----- Original Message ----- From: "Dan Minette" <[EMAIL PROTECTED]> To: "Killer Bs Discussion" <[email protected]> Sent: Saturday, August 13, 2005 10:54 PM Subject: Re: Physics question
> > ----- Original Message ----- > From: "Kevin Street" <[EMAIL PROTECTED]> > To: "'Killer Bs Discussion'" <[email protected]> > Sent: Friday, August 12, 2005 5:49 PM > Subject: RE: Physics question > > > >> A very good point, but I think The Fool is talking about the >> velocity (or >> maybe more interestingly, the momentum) with regard to the >> reference > frame >> of the place from which it was launched. i.e.: most likely Earth. >> By >> reducing the momentum to zero, you know exactly how much momentum >> the > craft >> has, which then introduces the uncertainty in the spacecraft's >> position. > > Well, our instruments are in an accelerating reference frame, so it > is not > clear that something thats, say, 200,000 miles away as it > approaches zero > velocity with respect to the earth's center of mass can have it's > velocity > measured all that easy. If you think of any measurement technique, > you > quickly intereact with the spacecraft enough to ensure that dpdx is > greater > than h-bar. > >> But I think there's one major problem here, beside the definitional > problem >> Warren mentioned: >> >> The thing is, Heisenberg's Uncertainty Principles are really meant >> to >> describe quantum mechanical or microscopic phenomena. Look at the > equation: >> >> [(Uncertainty in momentum) times (uncertainty in position)] is >> greater > than >> or equal to [(Planck's Constant) divided by (two times Pi)] >> >> (dp X dX) >= (h/2Pi) >> >> Planck's Constant is incredibly tiny, so the delta x of any >> macroscopic >> object would also be very small. >> >> And if delta p was zero, you'd end up with a singularity on the >> right > hand >> side of the equation when trying to determine delta x, which is > impossible. > > Actually, it means that as dp-> 0, dx-> infinity...which is what the > Fool > is thinking about. > > The real thing we can look at is the magnitude of dp and dx that we > are > talking about. h-bar is, roughly, 10^-34 J-s or 10^-34 kg m^2/s. > Let's > assume we have a 100 kg spacecraft. That gives us, dv*dx = 10^-36 > m*(m/s). > For the indeterminacy in the position to be 1 km, then the velocity > would > have to be restricted to within 10^-37 m/s. Just thinking about the > quantum variation in all the electrons, we can eliminate this > possibility. > > IMHO, the Fool's idea is a good example of the problem of trying to > force a > QM peg into a classical hole. > As I understand Relativity, it states that there are no privileged frames of reference. Therefore, there can be no "absolute velocity" since the velocity of an object is completely dependent on the frame of reference of an observer. What I see in The Fools question is an attempt to foist a privileged frame upon a relativistic universe, which is a no/go. Further, it seems questionable to me to relate the expansion of "space" to the velocity of objects. The expansion of space is not the same thing as the movement of objects within it (though related), and while space expands the objects within it (atoms and galaxies) do not. So I also see a conflation of expansion and velocity. Perhaps this is an artifact of the kinds of metaphors used to explain relativity? This causes me to pose a question: What causes space? (This is the simplest way to state the question I can think of) If I am misunderstanding The Fool or relativity, feel free to add to my understanding.<G> xponent A Defined Conglomerate Of Spatial Points Maru rob _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
