[Don Kingsley]

Don Kingsley asked the Naked Scientists:

Do all things in the universe, stars galaxies, planets, rotate relative to each
other, and does the universe rotate in reference to a theoretical
"fixed"point?

What do you think?

[ukmicky]

Hi Kingsley


Their are no fixed points in space.

[lyner]

And no overall rotation, either.

[graham.d]

I think if there were a rotation it would define a special point in the universe about which rotation occurred. None has ben found and is not in any models as far as I know. Also the universe would have an "up" and a "down" along any rotational axis so would no longer be isotropic. This would not fit in with observation but it is possible it is sufficiently small that it is unnoticable. The concept of an overall net rotation would mean that there was a very large angular momentum just after the Big Bang and I suppose an infinite one at some point. I don't think any models have this and it maybe not even a practical supposition, I'm not sure. Certainly the models that already exist are complicated enough without one that introduces artifacts to cope with a possible feature that has never been observed.

[Alan McDougall]

No it is more like a loaf of raison bread, with the galaxies imbedded like raison in the fabric of the universe (dark matter)

[syhprum]

If the universe is all there is which surely by definition it must be then rotation is a meaningless concept, rotation can only be relative to something else.

[graham.d]

Actually that is not so I believe. Rotation is not relative in that sense. It can be sensed as an absolute motion. Rotation of a roundabout is not equivalent to the universe rotating the other way for example whereas linear motion can be considered from any reference frame.

[syhprum]

In have always believed that the centripetal force that one experiences on a roundabout is due to ones movement relative to the general gravitational field of the whole universe.
What would cause centripetal forces if the whole universe was rotating?.

[lightarrow]

In have always believed that the centripetal force that one experiences on a roundabout is due to ones movement relative to the general gravitational field of the whole universe.
What would cause centripetal forces if the whole universe was rotating?.

Right. You must move with respect to something to measure inertial forces, that is, to be in an acceleretad frame of ref.
Example: we measure such inertial forces on Earth, because it's spinning with respect to the fixed stars; but what if were the fixed stars to be rotating around us? Nothing different, the result would be the same (Mach's principle).
If all the universe were spinning, it should have to do it with respect to a bigger universe which containes it.
What does it mean that our universe is contained in a bigger one?

[lyner]

This Mach Principle is quite hard to get your head around. If you are out on your own in deep space and you fire two rockets, tangentially, on two opposite sides of the ship it is hard to realise that the resulting rotational effects are only experienced because of the  rest of the Universe.
But the notion of doing the experiment in a Universe which consists only of the rocket is probably not valid.

[lightarrow]

This Mach Principle is quite hard to get your head around. If you are out on your own in deep space and you fire two rockets, tangentially, on two opposite sides of the ship it is hard to realise that the resulting rotational effects are only experienced because of the  rest of the Universe.

Never said it's simple, not even for physicists. The way you put in rotation an object (the ship in this case) doesn't change anything; once the object is in rotation, or even during the rotation, the laws of mechanics (Newton's Laws) are valid also because there is mass out there.

[graham.d]

Rotation generates forces that can be measured. The idea that Mach's principle applies here is not correct I think. If you are on a ship and fire rockets to rotate the ship you will feel a gravitational effect thought to be (according to general relativity) equal to the required cetripetal force exerted by the outer walls of your ship. You can tell you are moving without reference to the exterior world or you could posulate a gravitational field acting outwards. The force you feel varies with your distance from the centre of rotation. If you say that in an empty universe you would not experience these forces, which may or may not be the case as it is unclear what is meant by an empty universe (less clear than in Mach's day) and then say that the universe as a whole may be rotating but also not experience such forces, then surely the concept of rotation has no meaning in this case. 

[lyner]

I had a thought; perhaps you can regard rotation of any body by looking at the individual ('point') particles. Each particle of a spinning wheel can be regarded as performing independent sinusoidal, linear, motions, in phase quadrature. This gives a circular motion for each part of the wheel. The accelerations can be treated in the same way as for linear motion and Mach's principle can be said to apply. I imagine that the effects of centrifugal and coriolis forces would come out from any calculations.

What about the individual points? They must be spinning also but, by the time the scale is very small, QM may start to apply.

Here is a sort of classical version of the situation; First, imagine a rigid, light disc with some masses distributed around the edge. Spin the disc. The moment of inertia, angular momentum, etc are easy to calculate. You could predict the effect of moving the masses about and the resulting forces.

Now, imagine that all the masses are on friction free pivots. When you spin the disc, the masses are all moving round the axis of the disc but they are not, themselves spinning, because of their perfect bearings. When you start to move the masses around, how is the situation different from the first case?

An even simpler model would be to have two (not point) masses at the ends of a light rod. What  is the difference between the moment of inertia of the system with fixed masses and how would that change if the masses were freely pivoted? The r² term in the M.I. formula
Integral (m r²)
 would presumably change things because, for the fixed masses, the r values are different for each piece but, for the freely rotating masses, the r value for each mass would be that of its centre of mass. Can you actually add up the MI values for the overall system and for each mass?
This must be a bit of book work which someone may know already.

My main point is that you can separate the rotation into non-rotational (macroscopic) motion and rotational (microscopic) motion. Afaik, GR doesn't deal adequately with microscopic situations. Does this really call for more than GR?

