Poss. Symmetry Break?

[Wubbly]

The hyperphysics page on angular momentum is a little confusing because it shows omega twice. Once for the sense of rotation of the wheel, and once for the vector representing the angular velocity. The angular velocity vector is the one pointing in the direction of the thumb. The other omega is not the vector, it just shows you which direction the wheel is turning and only adds confusion by being labeled as omega.

You can get there by: Hyperphysics website -> Mechanics -> Conservation of angular momentum -> Angular momentum

If the wheel is rotating about the Z axis, then each little point particle on the wheel has a different radius and a different velocity in the X-Y plane. If you sum them, they would all add to zero (linear velocity, and linear momentum) and hence the need for Angular Momentum. The axis of rotation (Z axis in this case) is chosen for the angular momentum so a single number and a single direction can be used to define the motion.

If the wheel is rotating in the X-Y plane, the angular acceleration vector, angular velocity vector, and angular momentum vector will be perpendicular to that plane, along the Z axis in this case. The right hand rule is used to determine whether it is pointing toward the positive Z or negative Z direction.

If you look at MrV's post from here:
http://www.besslerwheel.com/forum/viewt ... 509#143509
you are looking at the X-Y plane. You can see that when a disk rotates CCW, the angular momentum is positive (positive Z in this case - out of the screen), and when a disk rotates CW, the angular momentum is negative (negative Z in this case - into the screen).

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If you look at the wikipedia link for angular momentum in 3 dimentions:
https://en.wikipedia.org/wiki/Angular_m ... dimensions
you can see that the angular momentum vector "L" is perpendicular to the plane of motion.
If you assume that the ball on the end of the rod is travelling in the X-Y plane, then the light green vector "L" is along the Z axis. You can use the right hand rule and see that when the ball is rotating CCW, the "L" is along the positive Z axis, and when the ball is rotationg CW, the "L" is pointing toward the negative Z axis.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Angular momentum is a 3D vector, but sometimes the vector notion is dropped to make it simpler to understand or explain. Or maybe it depends on the problem? If you have a uniform disk spinning, it has angular momentum about the axis of rotation, but if you have a mass moving along some trajectory and you choose an arbitrary reference point, you can have angular momentum about that arbitrary reference point.

Vector, pseudo vector - It's all a mathematical concept invented by mankind. You have this wheel spinning about an axis, and you have to come up with a way to describe it mathematically. Since the individual velocities and momentums all sum to zero (linear), man invented angular velocity and angular momentum as a mathematical tool to describe the motion. It was maybe arbitrary, but also somewhat logical to choose the axis of rotation as the reference point. The right hand rule was probably chosen since most of the population is right handed and they feel comfortable curling the fingers of their right hand in the direction of rotation. If we lived on a planet where most of the intelligent population was left handed (assuming we had hands), then maybe we would have invented the left hand rule (also assuming we had thumbs).



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Added:
Just looked at "Angular Momentum of a Particle" on the hyperphysics website. They describe the angular momentum of a particle as a vector product. Angular momentum is the vector "Cross Product" of the radius vector and the momentum vector. L = r x p. The "x" between "r" and "p" does not mean multiply. It means vector cross product.

If you look up "Cross Product" on wikipedia, you can see the resultant vector is perpendicular to the plane of the original two vectors.
https://en.wikipedia.org/wiki/Cross_product

Other than precessional effects is there any other significance to Angular Momentum that would make it substantially different from a linear momentum vector for comparison purposes ?

The Linear Momentum vector is in the plane of motion.
The Angular Momentum vector is perpendicular to the plane of motion.

[Mark]

I understand the concept of "handedness", as well as I understand that "peak" and "average" are two different (usually) measurements for the same exact thing.

But before a change of subject matter deviates the thread, I feel compelled to ask a question. In regards to;

MrV's experiment - Two unconnected counter-rotating discs, one unloads (completely?) into a rotary spring, time is stopped by the wizard, the spring is transferred to the other disc, time is resumed and the spring is unleashed.

and Wubbly's question - Shouldn't it nullify, instead of sum?

I ask, because I really don't know - Does a coiled spring retain handedness?

