Inegnuity v's Entropy 2 - Observations & Questions

[Ed]

James, in your typical colorful style of writing, I took "needling Greenie lessers" as people who where needling you with the intent of lessing [sic] your greenies. If that was not the intent of that phrase, then please except my apology.

Anyway, I could tell you who my 'partner' is likely to side with, but let's just say that if you were to start a "you're always against me" conspiracy rant with his name on it, that he would very likely behave similar. :-)

Please, just think twice about airing your dirty laundry in public. You did the same thing with me when you were PMing me. Abruptly stopped and then started posting flaming messages about me in public. Could you not have tried even once to explain what might be setting you off first? Instead of doing so in public? This is what seems to have happened here with you and Fletcher. No?

Here is a question for you, similar to your question to Bill. Have you ever donated to this forum? Relax, it's just a question. :-)

[Silvertiger]

Hi Silvertiger,

Just out of interest, how would you describe your own profile?

TLW

My personal preference is to build others up and encourage them and their success rather than spending my time finding clever ways to tear them down. If they want to have pissing contests, they shouldn't bring it to other people's threads. It's very rude, and shows they have no consideration for anyone but themselves. The dog-eat-dog mentality, if I were to adopt it, would probably give me acid reflux. Nobody ever wins; they keep going back and forth and on and on and on...the only thing that is perpetual here: the incessant ongoing blathering that interrupts everything productive. They should be both disgusted and embarrassed with themselves - old men acting like 14-year-old drama queens. If they have a disagreement, it should be polite, held in check, and pertaining to Fletcher's topic, since he's clearly invested much of his personal time and resources into compiling this research and, moreover, has been considerate enough to share his results with anyone who is interested. But, far-be-it, and pointless, for me to defend proper etiquette since these members would, no doubt, consider themselves to be above reproach. Right now, I am interrupting this thread, and I apologize.

[Trevor Lyn Whatford]

Hi Silvertiger,

I would agree with you above post, so I also apologizes for interrupting Fletcher's thread.

Given that this is Bessler's Wheel forum, do not be surprised if it attracts people of a similar profile to Bessler's.

Hopefully this will all have blown over by the time Fletcher comes back, I just hope he is enjoying him self so much, that he takes 3 weeks away from here.

Silvertiger quote,

and the intellectually challenged yet deluded ones who don't know it will never thirst for real knowledge since their cup is already full. If anyone thinks that they know someone who fits the aforementioned profiling statement, do not talk to them. Ignore them. It's that simple. Nuff said.

I hope this excludes anyone who builds (experiments), by this very act, of putting their theory's to the test, must show thirst for real knowledge.

We are looking for a anomaly in known physics, it is more likely to be found by accident, and may even be found in similar experiments that disprove it.

TLW

[MrVibrating]

Today a doubt arose in my mind about something i'd claimed previously:

I objected that a mass in motion will not accelerate if its mass is reduced - my point being to highlight that momentum is not conserved by a conjugate relationship between m and V, which would be a requirement for the momentum transfer hypothesis.

To support this contention i did a quick WM2D sim - set a mass in motion without gravity, paused, lowered the mass and continued the sim, finding that V remained constant as predicted.

I further supposed that if such an acceleration were to occur, it would be in violation of Newton's 1st.

However today i again considered the 'figure skater effect', whereby retracting one's outstretched arms while spinning causes angular acceleration, raising the RKE. Decreasing the radius of the mass distribution this way reduces angular inertia.

And of course, decreasing the angular inertia could be construed as functionally equivalent to reducing the net mass, since it is a mass's inertia that provides the scalar component of its momentum - the mass just dictates the amount of inertia.

So this seems at odds with my conclusion about linear momentum.

Furthermore, on second thoughts i'm not sure the WM2D test was valid - is it truly realistic? Maybe P is conserved and WM2D simply hasn't been designed to model it under these conditions, since changing the value of a mass in flight is itself a non-realistic proposal?

The rotational acceleration of the figure skater obviously doesn't violate Newton's 1st. Yet the rotational system is undergoing constant acceleration anyway, for a constant angular velocity, hence its moment of inertia remains in play, and hence why reducing it causes angular acceleration.

The linear flying object however is not subject to inertia when at constant velocity under normal circumstances. So in order for it to similarly accelerate when its mass (and thus inertia) is reduced, the act of reducing the mass must invoke an inertial reaction. If F=ma, then m=a/F, so does this imply that a spontaneous reduction in mass causes a corresponding counter-force and acceleration? Am i interpreting that correctly?

