Re: Who will stun the world as next Einstein?

From: Bill Hobba (bhobba_at_rubbish.net.au)
Date: 02/02/05 

Date: Wed, 02 Feb 2005 06:04:35 GMT

"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:2dg0011oc8h6u0oi1mkd65ft5fmm4h6vlc@4ax.com...
> On Tue, 01 Feb 2005 23:55:39 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
> wrote:
>
> >
> >"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
> >news:bh4vv0heo1cc9r74mq621ccpmnnlo76s6o@4ax.com...
> >> On Tue, 01 Feb 2005 05:32:30 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
> >> wrote:
> >>
> >> >
> >> >"Bill Hobba" <bhobba@rubbish.net.au> wrote in message
> >> >news:UoELd.143475$K7.17670@news-server.bigpond.net.au...
> >> >> Bill Hobba, being careless as usual, incorrectly wrote:
> >> >> > Thus the equation becomes m dvx/dx = part deriv (G M m
> >> >> > /x)/part deriv. dvx is the component of the velocity in the x
> >> >direction -
> >> >> > all other components are zero due to the coordinate system chosen.
> >> >Hence
> >> >> we
> >> >> > end up with m dvx/dx = - G M m /x^2. Or going to vector notation
m
> >> >dv/dx
> >> >> =
> >> >> > G M m Rhat /r^2 where Rhat is a unit vector in the line joining m
to
> >M.
> >> >>
> >> >> Of course that should not be dvx/dx or dv/dx it should be dvx/dt and
> >> >dv/dt.
> >> >>
> >> >> Thanks
> >> >> Bill
> >> >
> >> >It just occurred to me that I make this silly stupid obvious error and
> >these
> >> >guys who recon I have no idea what I am talking about miss it and
argue
> >> >about things like the derivative of a scalar function being a vector.
It
> >> >really makes you wonder, it really does.
> >> >
> >> >Bill
> >> >
> >> Even 3 consecutive messages to yourself are not enough to make a wrong
> >> into a right.
> >>
> >> It is well understood in gravity and potential theory that mg/r
> >> represents an equipotential surface quite like that at the metal
> >> surface of a charged sphere.
> >
> >That is one way of looking at the surface traced out as the vector that
> >function depends on and mg/r remains constant. But again that does not
> >change facts - a scalar is a scalar and a vector is a vector.
> >
> >>
> >> Grad mg/r has 3 partial derivative components each accompanied by the
> >> unit vectors i, j and k in spherical coordinates.
> >>
> >
> >Grad is not derivative.
> >
> >> In this case by
> >> symmetry it is obvious that d/dr will do the whole job and put a u_r
> >> on it if you want, it's a matter of notation.
> >
> >Your semantic ramblings will not change the facts - the derivative of a
> >scalar is a scalar. The grad of a scalar is a vector. Vectors must be
> >equated to vectors, scalars to scalars.
> >
> >>
> >> Grad mg/r = -mg/r^2 is a vector field extending like a porcupine from
> >> the equipotential surfaces.
> >>
> >
> >Grad mg/r = -mg/r^2 is a nonsense statement because you are equating a
> >vector to a scalar.
> >
> >>
> >> Why struggle with this?
> >>
> >
> >Because it does not matter how you cut and dice it a vector equals a
vector
> >and a scalar equals a scalar.
> >
> >> As you say, " It really makes you wonder, it
> >> really does" and that's a waste. It might even make a person stop and
> >> think.
> >>
> >> The OP has been lost in the dust, but out of respect for him, why not
> >> a tiny apology?
> >
> >Because a vector equals a vector and a scalar equals a scalar. But what
can
> >I expect from Mr Dual Space - you can not even get basic facts about the
> >history of the positron and Dirac correct. I notice you never apologized
to
> >me about that. Like most cranks you live in a dream world and refuse to
> >face facts.
> >
> >Bill
> >
> >
> >>
> >> John Polasek.
> Hobba you are a consummate fool.
> You say you are a mathematician; if you want to dip into physics study
> some physics.

As both a mathematician and a person interested in physics I know a scalar
is a scalar and a vector is a vector.

> There is just a whole bunch of field theory that hinges
> on equipotentials and the gradient thereof and of which you are
> totally unaware. You keep coming back with your insipid arguments.

Most would judge it is not an insipid argument to note -mg/r^2 is a scalar
and can not equal a vector. But each to his heirachy of importance - the
fact you consider such things not relevant is consistent with the rest of
the tripe you post. Indeed muddying the waters on such fundamental issues
is a well known crank tactic from crank behavior 101.

>
> I pointed out to you the 3 vector components of grad, and you should
> by now also be congizant of the fact that with spherical symmetry, the
> only derivative is with r, the other three are nil. So by not
> mentioning them does not demote my gradient to your profane
> deriviative.
> Don't go on with your sophomoric rantings about derivatives not
> making a vector. The derivative with r is what's left of the operator.

Are you consignment of the fact -mg/r^2 has one component so can not equal
the grad of anything which has 3 components? Obviously not by your
irrelevant ramblings.

Bill

>
> Remember what you are doing with your pustulations. You are defending
> an insult you made to an inquirer 3 days ago. Nothing has been gained
> by this exchange. Write this down; the gradient of a scalar potential
> is a vector no matter how denominated.
>
> John Polasek
>

Relevant Pages

  • Re: Derivative of the Determinant Function
    ... function as a function of the linear space of n x n matrices, ... In one sense the determinant is taken as a scalar ... One can ask for "directional derivatives" ...
    (sci.math)
  • Re: Who will stun the world as next Einstein?
    ... >>scalar is a scalar. ... The grad of a scalar is a vector. ... >>and a scalar equals a scalar. ... > Don't go on with your sophomoric rantings about derivatives not ...
    (sci.physics.relativity)
  • Re: Who will stun the world as next Einstein?
    ... >> represents an equipotential surface quite like that at the metal ... >change facts - a scalar is a scalar and a vector is a vector. ... Don't go on with your sophomoric rantings about derivatives not ...
    (sci.physics.relativity)
  • Re: Derivative of the Determinant Function
    ... function as a function of the linear space of n x n matrices, ... In one sense the determinant is taken as a scalar ... One can ask for "directional derivatives" ...
    (sci.math)
  • Re: Who will stun the world as next Einstein?
    ... >> represents an equipotential surface quite like that at the metal ... >change facts - a scalar is a scalar and a vector is a vector. ... Don't go on with your sophomoric rantings about derivatives not ...
    (sci.physics)
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