Motion
in [Physics]
|
Prev: Free fall
Next: 50% OF POPULATION BELOW AVG IQ!
From: Schoenfeld on 22 Sep 2005 05:02 jowr...(a)gmail.com wrote: > Schoenfeld wrote: > > Eric Gisse wrote: > > > Schoenfeld wrote: > > > > > > [snip] > > > > > > > > > > > s(t) = SUM{n=0}^+INF [ s^n(0) t^n / n! ] > > > > > > > > Your assumption that dg/dt = 0 reduces s(t) to: > > > > s(t) = s(0) + s'(0)t + 1/2 s''(0)t^2 > > > > > > > > > > > > Much simpler. > > > > > > I derived the solution, you reverse-engineered the solution from an > > > arbitrary equation. > > > > Not an arbitrary equation and no reverse engineering, just trivially > > true by definition. > > You missed the point. Doubtful. > I derived the equation that predicts your distance from Newton's 2nd > law and knowing two initial conditions. Correct, but you don't need Newtonian mechanics or those initial conditions whatsoever to derive the final result. That's why it's simpler to do it the way I showed. > You "derived" the same equation > by picking an arbitrary series that when it is differentiated twice, is > the same as the solution to a particular 2nd order ODE. Incorrect. All you need to know is if s(t) is analytic for all t, then by definition alone: s(t) = SUM{n=0}^+INF [ s^n(0) t^n / n! ] which for condition d^n s / dt^n = 0 for all n >= 2 gives the result: s(t) = s(0) + s'(0)t + 1/2 s''(0)t^2 or as commonly written: s = s0 + ut + 1/2 at^2 > They are NOT the same things. Indeed they are not. The point is that you can derive that result from definition alone without invoking Newtonian Mechanics at all. [...]
From: PD on 22 Sep 2005 05:08 Herman Trivilino wrote: > "PD" <TheDraperFamily(a)gmail.com> wrote ... > > > TomGee has opined that a body steadily loses kinetic energy due to drag > > from passage through dark matter, and that therefore an internal force > > is required to continue to supply energy to keep a body moving. > > There is no difference between a state of rest and a state of uniform > motion. If TomGee (or anyone else) has an explanation for what it is that > keeps an object in uniform motion (an internal force, or whatever) the same > explanation MUST apply to an object at rest. In other words, this same > internal force that keeps an object in uniform motion also keeps it at rest. > The two states are equivalent. > > Do you suppose that TomGee has an experimental result that distinguishes > beetween a state of rest and a state of uniform motion? No. In fact, TomGee believes that all of the predictions of his model are identical to those of the current model, just explained better. However, I think he has overlooked that this is a prediction of the current model, that there is *nothing* physically different between an object at rest and an object in uniform motion. Furthermore, TomGee has stated that *nothing* is stationary, so that whatever he is saying about an object at rest is spurious anyway. TomGee also seems to think that a reference frame has to have a physical object at rest in it to make sense as a concept, as evidenced by how he counts reference frames according to physical objects. > > > > ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- > http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups > ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
From: Randy Poe on 22 Sep 2005 06:16 TomGee wrote: > Randy Poe wrote: > > TomGee wrote: > > > > > [...], then we can safely and simply assume > > > > > that my internal force is nothing more than the kinetic energy of the > > > > > body. > > > > If you are using the term "internal force" to mean "kinetic > > energy" then what you are saying is that "without kinetic > > energy, a body would not move". I can accept that, though > > it is vacuous, since to me "kinetic energy" is just the term > > for the energy which is stored in the motion. > > > > > No, I am not saying that. I am saying that the isolated object moves > using the energy of its mass to move. Which to you is its kinetic energy. So you are making the statement that to you "kinetic energy" is a thing separate from motion, but when its there, a body uses it to move. So long as we agree that motion and KE always go together, that KE has the mathematical form that I would assign it (0.5mv^2 for a body at moderate velocity, E - mc^2 in general), we won't disagree on the relation between KE and motion. > There are no bodies without > energy visible to us and all mass contains relatively vast amounts of > energy. A moving body has energy due to its motion in addition to the > energy in its mass. Oh dear. You seem to be saying something different, that mc^2 is part of the energy of motion. Well, I'm afraid I'll have to disagree. A body with no motion has E = mc^2, and it won't spontaneously start moving unless there is an external force (Newton's First Law). And when it does, the KE will be present, in exactly the amount E - mc^2. And only that KE will be recoverable in collisions, or in the form of other energy such as PE or useful work. The energy stored in mc^2 is not generally available to us outside in mechanical processes. > The total energy of a mass is measurable as > momentum No, incomparable quantities. They go together, but they aren't the same thing. Bodies with KE also have momentum. Both are separately conserved. In inelastic collisions momentum is conserved but KE is not. > and it can be applied as a force as well, No, energy can't be "used as a force". Length can't be "used as an area" or as a time. Energy can't be used as a length. > as in the 3rd law of inertial motion. That doesn't tell use "energy can be used as a force". It does relate force and