Pmb wrote:
: "Joe Fischer" wrote
: Do you not agree that the precision balance
: will work on a merry-go-round where the sum of the
: centrifugal force and gravity are greater than 1.0 g?
:
: Obviously.

Haha.

: : But this thread has always been about inertial mass and the velocity
: : dependance thereof.
:
: As far as I am concerned, invariant mass means
: rest mass, and if mass is invariant, i.e., does _NOT_
: vary with velocity, then inertial mass equals rest mass.
:
: Then whyh do you refuse to address the subject of this post?

I have, you have learned physics according
to historical concepts without being able to remove
the old concepts when new knowledge becomes obvious.

: You've
: constantly stated your position on how to measure the mass of an object
: **which is not moving** by placing it at rest on a balance etc.

Of course it is not moving, of course I place
it on a balance, that is what you should have learned
in physics lab, or did you just take the written course
and skip the lab?

: The fact remains that, in principle, the weight of the object will increase
: as the body's speed increases.

That is false, I am moving at more that 0.9 c
relative to a very distant galaxy, and I still weigh
the same 200 pounds.

: That's never been questioned in relativity -

Wheeler did say it is false.

: It's a fact. A fact that cannot be gotten rid of by
: renaming quantities to fit your needs.

Other than for use in a particle accelerator,
it is nonsense.

: And let me remind you - This is the topic of this thread.

Did you just learn how to read? Too bad the
title of the thread doesn't make any sense at all,
rest mass is invariant at constant temperature in
a vacuum to avoid bouyancy.

: So do you want to address the topic of this thread?

Of course I do, the Divergent Matter model
of gravity would not be what it is without rest
mass being invariant.
That means it does not change with speed,
and not changing with speed means it does not
vary with speed, it means mass is not dependent
on speed.

: I.e. do you want to tell us how to measure the mass of moving body?

Sure, you slow it down, and I will put it
on one pan of a balance and then place known mass
standards on the other pan until they balance.

: Do you want to address the
: question "Does the weight of a body increase with speed?"
: Pmb

I think "Spacetime Physics" does that quite
well, but the style of arguing in teaching instead
of direct statements or rules makes it difficult
for some students.

Mass is invariant, it never changes without
absorbing thermal energy or being reconstructed in
a collision.

Your problem is that you think the energy
of a collision exists as mass before the collision
and that is obviously false.

While relativity is a study of physics in
which an observer tries to remove his prejudice
and learn nature, there is no reason the observer
should think his observations are correct without
relativity.

The title of the thread is obviously meant
to cause trouble, because rest mass must be identical
to inertial mass if "mass" is invariant.

People who want to teach do define terms and
symbols either in the text, or in an appendix as
in Weinberg "Gravitation and Cosmology".

Joe Fischer

--
3

"Joe Fischer" wrote in message
...
Pmb wrote:
: "Joe Fischer" wrote
: Do you not agree that the precision balance
: will work on a merry-go-round where the sum of the
: centrifugal force and gravity are greater than 1.0 g?
:
: Obviously.

Haha.

What's so funny? Would you laugh if I said 1 + 1 = 2?


I have, you have learned physics according
to historical concepts without being able to remove
the old concepts when new knowledge becomes obvious.

What in the world are you talking about "old concepts" - There's nothing
"old" about it. And if it was then you're wrong for thinking "old" means
"wrong" or "bad" or whatever. There's no new knowledge that has removed
relativity from physics - And the fact that you refuse to stay on topic
seems to give support to the fact that your notions are wrong.

You're like so many people here who are only able to say something is wrong
without supporting your arguement.

Here's what's wrong with your notion about mass. You have the erroneous idea
that conservtion of momentum is a newtonian concept and is wrong in some
way. You see an equations with m's and v's and all you are able to do is
think "Oooo! Newton!! Wrong!! Bad Physics!! Ooo. !"

The fact is that you have a severe lack of the most rudimentary concepts of
physics. And then you have the nerve to insuilt me when I explain newer
theoretical notions to you. Come on fischer - open your mind a tad,. Sheesh!


: You've
: constantly stated your position on how to measure the mass of an object
: **which is not moving** by placing it at rest on a balance etc.

Of course it is not moving, of course I place
it on a balance,

There you have it folks - fischer not only can't fathom what the weight of a
moving body is but he refuses to learn it.