[edited to change position of "(usually)", for clarity]

[MrVibrating]

I've adressed this same point repeatedly - and we could say the same thing about their RKE (equal and opposite, summing to zero).

Energy is a scalar. Is RKE a vector or a scalar? I believe it is a scalar. So No, you cannot say the same thing about RKE if the sign is always positive. You cannot sum equal and opposite RKEs to zero because both their signs are always positive. If you summed them, you would get double the energy, not zero. Is there any such thing as negative KE or negative RKE? I don't think so. Is there any such thing as negative momentum or negative angular momentum? Yes, because they are vectors and have a direction.

Taking that exact same point slightly further, the P and KE vectors of phonons in an isotropic body of heat are ...

I've never heard of a KE vector before. Is this a new physics concept?

RKE is signed positive or negative (CW or CCW).

Likewise, the angular momentum those RKE's pertain to.

KE has magnitude and direction, and phonons are quasi-particles carrying both quantised momentum and KE. But dimensionality (scalar or vector) has little to do with whether the P and KE in question are equal and opposite.

According to your thinking, a system of equal counter-rotating rotors has zero momentum yet positive KE... surely the only way to reconcile such a paradox is to recognise that both AM and RKE are actually signed...

Likewise, (as already laboured) charge can be positive or negative, and the net energy of equal opposite charges is zero.

So yes, all new physics indeed, for some apparently.

TBH all you seem to be trying to offer here is petty pedantry, so while i always appreciate necessary corrections or clarifications, and balance, etc... that's not really the vibe i'm getting from you.. more snarky, sardonic and snidey pedantics, which you don't seem to have thought through yourself yet..

Sorry if i'm misinterpreting you bud.. :|

[Wubbly]

In your animation from this post here:
http://www.besslerwheel.com/forum/viewt ... 509#143509
The angular momentum is negative during a CCW rotation, and it is positive during a CW rotation, but Rotational KE is positive in each case. Your own animation contradicts what you are saying.

KE has magnitude and direction.

Can you provide a link to some physics site that shows where KE or RKE of a macroscopic object has a negative sign or a direction?

According to your thinking, a system of equal counter-rotating rotors has zero momentum yet positive KE...

Positive RKE and zero angular momentum. Yes. This is accurate because energy is a scalar (sum) and angular momentum is a vector (sum).

...surely the only way to reconcile such a paradox is to ...

This is not a paradox. Momentum and energy are two completely different concepts. Is it a paradox that mv and 1/2 mv² can both be conserved at the same time? No.

But dimensionality (scalar or vector) has little to do with whether the P and KE in question are equal and opposite

Scalar or vector has everything to do with whether P in question is equal and opposite because equal and opposite vectors sum to zero. Technically there is no such things as "opposite" energy, since it is a scalar and is always positive with no direction. We are talking about macroscopic rotating disks here, not charged particles.


You seem to be confused about the difference between a vector and a scalar when dealing with macroscopic objects. You can't invent energy of a macroscopic object to be a vector just because you wish it to be so.

[MrVibrating]

I understand the concept of "handedness", as well as I understand that "peak" and "average" are two different (usually) measurements for the same exact thing.

But before a change of subject matter deviates the thread, I feel compelled to ask a question. In regards to;

MrV's experiment - Two unconnected counter-rotating discs, one unloads (completely?) into a rotary spring, time is stopped by the wizard, the spring is transferred to the other disc, time is resumed and the spring is unleashed.

and Wubbly's question - Shouldn't it nullify, instead of sum?

I ask, because I really don't know - Does a coiled spring retain handedness?

[edited to change position of "(usually)", for clarity]

In that particular test, the blue disc's momentum was positive (CCW), all of its energy is transferred to the spring (near as i could time it anyway), and so the spring is wound CCW, and wants to unwind in the CW direction.

So when it's swapped over to the red disc, with its negative (CW) momentum, it unloads in the same direction that the red disc is already spinning.

Obviously, we could've begun with both discs spinning in the same direction, or even flipped the rotary spring back-to-front, letting it unwind in the same relative direction it wound up..

