# Flywheels and gravity in GR

1. Jul 4, 2011

### yuiop

This is a diversion from the "Interesting energy problem" thread. That thread lost its direction a bit lost due to implications of Perpetual motion machines or over unity systems, but in this thread it is my intention to avoid those complications so that maybe we can have an sensible discussion.

I have two simple scenarios.

First scenario.

We have two identical flywheel devices. When spun up to speed they retain their angular momentum for a long time due to very efficient bearings. Once per revolution they emit a light signal giving an indication of their rotation speed. They are both spun up to X rpm and their flashing beacons flash at Y Hertz. One is slowly lowered down a shaft into a gravitational well, to a depth such that the gravitational red shift factor is 4.

Q1) What is the rpm of the flywheel as measured by a local observer at the bottom of the shaft?
Q2) What is the flashing rate in Hertz of the flywheel at the bottom according to an observer at the top?

Second scenario.

This time one flywheel is stationary at the bottom and the other is stationary at the top. Both flywheels initially have zero angular momentum. The flywheel at the top is spun up to X rpm. The flywheels have a radius of 1 metre and mass of 4 kg, so the moment of inertia of the flywheels is 2. Assume the energy required to spin up the top flywheel to X rpm is Z joules. The flywheel is connected to a generator and then to a light source that is transmitted downwards. The light energy is collected by a transducer and the bottom and powers a motor to spin up the lower flywheel. Assuming negligible power losses:

3) What will the rpm of the lower flywheel be according to local observer when all the energy from the top flywheel is transferred to the lower flywheel?

4) What is the flashing rate in Hertz of the flywheel at the bottom according to an observer at the top, when all the energy is transferred?

This is not a homework or textbook question. It is a direct extension of the previous thread here: https://www.physicsforums.com/showthread.php?p=3387677#post3387677

2. Jul 4, 2011

### Q-reeus

Not that it should matter, but as you didn't specify spin orientation, I will assume both spin axes are along the direction of g. Thence:

A1) Unchanged - still X rpm. Local physics doesn't change on descent, and there is zero gravitational torque acting at any stage.
A2) (X/60)/4 = X/240 (or Y/4 Hz). Why must you use mixed units here - rpm and Hz! Redshifted according to factor 4. It is after all acting as a clock.

[EDIT: One could argue differently to my A1) as follows: Both rest energy and rotational KE should equally redshift on slow descent. Hence, given that flywheel radius is unaltered, and rest mass redshifts by the factor 4, for a corresponding rotational KE redshift, rotation speed should drop only by the factor sqrt(redshift) = 2. This in turn implies the locally measured speed has risen to 2X. But I will stick with A1) as given. As I recall it from Bowler, coordinate values of inertial and gravitational mass go up to offset that scenario.]

This one-way transfer boosts energy (locally measured) by the blueshift factor 4, but since KE goes as rpm squared, rpm only goes up (locally) as the square root of 4, hence is 2X.
Twice the answer given in A2) - ie X/120 Hz (or Y/2), for the reason given in last answer - speed was boosted by a factor of 2 relative to first scenario.
Right yuiop, and btw care to answer whether my entry #56 there was correct iyo?

Last edited by a moderator: May 5, 2017
3. Jul 4, 2011

### bcrowell

Staff Emeritus
Q1: It's still X. A clock doesn't change its rate as measured by a local observer just because you transport it somewhere else. (If it did, it would violate the equivalence principle.)
Q2: Y/4
Q3: Light energy Z is emitted at the top, and it becomes 4Z at the bottom. Since the kinetic energy is proportional to the square of the angular velocity, the bottom flywheel spins up to 2X.
Q4: To a local observer at the bottom, the flashes are at frequency 2Y. This becomes Y/2 as measured optically at the top.

[EDIT] I posted the above answers without peeking at Q-reeus's post, then checked them against each other, and we seem to agree on all four.

4. Jul 4, 2011

### yuiop

Agree, the flywheel is effectively a clock.

Agree.
Agree.
Agree.

So we are all agreed on the answers, but it raises some further questions for me. I find it curious that when we transmit the energy downwards we get twice as much energy at the bottom than when we transport the flywheel downwards. Where has the "missing energy" of the transported flywheel gone?