: The fact remains that, in principle, the weight of the object will
increase
: as the body's speed increases.

That is false, I am moving at more that 0.9 c
relative to a very distant galaxy, and I still weigh
the same 200 pounds.

Still wrong. You're measuring your weight in the rest frame of the
gravitational field and you're measuruing your weight in the rest frame of
the scale.

Tell us what time dilation is so we know you know at least one fact of
relativity

[nonsense snipped]


When you show me that you understand at least one relativistic concept is
then we can continue. Otherwise your refusal to learn is getting extremely
boring fischer

"Joe Fischer" wrote

That is false, I am moving at more that 0.9 c
relative to a very distant galaxy, and I still weigh
the same 200 pounds.

Obviously you don't understand the question. The question "What is the
weight of a moving body" means that you choose a frame of referance to
measure the weight. Call that frame S. Then you measure the wieght of the
body when it is at rest. Call that W. The mass is defined as m = W/g where g
= acceleration due to gravity. Then let the body move in frame S, e.g. like
a car driving along the road - it still has weight but now it's moving in
the gravitational field (or accelerating frame of referance - it doesn't
matter since the results are identical). Then measure the weight while the
object is moving - impractical to do in real life but it has a definite
physical meaning. Le W' be the weight of the moving body. Then m' = W'/g is
the mass of the moving body. And m' = m/sqrt[1-(v/c)^2]





: That's never been questioned in relativity -

Wheeler did say it is false.

That's totally wong. Wheeler NEVER said that. If you claim that he did then
prove it. Strate exactly where he said that the weight of a moving body does
not depend on velocity.

If it's in the "use and abuse" section of Taylor and Wheeler's text
"Spacetime Physics -2nd Ed" then state which page. That section is now
online at

http://www.geocities.com/physics_world/stp/stp.htm

The pages and site were put online with the authors permission.


: It's a fact. A fact that cannot be gotten rid of by
: renaming quantities to fit your needs.

Other than for use in a particle accelerator,
it is nonsense.

: And let me remind you - This is the topic of this thread.

Did you just learn how to read?


Seems like you did. You haven't figured out that inertial mass is velocity
dependant yet.

[snipped rest of fischer's nonsense]

Pmb

Gauge:
Bilge:

(2) the
field in the center of momentum frame won't be a pure B-field since in
general, the center of momentum won't be at rest in the same frame in
which the field is purely magnetic and (3) I qualified the frame to be
valid at any instant and therefore at any instant, the center of momentum
is at rest. Apparently, the concept of an instantaneous rest frame goes
over your head.

So what's your point??

The point was that it apparently went over your head. If it hadn't,
you wouldn't have made a point of saying the answer was meaningless
and I wouldn't have needed to explain it.

[...]

My point was that relativistic mass always has a meaning.

You don't have a point, except perhaps whining because everyone
keeps telling you that relativistic mass is not really a fundamental
construct in relativity. I used to wonder why some of the physicists
on this newsgroup were so negative about the concept of relativistic
mass, since I've founnd it to be useful occasionally. Now, I know.
It's because people like you get carried away with it and try to
incorporate it as basic part of relativity and confuse others with
it because you are confused.


What you've
just given in relativistic mass in the zero momentum frame. But that
quantity can't meaningfully be measured in all frames of referance.

Don't be an idiot. The center-of-momentum frame is useful precisely
because it's usually very simple to use for calculations and the result
can be transformed into any other frame. What would you do, find the
frame in which the calculation is impossible as your choice of frames?


But I don't see that you're showing any understanding of what I've
explained to you. **** bilge - I;ve even given you the words straight
out of Ohanian's text and you still don't get it! Sheesh! You should
get your money back if you ever really got a degree in anything.

yada yada yada...


I'll explain it one more time for bilge - The problem is stated as
follows.

There are two electrons moving in an xy-plane in a uniform magnetic
field B
= B_o k (i.e. the B-field is in the +z direction.)

What is the total mass of the two electrons?

I already gave you that answer and in fact you referred to it when
you posted the first time. Are you stoned or just stupid?

[...]

[...], no one is arguing that four-momentum isn't conserved,
(except perhaps you - it's rather hard to tell, since you bugger
up everything).

