Subject: Rest mass or inertial mass?
From:
Rest mass of an elementary particle is an intrinsic property of the
elementary particle.

Ask yourself, if it is intrinsic then why do you refer to it by "rest". If it
is intrinsic than it is the same value according to every frame, invariant, and
as such with out need of reference to any particular frame.

But an accelerator accelerates mass, i.e.
inertial mass of an elementary particle. An accelerator accelerates
mass which changes with speed.

No. The mass is accelerated in accordance with
F = mA
and does not change with speed.

But
4-vectors are unobservable.

Nonsense, unobservables(i.e. hidden variables) have no place in modern physics.

Only time and space components of
4-vectors are observable.

If the components of an entity are observable then you can observe the entity
by its components.

So, invariant mass is observable only if the
4-vectors are equal to their time components, i.e. at rest.

That is not a logical inference following from your previous statement.

(invariant, 4-scalar) mass is not additive

Right, so?

and is not conserved in
nuclear reactions.

Wrong. The sum of masses is not conserved, but as you just stated the system
mass is not additive. The system mass is its center of momentum frame energy or
is the length of the system momentum four vector which is conserved. Energy is
conserved according to every frame, of course including the center of momentum
frame! see last paragraph-
http://www.geocities.com/physics_world/stp/pg_248.htm

Taylor and Wheeler wrote, {rest mass of final
system increases in an inelastic encounter} (p.121 of Spacetime
Physics, 1966).


See the second edition with corrections-
http://www.geocities.com/physics_world/stp/pg_247.htm
concerning the inelastic collision-"In all three examples the system momenergy
and the system mass are the same before as after."

[this topic is getting old so I sniped the rest of the misconceptions]

(WaiteDavid137) wrote in message ...
Subject: Rest mass or inertial mass?
From:
Rest mass of an elementary particle is an intrinsic property of the
elementary particle.

Ask yourself, if it is intrinsic then why do you refer to it by "rest".

Simple. Even trivial. The intrinsic properties don't determine its
inertia.

No. The mass is accelerated in accordance with
F = mA
and does not change with speed.

The parrot is back.

Why do you keep repeat yourself. Everyone here is accutely aware that
proper mass is invariant and you've been overly repetitive with your
identification of proper mass as just "mass"

That proper mass is invariant is NOT debtated. So why do you keep
repeating what everyone knows extremely well?


Wrong. The sum of masses is not conserved, ...

The sum of inertial mass (aka "relativistic mass" aka "mass") is
conserved where inertial mass is defined as the m in p = mv where v =
coordinate velocity


Pmb

(WaiteDavid137) wrote in message ...
Subject: Rest mass or inertial mass?
From:
Rest mass of an elementary particle is an intrinsic property of the
elementary particle.

Ask yourself, if it is intrinsic then why do you refer to it by "rest". If it
is intrinsic than it is the same value according to every frame, invariant, and
as such with out need of reference to any particular frame.
Radi to Wait 151917 (208)

I refer to the intrinsic property by {rest mass} because this property
is intrinsic, we do not observe rest mass of a moving object.
I refer to it by "rest" because I observe inertial mass. And the
observable mass I call {mass}.
I use the term {observe} in the same sense as Wheeler uses it at p.
248: (but only as observed in the frame…)


But an accelerator accelerates mass, i.e.
inertial mass of an elementary particle. An accelerator accelerates
mass which changes with speed.

No. The mass is accelerated in accordance with
F = mA
and does not change with speed.
You are mistaken.
An accelerator acts on a particle by the force f=e(E+vxB) rather than
by {F}.
The acceleration of the particle is {a} rather than {A}.
The inertia of a particle is important when the particle is under
acceleration.
Are you and Wheeler familiar with concept of inertia?


But
4-vectors are unobservable.

Nonsense, unobservables(i.e. hidden variables) have no place in modern physics.
We do not observe a 4-vectors in the Wheeler sense. We observe space
and time components of a 4-vector only.

Only time and space components of
4-vectors are observable.

If the components of an entity are observable then you can observe the entity
by its components.

So, invariant mass is observable only if the
4-vectors are equal to their time components, i.e. at rest.

That is not a logical inference following from your previous statement.