This is hard to think about because varying the value of mass is so unintuitive (because it's impossible). Again though, the whole point of this thought experiment is that the "full momentum transfer" hypothesis proposes that velocity will increase to compensate a reduction in mass, yielding a net rise in KE. So in this simplified scenario we're attempting to reduce the system to a single variable mass, dispensing with the collision for now, and "transfer" (or rather, conserve) the full momentum of a large moving mass as it hypothetically diminishes in mid-flight.

So whad'ya reckon - if m=F/a then a change in m causes a change in F and a, yay or nay?

[MrVibrating]

The first problem i have with this is the relative nature of motion. If a mass is moving from your point of view, but stationary from mine, and we then zap it with a shrink ray, which direction will it accelerate in from my point of view? Or would it simply speed up in the current direction from yours?

This is the bit i can't understand - what if i don't know that you're also observing it, and thus part of the system? From my point of view the shrink ray makes it accelerate in a preferred direction? To me, a stationary mass with six degrees of freedom has self-selected a consistent direction when i shrink it, apparently quite arbitrarily? Just because you're spying on me through a telescope from a moving frame a mile away?

I think this may falsify the hypothesis. If, from my point of view, the object has no momentum to conserve or transfer in the first place, then there's nothing to determine the vector of any force generated by reducing its mass. Therefore, no such force or attendant acceleration is viable.

If on the other hand it had positive momentum relative to me, but none relative to another observer, the same limitation must apply and no rise in KE is possible.

Ergo, the concept of full momentum transfer between unequal masses is unpossiblé - the smaller mass cannot accelerate to compensate its lower mass relative to that of a larger colliding body; the products of mV1 and mV2 can never be equal, and so KE must be conserved.

Fair assessment or false logic?

[Trevor Lyn Whatford]

Hi Mr V,

the way I see it, the ice skater increases the energy in the system with muscle energy input, as with all the angular motion experiments, they need a added input of energy to pull against CF to make them happen, but they do not have the effect of losing weight only make the circumference small, but increase their RPM so the kinetic energy should still be the same, conserved, but is the skater adding to the kinetic energy when the arms are pulled in? (Edit, it is more likely the energy input is equal to the pull against CF and the skater leaves the spin with the same energy, less friction. IMO there needs some more experiments with AM to find out for sure.)

If you reduce the mass in motion by half, the velocity should be the same, but you would lose half of your kinetic energy.

As for the smaller mass and bigger mass collisions it maybe worth your while putting a smaller mass on first drop of a Newtons Cradle and see if it is knocked back to the same height, I would guess it would, less friction though.

TLW

[Grimer]

The first problem i have with this is the relative nature of motion. ...

Forget about linear momentum. We don't have a datum for it therefore we are bound to get the confusion you describe.

Think in terms of angular momentum.

We do have a datum for that as can be easily shown by rotating a bucket of water.

Thus one can see that linear momentum is simply the boundary case of angular momentum where the radius is very large.

Of course there must be a datum for linear momentum but we don't know what it is. A moment's thought shows that.

[MrVibrating]

@Trevor

Yes, totally, the skater is performing work on the system to increase the RKE, and the system performs work on her when she lets her arms back out. I've done the sims on this previously and it all squares up.

And yes i agree, reducing mass by half should reduce the KE by half, while maintaining net velocity. However this is functionally identical to the funky collision suggested by Fletcher's line of enquiry here - instead of mass1 transferring its momentum to a separate mass2, we can simplify and consider what happens if mass1 simply becomes mass2, using a magic wand or gypsy curse etc. If the hypothesis were correct then we would expect KE to increase sixteen-fold. And if this were not to violate Newton's first then we'd need a force to apply an acceleration. And that force would need to be proportionate to the difference in mass between mV1 and mV2 - which, if it DID arise, would seem consistent with delta mass equal to force over acceleration; m=F/a, per F=ma. But the relativity thought experiment shows the absurdities that would result, if true.

Likewise i agree about Newton's cradle - if the outermost masses were unequal then we wouldn't expect a KE rise in the smaller mass. I'm probably missing some important aspect Fletcher's seen in scissorjacks, but for now i can't see them as anything other than an elaborate conduit for regular elastic collisions.




@Grimer


I think what falls out of this is that for anything more complicated to arise, inertia needs to be kept in play, as it is in the rotating system. Perhaps this is what Fletcher sees in the scissorjack collision - protracted acceleration and thus inertia, i don't know... But non-inertial frames seem ruled out.

So from hereon i'll try to consider whether rotating or accelerating linear systems might accommodate this full momentum transfer principle. For starters, going back to the ice skater effect, we'd need some way of retracting the moment of inertia for free, perhaps using gravity... but then even if that's possible the trick will be doing it cyclically, without losing the RKE gain when the MoI inevitably has to be extended again in order to repeat the process. For a linear accelerating system i currently don't even know where to start...

