Paddlewheels in a gravity well and angular momentum

1. May 8, 2012

jartsa

Let's shine two identical light beams onto two identical light absorbing paddlewheels in a gravity well at different heights.

The angular momentum of both paddlewheels increases.

Then we transfer the paddlewheels into same position and compare their angular momentums.

What will we observe?

(the light is aimed at one side of the paddlewheel)
(paddlewheels are vertical (standing up in the gravity field) )
(light sources are above the paddlewheels at the same position)
(after the experiment the paddlewheels are painted white, then the experiment is repeated with these perfectly reflecting paddlewheels)

Last edited: May 8, 2012
2. May 8, 2012

Q-reeus

jartsa wrote:
It would be helpful to specify the spatial relalation between each light source and paddlewheel, and the axial orientation of said paddlewheels wrt gravity source.

3. May 8, 2012

jartsa

I edited the OP

4. May 8, 2012

Q-reeus

We're getting closer to a tight spec, but not there yet:
Which, given the next spec of radial spin axes, initially suggests the light source is at the same potential as the wheel - i.e. aimed horizontally. But then...
Which now makes it hard to figure the geometry for me at least - seems to require a different oblique aiming angle for the two light sources - a complicated setup to analyze, even if precise figures (needed) are given as to relative heights - without which the beam angles cannot be known. Unless perhaps you have in mind that paddlewheel blades are angled at some oblique angle to the shaft axes (a sort of axial turbine) and the beams run parallel to the shaft axes but offset a given distance? But that then doesn't sound like a 'standard' paddlewheel where I would expect beam direction at right angles to shaft axes.
That part at least is easy - simple conservation of linear momentum tells us the white painted run will show twice the angular momentum change relative to when black (absorbing). The other easy thing imo is that qualitatively there is a gain in angular momentum simply by hauling the lower paddlewheel up to the potential level of the other - provided the spin axis is oriented radially during that lifting.
Back much later. :zzz:

5. May 8, 2012

A.T.

My guess: After switching off the laser we will observe both spinning at the same rate. When bought together the originally lower one will be spinning faster.

6. May 9, 2012

yuiop

I can see the source of your confusion. Just about every sentence in the OP is ambiguous.
That was not actually specified. He said the "paddlewheels are vertical" but did not specify which parts of the paddlewheels are vertical. I believe he meant the spin axes are horizontal. Most people would describe the orientation of the wheels on a typical car as vertical, instinctively referring to the plane of rotation parallel to the side walls of the tyres, rather than to the axes of rotation.
... but then again he did say "light sources are above the paddlewheels" suggesting they are not at the same potential as the paddlewheels, but then went on to muddy the waters again by adding "at the same position" ... :uhh: ... presumably meaning the light sources are at the same altitude as each other.

OK, here is what I believe he had in mind. There are two light sources at the same potential, that are aimed vertically downwards. Located in the path of each light beam is a paddlewheel. The paddlewheel axles are slightly offset horizontally (to the side) from the path of the beams so that the light strikes a paddle. If the light beams were not aimed to one side of the paddlewheels, they would strike the axle and not impart any rotation. Both the paddlewheels are below their respective light sources, but one is at a lower potential than the other. Now assuming that is what the OP meant, let's continue with rest of the his question:
OK, the light striking the lowest paddlewheel is blue shifted by the greatest amount and has the highest energy (all else being equal). By local measurements, the lower paddlewheel will have the greatest angular velocity. When we turn off the light sources and raise the lower paddlewheel to the height of the upper paddlewheel, the wheel that was initially lower will have the higher angular velocity and momentum when they are compared side by side (assuming no frictional losses). The ratio of their angular momentums will be proportional to the square root of the ratio of the gravitational redshift factors at their original altitudes.

7. May 9, 2012

Q-reeus

Splendid job of deduction yuiop - I'd say that assessment of the OP's intent is correct. Taking then the spin axes as horizontal, the relative gain in paddlewheel angular momentum is just about as you say, but I believe there is a subtle extra factor, which follows from what I wrote here: https://www.physicsforums.com/showpost.php?p=3901776&postcount=60 (unspecified there was that radial spin axis was assumed - spin-spin coupling)

Basically, for transverse spin axis, we should expect a spin-orbit coupling such that raising the lower paddlewheel=flywheel induces an orbital couple (equal and opposite linear impulses [transverse direction to both radial and spin axis] acting on the raised paddlewheel, and central mass, respectively). This must, to conserve overall linear and angular momentum, be balanced by a change in paddlewheel spin angular momentum, over and above the differential owing to beam blue shift you gave. I haven't attempted detailed calcs, but imo it almost certainly leads ultimately to adopting a truly isotropic metric if full consistency is demanded, as was angling at in that other thread.

[EDIT: There is the thorny issue of just where momentum resides here. In EM a physically real 'stored field momentum' is posited such that e.g. in the Feynman disk paradox, where a mechanical angular momentum imbalance exists, compensating angular momentum resides in the crossed static E and B fields. Formally it works. And it's needed there because acceleration is absolute not relative wrt electrodynamic effect. Unlike EM though, inertial interaction betweem matter distributions assumes reciprocity of relative acceleration - at least in a Machian scheme. So whether 'stored field momentum' is ever neeeded to be invoked in a gravitational setting may be doubtful, but I'm not sure. ]

Last edited: May 9, 2012
8. May 12, 2012

jartsa

If the wheels have different angular momentums at the end, where did the extra angular momentum come from, or where did the lost angular momentum go??