Also, I am curious about the conservation of angular momentum in a gravitational field. Locally, the angular momentum is conserved, but what about from a coordinate point of view?

Now we agree when the flywheel is lowered in the gravitational well, its velocity slows down by the gravitational gamma factor and assuming only sub relativistic angular velocities, this would suggest:

$$\omega = \frac{2L}{mR^2} \left(1-\frac{2GM}{rc^2} \right)$$

where $\omega$ is the coordinate angular velocity, L is the coordinate angular momentum, m is the mass of the flywheel and R is the radius of the flywheel.

Rearranging this gives:

$$L = \frac{mR^2\omega}{2} \frac{1}{\left(1-2M/(rc^2) \right)}$$

If L is to be constant, this suggests that the coordinate increase in mass causes the decrease in angular velocity of the descending flywheel, in order to conserve momentum. Does the above seem reasonable?

Yes, I meant and should have specified that the flywheels rotate around a vertical axis.
I will have to think about your alternative argument. It seems that it is the angular velocity that redshifts by a factor of 4 rather than the kinetic energy. Not sure though.

Tend to agree, but something is still bothering me. In the other thread, it seemed to make sense that the bottom of a vertical rotating rod directly connecting the bottom to the top would be turning faster at the bottom by the redshift factor and not the square root of the redshift factor. Clearly I need to think about this some more. Any further input would be welcome.

I agree that the power at the bottom is the square of the power transmitted from the top for the reasons you gave. However, I was concentrating on energy as it is energy that is generally conserved, while power is not a conserved quantity.

Last edited: Jul 5, 2011
5. Jul 4, 2011

### bcrowell

Staff Emeritus
If you're comparing locally measured KE of the flywheel at the bottom against locally measured energy of the light at the bottom, don't you get a factor of 4, not 2?

Assuming that this is an asymptotically flat spacetime, we have conserved scalar measures of mass-energy like Bondi mass and ADM mass. Presumably in the weak-field limit these would approximately equal something like rest mass plus KE plus PE plus energy of electromagnetic waves. But this isn't the weak-field limit, and even if it was, it seems to me that you're incorrectly omitting things like PE.

I believe that in an asymptotically flat spacetime, there is a conserved angular momentum. I don't see anything in this example that seems to violate that. I guess one thing to play with, if you're interested, would be to invent some cyclical processes and convince yourself that there is no net gain or loss of angular momentum in each cycle.

6. Jul 5, 2011

### yuiop

Yes, I think you are right, 4 not 2. That means we can get 4 times more energy to a lower station by converting and transmitting the energy rather than transporting the flywheel (assuming negligible energy losses in the transducer devices). I can partly mitigate this loss by raising a spent flywheel using a pulley device, while lowering the charged flywheel, but that does not explain all the difference.
Sorry, but Bondi or ADM mass is beyond me. Unlike pervect I am just a mere mortal. :tongue2: However PE is an interesting subject. Is it just a book keeping exercise? Do we just sum up all the energy and call any missing or additional energy the change in potential energy or can we localise and identify where this potential energy is?

Here is an very interesting cyclic process that is worthy of closer inspection. Let us say we have the same basic set up as before with a charged flywheel at the top and and a discharged flywheel at the bottom. The top flywheel initially has Z joules of energy stored in it. We transmit the energy of the flywheel to the lower flywheel using the light source. The lower flywheel is now rotating 2 times faster and 4 time the energy of the the top flywheel in its initial state as measured locally. We now raise the lower flywheel and lower the top flywheel using a pulley system and we end up with a flywheel at the top that is spinning twice as fast and has 4 times the energy as the original top flywheel and a spent flywheel at the bottom. We agree the rotation rate and the energy of the flywheel as measured locally does not change as it transported up or down. According to this analysis (which is obviously flawed), we have made a net gain of energy for free. Can anyone spot where the mistake is?
Using this equation I gave earlier:

$$L = \frac{mR^2\omega}{2} \frac{1}{\left(1-2M/(rc^2) \right)}$$
which no one has objected to yet, I have convinced myself that angular momentum is indeed conserved locally and on a coordinate scale. This is very useful, as it at least gives us one conserved quantity that clearly conserved in GR whereas energy conservation in GR is not at all clear (to me anyway). Did not Noether's theorem demonstrate that energy is not conserved in GR? As I recall she said something like there are proper theories where energy is conserved and improper theories where energy is not conserved and and GR is an improper theory?