So whether the momenta interact additively or subtractively is entirely up to us. "Handedness" usually refers to things like chiralty, which can't really be ambidextrous in the same way that we can easily flip over a loaded rotary spring. The momenta and KE's are simply signed - the standard engineering convention is that CCW rotation is positive, CW negative, and so we have positive and negative momenta and KE's, but we can flip their signs at will


The key points, such that they are, from the last round of tests:

- i was investigating how nature 'knows' whether a relative velocity of 20 m/s between two 1 kg masses is worth 100 J, or 200 J, in the hope of maybe exploiting any ambiguities there.

- this led to the unanticipated discovery that we can effectively change the net momentum of an interaction by 1:1.414, rising by this much for successive divisions of momentum, and decreasing by the same amount when amalgamating momenta.


- that particular ratio crops up in may inter-related fields, but some relevant examples here might be a sum of orthogonal vectors of equal magnitude, and/or the square root of two. Still lacking a conclusive answer tho..


- it's probably trivial and useless; i can't think of any good use for more or less momentum, unless we could somehow trade it up for more energy, which so far doesn't seem likely.


- we do have a non-dissipative loss mechanism though - which is exactly the same kind of asymmetry as a gain... just backwards. We can divide one angular momentum into two (which could be equal or opposite), conserving net momentum while halving net energy (quartering the energy per disc).

- unfortunately though, we don't seem to be able to reverse that process; when we re-combine child RKE's they merely sum, not quadrupling back up.

- so we seem to be running up against the second law of thermodynamics; we can un-dissipate momentum, but only conserving whatever fraction of net system energy remained at the last division. Moreover, half of the lost system energy hasn't even been dissipated, but destroyed, and so it can't be 'undissipated', even if we could undissipate the rest, which we can't...

- i'm still fascinated by this asymmetry. RKE is different to heat in many respects (tho obviously related in others)... I understand why RKE's sum linearly. I also understand why RKE quarters when divided into a second angular momentum. But i'm still lacking a conclusive understanding of why either case isn't time-symmetrical - ie. why we can't recombine two child RKE's back into their original four-fold sum.


- plus we have Wubbly's helpful interjections; that momentum can't be converted into KE, and that child momenta always sum to zero, while their corresponding energies always add cumulatively (because scalars can't be signed, apprently), and thus we can have mechanical energy in the absence of mechanical momentum! Similarly, if we then re-unify the momentums back into one, we must've created it all from zero. Besides, angular momentum is a pseudo-vector's in the z-plane, so there, or something..

Although to be fair, i may have missed the full expediency of his points..


But whichever way you look at it, there's one or two interesting things going on here... especially the bit about under-unity RKE division, IMHO.. and also this consistent 1:1.414 net momentum change (remembering Fletcher's point that these interactions are actually all menagé á troix with the earth's inertia, rather than strictly closed systems.. apart from the aformentioned under-unity result, which is exclusively between two rotors).

Doing other stuff ATM, but will get back into this shortly..

[MrVibrating]

In your animation from this post here:
http://www.besslerwheel.com/forum/viewt ... 509#143509
The angular momentum is negative during a CCW rotation, and it is positive during a CW rotation, but Rotational KE is positive in each case. Your own animation contradicts what you are saying.

KE has magnitude and direction.

Can you provide a link to some physics site that shows where KE or RKE of a macroscopic object has a negative sign or a direction?

Why? I'm not arguing that it's a vector, simply that it shares the sign of the momentum. If two momenta have a zero sum but positive KE, then how is that KE manifest? KE=1/2mV^2... if the system retains mass but zero momentum, then it has no velocity, and thus whence the KE?

According to your thinking, a system of equal counter-rotating rotors has zero momentum yet positive KE...

Positive RKE and zero angular momentum. Yes. This is accurate because energy is a scalar (sum) and angular momentum is a vector (sum).

So then where does the momentum come from when we combine two opposing motions back into one? If it's been conserved, then it was never actually zero.

...surely the only way to reconcile such a paradox is to ...

This is not a paradox. Momentum and energy are two completely different concepts. Is it a paradox that mv and 1/2 mv² can both be conserved at the same time? No.

How can you have non-zero 1/2mV^2 if m * V = 0? This is the very pinnacle of the nexus of the epitome of the quintessential essence of a "paradox", sir. It is a non-sequitir; your conclusions do not follow from your predicates. You're mis-applying concepts in a confused and self-contradictory attempt at pedantry, instead of offering the solution you think you are. I'm asking why net momentum varies by 1:1.414, and your 'answer' is that there is no momentum!