In the annihilation of a positron and an electron 4-momentum IS
conserved.

That's what I said in the paragraph to which you are blabbering
about. Are you stoned or just stupid?

[...]
The four-momentum is conserved. Here,
use your concept for the "relativistic mass" of a photon, m = h\nu/c^2,
with four-momentum q and an electron with four-momentum, p (and final
four-momentum, p') conserves mass (relativistic or otherwise):

?
(p + q)^2 = p'2

So what's the problem?

Let
P_e = 4-momentum of electron
= (E_e/c, p_e) = ("energy of electron"/c,momentum of electron)

P_p = 4-momentum of photon
= (E_p/c, p_p) = ("energy of photon"/c,momentum of photon)

P = total 4-momentum = ("total energy"/c, total momentum) = (E/c,p)

The magnitude of P is M = invariant mass of the system

M^2 = E^2 - p^2 = constant

If we're speaking total relativistic mass = m, then since E =
constant and m = E/c^2 it follows that mass is constant

This is so easy a child could understand it.

That explains why you didn't understand it. That reaction is
impossible. It doesn't conserve momentum and energy. It's the
reason that radiation processes are second order, not first order.
It also points out the problem with your relativistic mass fetish.


Here. I'll dumb it down so that maybe even you might understand

Try dumbing it down so you can understand it.

E = Energy = Consant

m = E/c^2 = constnat

So what problem are you having understanding this now that I've
explained this to you for the 3rd time?

Perfectly. You've just shown how to get the wrong answer
by using your misguided notion of relativistic mass. Using the
standard way everyone else does things:

(p + q)^2 = p^2 + q^2 + 2p.q = p'^2

since p^2 = p'^2 = m^2 in any frame,

2p.q = 0 = mw - mc k.\beta = 0

w - ck\beta cos(A) = 0

w = ck \beta cos(A) and since w = ck,

\beta cos(A) = 1 = v cos(A) = c

which can never be true.

[*further garbage snipped*]

Pmb wrote:
: "Joe Fischer" wrote
: That is false, I am moving at more that 0.9 c
: relative to a very distant galaxy, and I still weigh
: the same 200 pounds.
:
: Obviously you don't understand the question. The question "What is the
: weight of a moving body" means that you choose a frame of referance to
: measure the weight.

Sometimes it is ok to use the terms "mass" and
"weight" to mean the same thing, but we are talking
about mass, not weight.

: Call that frame S. Then you measure the wieght of the
: body when it is at rest. Call that W.

Why not "mass"?

: The mass is defined as m = W/g where g = acceleration
: due to gravity.

But the mass being "weighed" is sitting on
one of the pans, it is not accelerating.

And I want to compare the mass with a known
mass standard on a centrifuge rotating to produce
a centrifugal force of one kilogram on the mass
being "weighed".

Now are you going to tell me the centrifugal
force is active gravitational mass, or that it does
something to make the gravitational force stronger,
or that the mass being weighed gains passive
gravitational mass because of the rotation?

Where and how does g enter into this?

: Then let the body move in frame S, e.g. like
: a car driving along the road - it still has weight but
: now it's moving in the gravitational field (or
: accelerating frame of referance - it doesn't
: matter since the results are identical).
: Then measure the weight while the object is moving -
: impractical to do in real life but it has a definite
: physical meaning.

I don't know where you are going with this,
if you mean the mass of the car increases because
it is moving, and at 60 MPH, even your roundabout
way would give the same mass within hundreds of
significant digits of precision.

: Le W' be the weight of the moving body. Then m' = W'/g is
: the mass of the moving body. And m' = m/sqrt[1-(v/c)^2]

You are doing a little better than Don Shead,
but not much, he never got past m = w/g.
There is no reason for an object to gain mass,
just because something is moving relative to it and
the observers are on the something.

: : That's never been questioned in relativity -
:
: Wheeler did say it is false.
:
: That's totally wong. Wheeler NEVER said that.

Of course he did, "Spacetime Physics"
says that mass is invariant, and that mass does NOT
vary with velocity, and is NOT dependent ob velocity.

: If you claim that he did then prove it.

Maybe you could ask Dr. Taylor if Wheeler
is not available.

: Strate exactly where he said that the weight of a moving body does
: not depend on velocity.

