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From: Randy Poe on 23 Sep 2005 11:12 TomGee wrote: > Herman Trivilino wrote: > > "TomGee" <lvlus(a)hotmail.com> wrote ... > > > > >> Where did I say that? I would have snipped more, but I wanted > > >> to leave all the stuff you quoted from me so you can point > > >> me to the passage where you think I said F = p. > > >> > > > If you think you did not say it. then just what do you mean by F=dp/dt? > > > > Wow! Don't you know the difference between a quantity and its time rate of > > change? > > > > We can write v=dx/dt, where v is velocity and x is position. Do you think > > this means that velocity and position are equal? And that we can write > > v=x?! > > > So you don't know what he means by that either, eh? He just told you what I meant. I didn't say F = p, I said F is the time rate of change of p. Did you somehow miss this sentence in Herman's short post? "Don't you know the difference between a quantity and its time rate of change?" Since he obviously understood that dp/dt means "time rate of change of p" and TOLD you that's what it means and how it differs from saying F=p, what in the above gave you the impression he didn't know what I meant by F = dp/dt? - Randy
From: PD on 23 Sep 2005 14:05 Herman Trivilino wrote: > "TomGee" <lvlus(a)hotmail.com> wrote ... > > > Then, Herman, what does E=mc^2 mean to you? > > It means that mass is a form of energy. This is possibly where the confusion arises. The statement above gives the uninitiated the impression that mass is some sort of stuff and energy is another form of stuff, and that E=mc^2 means turning one kind of stuff into another kind of stuff. This very quickly leads to student confusion when told that a two-photon system, where each of the photons do not have mass stuff in them, can have a non-zero mass. The student is going to immediately guess that some of the stuff came from somewhere, possibly from the conversion of some of the photon energy stuff into mass stuff. Nothing could be further from the truth. It also leads to confusion where one is looking at an object from two different reference frames, where (depending on the definition) either the value of the mass or the value of the energy changes merely in going from one reference frame to another, and one has done *physically nothing* to change the amount of either kind of "stuff". Mass and energy are both quantifiable *properties*. Those properties have an operational definition. For example, energy is defined as what you can predictably attribute to an object such that, when the object is part of an isolated system, the system's total energy remains conserved regardless of interactions within the system. Read Feynman's "The Character of Physical Law" for a nice little chapter about this. What E=mc^2 means is that the operational definition that we apply to the property we call mass can also be applied to the property we call energy, and vice versa, with the quantitative conversion specified in the equation. > > When you are co-moving with respect to a composite body, the mass m that you > measure of the composite body consists of two parts. It is the sum of the > masses of each of the objects within the composite body PLUS the total > energy (divided by c²) of those objects. > > In other words, when we measure the mass of a composite body we find that it > consists not only of the masses of the components, but also the total of > their energies. Thus we see that mass is just one of the many different > forms of energy. > > That is what it means to me. Thanks for asking. That is one of my favorite > questions to answer! > And I hope that my comments above make a more careful answer even more enjoyable to give. Of course, my answer above will require more expansion to explain what those operational definitions of mass and energy really mean, and the condensed version above does no good to the unitiated like TomGee, but I'm sure you can do more with it as needed. PD
From: TomGee on 23 Sep 2005 14:37 Herman Trivilino wrote: > "TomGee" <lvlus(a)hotmail.com> wrote ... > > >> > If you think you did not say it. then just what do you mean by F=dp/dt? > >> > >> Wow! Don't you know the difference between a quantity and its time rate > >> of > >> change? > >> > >> We can write v=dx/dt, where v is velocity and x is position. Do you > >> think > >> this means that velocity and position are equal? And that we can write > >> v=x?! > >> > > So you don't know what he means by that either, eh? > > Yes, I do. He's talking about a derivative, or rate of change. In this > case, its a rate of change with respect to time. > > Yes I know that, but what is the relevance of that to the topic? > > > p is the momentum. > > Yes I know that, but what is the relevance of that to the topic? > > > dp/dt is the rate at which p changes with time. > > Yes I know that, but what is the relevance of that to the topic? > > > F is the net force. > > Yes I know that, but what is the relevance of that to the topic? > > > He's claiming that F=dp/dt, a valid claim. > > 2+2=4 is a valid claim too, but (You guessed it!): Yes I know that, but what is the relevance of that to the topic? > > > You're claiming that F=p, an invalid claim. > > No, I'm claiming no such thing. You just making a gross assumption. Gross as in "yucckkk!!"
From: TomGee on 23 Sep 2005 14:41 Randy Poe wrote: > TomGee wrote: > > Herman Trivilino wrote: > > > "TomGee" <lvlus(a)hotmail.com> wrote ... > > > > > > >> Where did I say that? I would have snipped more, but I wanted > > > >> to leave all the stuff you quoted from me so you can point > > > >> me to the passage where you think I said F = p. > > > >> > > > > If you think you did not say it. then just what do you mean by F=dp/dt? > > > > > > Wow! Don't you know the difference between a quantity and its time rate of > > > change? > > > > > > We can write v=dx/dt, where v is velocity and x is position. Do you think > > > this means that velocity and position are equal? And that we can write > > > v=x?! > > > > > So you don't know what he means by that either, eh? > > He just told you what I meant. I didn't say F = p, I said F is the > time rate of change of p. > > And the relevance of that to out topic is...? > > > Did you somehow miss this sentence in Herman's short post? > "Don't you know the difference between a quantity and > its time rate of change?" Since he obviously understood that > dp/dt means "time rate of change of p" and TOLD you that's what > it means and how it differs from saying F=p, what in the above gave > you the impression he didn't know what I meant by F = dp/dt? > > So you both can read a dictionary, so what? Obviously you can read; you can't comprehend much of what you read, however, since apparently you can only respond with nonrelevant answers.
From: TomGee on 23 Sep 2005 14:46
Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > TomGee wrote: > > > > It would be a pity if you were right, but you contradict yourself > > > > immediately by saying above that force and momentum are equivalent. > > > > > > Where did I say that? I would have snipped more, but I wanted > > > to leave all the stuff you quoted from me so you can point > > > me to the passage where you think I said F = p. > > > > > If you think you did not say it. then just what do you mean by F=dp/dt? > > That force is the derivative of momentum with respect to time. > > I see. And the relevance of that to our topic is ...? > > > For instance suppose p is a constant in time, say p = 5. > The F = dp/dt = 0. Five and zero are not the same. > > Suppose p is increasing linearly in time, p = p_0 + b*t > > Then F = dp/dt = b, a constant. The constant b and the > linearly increasing function p_0 + b*t are not the same. > > Okay, so what? |