In article ,
Y. T. wrote:
(John Schoenfeld) wrote in message
. com...
POSTULATES OF A MODIFIED CLASSICAL PHYSICS

Inertia law:
Mass is a scalar velocity field:
Scalar form: m(v) = m0 + int(0,v) (dm/dv) . dv


This has me puzzled already. You appear to be saying that there is a
quantity "m" that is an integrable function of some other quantity
"v". You don't define these letters, however.

The context seems to imply that "m" refers to a mass (of what?, if I
may dare ask). I appears further that "v" refers to a velocity but as
there is only a single "m" I fail to see what velocity is meant here;
i.e. velocity of what with respect to what else?

Of course if I am interpreting the letters right, then this would only
say anything new if dm/dv was not equal to zero and since there's no
observation ever made by a human being that would lead to such a
non-zero derivative I fail to see what the purpose of this exercise
might be. You could just as well have written dm/dT to include some
temperature dependence of mass (which would also be zero). So I'm
cautiously guessing that I'm misinterpreting the physical meaning of
the variables in that equation.

Care to eludicate your thoughts to the casual reader?

What became called relativistic mass was used since the 19th century, in
studies of electrodynamics. m(v)=m/sqrt(1-v^2/c^2) It's a little
old-fashioned today because in special relativity it's the kinematics, the
geometry, that leads to the extraordinary behaviors at high speeds, and
not anything special that's happening with forces and masses. But
relativistic mass can still get you through a lot of lab-frame problems
like the design of a particle accelerator.

--
"The polhode rolls without slipping on the herpolhode lying in the
invariable plane." -- Goldstein, Classical Mechanics 2nd. ed., p207.

"Franz Heymann" wrote:

Velocity is always a vector. It is permanently and uniquely
defined
as dx/dt. And dx is a vector and dt is a scalar, since we are not
talknig SR here.

Wrong. It could very well be a tensor of rank 3 or n, couldn't it?

You are waffling beyond the limitsof your competence.

So then velocity is defined "permanently and uniquely" as a rank-1
tensor? I hope you don't teach your students such nonsense.

"John Schoenfeld" wrote in message
om...
"Franz Heymann" wrote:

Velocity is always a vector. It is permanently and uniquely
defined
as dx/dt. And dx is a vector and dt is a scalar, since we are
not
talknig SR here.

Wrong. It could very well be a tensor of rank 3 or n, couldn't
it?

You are waffling beyond the limitsof your competence.

So then velocity is defined "permanently and uniquely" as a rank-1
tensor? I hope you don't teach your students such nonsense.

You are still waffling.

Franz

"Franz Heymann" wrote in message ...
"John Schoenfeld" wrote in message
om...
"Franz Heymann" wrote:

Velocity is always a vector. It is permanently and uniquely
defined
as dx/dt. And dx is a vector and dt is a scalar, since we are
not
talknig SR here.

Wrong. It could very well be a tensor of rank 3 or n, couldn't
it?

You are waffling beyond the limitsof your competence.

So then velocity is defined "permanently and uniquely" as a rank-1
tensor? I hope you don't teach your students such nonsense.

You are still waffling.

Such a simple question, and you can't answer it.

Franz