But i'll keep thinking it over to see if i can get some kind of handle on it...

[Trevor Lyn Whatford]

Hi Mr V,

What if the ice skater holds 1 kilo weights in each hand before going into the spin and then drops them when spinning? the mass acceleration should still be the same as a skater without the extra mass, when mass density and positions are allowed for, the skater should still get the normal spin rate, I would guess that the skater would only lose kinetic energy equal to the drop in mass, so KE is conserved again, it looks like you are correct.

I would like to see a graph of Angular Momentum experiments, and the forces required to pull against CF at different RPM's, because I should think that CF and the pull against it would be climbing a lot higher and faster than the other elements on the graph.

[Silvertiger]

To support this contention i did a quick WM2D sim - set a mass in motion without gravity, paused, lowered the mass and continued the sim, finding that V remained constant as predicted.

Mr. Vibrating, it isn't a prediction - it's just physics. The velocity is not supposed to change since P=mv. The momentum works just fine. You can change the mass all you want and the velocity will remain exactly the same...but the momentum will change. Next time you run the sim, create a generic controller and set the range from 1 to 10 and assign that input to the mass value box in the Properties window. Set the Vx value to whatever velocity you want. Select your mass, go to your top menu and click on Measure, and go down and click on Momentum. Run your sim. Slide your controller up and down to change the mass between 1 and 10 pounds and watch the linear momentum value change with it.

[ME]

Combined Momentum of 'unstable' mass: P=(m[1]*v[1]+m[2]*v[1])
Then it will loose mass 2. Without effort the resulting P will be (m1*v[1]), thus with the same velocity.
But where did mass 2 go? I guess the combined P is still (m[1]*v[1] + m[2]*v[1]). v[1] remains the same even when m[2] is reduced to 0.
And why is mass 2 gone (seems like a valid question)?
Perhaps there was a system where P[total]=(m[1]*v[1]+m[2]*v[1])+(m[3]*v[2]) and became P[total]=(m[1]*v[3])+(m[2]*v[4])+(m[3]*v[5])? With only this information, little can be said about the new velocities.

Marchello E.

[Trevor Lyn Whatford]

Hi Silvertiger,

Momentum is the masses kinetic energy, reduce the mass in motion reduces the kinetic energy, its just another way of saying, reduce the mass in motion reduces the masses momentum.

Its a impression thing, if someone says mass is losing its momentum, I tend to think of it slowing down, and not that it has lost mass, but then again it does happen often, you could say it happen with the Sun, wherein it is losing its mass by burning it up.

[Silvertiger]

You can change the mass all you want and the velocity will remain exactly the same...but the momentum will change.

Momentum is the masses kinetic energy, reduce the mass in motion reduces the kinetic energy, its just another way of saying, reduce the mass in motion reduces the masses momentum.

Umm...you just repeated what I said, just fyi. ;)

...I tend to think of it slowing down...

Why do you think that? It does not slow down; nor does it speed up. This is linear momentum, not angular. P=mv, Trevor. Look at the equation. Let's say m=1 Lb, and that v=1 in/s. That gives the mass a momentum of 1 Lb-in/s. Now change the mass to 10 Lbs. P=mv=(10 Lbs)(1 in/s)=10 Lb-in/s. In both cases, the velocity did not change - it was still 1 in/s.

[Trevor Lyn Whatford]

Hi Silvertiger,

I tend to think losing momentum as slowing down and not as losing mass, because that is just how the word is generally used, if something is speeding up it gains momentum, and if something slows down it is losing momentum, that is how I always thought of it, Its a impression thing, that was my impression of momentum, I never gave it much thought of mass in motion losing mass, until now.
Edit,

If you reduce the mass in motion by half, the velocity should be the same, but you would lose half of your kinetic energy.

At least I was not miles away.

TLW

[Silvertiger]

Well I would say that people will and do think of it that way because it's true, and the fact that since mass not change unless you're dumping cargo while moving, the only thing left to change that changes momentum is velocity. If you change the mass, the momentum changes; if you change the velocity, the momentum changes. But if you do change the mass instead of the velocity, the momentum changes, and the velocity remains unchanged. An airplane needs to get to an island but halfway there, over the ocean, they discover they don't have enough fuel to quite get them there at the minimum speed to maintain a minimum altitude. Sooo, what they wind up doing is dumping all their non-essential cargo to lighten the plane so it can maintain the same minimum speed. Same principle; real-world application.