Again, we needn't divide a momentum into opposing child momentums - both could share the same sign... and the net change in momentum would still[ be 1:1.414, and your argument that there wasn't actually any momentum at all wouldn't have a leg to stand on.. and so it doesn't here - if the momentums are equal or opposite, it makes no substantive diference whatsoever to the net reserves of KE or momentum, or our options for dividing or combining them. Obviously we can't create momentum from nothing, in the latter case, hence it was self-evidently conserved as equal opposite quantities, not cancelled.

But dimensionality (scalar or vector) has little to do with whether the P and KE in question are equal and opposite

Scalar or vector has everything to do with whether P in question is equal and opposite because equal and opposite vectors sum to zero. Technically there is no such things as "opposite" energy, since it is a scalar and is always positive with no direction. We are talking about macroscopic rotating disks here, not charged particles.


You seem to be confused about the difference between a vector and a scalar when dealing with macroscopic objects. You can't invent energy of a macroscopic object to be a vector just because you wish it to be so.

LOL where have i "invented" energy? Besides, you're creating and destroying momentum.

FYI, scalars can be signed (like temperature scales etc.), including RKE, and i have owned and operated several macroscopic objects. Trolling over irrelevant details is totally gay, please explain this 1.141 thing or why the under-unity RKE divisions can't be reversed, or else quit the feckless nitpicking? You're subverting the discussion without offering anything productive..?

[MrVibrating]

Please, everyone, can we draw a line under this whole "there is no spoon!" sophistry? I've repeatedly acknowledged it, and tried to put it to bed, it's a totally unproductive angle since the momentum self-evidently is conserved and constant and freely-transferable throughout.

What IS interesting is its corresponding energy value, and how it changes when dividing, vs recombining.

Please can we talk about this instead?

[Mark]

Okay, so.....

The spring loaded with one handedness, and unloaded with the opposite handedness.
And that is why the sim program showed a greater sum, rather than a zero sum.

-- -- -- -- --

I don't know why that didn't occur to me right away. Guess my old brain needs to think more, and speak less.

My apologies if I stalled anyone's train of thought.

[Mark]

What IS interesting is its corresponding energy value, and how it changes when dividing, vs recombining.

Please can we talk about this instead?

I would very much like to see you guys develop the topic of a Possible Symmetry Break further.

And MrV, I am particularly interested to see your posts regarding this comment:

When i initially jotted down this concept, the night before starting this thread, i also noted two other possible leads.. one of which involved transfering momentum from a wheel, via the counterforces induced by unbalanced orbiting rotations, to the axle, and back.

[MrVibrating]

I got to the root of the 1:1.414 thing, and it is indeed trivial..

I performed the previously mooted test wherein only half of a disc's RKE is drawn off into a spring, which is then transferred over to a second stationary disc, accelerating it with the stored PE.

Again, this resulted in energy unity, and this 1:1.414 momentum rise.

So then i tried a slight variation, instead using the spring to draw off half of one disc's momentum, rather than its RKE, and then transferred that over to a second static disc as before.

This resulted in a 1:3 RKE distribution - the 2nd disc having 3 times more RKE than that remaining on the first disc... and if i'd anticipated that the distribution of momentum would now be 50:50... it wasn't - as well as having three times the energy, the 2nd disc had 1:1.73 times more momentum than that remaining on the first..

So then i tried a number of quick variations in rapid succession - transferring all of one disc's RKE over to a larger, or smaller disc, quickly concluding that exchanging a given quantity of energy for absolutely any amount of momentum, large or small, is entirely trivial and limited only by practical concerns


I recalled encountering the sqrt2 signature earlier, and a quick check back thru the thread shows that it's reciprocal, 0.707, arose on page 7, as the change in radial distance corresponding to a halving of the respective MoI, following (MoI = m * r^2), so, as previously assumed, it's just an incidental consequence of dividng one mass's KE into two identical masses.

Bottom line is that we could put 1 J into a large mass accelerated to a low velocity, or a smaller mass at higher velocity - two different momentums, same energy, utterly trivial, something we all knew all allong..

