Re: Who will stun the world as next Einstein?

From: Tom Capizzi (etianshrdlu_at_verizon.net)
Date: 02/03/05 

Date: Thu, 03 Feb 2005 19:13:48 GMT

This is the complete text of your last post. Is something missing?

"Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
news:1107384256.335012.211010@z14g2000cwz.googlegroups.com...
>
> Tom Capizzi wrote:
>> "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
>> news:1107373965.215881.192060@o13g2000cwo.googlegroups.com...
>> > Tom Capizzi wrote:
>> >> "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
>> >> news:1107370962.408865.4520@o13g2000cwo.googlegroups.com...
>> >> >
>> >> > Tom Capizzi wrote:
>> >> >> "Ken S. Tucker" <dynamics@vianet.on.ca> wrote in message
>> >> >> news:1107303847.690945.56720@f14g2000cwb.googlegroups.com...
>> >> >> > Bill Hobba wrote:
>> >> >
>> >> >> >> Grad is not derivative.
>> >> >> >
>> >> >> > Say what???, last I learned, the Grad is the
>> >> >> > maximum slope (rate of change == derivative)
>> >> >> > at the point it's being done at. I was quite
>> >> >> > impressed by the Grad for that reason, but I
>> >> >> > was a HS punk at the time, and maybe missed
>> >> >> > something, please explain...
>> >> >> > TIA
>> >> >> > Ken
>> >> >
>> >> >> Grad is a vector and derivative is a scalar.
>> >> >
>> >> > That can't be right, derivatives don't
>> >> > automatically generate invariants.
>> >> > Tom, me thinks you've been Hobbatated.
>> >> > Ken
>> >>
>> >> Maybe I should have been more explicit. The Grad of
>> >> a scalar is a vector, but the derivative of a scalar is a scalar.
>> >
>> > Perhaps I was not clear.
>> > A scalar is an invariant ok?
>> > An invariant is not necessarily a constant ok?
>> > Denote "A" to be such a variable invariant.
>> >
>> > Then dA/dt =/= invariant because dt is not
>> > invariant and hence dA/dt is not a scalar.
>> >
>> > Something that's interesting though, is the
>> > relativistic fact that dA/dt is not a scalar.
>> >
>> > Maybe that's a good way to present dt as being
>> > a vector component, specifically with ct=x^0,
>> > then, dA/dt == dA/dx^0, and is also a vector
>> > component, but not a scalar (invariant).
>> >
>> > I just wanted to caution a derivative only
>> > in special cases produces a scalar, and
>> > certainly not generally in physics.
>> > Regards
>> > Ken S. Tucker
>> >
>> Interesting semantic quibble. I use the definition of scalar
>> as an entity with magnitude but no direction, and a vector
>> has both. To require a scalar to be an invariant is overly
>> restrictive, but I won't rule out the possibility that this is
>> customary usage in some circles. However, your argument
>> above is invalid because d/dt is not a component of the
>> Grad operator. Grad is a spatial vector derivative. And it
>> is a vector. The derivative of a scalar apparently may or
>> may not be a scalar (in the restricted sense of the word),
>> but it is never a vector. I am curious what you call an entity
>> with magnitude and no direction that is not an invariant.
>

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