7. Jul 5, 2011

### Q-reeus

This just addresses issues in entry #4, and I see there has been further correspondece since. Anyway....
This tends to be a tricky area. When it comes to assigning coordinate values to such things as mass and charge, turns out that in one consistent approach there is necessarily a 'splitting' process required. Mass is no longer just mass, charge is no longer just charge. In order to reconcile various relationships such as W = mc2, KE = 1/2 mv2, Fe = qE, Fm = qvxB, Fg = Gm1m2/r2, Fa = ma, etc, the *local* equivalence of active, passive, and inertial mass no longer works upon tranlation into coordinate measure, and similarly for charge. Bowler for the most part addresses all this in 'Gravitation and Relativity', ch6. Note though the approach there is to assign all values to matter, none to the gravitational field. Which works as a formal arrangement, but sidesteps the 'elephant in the living room' problem of gravitational field as source of an ill-defined enrgy-momentum density but not as source of further gravitation. A somewhat 'delicate' issue best not carried further here.
But it is turning faster by the blueshift factor! Thing is because of the squared power factor there is also an equal blueshift of torque. It all then balances nicely, up and down, not only with your motor generator setup, but the meshing gear arrangement of the OP in that other thread. No need to consider Born rigidity etc.
Umm..., over any interval where steady operation applies - assumed I would think in the setup there, the two are synonymous surely?

8. Jul 5, 2011

### Q-reeus

Just got around to reading this bit. Sure the raised flywheel has 4 times the energy of that in the original top flywheel. But the 'magic' is in the pulling up process - you have to work 4 times harder to get it's now increased energy-mass back up to the top! Do I get a first prize of a trip to DisneyLand?:tongue:
[EDIT: to round the above out a bit, note this is highlighting the difference between a dissipative process and a non-dissipative one. In merely lowering slowly the spinning flywheel, work is done that is transferred to the lowering winch - an effectively dissipative process. Hence at the bottom, there is reduced (redshifted) mass. A reverse winch back up simply reverses this process - original mass & KE is restored *via the winching process*. When alternately energy is *non-dissipatively* added to the bottom-dwelling flywheel, a winch up has to deal with that added mass-energy.]

Last edited: Jul 5, 2011
9. Jul 5, 2011

### bcrowell

Staff Emeritus
The equations for calculating them are complicated, but the concept is simple. Basically if you're in an asymptotically flat spacetime, then you can go far away from the sources, where the field is weak and everything is Newtonian. You measure the field out there, and use Newton's law of gravity to solve for the mass. The difference between Bondi mass and ADM mass is that Bondi mass doesn't include energy being radiated away to null infinity by gravitational waves, whereas ADM mass does.

For technical reasons, Noether's theorem doesn't apply to GR. The symmetry group is the group of diffeomorphisms (smooth changes of coordinates), and it doesn't produce anything useful when you put it in Noether's theorem.

10. Jul 5, 2011

### yuiop

Ah, OK. I was not aware Noether's theorem does not apply to GR. Thanks.

Now back to the original experiment. So far we all think that the lower flywheel will spin twice as fast and have 4 times as much energy as the top flywheel started with, after the energy is transmitted, as measured locally.

Now if we add a long vertical rod to the bottom flywheel (please humour me here) then what would the rotation speed of the top part of rod be measured to be by the top observer?

(Please assume greater than 95% efficiency in the system and the top part of the rod is not attached to anything other than perhaps a support bearing.)

11. Jul 5, 2011

### pervect

Staff Emeritus
It's a bit misleading to say that Noether's theorem doesn't apply to GR. If you have certain symmetries of the metric, like a metric that doesn't depend explicitly on time, or position, then you can use Noether's theorem to explicitly find a conserved quantity (Energy or momentum, or angular momentum, for instance, depending on whether the metric is time-invariant, position-invariant, or angle-invariant.

