"John Schoenfeld" wrote in message
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(Y. T.) wrote in message
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(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

"Franz Heymann" wrote in message ...
"John Schoenfeld" wrote in message
om...
(Gregory L. Hansen) wrote in message
...
In article ,
John Schoenfeld wrote:
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
Vector form: m(v) = m0 + intP(0,v) grad_v(m) . dv


1st Law of Motion:
The momentum of a body remains invariant unless otherwise acted
on by a Force.
Scalar form: p = m(v) v
Vector form: p = m(v) v


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


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


How are these modified postulates? They look like Newton's
postulates to
me, except you've specialized them to a mass that varies with
velocity in
an unspecified way (what's dm/dv?). It's still p=mv, F=dp/dt, and
conservation of momentum.

No, it's p = m(v)v. The different is important.

Then please explain this important difference between
p = mv
and
p = m(v)v

You can't see the major and fundamental difference?

p can remain constant while m(v) and v are varying, THAT'S the difference.

Franz

(John Schoenfeld) wrote in message . com...
"Franz Heymann" wrote in message ...
"John Schoenfeld" wrote in message
om...
(Gregory L. Hansen) wrote in message
...
In article ,
John Schoenfeld wrote:
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
Vector form: m(v) = m0 + intP(0,v) grad_v(m) . dv


1st Law of Motion:
The momentum of a body remains invariant unless otherwise acted
on by a Force.
Scalar form: p = m(v) v
Vector form: p = m(v) v


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


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


How are these modified postulates? They look like Newton's
postulates to
me, except you've specialized them to a mass that varies with
velocity in
an unspecified way (what's dm/dv?). It's still p=mv, F=dp/dt, and
conservation of momentum.

No, it's p = m(v)v. The different is important.

Then please explain this important difference between
p = mv
and
p = m(v)v

You can't see the major and fundamental difference?

p can remain constant while m(v) and v are varying, THAT'S the difference.

In F = d(mv)/dt, if F = 0, then mv = constant, whether m = constant or
some function of a variable x, including v. For instance, in a
rocket/exaust gas system, the total system momentum remains constant
although the rocket mass decreases as a function of time. You must
understand once for all that v = f(t) and hence m(v) is equivalent to
some function m(t).

Since your force law is integratable to yield F = d(mv)/dt, there is
no difference between what you have proposed and Newton's formulation,
as the latter is general enough to account for any functional
variations in the 'quantity of motion' (look up the term).

The reason momentum is conserved in dynamically isolated systems is
ultimately due to space translation symmetry postulation.

You have failed to provide an alternative force law to Newton's but
only a specific problem where m varies as a function of v. You have
called this special case a general law, which is not as Newton's law
already covers it.

As such, you got nothing but a bold speculation that non-relativistic
mass is a function of velocity. That is where your whole think reduces
to. A bold speculation.

Mike

"John Schoenfeld" wrote in message
om...
"Franz Heymann" wrote in message
...
"John Schoenfeld" wrote in message
om...
(Gregory L. Hansen) wrote in
message
...
In article ,
John Schoenfeld wrote:
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
Vector form: m(v) = m0 + intP(0,v) grad_v(m) . dv


1st Law of Motion:
The momentum of a body remains invariant unless otherwise
acted
on by a Force.
Scalar form: p = m(v) v
Vector form: p = m(v) v


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


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


How are these modified postulates? They look like Newton's
postulates to
me, except you've specialized them to a mass that varies with
velocity in
an unspecified way (what's dm/dv?). It's still p=mv, F=dp/dt,
and
conservation of momentum.

No, it's p = m(v)v. The different is important.

Then please explain this important difference between
p = mv
and
p = m(v)v

You can't see the major and fundamental difference?

p can remain constant while m(v) and v are varying, THAT'S the
difference.

But since you have already said that your m is a function of v, it
follows that
p = mv
is identical with p = m(v) v

Franz

"Franz Heymann" wrote in message ...
"John Schoenfeld" wrote in message
om...
"Franz Heymann" wrote in message
...
"John Schoenfeld" wrote in message
om...
(Gregory L. Hansen) wrote in
message
...
In article ,
John Schoenfeld wrote:
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
Vector form: m(v) = m0 + intP(0,v) grad_v(m) . dv


1st Law of Motion:
The momentum of a body remains invariant unless otherwise
acted
on by a Force.
Scalar form: p = m(v) v
Vector form: p = m(v) v


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


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


How are these modified postulates? They look like Newton's
postulates to
me, except you've specialized them to a mass that varies with
velocity in
an unspecified way (what's dm/dv?). It's still p=mv, F=dp/dt,
and
conservation of momentum.

No, it's p = m(v)v. The different is important.

Then please explain this important difference between
p = mv
and
p = m(v)v

You can't see the major and fundamental difference?

p can remain constant while m(v) and v are varying, THAT'S the
difference.

But since you have already said that your m is a function of v, it
follows that
p = mv
is identical with p = m(v) v

ha ha ha

(Mike) wrote in message
You can't see the major and fundamental difference?

p can remain constant while m(v) and v are varying, THAT'S the difference.

In F = d(mv)/dt, if F = 0, then mv = constant, whether m = constant or
some function of a variable x, including v.

Wrong.

For instance, in a
rocket/exaust gas system, the total system momentum remains constant
although the rocket mass decreases as a function of time. You must
understand once for all that v = f(t) and hence m(v) is equivalent to
some function m(t).

Wrong.

dp/dt = 0 does not imply (dm/dt = 0 and dv/dt = 0).

Since your force law is integratable to yield F = d(mv)/dt, there is
no difference between what you have proposed and Newton's formulation,
as the latter is general enough to account for any functional
variations in the 'quantity of motion' (look up the term).

Wrong.

F = d(m(v)v)/dt = ma + grad_v(m) . a

The reason momentum is conserved in dynamically isolated systems is
ultimately due to space translation symmetry postulation.

Irrelevant.

You have failed to provide an alternative force law to Newton's but
only a specific problem where m varies as a function of v. You have
called this special case a general law, which is not as Newton's law
already covers it.

Wrong.

As such, you got nothing but a bold speculation that non-relativistic
mass is a function of velocity. That is where your whole think reduces
to. A bold speculation.

Wrong.

Mike

woops..

that was .. + v (grad_v(m) . a)

;-)

"Franz Heymann" wrote in message ...
"John Schoenfeld" wrote in message
om...
(Y. T.) wrote in message
om...
(John Schoenfeld) wrote in message
om...
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.

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


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

"John Schoenfeld" wrote in message
om...
"Franz Heymann" wrote in message
...
"John Schoenfeld" wrote in message
om...
(Y. T.) wrote in message
om...
(John Schoenfeld) wrote in message
om...
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.

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

You are waffling beyond the limitsof your competence.


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

Franz

In article ,
John Schoenfeld wrote:
"Franz Heymann" wrote in message
...

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?

Only if position was a tensor of rank 3 or n.

--
"The main, if not the only, function of the word aether has been to
furnish a nominative case to the verb 'to undulate'."
-- the Earl of Salisbury, 1894