Poss. Symmetry Break?

[MrVibrating]

Sooo, here's a quck linear test:



..as you can see, two 1 kg blocks, initial velocity 20 m/s, net P = 20 kg-m/s, net KE = 200 J, and the spring constant is 1 N/m^2.

As before, the blue mass's KE unloads into the spring, then at the simulation frame where its energy is lowest, it pauses for a few seconds (so you can read it), before releasing the rigid joint securing the red block... Pausing again at the next peak energy frame.

Full energy conservation, peak P for each mass is 14.14 kg-m/s, thus summing to a net P of 28.8 kg-m/s, which, divided by our initial 20 kg-m/s, gives... a certain ratio. I'm not going to dignify it with an enumeration, but i think we all know what ratio i'm on about...

[Fletcher]

Ok .. so when an energy storage device, like a perfectly elastic spring, is used as the intervention method between the two identical masses (where one is stationary) then we get a different result from a 'true' collision between the two masses (without the spring).

In the spring example the KE is conserved, and stored in the spring as SPE, before being imparted equally to both masses thru the spring action.

Since both masses will now be traveling in opposite directions they each have 100 J's (scalar) of KE at maximum, and the spring has zero SPE. To make the sums balance then the initial P of 20 units (vector) must now be positive 14.14 units and the other negative 14.14 units.

Total P is the sum of both which is zero. Where did this P go ? It went to ground/earth because the spring is anchored (thru the pinned mass 'B') and released - IOW's, the earth moved and gained 20 units of P. But the sim can't show that.

ETA: maybe it can if you use 'Player Mode' - I've never looked at it.

N.B. in a regular perfectly elastic collision example you will find that although it is a perfectly elastic collision the original moving mass 'A' with 200 J's of KE and 20 units of mv has come to a complete stop. Mass 'B' shoots off with all the KE and mv that 'A' had. Both KE and mv (P) are conserved. This is because there is no spring intervention where the mechanical medium is 'grounded'.

So, lets apply that to the dual identical disks scenarios, in terms of Angular Velocity and Angular Momentum ?

What happens when the rotary spring was used ? Where they turn in opposite directions.

What happens when a rod connection was used between displaced disks ? Where they turn in the same direction.

Are there parallels to the linear examples because the 1.414 ratio shows up in one situation ?

[ME]

Obviously momentum cannot literally be 'transferred', but merely induced by an exchange of energy

Interesting.
Perhaps I'm just rocking your boat - if that's the case then don't mind me and skip the rest.
---
According to my current understanding I think there only exists a literal transfer of Momentum, while 'energy' is a way of predicting the outcome in relation to something else.

When there is a constant Force (by a spring) then there is a constant exchange in Momentum over time (F=∆p/∆t=m·∆v/∆t =m·∆a or F=∆m·∆v/∆t), meaning a constant exchange in Velocities over time: The integral of ∆v over time (or the result of this all) is Energy (F=E/∆x).

In case of a collision there's no Force to begin with and there's no Force afterwards. It only exists during collision, and things get slightly different (F·∆t=∆p=∆m·∆v).
While ∆t is small, Momentum will only be exchanged in that instant.

[MrVibrating]

Quick thought: if energy was merely dissipated, not destroyed, in the direct-transmission test - say, by the friction of the chain and sprocket - despite the fact that the discs end up with the right energy for their momentum, which remember is 1.414 times more than we began with, then a perfectly lossless chain and sprocket would have left us with 2.828 times more momentum than we started with, at the same exchange rate.

This seems to weigh significantly against the possibility the missing 100 J has escaped, somehow, rather than being annihilated..

[MrVibrating]

Obviously momentum cannot literally be 'transferred', but merely induced by an exchange of energy

Interesting.
Perhaps I'm just rocking your boat - if that's the case then don't mind me and skip the rest.
---
According to my current understanding I think there only exists a literal transfer of Momentum, while 'energy' is a way of predicting the outcome in relation to something else.

When there is a constant Force (by a spring) then there is a constant exchange in Momentum over time (F=∆p/∆t=m·∆v/∆t =m·∆a or F=∆m·∆v/∆t), meaning a constant exchange in Velocities over time: The integral of ∆v over time (or the result of this all) is Energy (F=E/∆x).

In case of a collision there's no Force to begin with and there's no Force afterwards. It only exists during collision, and things get slightly different (F·∆t=∆p=∆m·∆v).
While ∆t is small, Momentum will only be exchanged in that instant.