[MrVibrating]

What IS interesting is its corresponding energy value, and how it changes when dividing, vs recombining.

Please can we talk about this instead?

I would very much like to see you guys develop the topic of a Possible Symmetry Break further.

Yesterday i replicated the result without using the chain and sprocket transmission, instead simply letting the two discs roll against eachother, edges in contact, switching collision detection on and off (which is magic, but sufficiently equivalent to opening up a small airgap between them).

Again, this resulted in a perfectly equal division of momentum, all the original momentum conserved.. but with only a quarter the original energy residing on each disc.. half of the system energy disappeared.

Again, both discs have exactly the right amount of energy for their respective MoI and velocity. As such, no energy is strictly "missing". We still have all of our original momentum, so the absent energy hasn't been dissipated to heat. Everything is perfectly consistent... half of our system's KE has simply vanished without trace.


In principle, then, we can repeat this process, repeatedly bifurcating the angular momentum in successive stages, first reducing it by 50%, then destroying half of that remainder, so now down to 75%, and so on... converging to an infinitesimal after a few steps.

In principle, we could use such a system to brake a freight train, destroying megajoules without dissipating any of it as heat...

Calorimetry would just confirm the maths - that the 50% energy shed at each stage no longer exists - its reason for being, obviated - the mass and velocity it represented no longer applying.


And so this is a classical symmetry break - an asymmetry between the net input vs net output energies.

It is exactly the same beast as a gain. Its opposite twin.

And MrV, I am particularly interested to see your posts regarding this comment:

When i initially jotted down this concept, the night before starting this thread, i also noted two other possible leads.. one of which involved transfering momentum from a wheel, via the counterforces induced by unbalanced orbiting rotations, to the axle, and back.

Well, the thought was, that moving a mass on a wheel induces a counter-torque due to Newton's 3rd law, whereas if it can alternately push against the earth instead, then a complete interaction or cycle of said mass could remain open instead of closed, momentum-wise.

The objective would be to try to thwart the normal balance of counter-momenta that otherwise enforces momentum symmetry - ie. to sidestep Newton's 3rd law. As shown previously, N3 breaks can generate energy in subtle and unintuitive ways - such as circumventing the usual requirement for acceleration costs to scale exponentially with velocity; so the internal cost per unit of acceleration could remain constant (much like momentum), while externally, net energy still equals 1/2mV^2.

The basic concept is buying torque from within a rotating frame without applying counter-torque to a physical stator or the wheel itself.

If Bessler was doing something like this then the weights heard landing on the descending side - which Wolff had speculated must have somehow gained additional energy as they fell - might have been 'fly-weights' rather than 'drop weights' - their purpose, to transfer RKE / momentum invested by some other falling mass, having side-stepped the usual counter-torques in accerating it... perhaps by something as simple as a weight spooling off the axle, or something..


Not a particularly solid hypothesis yet, just looking for one really..

Something i may try shortly is loading the RKE of orbiting discs into springs etc. and swapping between the rim and common axis between loading and unloading... not sure where it might lead, just an investigative angle, to see how P's and KE's stack up..

[MrVibrating]

Trying to understand why this non-dissipative loss resists time-symmetry.

I suspect it may be because the MoI of the system doesn't instantaneously halve when using a spring, as it doubles when using a sudden direct drive..

Not a particularly pithy summary, and like i say i don't currently fully understand it.

If we run the experiment in reverse, it looks like this:

- Two equal discs spinning equally, via direct drive (meshed or rolling or belt-drive etc.)

- One stops, and all of its momentum instantly transfers to the other.

- The spinning one now has four times its initial energy, and net system energy has doubled.


Works like a charm in the forwards direction, but something's non-reversible in step two...





This is presumably due to the second law doing its thang, but just how it applies - and thus how it may or may not be circumvented - eludes me.

The deference of CoE to CoM on the way down, is reversed on the way back up - we can reunify all the system momentum back into a single rotor, yet eviscerated of its original energy value.


The reason i've been banging on about non-dissipative losses for the last few years is that i learned from Steorn that they're the oft-overlooked corollary of the much-sought gain - being "under-unity" in precisely the same manner as their inverted twins, yet easily dismissed as worthless, since energy is lost instead of gained.