(invariant, 4-scalar) mass is not additive

Right, so?

and is not conserved in
nuclear reactions.

Wrong. The sum of masses is not conserved, but as you just stated the system
mass is not additive. The system mass is its center of momentum frame energy or
is the length of the system momentum four vector which is conserved. Energy is
conserved according to every frame, of course including the center of momentum
frame! see last paragraph-
http://www.geocities.com/physics_world/stp/pg_248.htm

Taylor and Wheeler wrote, {rest mass of final
system increases in an inelastic encounter} (p.121 of Spacetime
Physics, 1966).


See the second edition with corrections-
http://www.geocities.com/physics_world/stp/pg_247.htm
concerning the inelastic collision-"In all three examples the system momenergy
and the system mass are the same before as after."

[this topic is getting old so I sniped the rest of the misconceptions]
Dear Waite,
I present a method to learn if mass changes with velocity.
Let a ball move in a box. Does its gravitational pull change with its
velocity?

Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/16/03 10:36 AM US Mountain Standard Time
Message-id:

(WaiteDavid137) wrote in message
...
Subject: Rest mass or inertial mass?
From:
Rest mass of an elementary particle is an intrinsic property of the
elementary particle.

Ask yourself, if it is intrinsic then why do you refer to it by "rest". If
it
is intrinsic than it is the same value according to every frame, invariant,
and
as such with out need of reference to any particular frame.
Radi to Wait 151917 (208)

I refer to the intrinsic property by {rest mass} because this property
is intrinsic, we do not observe rest mass of a moving object.

BS

I refer to it by "rest" because I observe inertial mass.

You still don't get it. The inertia or the m in
F = mA is invariant so there is no need to refer to it with the rest frame in
mind.

And the
observable mass I call {mass}.
I use the term {observe} in the same sense as Wheeler uses it at p.

No you don't. You misconstrued what he said.

But an accelerator accelerates mass, i.e.
inertial mass of an elementary particle. An accelerator accelerates
mass which changes with speed.

No. The mass is accelerated in accordance with
F = mA
and does not change with speed.
You are mistaken.
An accelerator acts on a particle by the force f=e(E+vxB) rather than
by {F}.

No, the electic and magnetic fields are not vectors. They are incomplete parts
of the actual field acting. The actual field is the unification of the two into
the electromagnetic field tensor. The force an accelerator actually acts on a
particle by is then equation 7.1.9 at
http://www.geocities.com/zcphysicsms/chap7.htm#BM84
whisch is F, not f.

The acceleration of the particle is {a} rather than {A}.

No, the acceleration in term of proper time derivatives instead of coordinate
time derivatives is A, not a.

The inertia of a particle is important when the particle is under
acceleration.
Are you and Wheeler familiar with concept of inertia?

Yes it is the m in
F = mA.
Apparently you are not.

But
4-vectors are unobservable.

Nonsense, unobservables(i.e. hidden variables) have no place in modern
physics.
We do not observe a 4-vectors in the Wheeler sense.

Sure we do. You misconstrued what he said.

We observe space
and time components of a 4-vector only.


Fine, then you observe the 4-vector through its components.

Only time and space components of
4-vectors are observable.

If the components of an entity are observable then you can observe the
entity
by its components.

So, invariant mass is observable only if the
4-vectors are equal to their time components, i.e. at rest.

That is not a logical inference following from your previous statement.

(invariant, 4-scalar) mass is not additive

Right, so?

and is not conserved in
nuclear reactions.

Wrong. The sum of masses is not conserved, but as you just stated the
system
mass is not additive. The system mass is its center of momentum frame
energy or
is the length of the system momentum four vector which is conserved. Energy
is
conserved according to every frame, of course including the center of
momentum
frame! see last paragraph-
http://www.geocities.com/physics_world/stp/pg_248.htm

Taylor and Wheeler wrote, {rest mass of final
system increases in an inelastic encounter} (p.121 of Spacetime
Physics, 1966).


See the second edition with corrections-
http://www.geocities.com/physics_world/stp/pg_247.htm
concerning the inelastic collision-"In all three examples the system
momenergy
and the system mass are the same before as after."


