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 23:45:06 GMT

"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:aft101de7ttugrnup2r5nhrhvctoesq4ak@4ax.com...
> On Wed, 02 Feb 2005 06:04:35 GMT, "Bill Hobba" <bhobba@rubbish.net.au>
> wrote:
>
> >
> >"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:
> >>
> snip
> >> >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
> >> >
> snip
> >
> >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.
> >
> snip
> >>
> >> 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
> >
> snip
>
> OK Bill try this. I wrote a whole new law of gravity by adapting the
> Navier Stokes equation
>
> grad P = -rho g

Define rho - is it a vector? In normal usage rho would be a constant so I
must take g as a vector until you detail rho better.

>
> leading me to add 1 term to Newton's equation:
>
> d(c^2/2)/dr = MG/r^2 = -g

The above implies g is a scalar - back to the drawing borard.

>
> which you can see as my Eq. (1) of my new gravity theory on my website
> http://www.dualspace.net.

Why would I be interested in the ramblings of a crank who can not even
understand vectors equate to vectors and scalars to scalars or the correct
history if Dirac and the positron?

Bill

>
> There you can see, with some study, that this new derivative term by
> itself does all the work, in open handed fashion, of all the cryptic
> time-manglings of the timelike Schwarzschild metric.
>
> For example, take a look at how it computes, piece by piece, the 4
> separate corrections needed in the GPS system.
>
> I would not know how to proceed if I had to keep to the grad
> formulation, except to affix the Kronecker delta D_ir so we would
> have del_i D_ir = del_r being the derivative vs r.
>
> Instead of arguing about what the OP said the other day, you will find
> plenty of equations in my paper to mull over and cut to pieces, and,
> today only, I'll give you $10 apiece for each error in mathematics
> that you find, scrivener's errors aside.
>
> John Polasek
>
>

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