What keeps electrons spinning around their nucleus?
in [Relativity]
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From: Daryl McCullough on 29 Mar 2005 15:58 Gregory L. Hansen says... >>Well, in order to compare vectors at different times, you >>have to have some operational way to do parallel transport. >>For spatial vectors, you can use gyroscopes: Set the gyroscope >>into motion pointing along a particular vector A at time t1. >>At a later time, t2, you compare A at time t2 to the direction >>that the gyroscope is pointing. > >And then declare that it's not forces, but time-dependent basis >vectors, that make the gyroscope turn? What would we say about >a gyroscope precessing under the influence of gravity? I'm not exactly sure what you are arguing here. There is an operational way to perform parallel transport of vectors. Yes, you are right, that if there is an unknown force acting on our measurement devices, then that will screw things up. But so what? >>The coordinates of the black horse don't >>change with time, but the basis vectors do. > >The basis vectors change with time WITH RESPECT TO THE BASIS VECTORS OF >THE INERTIAL FRAME. You don't need to yell. Yes, that's essentially what parallel transport amounts to. >That's picking the inertial frame out as a special >frame, and referring all mechanics done in the accelerated frame to the >inertial frame. Inertial frames are special any way you look at it. Either they are the frames with no fictitious forces, or they are the frames in which the connection coefficients are all zero. >It requires the accelerated observer to have an inertial >frame to reference. No, it requires that the observer make observations. The connection coefficients are empirically deducible from observations involving gyroscopes, test particles, etc. >Or, if no inertial frame is observable (e.g. an >observer on a spinning planet with a cloudy sky), it requires the observer >with a Foucalt pendulum or a cannon to construct a hypothetical frame that >makes the inertial forces disappear, and then declare the hypothetical >frame to be the one that should really be used. I never said that you *should* use inertial frames. I just said that *if* you use noninertial frames, then the term for acceleration includes additional terms involving connection coefficients. >And that program has been followed before, in the general theory of >relativity. Except in that theory there's no such thing as an inertial >frame except as an approximation in a limited region. You don't need global inertial frames. All you need is parallel transport in order to compute the connection coefficients. -- Daryl McCullough Ithaca, NY
From: PD on 29 Mar 2005 16:32 TomGee wrote: > Well, Greg, even PD admits that the centripetal acceleration is a > force, Either you don't know how to read, or you don't know how to quote, or you just like to mangle as you please. Do not run with scissors, Tom, I beg you. > but he does not know the bases of such forces. In the case of > the electron, it is the em force and in the case of his examples, it is > the gravitational force. Obviously, if we decide to not call > centrifugal force a force any longer, the other should not be called > one either. The centripetal force is imposed as an inward pull while > centrifugal force is a function of inertia, which is that tendency of a > body to continue unaccelerated when no net force is acting upon it. > The inertia of a body is what keeps it wanting to leave its orbit but > the pull of gravity keeps it from doing that. > > He has denied that em is the force involved in an atom which binds the > electron into orbitals. He was taught that both centripetal force and > centrifugal force exist, but now he lemmingly goes over the cliff by > discarding centrifugal force but retaining centripetal force without > knowing why. Wormy is doing the same, of course, as a Chief Lemming. > > Is everyone here so full of knowledge that they think the basics no > longer apply? > > TomGee You are a piece of work, TomGee, I'll give you that. Not a piece of work that is particularly a source of pride, but a piece of work nonetheless. PD