Do I really have to recite a page, surely you
can find it easily, the book is pretty much dependent
on the fact that mass is an invariant and that energy
is frame dependent.

: If it's in the "use and abuse" section of Taylor and Wheeler's text
: "Spacetime Physics -2nd Ed" then state which page. That section is now
: online at
:
: http://www.geocities.com/physics_world/stp/stp.htm

It is some other place early in the text,
because it is an important concept, and because it
avoids confusion if everybody learns to mean the
invariant rest mass when the usen the unqualified
word mass.

: : It's a fact. A fact that cannot be gotten rid of by
: : renaming quantities to fit your needs.
:
: Other than for use in a particle accelerator,
: it is nonsense.
:
: : And let me remind you - This is the topic of this thread.
:
: Did you just learn how to read?
:
: Seems like you did. You haven't figured out that inertial mass
: is velocity dependant yet.

Why do you write mass one time, then write rest
mass the next, and now jump to inertial mass.

I told you I don't know what inertial mass is
if it is not rest mass.
My discussion is about what most people mean
when they say mass, which might be a bowling ball,
a chair, or a billiard ball.

The concept of an observer not in an object's
rest frame determining the mass of the object has
litte use.
Observers should be smart enough to use
relativity to determine how much their observations
are skewed by not being in the rest frame of the
object, and use values that would be measured in
the rest frame of the object.

This is needed to get every possible
observer to agree on the mass of the object.

My rotating platform with a precision
balance will compare masses without any problem,
removing the w/g from the equation.
There is only the mass that equals the
combination of mass standards placed on the
other pan.

What difference does it make to observers
other places what the object weighs?

Joe Fischer


--
3

"Joe Fischer" wrote [same old stuff]

I said that when you showed that you understood relativity that I'd
continue. It seems you don't want to do that. Fine. However if you simply
forgot - What is time dilation?

Pmb

ps - you're confusing passive gravitational mass with rest mass

(Bilge) wrote in message ...
Gauge:
Bilge:

(2) the
field in the center of momentum frame won't be a pure B-field since in
general, the center of momentum won't be at rest in the same frame in
which the field is purely magnetic and (3) I qualified the frame to be
valid at any instant and therefore at any instant, the center of momentum
is at rest. Apparently, the concept of an instantaneous rest frame goes
over your head.

So what's your point??

The point was that it apparently went over your head. If it hadn't,
you wouldn't have made a point of saying the answer was meaningless
and I wouldn't have needed to explain it.

Oh! I see. You still don't understand your mistake here. Well I never
expected you to.

I've told you already that you can choose to define mass in the zero
momentum frame by taking the energy in that frame and dividing it by
zero - However that can only be measured in that frame in a meaningful
way. The result you get is is not an invariant mass and you are
therefore redefining mass to fit your needs.

In relativity one should be able to measure any quantity from any
frame.

[...]

My point was that relativistic mass always has a meaning.

You don't have a point, except perhaps whining because everyone
keeps telling you that relativistic mass is not really a fundamental
construct in relativity.

Nope - sorry flamer - but all I see is a bunch of internet cranks anc
crackpots trying to force their opinion on me.


What you've
just given in relativistic mass in the zero momentum frame. But that
quantity can't meaningfully be measured in all frames of referance.

Don't be an idiot. The center-of-momentum frame is useful precisely
because it's usually very simple to use for calculations and the result
can be transformed into any other frame. What would you do, find the
frame in which the calculation is impossible as your choice of frames?

That's a clear indication that you don't know what you're talking
about. You're trying to add these 4-vectors and you think that the
resulant 4-vector is physically meaningful. It is not. The particles
in the system have forces exerted on them from outside the system. As
such both their energy and momentum will vary along the particle's
world line. No matter what frame of referance you choose to measure
the particles energy and momentum in the only meaningful physical
quantities are the total energy and total momentum. That is the sum or
the energy of each particle is the total energy.


