Re: Hansen discovers how to reason.

From: Timo Nieminen (timo_at_physics.uq.edu.au)
Date: 01/27/05 

Date: Thu, 27 Jan 2005 15:11:43 +1000

On Thu, 27 Jan 2005, Androcles wrote:

> "Timo Nieminen" <timo@physics.uq.edu.au> wrote:
> > On Thu, 27 Jan 2005, Androcles wrote:
> >
> >> 4th axiom...
> >>
> >> Now, is (x,y,z,t) a vector?
> >>
> >> Does there exist a -t such that -t + t = 0?
> >>
> >> Or, in simpler terms that you use as you teachiing method, do tell us
> >> what is the magnitude and direction of time?
> >
> > I'm a little surprised to see you asking this again, since you and I
> > discussed this so recently.
>
> Yes, but you haven't a clue and are a total waste of time.

Avoiding the issue by dealing insults. One of your typical tactics. So
mature and resourceful of you! Of course, logic might be more effective.

> > Anyway, given that you ask the question
> > again, perhaps this is a good place to summarise the conclusion:
> >
> > Choose a time to be the origin of a time coordinate axis. Times later
> > than
> > the origin will be positive, earlier times will be negative. The
> > magnitude
> > of the time coordinate is equal to the length of the time interval
> > from
> > the origin to the time represented by the time coordinate.
> >
> > Choose 4 pm to be the origin. At twenty past four, we have
> > t1 = t = 20 minutes. Twenty to four would then be t2 = (-t) = - 20
> > minutes.
>
> Learn what a scalar is, Nieminen, because you are juggling them as if
> they were vectors. Scalars are not vectors.
> The scalar -1 is not the vector -t. There is no vector -t.

Do you know the difference between a vector and a component of a vector?

Do you know that the definition of a vector space requires the vector
(-1)*t to be equal to (-t)?

If we choose 4pm to be the origin, later times to be positive, and earlier
times to be negative, then 20 past four is t = 20 minutes. Yes or no?

What is the time-coordinate assigned to twenty to four?

> If I have two opposing forces, as in a tug o' war, I have two opposing
> force vectors. I can yell "Heave" and increase the scalar on both teams.
> The team with the greatest scalar wins.

At this point, it seems like a good idea to ask you to define "scalar".
Are you (mis)using scalar to mean "magnitude of a vector quantity"?

> It is the vector itself, with unit scalar, that has to have an inverse.
> Time has NO inverse. You can go from 4:00 pm to 4:20 pm (in fact you
> cannot avoid it) but you can NEVER go from 4:00 pm to 3:40 pm. There is
> no inverse time. Saying that 3:40 exists or did once exist does NOT make
> time a vector.
>
> Let's try this. I'll use zero as my scalar. It's 4:00 pm. Time will
> march on to 4:20 pm. Apply the scalar. It is still 4:00 pm. Go ahead
> and do it. Prove you can stay at 4:00 pm.

Why does time travel have to do with? Time coordinates are just a
mathematical description of time. You don't have to be able to go to a
particular time for a mathematical description to label that time.

Is 6 hours earlier than 6 hours later than now a time other than now, or
is it now?

> > Since t1 + t2 = t + (-t) = 20 + -20 = 0, t2 is the additive inverse of
> > t1.
>
> No it isn't. You have not stopped time. You have only shown that 4:00 pm
> once was.

I'm not claiming to have stopped time. You might note that if we choose
the origin to be 4pm 3 days from now, *both* times lie in the future. How
can time travel into the past be possibly required for that?

The requirements for time to be labelable by a time coordinate that is the
component of a 1D vector are simple. No travel, whether physical travel or
time travel is required by the mathematical definition. Why do you
continually insist that it is? Please, point out in an accepted definition
of a vector that such travel must be possible!

> > Since twenty to four existed/will exist (depending on exactly which
> > 4pm is
> > chosen to be the origin) this is a physically meaningful time,
>
> No it isn't physically meaningful. Even if your chosen 4:00 pm will
> exist, it will pass and become 4:00 pm did exist, and you cannot stop it
> from happening.

So, are you saying that twenty to four is not a physically meaningful
time?

>From a physical perspective, the description of times by a vector simply
means that time later than the origin have a positive vector component,
and times earlier than the origin have a negative vector component.

Are you really trying to say that "3 hours ago" is not physically
meaningful?

Alas, it's a cornerstone of the great edifice that is Galilean-Newtonian
mechanics. I guess you must disagree with the results of that field of
physics, too.

Hint: if you want to be a successful anti-SR advocate, it might be
worthwhile attacking the differences between SR and Newtonian physics, not
the points that they have in common.

> There is nothing about making time a vector that is physically
> meaningful.
> Time does exist, yes, but it is NOT a vector, anymore than (apple,
> orange) is a vector. You cannot turn apples into oranges by rotating the
> supposed vector, but you can turn x into y by rotating the vector (x,y)
> into (y,x).

Oh my! You must be proud of that irrelevant smokescreen!

Alas, logical argument (as opposed to debate, where such things as
irrelevant diversions, lying, ignoring evidence, making up your own
definitions, and shouting down the opposition can actually be successful
tactics) is not the place for that kind of thing.

Since you have no reply of substance, merely an unsubstantiated claim that
the mathematical definition of a vector (or did you mean additive
inverse?) requires travel through space or time, I think it can be safely
assumed that you have no valid counterargument.

> > provided the time axis is infinite and unbounded, the same can be done
> > for
> > any time t, and thus the requirement of existence of an additive
> > inverse
> > is satisfied.
>
> No it isn't.

By the usual mathematical definition of an additive inverse, it is.

> You cannot stop time passing, or reverse it, or change it
> in any way.

And this requirement can be found exactly where in the mathematical
definition of an addition inverse, or for that matter, a vector space? 

-- 
Timo

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