Dear Waite,
I present a method to learn if mass changes with velocity.

I already told you the answer. It does not.

Let a ball move in a box. Does its gravitational pull change with its
velocity?


In relativity gravity doesn't pull at all.
A = 0 is the geodesic equation from F = 0.

If you instead asked if a ball has a different spacetime geometry when it is
spinning without translation than when it is rolling then the answer would
still be no. The spacetime geometry in either case is kerr geometry. g_mu_nu in
one case is just a constant velocity transformation of g_mu_nu in the other
case, but either way the spacetime it yields is kerr geometry.

In highly relativistic scenarios you can not use F = GMm/r^2 at all. Instead
you use Einstein's field equation which has T^mu^nu as the source term not m.
This tensor is frame covariant. Its "length" does not depend on speed.

Radi to Waite 161408 (157)

Please, confirm that, according to you and to like-minded persons:
(1) Inertia of a body equals mass of the body and the mass is mass of
the body at rest.
(2) A spring-balance does not show force which acts on a body from the
spring-balance.
(3) Speed of a body does not equal to length divided by time.
(4) Acceleration of a body does not equal to the second derivative
with respect to time.
(5) The electic field is not a vector field.

Dear Waite,
I had read at the page 248:
{How do we find out the mass of a particle?
- Weight it! Or determine its gravitational pull.}
So, I present a method to learn if mass changes with velocity.
Let a ball move in a box.
Does its gravitational pull change with its velocity?

Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/17/03 9:53 AM US Mountain Standard Time
Message-id:

Radi to Waite 161408 (157)

Please, confirm that, according to you and to like-minded persons:
(1) Inertia of a body equals mass of the body and the mass is mass of
the body at rest.

I already explained this. Go back and read it.

(2) A spring-balance does not show force which acts on a body from the
spring-balance.

I never mentioned any spring balance. But, since you mention one, once in
equalibrium it displays the length of the four-vector force.

(3) Speed of a body does not equal to length divided by time.

I never said that. Four-velocity is length divided by proper time. The "length"
of this vector is c. Coordinate velocity is length divided by coordinate time.
The magnitude of this pseudovector is the speed.

(4) Acceleration of a body does not equal to the second derivative
with respect to time.

I never said that. In SR four-acceleration is the second derivative of position
with respect to proper time.

(5) The electic field is not a vector field.

E by itself has only 3 components and therefor can not obey tensor
transformation. Something is a vector only when it obeys tensor transformation
properties. The electormagnetic field is the actual 16 element vector that one
has producing the four-force.

Dear Waite,
I had read at the page 248:
{How do we find out the mass of a particle?
- Weight it! Or determine its gravitational pull.}
So, I present a method to learn if mass changes with velocity.
Let a ball move in a box.
Does its gravitational pull change with its velocity?


I already explained this. Go back and read where I explained why gravity does
not pull at all and why the mass of the ball doesn't change in your scenario.

By the way, the weight you measure on a scale is not a measure of gravity force
down which four-force is zero. It is a measure of the normal force up on the
object being weighed. This force produces nonzero four-acceleration. Or
"really" what it displays is the equal magnitude reaction force down on the
scale. The scale reading is the length of this four-force, not the pull of
gravity. Once more, the four-force of gravity is F=0.

(WaiteDavid137) wrote in message ...
Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/17/03 9:53 AM US Mountain Standard Time
Message-id:

Radi to Waite 161408 (157)

Please, confirm that, according to you and to like-minded persons:
(1) Inertia of a body equals mass of the body and the mass is mass of
the body at rest.

I already explained this. Go back and read it.
So, you confirm that inertia of a body equals mass of the body and the
mass is mass of the body at rest.

(2) A spring-balance does not show force which acts on a body from the
spring-balance.

I never mentioned any spring balance. But, since you mention one, once in
equalibrium it displays the length of the four-vector force.
Do you confirm that a spring-balance does not show force which acts on
a moving body accelerating by the spring-balance?

(3) Speed of a body does not equal to length divided by time.

I never said that. Four-velocity is length divided by proper time. The "length"
of this vector is c. Coordinate velocity is length divided by coordinate time.
The magnitude of this pseudovector is the speed.
And what is {velocity} without an adjective?