These terms are not exclusive to momentum, but shared in common with energy. If momentum was directly transferred, rather than induced, then there'd be nothing stopping us transferring the momentum of a large slow mass to a smaller faster one, gaining energy, as Fletch and others have explained.

But what confounds these attempts - what is exchanged instead - is energy, from which momentum is induced uniquely and independently as a function of a specific body's inertia, on a case by case basis - just because it's an equal momentum doesn't imply it's one and the same entity, 'transferred' like some kind of metaphysical essence that leaps between bodies..

From the classical perspective it's usually OK to think in such terms, but the Higg's field is confirmed, we know mass is mediated by a gauge boson that interacts with half-spin particles, hence momentum is induced in much the same manner as relative motion between charges induces magnetic force. It's usually fine - and perfectly intuitive - to consider the magnetic field to be a property of permanent magnets. But it's not literally true, as Faraday's paradox reveals..

I think this same issue underlies the situation here - which is pretty much what you're getting at, i think - that the right amount of momentum has been induced for the available energy.

This i can't disagree with - it checks out perfectly, when cross-referencing the terms.

And so it's - apparently - trivial that the system's net momentum can rise. It's just surprising to me, i guess. 1 / 2 = 0.5, or 0.7, depending on whether we divide it directly or re-apportion its energy. We have a system of elementary simplicty - a simple transfer of P and KE between two identical bodies, yet we can destroy energy, create momentum, and even do both at the same time.. and it's all perfectly reasonable, nothing to get excited about, just basic maths.

Still more interesting than anything i've found in gravitational systems tho..

[MrVibrating]

Ok .. so when an energy storage device, like a perfectly elastic spring, is used as the intervention method between the two identical masses (where one is stationary) then we get a different result from a 'true' collision between the two masses (without the spring).

In the spring example the KE is conserved, and stored in the spring as SPE, before being imparted equally to both masses thru the spring action.

Since both masses will now be traveling in opposite directions they each have 100 J's (scalar) of KE at maximum, and the spring has zero SPE. To make the sums balance then the initial P of 20 units (vector) must now be positive 14.14 units and the other negative 14.14 units.

I'm sure i'm almost there, but could you just patronise me with the maths here? I saw straight away the relevance to the hypotenuse of a right triangle, and hence the signature of an orthogonal vector sum, yet here we have parallel vectors.. i just can't see how parallel momemta can yeild fractional sums. Doubles and quarters make sense, but 1.414..? Yet it's also the square root of two, and we're playing with squaring functions, and dividing them by two bodies... so that's where i was expecting to find resolution..

Total P is the sum of both which is zero. Where did this P go ? It went to ground/earth because the spring is anchored (thru the pinned mass 'B') and released - IOW's, the earth moved and gained 20 units of P. But the sim can't show that.

Yes! I'd been considering the role of the grounded momentums (acknowledging that the spring was compressing against earth when describing the tests) but you're absolutely right - the initial momentum hasn't been transferred or shared between the masses, hence, strictly speaking, it hasn't increased either...

Instead, we sunk the original momentum to earth, storing the corresponding energy in the spring, and then disconnected from earth, leaving it with that 20 units of P, and induced a new P of equivalent energy value divided by the doubled output mass.

I still have the same reservations about distinguishing objective momentum from equal opposite momentums (since such a system can likewise be described as having zero energy, even tho it may be conserved and available).

Still, now our net system momentum would be 20 units in the earth, plus 28 more between our masses, so technically we actually have 2.414 times more than we began with...

I think we're close to an understanding tho..

ETA: maybe it can if you use 'Player Mode' - I've never looked at it.

N.B. in a regular perfectly elastic collision example you will find that although it is a perfectly elastic collision the original moving mass 'A' with 200 J's of KE and 20 units of mv has come to a complete stop. Mass 'B' shoots off with all the KE and mv that 'A' had. Both KE and mv (P) are conserved. This is because there is no spring intervention where the mechanical medium is 'grounded'.

So, lets apply that to the dual identical disks scenarios, in terms of Angular Velocity and Angular Momentum ?

What happens when the rotary spring was used ? Where they turn in opposite directions.

What happens when a rod connection was used between displaced disks ? Where they turn in the same direction.

Are there parallels to the linear examples because the 1.414 ratio shows up in one situation ?

