Re: Who will stun the world as next Einstein?

From: John C. Polasek (jpolasek_at_cfl.rr.com)
Date: 02/02/05 

Date: Tue, 01 Feb 2005 22:07:11 -0500

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. 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.

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.

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: 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: Who will stun the world as next Einstein?
    ... >>about things like the derivative of a scalar function being a vector. ... > represents an equipotential surface quite like that at the metal ... change facts - a scalar is a scalar and a vector is a vector. ... Because a vector equals a vector and a scalar equals a scalar. ...
    (sci.physics.relativity)
  • Re: Who will stun the world as next Einstein?
    ... >>about things like the derivative of a scalar function being a vector. ... > represents an equipotential surface quite like that at the metal ... change facts - a scalar is a scalar and a vector is a vector. ... Because a vector equals a vector and a scalar equals a scalar. ...
    (sci.physics)
  • 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)