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Why does an object start to fall under gravity 
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#19
Jan2013, 10:06 AM

P: 21

Heisenberg uncertainty relation. ΔE.ΔT <h/2∏
Surely this allows causality as I suggest as Δt can be made as small as you wish 


#20
Jan2013, 10:21 AM

P: 21

An electron to be not moving would have to be at absolute zero as if not it will be moving by the momentum change caused by infrared photons hitting it a little like brownian motion. sorry if that sounds a little patronizing it was not meant to be Is there any experimental evidence that such a electron starts to move.? 


#21
Jan2013, 10:36 AM

P: 21

I am trying to explain motion from energy only and it works for every observable situation. (by works it agrees with forces produced by interaction with fields.) From energy only, it can only accelerate if it has a velocity ie it moves from one potential to another hence loses PE and gains KE hence it accelerates This is always observed as Δx<>0. Δx=0 is never attainable although classical theory I think says this would happen at absolute zero. When if Δx= 0 then Δpe =0 ∴Δke =0 ∴Δv=0 ∴acceleration = 0 In this theoretical situation there would be no acceleration and hence it implies no gravitational force 


#22
Jan2013, 11:31 AM

P: 21

I have been warned by the moderator,
So can I apologize I have " introduced a number of offthewall concepts that I probably made up on my own. Yet, I am USING them as if these are unambiguous physics concepts that are welldefined" and for trying can I apologize for trying " to educate all of us of simple, basic mechanics." it was not meant and can I appologize for not knowing my audience. I am sorry but find it difficult to communicate possibly as I have aspergers 


#23
Jan2013, 11:33 AM

Sci Advisor
Thanks
P: 3,757




#24
Jan2013, 11:42 AM

P: 446

You don't even need to consider photons colliding with an electron. If the electron is in a position eigenstate, then it doesn't have a well defined momentum. You can't have a localized particle that is not moving in quantum mechanics, so your question is meaningless in that circumstance. In classical mechanics, it is perfectly acceptable and your claim is wrong. You have chosen rules that do not describe the world. I checked out your blog before the mods pulled the link down. It is good that you are trying to think about such interesting questions, but you would serve yourself much better if you spent your time learning about the well studied theory of quantum mechanics instead of making up your own theory of nonnewtonian mechanics. 


#25
Jan2013, 11:44 AM

P: 446




#26
Jan2013, 11:45 AM

P: 38

[tex] \frac {1}{2} m \dot{q}^2 + mgq = E [/tex] contains all the information contained in [tex] m \ddot{q}^2 + mg = 0, [/tex] which i think we both agree gives the trajectory uniquely given an initial velocity and initial position. The former is just the "first integral" form of the latter. It can be solved according to (apologising in advance for potential algebraic slips but the principle is sound): [tex] \int_{q(0)}^{q(t)}{\frac{dq'}{\sqrt{2(\frac{E}{m}gq')}}}=\int_{0}^{t}{dt'}. [/tex] I think the solution is unique up to throwing away an unphysical solution in which the object accelerates upwards  certainly there is no room here for the object to stay stationary. I can't deny that the object staying stationary forever satisfies conservation of energy though. I think the reason why the maths works is to do with the fact that in the above integral over q', although there is a singularity in the integrand, the actual integral itself is wellbehaved. I reckon Zeno's paradox is also absorbed into this observation: the time stays finite because the q'integral stays finite. Certainly an interesting one... i agree that in general conservation of energy is not enough to work out dynamics. 


#27
Jan2013, 11:46 AM

P: 21




#28
Jan2013, 01:09 PM

P: 21

It works well if [itex]\dot{q}[/itex]>0 


#29
Jan2013, 01:21 PM

P: 38

Yeah, i retract what i said before. Didn't think about it carefully enough the first time. The solution that needs to be rejected as 'unphysical' is precisely that which we are trying to avoid in the first place, i.e. staying still at q(0). Apologies, DrewD!
I just mentioned this curiosity to a scientist friend. She suggested that the existence of the stationary solution could be the mechanism by which clouds stay up. Maybe the OP could work on turning this into a viable theory? :P 


#30
Jan2013, 01:57 PM

Mentor
P: 17,330

In any case, all of QM is based on the Lagrangian and the closely related Hamiltonian approaches, using energy principles. The mechanism is what I described before, so if you want to reason from energy principles, that is the way to do it. 


#31
Jan2013, 02:04 PM

Mentor
P: 17,330

For example, given an object of 1 kg mass in a 1 g uniform field, what is the trajectory if it has KE = 100 J and PE = 0 J initially? If that is all that you are given then you cannot solve for the trajectory. Even if you are also given an initial velocity, how does simple conservation of energy tell you that you get a parabolic path. Why not just a straight line path with the speed determined by energy considerations? You need something additional, the principle of least action. Since your explanation doesn't work for the moving case, you shouldn't be surprised that it doesn't work for the stationary case either. 


#32
Jan2113, 02:33 AM

P: 21

My apology my opening statement is misleading I agree 100% with what you say I am not trying to replace mechanics (forces) with potential changes for moving objects. Therefore I am not trying to prove a path or trajectory mechanics both works and is very sophisticated at this. My only concern is if a particle starts moving from rest ie stops remaining stationary. My model predicts it does not move, Any path would prove the model wrong. I am only trying to prove my model wrong. Any path would do. To try to answer your point I feel the Principle of minimum energy or 2nd law of thermodynamics constrain the particle to move to a lower potential. If the field is linear this will be perpendicular to the field (thus in the direction of the force). Such would be sufficient to prove my model wrong. again I apologise if my comment seems condescending it is just that I am working from first principles only. 


#33
Jan2113, 03:46 AM

P: 21

Not sure if I am allowed to do this but there is a thread which although it is inconclusive indicates that conservation of energy cannot be observed if Δt.ΔE is small enough. http://www.physicsforums.com/showthread.php?t=120224 Therefore I feel my causality which you object to is at least feasible or more importantly not yet shown to be false. Thanks very much for your thoughts 


#34
Jan2113, 04:14 AM

P: 21

I agree work done is done as there is a change in energy. and I agree mechanics use F.X=Work Done To answer your second question what is new. First I do not know if what I suggest is new. However it seems mechanics has been built on theories which are then proved true or false by observation. Hence mechanics is an excellent model for observable objects. I believe as it is impossible to observe an object with velocity = 0 only objects whose average velocity = 0 by this I mean all observed objects have internal energy. I think of them as having little bits vibrating around ( quarks molecules ...whatever) However mechanics takes these observations and applies them to a theoretical unobservable particle and assigns such particle with velocity =0 In short I am saying mechanics has not been validated for this theoretical state an unobservable state of a particle. I am unsure of the ramifications of this. My apologies again for lack of scientific terminolgy 


#35
Jan2113, 06:35 AM

Mentor
P: 17,330

It is possible to use energy principles only to correctly describe a particle in general, you just have to use the Lagrangian approach. You should hold off trying to invent your own approach until you have actually learned Lagrangian mechanics. 


#36
Jan2113, 09:20 AM

P: 21

I apologise if Lagrangian mechanics shows this. I leave you to judge. However I conclude that gravitaional force is not equal to mg but ma where a is the actual acceleration of the actual particle not a theoretical value. To be precise f=mg does not work for a stationary particle but does at all other times. 


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