What keeps electrons spinning around their nucleus?
in [Relativity]
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From: Daryl McCullough on 29 Mar 2005 09:28 Gregory L. Hansen says... >A force makes something accelerate. If you're at rest in an accelerate >frame you might see a ball accelerate spontaneously. If it accelerated, a >force must have acted on it. "Inertial force" might not be the best name >for that because they don't appear in an inertial reference frame. I guess whether or not the ball accelerates when you let go of it depends on how you define "acceleration". Here's the way I see things, mathematically: Letting V be the velocity vector, and A be the acceleration vector, we have: (where e_k = the kth basis vector) V = V^k e_k A = (d/dt V^k) e_k + V^k (d/dt e_k) = (d/dt V^k) e_k + V^i (d/dt e_i) (changing the dummy index k to i) Now, we can rewrite d/dt e_i as follows: d/dt e_i = dx^j/dt d/dx^j e_i = V^j (d/dx^j e_i) = V^j G^k_ji e_k where the matrix G^k_ji is defined via d/dx^j e_i = G^k_ji e_k Putting this altogether, A = (d/dt V^k) e_k + V^i V^j G^k_ji e_k Or in component form, A^k = (d/dt V^k) + V^i V^j G^k_ji If you release a ball while sitting on a spinning carousel, the first term (d/dt V^k) is nonzero, but the second term V^i V^j G^k_ji exactly cancels it, to give zero acceleration. -- Daryl McCullough Ithaca, NY
From: Gregory L. Hansen on 29 Mar 2005 10:03 In article <d2bom40jkc(a)drn.newsguy.com>, Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: >Gregory L. Hansen says... > >>A force makes something accelerate. If you're at rest in an accelerate >>frame you might see a ball accelerate spontaneously. If it accelerated, a >>force must have acted on it. "Inertial force" might not be the best name >>for that because they don't appear in an inertial reference frame. > >I guess whether or not the ball accelerates when you let go of it >depends on how you define "acceleration". Here's the way I see >things, mathematically: Letting V be the velocity vector, and >A be the acceleration vector, we have: (where e_k = the kth basis >vector) > > V = V^k e_k > A = (d/dt V^k) e_k + V^k (d/dt e_k) > = (d/dt V^k) e_k + V^i (d/dt e_i) (changing the dummy index k to i) > >Now, we can rewrite d/dt e_i as follows: > > d/dt e_i = dx^j/dt d/dx^j e_i > = V^j (d/dx^j e_i) > = V^j G^k_ji e_k > >where the matrix G^k_ji is defined via > > d/dx^j e_i = G^k_ji e_k > >Putting this altogether, > > A = (d/dt V^k) e_k + V^i V^j G^k_ji e_k > >Or in component form, > > A^k = (d/dt V^k) + V^i V^j G^k_ji > >If you release a ball while sitting on a spinning carousel, >the first term (d/dt V^k) is nonzero, but the second term >V^i V^j G^k_ji exactly cancels it, to give zero acceleration. The way I see that, you have declared "There shall be no mechanics performed in an accelerated frame!" What makes you think the e_i have a time dependence? On a spinning carousel, if I define e_1 by a line extending from the center to the black horse, and e_2 from the black horse to the red one just ahead, then they're not going to change. I could sit on the black horse and stare at the red horse, and I won't have to move my head around and around because the red horse and the basis vectors are stationary with respect to myself. Of course, relative to the snack stand we might have e1=(cos(wt),sin(wt)) and e2=(-sin(wt),cos(wt)). And then it has a time dependence if we assume that "real" mechanics is done from an inertial frame, and that the e1 which is constant for the observer on the carousel isn't really constant because it's not constant for the observer at the snack stand. But the observer on the carousel is still at rest with respect to himself, and he can still define a coordinate system relative to himself. He's not required to reference a customer on a horsie to the snack stand, he can reference her to himself and just walk towards her. To calculate inertial forces we usually do reference an inertial frame. That's because we know what natural motion looks like in an inertial frame-- a straight line. But once we know the centrifugal and coriolis forces in a rotating frame we could use that as our reference for a third accelerated frame, like "The Scrambler". We just have to know what the natural motion looks like in any frame, and then relate it to a frame of interest if that's not the one we know. -- "In any case, don't stress too much--cortisol inhibits muscular hypertrophy. " -- Eric Dodd