Compelling questions, and i'll come back and work through them tomorrow, but for now you've just reminded me of something..

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. So for example, an orbiting mass is rotated, and / or a spring wound, causing a counter-torque on the wheel, and then the mass / spring or whatver is moved to the axis to reset, earthing the reaction forces.

Hadn't made any kind of start on it, was just a one-line idea on a 'to do' list.. but it looks like things are going that way anyway.. lots of ideas to sift through here, will come back to it tomorrow and over the w/e..

[MrVibrating]

Quick thought: if energy was merely dissipated, not destroyed, in the direct-transmission test - say, by the friction of the chain and sprocket - despite the fact that the discs end up with the right energy for their momentum, which remember is 1.414 times more than we began with, then a perfectly lossless chain and sprocket would have left us with 2.828 times more momentum than we started with, at the same exchange rate.

This seems to weigh significantly against the possibility the missing 100 J has escaped, somehow, rather than being annihilated..

It was late, made a mistake there - the test of was thinking of was 'discs1.3' where a chain and sprocket suddenly becomes active, dividing the motion of one disc into two.. in which the momentum was conserved, but half the energy disappeared (so no 1.414* momentum increase in this case).

However the same point stands - if we still have all of our momentum, yet half the energy's missing, then it can't have been a dissipative loss.

Hence if we can run the same sequence of actions in reverse, we'll definitely gain twice our input energy...

[MrVibrating]

Just repeated the linear test, but with a slight difference - once the spring was compressed, i replaced the 1 kg mass with 2 * 500 gram masses, leaving the other end of the spring still attached to earth.

Upon releasing the spring, the 2 half kilo masses were both accelerated up to 20 m/s, with 10 kg-m/s each, so P and KE were conserved, with no funky ratio this time.

Obvious outcome, but just nailing down exactly which conditions cause the rise in P, and which don't, so had to get this one outa the way....

[MrVibrating]

Further extrapolating from 'discs1.3' - we could continually divide the conserved angular momentum - from two discs into four, then from four into eight and so on, each time halving our remaining energy, while keeping all of the momentum..

On the one hand, this looks a lot like dissipative loss, yet it is only momentum that is dissipating, along with the remaining half the the energy at each succesive step - the other half of which keeps disappearing...

So at the first division we go from 20 J down to 10 J, then 5 J, then 2.5 J and so on, maintaining our full system momentum throughout. We end up with a lot of mass going very slowly, instead of a little bit of mass going really fast.

So, in the interest of accurate reporting, it is a semi-dissipative loss mechanism. The portion of the energy of interest is the 50% that keeps disappearing at each division - the bit that isn't conserved, rather than the part that is.

Yet in this case, because the two quantitites change hand-in-hand, in order to reverse the non-dissipated portion (creating energy), we'd need to reverse the disspiated portion it is twinned with, and context-dependent upon...

And this is where we come up against 2LoT - not in trying to re-create the destroyed energy, but simply in trying to consolidate the portion that hasn't disappeared, but has nonetheless dissipated along with the momentum...


In other words we'd need a Maxwell's daemon to make this work. If we had one, it'd pay out double... a real double-whammy... but by the same token, i've heard that invisible pink unicorns pay out triple.. :|

[MrVibrating]

Moving on, i've been wracking my brains trying to come up with some useful application of a momentum gain, for a conserved energy.

Some way to trade it up for an energy profit...

Being able to buy and sell two different quantities of net momentum for the same net energy seems to be suggestive a gain margin, somewhere... and if we can cyclically gain momentum, violating CoM, then CoE should follow it off the cliff..

Just chewing it over, no firm conclusions yet..

[MrVibrating]

New idea - Slow a disc to half its energy using a rot. spring, which is then discharged into a 2nd disc.

This seems a good counterpart to the direct-drive 1-into-2 test ('discs1.3').

It's also easily reversible and presumably energy symmetrical in either direction.


The thing is, despite my earlier deference to 2LoT, there's still something i can't get my head around - why does energy rise by the sum when the KE's of two discs are added together, but drop by the square when the KE of one disc is divided into two?

Surely this is due to differences in the ways we're achieving these ends...

I've already shown that it's trivial to halt one disc and add its energy to a second via a spring, and 2LoT didn't seem to mind in the slightest. So, in retort to my previous assertion, we could keep transferring the momentum and KE of many slow discs, into progressively fewer ones, spinning ever faster... yet, unlike what happened on the way down, coming back up our net energy remains constant, along with net momentum.