(4) Acceleration of a body does not equal to the second derivative
with respect to time.

I never said that. In SR four-acceleration is the second derivative of position
with respect to proper time.
I claimed: The acceleration of the particle is {a} rather than {A}.
You wrote: No.
So, according to you, {a}, i.e. the second derivative with respect to
time is not the acceleration.


(5) The electic field is not a vector field.

E by itself has only 3 components and therefor can not obey tensor
transformation. Something is a vector only when it obeys tensor transformation
properties. The electormagnetic field is the actual 16 element vector that one
has producing the four-force.
So, you confirm that the electic field is not a vector field and the
magnetic field is not a phseudovector field.

Dear Waite,
I had read at the page 248:
{How do we find out the mass of a particle?
- Weight it! Or determine its gravitational pull.}
So, I present a method to learn if mass changes with velocity.
Let a ball move in a box.
Does its gravitational pull change with its velocity?


I already explained this. Go back and read where I explained why gravity does
not pull at all and why the mass of the ball doesn't change in your scenario.
Do you words mean that Wheeler made a mistake? One can learn how to
find out the mass of a body at the page 248.

By the way, the weight you measure on a scale is not a measure of gravity force
down which four-force is zero. It is a measure of the normal force up on the
object being weighed. This force produces nonzero four-acceleration. Or
"really" what it displays is the equal magnitude reaction force down on the
scale. The scale reading is the length of this four-force, not the pull of
gravity. Once more, the four-force of gravity is F=0.
Confirm that {normal force without other adjective produces nonzero
four-acceleration}
Confirm that {the scale reading is the length of four-force}. I think
that the scale reading is the modulus of the 3-force.

And what about a definition of mass?
My opponents write about mass all time.
But they do not know what is mass.
I asked my opponents,
Joe Fischer, Jeff Kimmel, Waite David, Vetwannabel, Tom Roberts,
Bilge,
to define the mass in the spirit of operationism.
I asked for an operation to determine the mass.
I have no answer.

Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/20/03 11:47 AM US Mountain Standard Time
Message-id:

(WaiteDavid137) wrote in message
...
Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/17/03 9:53 AM US Mountain Standard Time
Message-id:

Radi to Waite 161408 (157)

Please, confirm that, according to you and to like-minded persons:
(1) Inertia of a body equals mass of the body and the mass is mass of
the body at rest.

I already explained this. Go back and read it.
So, you confirm that inertia of a body equals mass of the body and the
mass is mass of the body at rest.

And the mass m of the body, the inertia term in F = mA is also the mass of the
body moving. It is invariant.


(2) A spring-balance does not show force which acts on a body from the
spring-balance.

I never mentioned any spring balance. But, since you mention one, once in
equalibrium it displays the length of the four-vector force.
Do you confirm that a spring-balance does not show force which acts on
a moving body accelerating by the spring-balance?

Since when is a spring balance of any use at all at relativistic speeds with
respect to the test mass? Good luck keeping such a scale in one piece. All I am
saying is that they are designed in such a way that they display the length of
the four-force when in equalibrium with the test mass.
|F| = (-g_mu_nuF^muF^nu)^(1/2)
Lengths of four-vectors are invariant. This value is independant of frame.


(3) Speed of a body does not equal to length divided by time.

I never said that. Four-velocity is length divided by proper time. The
"length"
of this vector is c. Coordinate velocity is length divided by coordinate
time.
The magnitude of this pseudovector is the speed.
And what is {velocity} without an adjective?

In modern relativity when the context does not make it clear one should always
use an adjective.


(4) Acceleration of a body does not equal to the second derivative
with respect to time.

I never said that. In SR four-acceleration is the second derivative of
position
with respect to proper time.
I claimed: The acceleration of the particle is {a} rather than {A}.
You wrote: No.
So, according to you, {a}, i.e. the second derivative with respect to
time is not the acceleration.


No, You sniped what I actually said.


(5) The electic field is not a vector field.

E by itself has only 3 components and therefor can not obey tensor
transformation. Something is a vector only when it obeys tensor
transformation
properties. The electormagnetic field is the actual 16 element vector that
one
has producing the four-force.
So, you confirm that the electic field is not a vector field and the
magnetic field is not a phseudovector field.