From: jmfbahciv on 29 Mar 2005 09:12 In article <7K52e.21$45.3808(a)news.uchicago.edu>, mmeron(a)cars3.uchicago.edu wrote: >In article <Xns9627C5AEB62D6WQAHBGMXSZHVspammote(a)130.39.198.139>, bz <bz+sp(a)ch100-5.chem.lsu.edu> writes: >>mmeron(a)cars3.uchicago.edu wrote in >>news:8B12e.18$45.3391(a)news.uchicago.edu: >> >>> In article <Mo12e.16031$C7.902(a)news-server.bigpond.net.au>, "Bill Hobba" >>> <bhobba(a)rubbish.net.au> writes: >>>> >>>><mmeron(a)cars3.uchicago.edu> wrote in message >>>>news:XQ02e.15$45.3352(a)news.uchicago.edu... >>>>> >coordinates, but I think what is misleading is to call the correction >>>>> >terms "forces". >>>>> >>>>> Only if you attach more meaning to the term "force" than it deserves. >>>> >>>>Ahhhhhh. Yes. As Feynman says it is half a law. It gains its full >>>>meaning when combined with other laws and/or concepts such as Coulombs >>>>law or the introduction of non inertial reference frames. >>>> >>> Yes, it is a rather complex issue. I wrote some stuff about it in the >>> past, here, but I never kept a copy. But it certtainly needs some >>> sort of broad framework, to make sense. >> >>Especially when someone keeps insisting that force is always the result of >>acceleration > >Cause, not result. > >> and that without acceleration (as for example when a gyro >>precesses at a constant rate, or when a mass moves at a constant velocity >>because it is overcoming drag or friction) there is no force and no work. >> >Where there is net force, there is acceleration. The F in Newton's >law is the total (i.e.) net force acting. Since forces are vectors, >it is perfectly possible to have different non-zero forces to sum up >to a zero net force. This is why using algebra in first physics courses hurts learning more than it helps. > >If I put your finger in a vise and squeeze, your finger is being acted >upon by two forces, equal and opposite. The net force is zero and >your finger is going nowhere. Which by no means mean that since the >net is zero, there are no observable (or, for that matter, audible, >i.e. loud screams) effects present. I always have problems drawing that force diagram. The arrows never "matched" my common sense. And torque always confused me and, as a result, I always reverted to "memorized" formulas whenever I did those problems. /BAH Subtract a hundred and four for e-mail.
From: Bilge on 29 Mar 2005 10:35 mmeron(a)cars3.uchicago.edu: >In article <d2ac7c$65m$5(a)rainier.uits.indiana.edu>, >glhansen(a)steel.ucs.indiana.edu (Gregory L. Hansen) writes: >>In article <slrnd4h9dg.6h5.dubious(a)radioactivex.lebesque-al.net>, >>Bilge <cranks(a)fghfgigtu.com> wrote: >>> mmeron(a)cars3.uchicago.edu: >>> >>> >I don't see anything in the formulation of newton's laws saying that >>> >forces must be traceable to fundamental interactions. What should be >>> >taught is that there are "physical forces" which are, indeed, a >>> >measure of interactions between objects, and there may be additional >>> >"inertial forces" which are an artifact of the choice of reference >>> >frame. But being an artifact, does not mean that they've no >>> >>> ``Inertial force'' is an oxymoron. If ``inertial forces'' are >>>to be considered forces, then the word ``force'' is nothing but >>>a placeholder for a noun, since anything is then a force. >> >>A force makes something accelerate. If you're at rest in an accelerate >>frame you might see a ball accelerate spontaneously. If it accelerated, a >>force must have acted on it. "Inertial force" might not be the best name >>for that because they don't appear in an inertial reference frame. >>-- >Got a point. Lets call them "non-inertial forces". That would be redundant.
From: David Cross on 29 Mar 2005 11:33
On Tue, 29 Mar 05 14:12:53 GMT, jmfbahciv(a)aol.com wrote: >I always have problems drawing that force diagram. The arrows >never "matched" my common sense. And torque always confused me >and, as a result, I always reverted to "memorized" formulas >whenever I did those problems. I think I was lucky; I learned about torques after I'd learned how to apply cross products. :) Therefore all I needed was the radial distance away from the center and the applied force. :) --- David Cross dcross1 AT shaw DOT ca |