So why is it, when descending through the same sequence of outcomes, that our net energy keeps reducing?

Why's net energy decrease as we keep dividing a given momentum into ever-more bodies, but not increase when combining an aggregate momentum into one body?

These are the questions, people! If 2LoT isn't physically preventing us from freely consolidating momenta, then we don't need a Maxwell's daemon.. the door's wide open and we're free to re-arrange the furniture as anally as we like.

So dissipation doesn't seem to be our problem. We can 'un-dissipate' every last crumb of momentum, back into a full hearty loaf.. of momentum. Type stuff.

Trouble is, the finished article only has the calorific value of the smallest crumb it was reduced to.

As if un-slicing a loaf reduces its nutritional value to that of a single slice.

I'll start a new round of tests tomorrow, halving the energy of discs instead of fully-draining them, to see if this helps zoom in on the cause of this asymmetry.

Right now i think i need a sandwich..

[Wubbly]

OK so obvious idea - i gave an inital velocity of 60°/s (10 RPM) to each disc, spinning in opposite directions, CW & CCW.

- we began with 244200 kg-m²-°/s, on each wheel, so 488400 in total.

Isn't Angular Momentum a 3-D vector that follows the Right Hand Rule rule? If you have equal and opposite angular momentum vectors, wouldn't they sum to zero kg-m²-°/s, not 488400 kg-m²-°/s ?

[MrVibrating]

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

Whether either are zero or non-zero is entirely context dependent:

- it is trivial to arrange for both discs to spin in the same direction (use a 'figure 8' transmission loop, or flip the rot. spring back-to-front when swapping it between discs, or whatever) - in which case the P's and KE's are no longer opposing..

- but even if the motions are equal and opposite, we can still transfer the P and KE of one disc back over to the other, which now has twice the P and KE it had previously... yet if we assumed the position that there was no net P or KE, then both have just been created from nothing!

- in a perfectly-elastic collision between equal and opposite momenta, P and KE is conserved - the masses will keep oscillating back and forth forever. Each half-cycle is a discrete interaction, and a system with latent activity has latent energy / momentum.

- if we wanted to be consistent with our glass-half-empty approach, a voltage accumulator contains an equal and opposite charge separation... yet if we just sum ions to anions, a fully-charged battery has zero energy..


Whether such systems have positive PE or zero depends on the properties of the net system with respect to some additional reference frame - for instance, a system of equal and opposite momenta and KE's has constant energy with respect to some other body or system - it's not accelerating or decelerating in relation to this other frame.

But that doesn't mean that system can't have positive PE in relation to its own internal reference frames.

For example a pair of oscillating masses - a pair of rubber balls connected by an elastic band, say - in geostationary orbit, has zero energy relative to the ground, but positive internal energy.

Taking that exact same point slightly further, the P and KE vectors of phonons in an isotropic body of heat are randomly distributed and amorphous, so could be summed to zero in much the same manner... ie. a lump of molten metal might not be accelerating... but that doesn't mean it's safe to stick in your pocket...!


Sooo, in these current systems here, we can get the energy back out (or half of it, anyway - we can un-dissipate the momentum, and at least half the original energy along with it), so our equal-opposing P's and KE's remain objective reserves, rather than subjective zero sums.


I still wanna work thru Fletcher's previous questions, and the last round of tests i mentioned, but been doing other stuff last couple of days, hopefully back to it later tonight..

[Wubbly]

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?

[Fletcher]

Hi Wubbly .. after your last post and diagram of the right hand rule for Angular Momentum I read up the Hyperphysics page on Angular Momentum and the wikipedia page.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

You have to go thru Mechanics > Angular Momentum.

Angular Momentum of a Particle ; Angular Momentum ; Angular and Linear Momentum

Rotation Vectors ; Direction of Angular Quantities ; Angular Momentum Change

https://en.wikipedia.org/wiki/Angular_momentum

Anyways, the upshot being that a rotating object has precession about the axis of rotation (as we know), presumably because of the right hand rule - so any torque applied to the axis is resisted. Most of us have seen a gyroscope or spinning top stand up etc.

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 ?

I note that in the wiki page it is described as a vector, a scalar, and a pseudovector in 3 dimensions.

I'm a little confused by how it is treated, your comments appreciated.