overlook this issue myself and continued talking about "the planet" as if it will still be something like our planet, just to keep the discussion as simple as possible.  However, at least I am mentioning that space changes as you start eliminating matter from it.   Meanwhile, you ( @alancalverd ) are simply not doing this but instead just ploughing on as if conventional rotational dynamics derived in flat Minkowski space will somehow always continue to apply.    For example, the whole spacetime manifold has probably degenerated into something vaguely like the Schwarzschild solution (or the Kerr Solution if we maintain that the planet-like object was rotating) and so a small change in the r co-ordinate,  dr,  doesn't even have to describe a change in an amount of space.   Where r < r_s (the Schwarzschild radius), then the r co-ordinate has become time-like and not space-like.   Pause for a moment and recognise that the underlying geometry of this space is thoroughly messed up and conventional formulae for rotational dynamics like mω²r (that were derived for circular motion in flat space) have long since "gone out of the window".

    Let's just approach the situation the other way around, start from the very specific for which we have an exact solution to the EFE and then generalise to our situation.   What is it that happens in the Kerr solution that describes the spacetime around a rotating black hole?   Can we generalise this to a universe where the only thing in it is a rotating planet (that hasn't necessarily collapsed to a proper BH)?

The Kerr metric is a generalization to a rotating body of the Schwarzschild metric ......
According to the Kerr metric, a rotating body should exhibit frame-dragging (also known as Lense?Thirring precession), a distinctive prediction of general relativity. ....... Roughly speaking, this effect predicts that objects coming close to a rotating mass will be entrained to participate in its rotation, not because of any applied force or torque that can be felt, but rather because of the swirling curvature of spacetime itself associated with rotating bodies.

[Extracts taken from  Wikipedia.  https://en.wikipedia.org/wiki/Kerr_metric ]

    There you go.   That's how a theoretical scientist does it.   There may be some practical difficulties when trying to delete all other matter from the universe but that's a separate issue.   If we assume that we can delete all the other matter in the universe and leave only the (previously spinning) planet, then local inertial frames centred at some point on the surface of the planet should do as previously discussed.  In particular, test masses on or just above the surface of the planet are "entrained" to follow the rotation of the planet without needing to have any centripetal or centrifugal forces included in the equations that will govern their motion.

Best Wishes.

[paul cotter]

I am not in a position to argue with many of your points as my level of GR understanding is abysmal(tensor math with all it's sub/super scripts just confuses me). However, space as it currently exists is locally close to flat Minkowski space and deletion of all other mass in the universe apart from our test planet would surely make it even flatter?  NB: engineering argument as opposed to a physicist's argument.

[alancalverd]

In particular, it's not clear that the planet would still look spherical or have properties like a sphere in Euclidean space.

It will because, like a good mathematician, I said it does, in reply #6

a completely spherical and uniform planet

In particular, test masses on or just above

I said on, not just above. And the entrainment, whether by frame dragging or a piece of string, means that it will be accelerating towards the center of the planet, so the gravitational force will be offset by the fact that it is in permanent free fall.   

[Bored chemist]

I'm curious...
Unfortunately, I'm a curious chemist, rather than a physicist.

So...
Imagine I'm on a perfectly spherical planet which is alone in the whole infinite universe.
I can check it's a sphere. I just need a big one of these.
https://en.wikipedia.org/wiki/Spherometer

And I can choose a planet which is made from a single practically rigid material.
I can get a pendulum clock and a quartz clock.
I can put them next to each other and compare how fast they run.
If I plot the rate of change of difference between the two clocks as a function of position, I think I should get a pattern where the pendulum clock runs faster at the poles and slower at the equator.
Not some strange relativistic thing, just the effect of centrifugal force meaning the pendulum isn't pulled down so hard at the equator.

If this
https://www.engineeringtoolbox.com/acceleration-gravity-latitude-d_1554.html
is to be believed, the effect is such that g changes from 9.78 to 9.83 m/s/s here on earth, and the error in timing is comparable (About 0.25% if I have got the arithmetic correct),

There will be some tiny effect of local gravity on the quartz clock, but I can pretty much ignore that if all I want to do is find the equator.

If the two clocks are consistent over the whole of the sphere's surface I can say it's not rotating.

Or doesn't that work?

[alancalverd]

It's a pretty basic form of gravimeter, but an even simpler one just* uses a helical spring and a weight.

http://www.cleonis.nl/physics/phys256/eotvos.php  discusses some interesting experiments where a gravimeter in an airship flying east to west measured a different value from one flying west-east, the difference being due to the difference in centrifugal force.


*physicist's shorthand for hundreds of hours in an instrumentmaker's workshop.

[Bored chemist]

I just want the excuse to have a good pendulum clock.
(Even if it's just  in a thought experiment)

https://en.wikipedia.org/wiki/Shortt%E2%80%93Synchronome_clock

The great advantage to using clocks is that, if they are reasonably well built, I can improve the accuracy of my measurement by simply waiting longer.

[geordief]

So if we have two rotating objects with mass (obviously in a vacuum)  does the rotation of  the one affect the rotation of the other?