momentum: F = dp/dt. > Thus, momentum is both a measurement and a force, > like it or not. Your "thus" does not follow. The fact that F and p are related by F = dp/dt does not mean p "can be used as F". > > And finally I guess you'd say that a body with KE will maintain > > a constant velocity forever if there are no external forces. > > And again I'd agree: if there are no external forces, KE > > is conserved, and so motion is unchanged. > > > Oh, wonderful. I am surprised, but elated too at having been able to > get my point across to you. However, I would not say the constant > velocity can be maintained forever. Well then I'm not sure why you're elated since there is no part of the above paragraph on which we agree. > Remember, "Any system will run > down when left to itself, through various forms of attrition." Incorrect. Systems in which friction are present will run down due to energy losses. Is this another quote from Encarta? Give it up. > > They are OF different dimensions. They HAVE different dimensions. > > They are different things. That's why we don't like to use a word > > like "force" to describe a thing like KE which doesn't have > > the dimensions of force. > > > Yes, we don't like to use force in that way, I agree, but we need to No, we don't "need to" mix our units, since doing so gives different answers. When I said we don't "like to use force that way" I meant the semantic usage. We don't like to apply the same name to things which can't be used in the same equations. It confuses things. In the equation F = dp/dt, it is just as correct if we use the word "force", as you are wont to in your Encarta-motivated vocabulary, for F as well as for p, so long as one is careful never to use the p-type "force" on the left or the F-type "force" on the right. If one uses the word "force" to mean momentum and uses momentum units to mean this type of "force", you'll confuse everyone you talk to, but you can still have correct equations. So we don't like this abuse of language and no we don't "have" to abuse our language. It might lead us to do stupid mistakes, like use "A force" in equations where we can only use "B force". > Yes, that's true. But we know energy comes in various forms and can be > measured in various ways. All of which are subject to conservation of energy and all of which are measured in the same units. > My model contends that energy is the fifth > force Well, that language makes it sound like you're using it as the kind of "force" that can be used where we use F. For instance, you would use "energy = ma". And that's just wrong. If it's a "force", then it will accelerate masses, because that's what everything we call a force does. In other words, saying its the "fifth force" while it doesn't obey a single equation that the other four forces do, is wrong-o. > and that as such it will be the cornerstone of the GUT Well, that will be hard to do for something which fails on the grounds that F = ma. > > Energy is not, of course, a force. Trying to equate energy to > > ma will lead to nonsense. > > > Is that what you thought about mass and energy? No. That was introduced a half-century before I was born. > Were you one of those > who would burn at the stake anyone claiming that mass and energy are > equivalent? No. Do you have some evidence that anybody ever did that? > And now that everyone agrees with that, are you now saying > the same thing about energy and force. No. Einstein's theories were radical but don't fail on elementary dimensional grounds. Remember, they laughed at Galileo, but they also laughed at Bozo the clown. Being "different from existing physics" doesn't automatically mean you're right. > Let's examine the possibility that my wish to unite the two could work. OK, I examine it. I try to put energy on the left hand side of F = ma. Oops, it fails out the starting gate. End of story. - Randy
From: TomGee on 22 Sep 2005 15:06 No it wasn't, Matthew.
From: Herman Trivilino on 22 Sep 2005 15:15
"Schoenfeld" <schoenfeld1(a)gmail.com> wrote ... >> I derived the equation that predicts your distance from Newton's 2nd >> law and knowing two initial conditions. > > Correct, but you don't need Newtonian mechanics or those initial > conditions whatsoever to derive the final result. That's why it's > simpler to do it the way I showed. The difference is that you didn't use ANY mechanics at all. Only kinematics. > All you need to know is if s(t) is analytic for all t, then > by definition alone: > s(t) = SUM{n=0}^+INF [ s^n(0) t^n / n! ] > > which for condition > d^n s / dt^n = 0 for all n >= 2 > > gives the result: > s(t) = s(0) + s'(0)t + 1/2 s''(0)t^2 As soon as you say that the displacement s is a function of time t, you are doing kinematics. > Indeed they are not. The point is that you can derive that result from > definition alone without invoking Newtonian Mechanics at all. You can derive the result from kinematical considerations alone, and that is what you did. Without mechanics you have no physical justification for truncating your series with the second order term, other than to perhaps claim that it's a good fit with what's observed. Essentially, this is what Galileo discovered for free fall, although he wasn't able to derive it in the elegant way that you did. When mechanics is brought into consideration, we gain more insight as to why your truncation at the second term is the right place to truncate. It's because a constant force implies a constant acceleration. And a constant acceleration, by definition, implies that all terms of order greater than 2 are zero. ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |