"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message

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"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
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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

That is brand new. Why did you change your old approach?

Wrong. It's a direct consequence from the fundamental theoreom of
calculus.
m(v) - m(u) = int(0,v) (dm/dv).dv
in vector form:
m(v) - m(u) intP(u,v) grad_v(m).dv

Vector form: m(v) = m0 + intP(0,v) grad_v(m) . dv

So which one are we supposed to choose, vector or scalar?

Newton's Laws can be expressed in Scalar form or Vector form, what's
your problem with what i have posted above?

My problem is that Newton's Laws refer only to the vector quantities
acceleration, force, momentum, etc.

1st Law of Motion:
The momentum of a body remains invariant

I think you mean constant. Invariant has to do with the behaviour

of

a quantity when changing frames of reference.

In my opinion, they are invariant and constant are equal.

No.. Before you came on the scene, the word "invariant had acquired a
quite specific usage. *Not* to be confused with "constant".
You are, of course, welcome to use exisging terms with meanings
privatew to you, but I would not advise it if you are trying to
convince folk that you are not just cheering from the side lines.

unless otherwise acted on by a Force.
Scalar form: p = m(v) v

That is not an expression for a force

That's right, it's the expression for momentum.You said it was an

expression for force. Reread what you said.

Vector form: p = m(v) v

That, also, is not an expression for a force

That's right, it's the expression for momentum.

That's not what you said.

And a force is *always* a vector.

For a lot of problems, Force can be treated as a scalar - |F|.

|F| is the magnitude of the froce vector F. Force is *always* a
vector.

2nd Law of Motion:
Force is the 1st time derivative of momentum.
Scalar form: F = (m + v dm/dv) a

There is no such thing as a scalar force.

Wrong. It's just the length of the Force vector, and v the length of
the velocity vector, which are scalars.

In that case you are still wrong, because the acceleration is a
vector.

You should learn to express yourself clearly, correctly and
unambiguously.

Vector form: F = (m + v . grad_v(m))a

These equations disagree with the definitions you gave higher up

for

force, where you called it "p" for some unknown reason.

That was a disastrous typo that I made. The "Vector" form is:

F = m(v) a + v(grad_v(m) . a)

Proof:
p = m(v) v
dp/dt = m(v) dv/dt + v(grad_v(m) . dv/dt)

Why not just keep it simple, and just say
dp/dt = m dv/dt + v dm/dt ?

3rd Law of motion:
Interacting bodies conserve net momentum through equal and

opposing

Forces.

That is not a law of motion. It is deducible from the 3rd law of
motion. (I hasten to say that I refer to Newton's 3rd law, not

your

crap.)

It is Newtons 3rd law.

No. At the stage at which newton's laws are initially presented,
momentum has not yet been defined.
Newton's 3rd law says

"An action is always opposed by an equal and opposite reaction"

There is no mention of momentum conservation there. The latter is
derived a page or two later, after "momentum" has been defined".

I do most sincerely hope that you are not just aiming to give us a
repeat performance of the crap you posted earlier.

Franz

John, I have a serious question for you. Why do you keep doing this to
yourself? What do you hope to gain by rederiving 300 year old equations
that have been demonstrated accurately time and again?

Before you go and throw a little temper tantrum about how I am just

another

cheerleading naysayer who isn't doing anything to prove you wrong or

show

mistakes in your math,

I had a typo, which none of you idiots picked off.

F = m a + v (grad_v(m).a)

Blah blah blah.

If you can show that my equation has been derived 300 years ago then I
will retract it, otherwise behold the astounding truth.

Of what?

I challenge you to prove yourself right in each and
every new theorem you have presented. Yes. You heard me correctly.

Prove

them right.

They are right. The "how" is purely irrelevant.

If they are right, you will have some evidence to show that they are right.
What does your postulates say that Newton's don't?

Sean

"John Schoenfeld" <j.schoenfeld@programmer.net> wrote in message
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ytyourclothes@p.zapto.org (Y. T.) wrote in message

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j.schoenfeld@programmer.net (John Schoenfeld) wrote in message

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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?

m is mass v is velocity.

m(v) is velocity-dependent mass.

dm/dv is rate of change of mass w.r.t speed.

grad_v(m) is the rate of change of mass w.r.t velocity.


From the fundamental theoreom of calculus,

m(v) - m(u) = int(u,v) (dm/dv)dv

m(v) = m0 + int(0,v) (dm/dv).dv

likewise, for vector velocity

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.

m(v) = m0 + intP(0,v) grad_v(m) . dv