No. I said the electric field and the magnetic field are not vectors, that they
are pseudovectors. I said that the electromagnetic field F^mu^nu is the actual
vector.


Dear Waite,
I had read at the page 248:
{How do we find out the mass of a particle?
- Weight it! Or determine its gravitational pull.}
So, I present a method to learn if mass changes with velocity.
Let a ball move in a box.
Does its gravitational pull change with its velocity?


I already explained this. Go back and read where I explained why gravity
does
not pull at all and why the mass of the ball doesn't change in your
scenario.
Do you words mean that Wheeler made a mistake?

He either made a mistake or you misconstrued what he said.

By the way, the weight you measure on a scale is not a measure of gravity
force
down which four-force is zero. It is a measure of the normal force up on
the
object being weighed. This force produces nonzero four-acceleration. Or
"really" what it displays is the equal magnitude reaction force down on the
scale. The scale reading is the length of this four-force, not the pull of
gravity. Once more, the four-force of gravity is F=0.
Confirm that {normal force without other adjective produces nonzero
four-acceleration}

Fine, but next time do your own homework.
The force of gravity is Fg = 0. The normal force is
Fn = DP/dtau
Fn=dP^lamda/dtau+G^lambda_mu_nuU^muP^nu
dP^lamda/dtau = 0
Fn^lamda = G^lambda_mu_nuU^muP^nu
U^i and P^i are 0.
Fn^lambda = G^lambda_0_0U^0P^0
G^lambda_0_0 = (1/2)g^lambda^rho(g_0_rho,0 +g_0_rho,0 - g_0_0,rho)
Consider Schwarzschild geometryThe only nonzero contribution is
G^r_0_0 = (1/2)g^r^r(- g_0_0,r)
So
Fn^r =(1/2)g^r^r(- g_0_0,r)U^0P^0
P^0 = mU^0 = mcdt/dtau = mc(1 - 2GM/rc^2)^(-1/2)
g^r^r = - (1 - 2GM/rc^2)
g_0_0,r = 2GM/r^2c^2
Fn^r = - (1/2)(1 - 2GM/rc^2)(-2GM/r^2c^2)mc^2(1 - 2GM/rc^2)^(-1)
Fn^r = GMm/r^2
This normal force produces a four acceleration according to F = mA so,
A = F/m
A^r = GM/r^2

Confirm that {the scale reading is the length of four-force}. I think
that the scale reading is the modulus of the 3-force.

Next time do your own homework
|F| = (-g_mu_nuF^muF^nu)^(1/2)
According to the scale frame this is
|F| = (-eta_mu_nuf^muf^nu)^(1/2)
But this yields
|F| = |f| so yes it is also the scale frame modulus of the 3 force, but since
the length of a four-vector is invariant it is also the length of F according
to every frame.


And what about a definition of mass?

I already gave it to you.

And what about a definition of mass?

I already gave it to you.
Radi to Waite 201418
Dear Waite, I have read all you wrote, and I cut our texts to shorten
our messages.
Please, repeat your definition of mass.
Do you agree with the Wheeler definition: To find out the mass of a
particle one must weight it! Or determine its gravitational pull?
And, please, explain what occurs when a constant force accelerates a
body: is the accelerate constant? is the velocity constant?

Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/21/2003 2:34 PM US Mountain Standard Time
Message-id:

And what about a definition of mass?

I already gave it to you.
Radi to Waite 201418
Dear Waite, I have read all you wrote, and I cut our texts to shorten
our messages.
Please, repeat your definition of mass.

See-
http://www.geocities.com/zcphysicsms/chap3.htm
pay psrticular attention to the paragraphs immediately after equation 3.1.5

Do you agree with the Wheeler definition: To find out the mass of a
particle one must weight it! Or determine its gravitational pull?

No, but I doubt that is precisely what he said.

And, please, explain what occurs when a constant force accelerates a
body: is the accelerate constant? is the velocity constant?


When a constant force F acts on a body the acceleration A is constant, but due
to time dilation as relating the time differentials of the two the coordinate
acceleration "a" is not. Why in the world would you ask if the velocity is
constant? Of course not.