Furthermore, historically, Noether's theorem was developed specifically to explain why GR had the problems it does with energy. So, saying it doesn't apply is a bit of a cognitive dissonance. Hilbert was puzzled by the issues he was having with finding that GR didn't have the sort of conserved energy he expected, and he asked Emily Noether to look into it. Her famous theorem was the result of that investigation. It's a negative result, but it's still significant, it pins down to some extent why one has so much trouble finding a conserved energy in GR.

12. Jul 5, 2011

### yuiop

I was thinking something along those lines but I certainly would not have had the language skills to come up with something like "cognitive dissonance". Anyway, is energy conserved as a pair along with something else like momentum in GR?

Well, maybe not the trip to disneyland, but a virtual cookie is on its way. To sum up your position, the additional relativistic mass of the charged flywheel due to its angular kinetic energy requires more energy to raise back up again then an uncharged flywheel, presumably accounting for the extra "free" energy.

Still, I would like to hear from anyone that would like to put a number to the question I posed in #10.

Last edited: Jul 5, 2011
13. Jul 5, 2011

### bcrowell

Staff Emeritus
I think we're talking about two different things here. I was talking about diffeomorphism invariance of the field equations. You're talking about Killing vectors of a fixed spacetime, and things like conserved quantities for world-lines of test particles.

14. Jul 6, 2011

### Q-reeus

Thanks for the vrtual cookie - virtually very tasty! But yes, your summary matches.
OK, to quote from that:
X/2 - since it is rotating twice the speed of that when the original X rpm flywheel was lowered down and redshifted by a factor 4. Redshift given by SC's must correspond to the 'physical' measurement up top otherwise there would be a continual twisting deformation going on, whereas we assume a steady state with a tortionally unstressed rod, yes?

I'm going to take this opportunity to correct some of my comments in #7. On taking a closer look at Bowler's approach to assigning coordinate values, potential dependence is typically attached to the 'constants' such as c, G, mu, epsilon, etc, rather than mass, charge. Even there, seems he was assuming isotropic SC's, where one can, at least for the most part, get away with using a scalar transformation for such entities. In standard SC's though this is not possible (light speed c and distance are explicitly orientation dependent), and tensor relationships would be mandatory in general. A consistent assignment of what to what could get quite tricky - I suppose it has been categorized in some standard textbooks. Just a general observation - I'm not 'into' tensor math. In your scenario where spin axis is vertical, one doesn't really encounter the difficulty, but if you had specified horizontal spin directions, the tensor nature in SC's would rear it's ugly head. You might be inspired by this to pose a situation, for instance, of a horizontally spinning 'dumbbell' type flywheel, and ask how angular momentum would in coordinate measure be invariant for any instantaneous orientation of the 'dumbbell'. (I have my own reasons for not wanting to work it all out in SC's. I'm terming it 'the SM DOG'. Seems I'm the only one hearing it barking loudly, but to be safe am keeping it muzzled and on a tight leash. Enough said!)

15. Jul 7, 2011

### yuiop

Well yes, we want a rod that is in a steady state of stress rather than increasing stress leading to structural failure, but I am still not convinced this has been adequately solved yet. If we turned the whole experiment upside down, we could take the energy from the charged flywheel low down and beam its energy up to the top where the photons have been redshifted down by a factor of 1/4 so that energy would be used to spin the top flywheel up to 1/2 the speed the lower flywheel had and then a vertical axle attached to the top flywheel would have a turn rate of 2 at the bottom due to the blue shift factor of 4. Doing it this way we have doubled the speed of rotation of the shaft (so we have invented the worlds first continuously variable gravitational gear box). Now the question is how much torque is developed at the bottom of the vertical shaft and what speed would this turn another flywheel attached to it at the bottom? Are we sure the energy and momentum figures are all accounted for correctly here?

16. Jul 7, 2011

### Q-reeus

Getting very late for me, so just a short answer. Basically what you propose is I believe just a time reversal of the first setup. Running the camera backwards so to speak should work just fine!:zzz: