
#37
Jan2113, 10:24 AM

P: 427

If the force due to gravity is the mass times the actual acceleration of the particle, then a particle at rest would have no weight. Again, this is pure silliness. If there were some observation of this, then maybe it would be interesting, but there hasn't been and this isn't. [itex]f=mg[/itex] does work for a stationary particle. The force do to gravity of a particle at rest is [itex]mg[/itex] which opposed by some normal force that is equal and opposite and therefore the sum of all forces is 0. Your "model" is not a model because it does not describe anything real. You are trying to find a way to solve a problem that arises because of a false assumption. The resolution should be obvious. I'm pretty much done with this thread since you appear to be sticking to your guns and uninterested in actual study. Your "model" predicts things that don't happen. Give it up. 



#38
Jan2113, 11:06 AM

P: 21

If as you say I predict things that at this moment in time can not be observed makes my theory wrong so be it. I am not, as I am sure you guess, a professional physicist . I am not trying to create a new model for the scientific community. I am only excising my enquiring mind You are right I am sure that it is pure silliness to suggest a particle not an object at rest has no weight. I was hoping for a more informative and reasoning other than silliness. Particularly as you can produce no experimental evidence to show me wrong You keep stating f=mg for a particle is correct you are right it is for an object stationary or moving, and a moving particle, but not a stationary particle as I have shown. So there are no challenges to mechanics there never was I always said the effect was probably not observable. I am still not sure if the effect can be observed and measured that is for better physicists than me to work out. Also my conclusion is only F=MA which seems highly satisfactory. Please before you go could you explain my false assumption. I have only used conservation of energy, Heisenberg's EnergyTime uncertainty principle and 2nd law of thermodynamics. oh and causality. Even if you cannot do this as you perceive me as a waste of space and do not wish to waste your time I really do thank you and the others for your efforts and time indulging my enquiring mind. I have learned a great deal If that sounds sarcastic it really is not. It is entirely genuine and I wish you all well. 



#39
Jan2113, 12:10 PM

Mentor
P: 16,476

[tex]x(t)=y(t)=\frac{E}{gm} \left( 1 e^{\sqrt{2}gt} \right)[/tex] This is not a correct solution, but it satisfies your model by maintaining a constant total energy with KE decreasing as the object moves to an area of higher PE. Your model is WRONG, even for moving particles. It does not have enough information to select the correct trajectory which is compatible with energy considerations. For that, you need the principle of least action. 



#40
Jan2113, 12:12 PM

Sci Advisor
PF Gold
P: 11,352

I don't think he's switched to 'receive' on this one.




#41
Jan2113, 12:50 PM

P: 21

It only says that when a particle not an object is stationary it will not experience a force. As I am lead to believe this is never observable therefore this cannot I believe be shown to be wrong. Clearly you could say it was irrelevant and you may be correct time will tell. You are correct the present models have been rigorously tested and work I have said this before. Theories on the present testable/observable world are not changed in my model in any way stationary or moving. I am not trying to improve the models science use it is a personal model of the world to aid my understanding. I thank you for help in refining my model and sorting out in my mind the problem of action at a distance which always seemed nonsense. I really thank you for your time and patience while I indulged my enquiring mind. I wish you well 



#42
Jan2113, 12:54 PM

Mentor
P: 16,476

If you disagree and believe that your model is correct, then you need to demonstrate in which way my posted trajectory violates your model. If my posted trajectory does not violate your model then your model is wrong, as I have claimed. 



#43
Jan2113, 12:55 PM

P: 21

I wish you well I really enjoyed it ps great signature 



#44
Jan2113, 12:59 PM

P: 21

Again thanks 



#45
Jan2113, 01:12 PM

Mentor
P: 16,476

The whole point of this thread is that you believe that the above model correctly describes the physics of a moving body. You were then concerned by the fact that this model admits a nonphysical solution for a stationary body, namely the solution where it doesn't move at all. I demonstrated that it also admits a nonphysical solution for a moving body. Therefore, since it admits nonphysical solutions for moving bodies it should come as no surprise that it admits nonphysical solutions for stationary bodies also. There is no difference wrt moving and stationary bodies, the model fails for both. 



#46
Jan2113, 03:26 PM

P: 427

I know I said I wouldn't come back, but this is starting become funny. Adeste's "model" predicts things that are counter to current theory, BUT since his/her model only applies to particles that are defined by the fact that they follow this model, there is no problem with the fact that current theory predicts different things. So we have a theory that ONLY applies to stationary particles that are not effected by Newton's laws and the theory's conclusion is that these particles are not effected by Newton's laws! Inevitably, they will also not be effected by QM because those are different particles for which this theory doesn't apply. I think this might actually turn out to be a good example to use in HS physics to show why energy conservation is not sufficient to describe reality and what the difference between science and pseudoscience is. 



#47
Jan2113, 07:04 PM

P: 177

The instantaneous acceleration of an object does not depend on the instantaneous velocity of the object (for uniform gravitational fields). Therefore it is not possible to obtain from the velocity of an object (at one point in time) any information about its acceleration (for uniform gravitational fields). 


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