9. May 12, 2012

jartsa

Hello again. I lost interest for a while.

But maybe there is something in these paddlewheels with horizontal spin axes things:

I would say that the wheels must have the same angular momentum at the end.

And it must look like the lower wheel is spinned up by a more powerful light beam.

So it must be so that when a wheel is lifted up, it slows down! And when a wheel is lowered down it speeds up!

Last edited: May 12, 2012
10. May 12, 2012

yuiop

The wheel that we raised up has extra angular momentum and energy, but it required energy to raise the wheel so no energy is gained or lost.

Sorry, but it is the other way around. When we lower clocks they slow down relative to a clock higher up. We could construct a clock based on an efficient flywheel and it will behave the same as other clocks and the flywheel slows down.

Relativistic mass is an unfashionable concept but it may be useful here. In SR when we accelerate an object the inertia of the mass increases. A similar thing happens when we lower a mass into a gravitational well. The inertia of the mass increases. In the case of the flywheel the moment of inertia of the flywheel increases as it is lowered and the rotation speed reduces due to conservation of angular momentum, so the total angular momentum of the flywheel does not change. The angular energy of the flywheel does appear to reduce as the flywheel is lowered, but if we attached a pulley and a generator to the lowering rope, we could recover the energy that the flywheel appears to be losing, so overall the total energy of the system is conserved.

11. May 12, 2012

jartsa

yuiop, Where does the extra __angular momentum__ come?

angular momentum, not energy

12. May 13, 2012

yuiop

There is no extra angular momentum. It is conserved. The angular momentum (L) is given by L = Iω where I is the moment of inertia and ω is the angular velocity. Any increase is ω is matched by an equivalent decrease in I and vice versa and L remains constant. If you put a sliding lead weight on the spoke of a spinning wheel, the weight moves outwards and the wheel slows down, but the moment of inertia of the wheel changes by a correspond amount and angular momentum is conserved.

Remember, it is the angular velocity or speed of the flywheel that is changing and not the angular momentum which is constant.

13. May 13, 2012

jartsa

When we are doing the comparison of the wheels and noticing that the angular momentums differ, how do we explain the difference? Where did the wheel with more angular momentum get the extra angular momentum?

14. May 13, 2012

yuiop

At which stage? When they are both back at the same level at the end?

15. May 13, 2012

jartsa

Yes, when they are both back at the same level at the end.

16. May 13, 2012

yuiop

I concede you may have a valid point here. Possibly I have got local and distant measurements mixed up earlier and arrived at the wrong conclusions. I will do some more analysis and see if I can figure out what should be happening.

17. May 13, 2012

yuiop

OK, I have given it some more thought and I am sticking with my original conclusion. The extra energy (and momentum) that the lower flywheel has when it raised to be alongside the higher flywheel comes from the extra energy required to raise it (eg the energy to power the electric winch that raises it up).

18. May 13, 2012

Q-reeus

Yuiop - that is not properly your earlier argument. The part, presumably dealing just with the effect of raising the paddlewheel, about change in angular KE is (and I agree there though probably not in fine detail), but not that wrt angular momentum L. Which was that the extra L is generated in situ - at the lower potential location, with no subsequent change upon raising the paddlewheel/flywheel. And I had also initially believed that to be the case too. But gave a subsequent perspective in #7 which evidently you have implicitly rejected. So you see no need for an action-reaction principle between flywheel and gravitating mass?

Last edited: May 13, 2012
19. May 13, 2012

jartsa

Hey Q-reeus, what do you think about a supermassive flywheel, that expands along the spin axis.

Well this is what I think:

Obviously spinning rate does bot change.

Observer in the flywheel says that spinning rate decreases. That's because the observer's brain speeds up, when gravitational time dilation decreases.

20. May 13, 2012

yuiop

That was my position and still is although I did not make it clear in the last post. When I said "The extra energy (and momentum) that the lower flywheel has when it raised.." I was referring to the fact that the raised flywheel has more angular momentum than the flywheel that was already at the raised position, but as you say, the angular momentum of the raised flywheel did not change during the raising and was already there, while the angular KE did increase during the raising (both as measured by a distant observer. A local observer that ascends with the raised flywheel, will not measure any change of rotation speed or energy or momentum of the flywheel as it is raised.

Well I did not implicitly reject it as far as I know, I simply did not understand what you were getting at in that post or what your final conclusion was. Also, you have never acknowledged that a flywheel rolling at relativistic speeds past an observer in SR, has different instantaneous momenta at different positions along its perimeter and that this messy feature of angular momentum in relativity is not just a feature of the Schwarzschild metric. You also suggested introducing the isotropic S metric, but I am not sure how that would help determine whether the flywheels have equal momentum or not at the final position.

As for the action-reaction principle involving the gravitating mass, I am not sure how that helps here either. For most practical purposes where we have huge gravitational mass and tiny test object, we only introduce a negligible error by ignoring the back reaction of the gravitational mass.

Last edited: May 13, 2012