[lightarrow]

Rotation generates forces that can be measured. The idea that Mach's principle applies here is not correct I think. If you are on a ship and fire rockets to rotate the ship you will feel a gravitational effect thought to be (according to general relativity) equal to the required cetripetal force exerted by the outer walls of your ship. You can tell you are moving without reference to the exterior world or you could posulate a gravitational field acting outwards. The force you feel varies with your distance from the centre of rotation. If you say that in an empty universe you would not experience these forces, which may or may not be the case as it is unclear what is meant by an empty universe (less clear than in Mach's day) and then say that the universe as a whole may be rotating but also not experience such forces, then surely the concept of rotation has no meaning in this case. 

<<You can tell you are moving without reference to the exterior world>>
How? You measure inertial forces because of Mach's principle. There wouldn't be any without the masses of the fixed stars.
The simple "principle of inertia" F = ma, comes from Mach's principle.
Dennis Sciama has made a semi-quantitative computation, postulating the existence of a new kind of force between two accelerated masse, which, like in the EM case, goes as 1/R instead of 1/R² (it is reasonable considering that such a force is so small with respect to the normal grav. force to be negligible, at small scales). He then integrated in dV of the entire universe, with tabulated values of average density, taking as the superior extreme of integration the "horizon of visibility". With such rough assumptions he find for F/ma the value ≈ 0.3. Which I consider very impressive.

[syhprum]

I would be interested to know if when we experience centripetal forces due to rotation it is due to our rotation relative to the local galaxy or due to the universe as a whole (the fixed stars as Mach put it), I can conceive of no way this could be measured.

[lightarrow]

I would be interested to know if when we experience centripetal forces due to rotation it is due to our rotation relative to the local galaxy or due to the universe as a whole (the fixed stars as Mach put it), I can conceive of no way this could be measured.

The second you wrote.

[lyner]

which, like in the EM case, goes as 1/R instead of 1/R2

Could you enlarge on that? The Coulomb force and radiation law drop off as 1/r², don't they?  Have I missed something? I agree that potential in both cases follows 1/r.

But a 1/r law, or at least a non-inverse square law, would be reasonable if the spherical symmetry of the situation no longer held and it wouldn't be a 'point source' situation because it would involve a very distributed cause.

[graham.d]

If you have a rotating universe in any meaningful sense of the word rotation, you have an axis hence anisotropy. In an "empty" universe you can postulate that you would not feel any forces when rotating because you are not rotating with respect to anything and angular momentum would not have any magnitude. So how would this apply to a rotating universe? For argument's sake, assuming a closed universe (which may not be the case), would it not be also difficult to say with what the universe is rotating with respect to? Would then the meaning of "rotation" be something different from our normal usage? Would any forces result? If not, what is the meaning of rotation in this case and if "yes" why is there no apparant anisotropy?

[Alan McDougall]

I agree rotation is relative, for instance take everything out of the universe, would it be revolving?

I say no,

[lightarrow]

which, like in the EM case, goes as 1/R instead of 1/R2

Could you enlarge on that? The Coulomb force and radiation law drop off as 1/r², don't they?  Have I missed something? I agree that potential in both cases follows 1/r.

No, at not very low distances, radiation drops as 1/r.
I only found this article on wikipedia, but the formula for the electric field from an oscillating charge is on many books, included "Feynman's Physics"
http://en.wikipedia.org/wiki/Antenna_measurement

Physical background.
The electric field created by an electric charge is:

E = (-q/4πε₀)[e_r'/r'² + (r'/c)(d/dt)(e_r'/r'²) + (1/c²)(d²/dt²)(e_r')].

where:
q is the accelerating charge
E is the electric field vector.
c is the speed of light in vacuum.
ε₀ is the permittivity of free space.
r' is the distance from the observation point (the place where E is evaluated) to the point where the charge was r'\c seconds before the time when the measure is done.
e_r' is the unit vector directed from the observation point (the place where E is evaluated) to the point where the charge was r'\c seconds before the time when the measure is done.

The "prime" in this formula appears because the electromagnetic signal travels at the speed of light. Signals are observed as coming from the point where they were emitted and not from the point where the emitter is at the time of observation. The stars that we see in the sky are no longer where we see them. We will see their current position years in the future; some of the stars that we see today no longer exist.

The first term in the formula is just the electrostatic field with retarded time.

The second term is as though nature were trying to allow for the fact that the effect is retarded (Feynman).

The third term is the only term that accounts for the far field of antennas.

The two first terms are proportional to 1\r^². Only the third is proportional to 1\r.

But a 1/r law, or at least a non-inverse square law, would be reasonable if the spherical symmetry of the situation no longer held and it wouldn't be a 'point source' situation because it would involve a very distributed cause.

Certainly a single oscillating charge (the field which it generates is proven to be equivalent to that of an oscillating dipole) has not spherical symmetry and the field is maximum, at far distances, around the axis of oscillation, in the perpendicular direction to it. I don't know how to imagine a perfect spherical simmetry for a system of accelerated charges; even an oscillating multiple will give rise to an almost, but not perfect, spherical symmetry. I don't know much of the subject, but the fact radiation at far distance drops as 1/R instead of 1/R² is one of first amazing things I studied at university on the course of "Physics II" and so I remember it very well.