On Thu, 22 Sep 2011 09:27:55 -0500, PD <thedraperfamily@gmail.com> wrote:

> On 9/20/2011 12:31 PM, Aetherist wrote:
> > On Tue, 20 Sep 2011 10:03:30 -0700 (PDT), PD<thedraperfamily@gmail.com>  wrote:
> > 
> > > On Sep 19, 11:19 am, Byron Forbes<chocol...@caramel.com.au>  wrote:
> > > > In article<j54t3v$n2...@speranza.aioe.org>, thedraperfam...@gmail.com says...
> > > > 
> > > > > On 9/17/2011 11:24 PM, Byron Forbes wrote:
> > > > 
> > > > > > So what you're trying to get away with here with this "world line"
> > > > > > crap is that the accelerations cannot be made
> > > > > > insignificant by long durations of constant v?
> > > > 
> > > > > > Just crap.
> > > > 
> > > > > Just saying "just crap" doesn't remove the kink from the world line.
> > > > 
> > > > > > Be as determined as you like, along with hoards of the scientific 
> > > > > > community, but you have been fools all these
> > > > > > years and that's a simple fact.
> > > > 
> > > > > If you wish to believe so, being as determined as you like and all.
> > > > 
> > > > > The point, remember, is not to convince you of anything. After all, one
> > > > > can't convince a mule or a lawn ornament, and there really isn't much
> > > > > point in trying, even if the mule or the lawn ornament is daring you to try.
> > > > 
> > > >          Observer A and B move with constant v to each other.
> > > >          Time dilates the same to each other.
> > > >          Whose clock "is" running slower?
> > > 
> > > I see those quotation marks around the "is". I gather you think there
> > > is a definitive, frame-independent answer. This is not the case.
> > > 
> > > Let me show you something related.
> > > 
> > > Observer A and B are both massive objects and move with constant v
> > > with respect to each other. To observer A, observer A is motionless
> > > and has zero kinetic energy, while observer B is moving and has
> > > nonzero kinetic energy. To observer B, observer A is moving and has
> > > nonzero kinetic energy, while observer B is motionless and has zero
> > > kinetic energy.
> > > 
> > > Which observer "is" the one with more kinetic energy?
> > > 
> > > Think about that, and in particular ask why it is you would think that
> > > this question even should have a single, frame-independent answer.
> > 
> > Because it does...  Namely the only way that 'energy' becomes
> > 'observable' is if, and only if, it is realized.  As long as
> > A moves at v and B moves at v no energy is 'measurable'.  If
> > A collides with B then the actual energy realized is the lesser
> > of of (1/2)A(dv)^2 and (1/2)B(dv)^2.  If each changes their speed by
> > v (as you say, each can consider themselve 'at rest') then
> > the energy required to do so is (1/2)[A or B]v^2.  So the answer
> > to your question is, the frame independent answer is, there is
> > NO! kinetic energy to either...
> > 
> 
> Please reread the case. Observers A and B are both massive objects
> moving with constant v with respect to each other.
> This is not a case of a change in v at all.
> 
> I simply love your statement that the frame independent answer is that
> no kinetic energy in either of the objects. Love it, love it, love it. LOL.

Hmmm, let's write the down in math.

      /v
E = m |  v dv = (1/2) mv^2
      /0

now let dv -> 0

What is E?

In fact, without 'frames' what the hell is v???

KE like PE is 'a concept' and cannot be 'measured' unless
and until you have an actual dv...

It's simply NOT! real.  Any actual energy realized is
(1/2)m[dv]^2 IN THE REST FRAME from which the dv is
measured!

What I find sad is your inability to think logically and
not depend upon rote, memorized knee-jerk responses.