From: Gregory L. Hansen on 29 Mar 2005 16:47 In article <d2cfih03101(a)drn.newsguy.com>, Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: >Gregory L. Hansen says... > >>>Well, in order to compare vectors at different times, you >>>have to have some operational way to do parallel transport. >>>For spatial vectors, you can use gyroscopes: Set the gyroscope >>>into motion pointing along a particular vector A at time t1. >>>At a later time, t2, you compare A at time t2 to the direction >>>that the gyroscope is pointing. >> >>And then declare that it's not forces, but time-dependent basis >>vectors, that make the gyroscope turn? What would we say about >>a gyroscope precessing under the influence of gravity? > >I'm not exactly sure what you are arguing here. There is an >operational way to perform parallel transport of vectors. Yes, >you are right, that if there is an unknown force acting on our >measurement devices, then that will screw things up. But so what? So you're arguing against the notion of inertial forces. If it could be an unknown force making the gyroscope turn, that doesn't really rule out forces. Or a formulation that treats it as a force. But I think we're arguing about the sort of thing that doesn't really matter. I certainly don't disagree with any of your physics. > >>>The coordinates of the black horse don't >>>change with time, but the basis vectors do. >> >>The basis vectors change with time WITH RESPECT TO THE BASIS VECTORS OF >>THE INERTIAL FRAME. > >You don't need to yell. Oops! Sorry. I'll use my indoor voice. > >Yes, that's essentially what parallel transport amounts to. > >>That's picking the inertial frame out as a special >>frame, and referring all mechanics done in the accelerated frame to the >>inertial frame. > >Inertial frames are special any way you look at it. Either they >are the frames with no fictitious forces, or they are the frames >in which the connection coefficients are all zero. Or they're frames whose inertial forces exactly cancel out the Coriolis and centrifugal forces? At any rate, I don't think anyone calculating connection coefficients will be confused as to the nature of an inertial force. > >>It requires the accelerated observer to have an inertial >>frame to reference. > >No, it requires that the observer make observations. The >connection coefficients are empirically deducible from >observations involving gyroscopes, test particles, etc. > >>Or, if no inertial frame is observable (e.g. an >>observer on a spinning planet with a cloudy sky), it requires the observer >>with a Foucalt pendulum or a cannon to construct a hypothetical frame that >>makes the inertial forces disappear, and then declare the hypothetical >>frame to be the one that should really be used. > >I never said that you *should* use inertial frames. I just said >that *if* you use noninertial frames, then the term for acceleration >includes additional terms involving connection coefficients. > >>And that program has been followed before, in the general theory of >>relativity. Except in that theory there's no such thing as an inertial >>frame except as an approximation in a limited region. > >You don't need global inertial frames. All you need is parallel transport >in order to compute the connection coefficients. > >-- >Daryl McCullough >Ithaca, NY > -- "Beer is living proof that God loves us and wants us to be happy." -- Benjamin Franklin
From: puppet_sock on 29 Mar 2005 17:09 dubious(a)radioactivex.lebesque-al.net (Bilge) wrote in message news:<slrnd4dc08.n0h.dubious(a)radioactivex.lebesque-al.net>... > Nick: > >What is the velocity of an electron in a shell? > > Velocity isn't a quantum mechanical observable. Of course it is. What ever do you think you get when you take the time derivative of position? Socks
From: Daryl McCullough on 29 Mar 2005 17:09
Gregory L. Hansen says... >Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: >>I'm not exactly sure what you are arguing here. There is an >>operational way to perform parallel transport of vectors. Yes, >>you are right, that if there is an unknown force acting on our >>measurement devices, then that will screw things up. But so what? > >So you're arguing against the notion of inertial forces. I'm arguing against *calling* them inertial forces. Instead, they are connection coefficients, which are part of the *acceleration* vector. >>Inertial frames are special any way you look at it. Either they >>are the frames with no fictitious forces, or they are the frames >>in which the connection coefficients are all zero. > >Or they're frames whose inertial forces exactly cancel out the Coriolis >and centrifugal forces? I'm not exactly sure what you mean here. I thought the Coriolis and centrifugal forces *were* the inertial forces. Somebody quoted something that Einstein wrote about the equivalence principle. He said that what was special about the freefalling frame in the presence of gravity is that the inertial force due to acceleration downwards (which would tend to throw you towards the top of a downwardly accelerating rocket) is exactly cancelled by the force of gravity. I think that's a confusing way to look at it. -- Daryl McCullough Ithaca, NY |