Magnetic entropy viscosity, in electromagnetic systems, causes dissipative energy loss due to Joule heating. Yet in passive magnetic systems, the same losses are non-dissipative - it is trivially easy to destroy energy using a permanent magnet and an old lump of iron, by simply prising them apart against a greater force after the induced field has reached saturation some time after the parts attracted together. The opposite of this effect is basically "Orbo" tech.

Yet in two long years of researching gravitating systems i never found a single non-dissipative loss. This would be the first mechanical example i've encountered, if it stands to scrutiny.

Turning a passive Sv loss into a gain isn't a straightforward process, yet it's possible. So the hope is that maybe the same will be true here - a successful implementation might look radically different to a simple pair of interacting masses... or may just end up seeming flat-out impossible. Whatever, for now this is a foot in a door previously locked out, bricked over, plastered up and wallpapered.. (and not in a 'fun night out' way.)

Any tangible example of energy destruction is a white rabbit we're compelled to follow..

[MrVibrating]

Just thought of a possible angle on this non-dissipative loss - work has been performed in decelerating a mass...

To undissipate the 100 J back onto disc 1, we'd need to spend 25 J un-accelerating it, and another 25 J un-decelerating disc 2 - at which point we'd have just 50 J left on disc 1, having spent the other 50 J doing work in changing the mass's speeds. All 100 J apparently accounted for, in terms of energy spent, and energy remaining.


So we start off with, say, 100 J on one disc, then share half it's momentum with another, and each ends up with 25 J, for a 50 J net remaining.

So we began with 100 J, and ended with 50 J left, 25 J on each disc.

Disc 1 has given 25 J to disc 2.

It has also performed 25 J of deceleration work upon itself.

So it's spent 50 J, of its original 100 J budget.

The 25 J it gave to disc 2 is still there, conserved as RKE.

Likewise the 25 J it kept for itself.

So that's 50 J all present and correct.

But it also performed 25 J of deceleration work upon itself - it spent energy, to change its speed.

And because that workload decelerated a mass, it is irreversible - we've spent KE to cancel (decelerate) KE, and now we can't un-decelerate it as that energy's not conserved in the form of motion - we invested it in negative momentum, instead of positive...

!

That still leaves another 25 J unaccounted for, but also points to a way out of the stalemate - we'd need to arrange a system wherein a deceleration with respect to one reference body, is an acceleration with regards to another... after all, whether a change in velocity is positive or negative is subjective - especially in a rotating frame..

In such a scenario, a disc that was stationary in our external frame would nonetheless effectively have positive RKE with respect to some other mass rotating about the same axis.. IOW a mass oscillates between two alternate FoR's, and when one FoR sees an acceleratory workload, the other sees a deceleratory workload, while objectively, both are simply spending equal energy to change the mass's rate of motion, and whether it has accelerated or decelerated is entirely subjective..

Then we'd still have a reserve of KE with which to undissipate the lost RKE by undecelerating the quasi-accelerated mass..



_____
TL:DR

Basically i'm still stuck on relative vs absolute motion.. And i guess since a static rotor experiences no CF, rotation is absolute in a way that linear motion isn't, since rotation is also acceleration..? In which case it seems unlikely a static disc could ever have positive KE compared something actually rotating..?

No idea what i'm on about.. just thrashing it out...

[MrVibrating]

The above notion is clearly gibberish - decelerating disc 1 is not a separate workload to accelerating disc 2, to be counted twice.. it's the same energy.

Work has only been performed by disc 1 upon disc 2 - it's completely extraneous to suppose that disc 1 has 'performed work upon itself' in decelerating, and would be counting the same energy twice.

I was trying to make the loss go away, the same way you would for a gain, but if this is indeed a real symmetry break - energy destruction - then it can't be explained away, without invoking another violation (intentionally or otherwise) somewhere else along the line.

I think the only valid point i might have been grasping at is that if we have some energy, we can use it to get rid of it.. Whereas, when we don't have some energy, we can't employ its absence to make it appear... that's basically why it's currently irreversible... i think.

[Fcdriver]

The acceleration would have to be very slow and steady, using less energy than it gave.