I'll explain this yet one more time for you (but you should really pay
attention this time).
----------------------------------------------------------
In frame S = zero momentum frame:

The energy of particle one, E_1, is measured **at the same time** the
energy of particle two, E_2, is measured. The momentum of particle
one, p_1, is measured **at the same time** the momentum of particle
two, p_2, is measured. This is the same time at which the energy
measurements were made. That means that you have choosen two events
which are spatially seperated which lie on the **non-straight**
worldline of the two particles. The observer in S forms the total
4-momentum as follows

E = E_1 + E_2 = total energy as measured in S at time t
p = p_1 + p_2 = total 3-momentum as measured in S at time t

P =(E/c, p)

Define M as follows

M = sqrt[ E^2 - (pc)^2]/c^2

----------------------------------------------------------
In frame S' = (frame moving relative to S:

Observe in S' follows same procedure -

The energy of particle one, E'_1, is measured **at the same time** the
energy of particle two, E'_2, is measured. The momentum of particle
one, p'_1, is measured **at the same time** the momentum of particle
two, p'_2, is measured. This is the same time at which the energy
measurements were made. The observer adds these 4-momenta together


E' = E'_1 + E'_2 = total energy as measured in S at time t
p' = p'_1 + p'_2 = total 3-momentum as measured in S at time t

P' =(E'/c, p')

Define N as follows

N = sqrt[ E^2 - (pc)^2]/c^2
----------------------------------------------------------

Fact #1: P is a 4-vector
Fact #2: P' is a 4-vector
Fact #3: In general: N does not equal M

Reason - When observers in each frame measure energy and momentum for
each particle they do so simultaneously. Since simultaneity is not
Lorentz invariant the resulting 4-vectors are not the same 4-vectors
since they pertain to different event. ****** When the particles are
free particles this problem does not arise!!!!*****.


And as I keep telling you a million times - that was what Ohanian
meant in his footnote in his text -
www.geocities.com/physics_world/ohanian.htm

----------------------------------------------------------
Since P^mu is the sum of two four-vectors (see Eq. [188]), at first
sight it seems obvious that it should be a four-vector. However there
is a catch: in Eq. [188] it is implicitly assumed that two
four-velocities are evaluated 'at the same time'. Since Lorentz
transformations do not preserve simultaniety, the meaning of 'at the
same time' is different in two frames, and this difference could
complicate the transformation law of the sum [188]. However, for the
special case of free particles, Eq. [189] is obviously correct since
in this case the four velocities are constant, and it is irrelavant
whether the velocities are evaluated simultaneously or not. It then
follows that E. [189] must be correct for particles that were free
(did not impact) at some time in their past: the momenta Pm and Pm are
constant by hypothesis, and if they had the transformation law [189]
at one time they must keep it forever."
----------------------------------------------------------

To get the same value for the invariant mass different observers would
need to measure energies and times at different times in their frame
of referance - While mathematically that is reasonbable - It's not
**physically meaningfull**.

Now what part of the above didn't you understand this time?



In the annihilation of a positron and an electron 4-momentum IS
conserved.

That's what I said in the paragraph to which you are blabbering
about. Are you stoned or just stupid?

I read it mr. flamer - and I explained to you that since it's
conserved it means that the invariant mass of the system is also
conserved - what part of that were you unable to follow??

That explains why you didn't understand it. That reaction is
impossible.

I assumed that when you wrote

---------------------------------------------------------------------
...,use your concept for the "relativistic mass" of a photon, m =
h\nu/c^2,
with four-momentum q and an electron with four-momentum, p (and final
four-momentum, p') conserves mass (relativistic or otherwise):
?
(p + q)^2 = p'2

---------------------------------------------------------------------

That you meant that the *total* 4-momentum (photon + electron) was p'
and is a conserved quantity when the photon scatters off the electron.
Since you claim that is not what you meant then disrgard the rest.


So - You wrote an expression that can't happen - big deal. Then you
tell me to use the concept of relativistic mass on an interaction that
can't occur. You're such a fruitcake bilge - So what's your point?




Back to the topic sparky - Mass is conserved. E.g. electron positron
annihilation - You acknowledge the fact that 4-momentum is conserved.
Since that's true it follows that

(1) invariant mass = magnitude of P (P = total 4-momentum) is
conserved
(2) relativistic mass = P^0/c = conserved

I see you snipped that part because I've just proved you wrong.

I also see that you snipped the part about the nuclear fission

http://www.geocities.com/physics_world/mass_energy.htm

I guess you snip all parts that prove you wrong. Expected at this
point.


You claim that mass is not conserved such reactions like a fission of
a nucleus - I've proven that it is.


Are you going to directly state what you claim is wrong regarding the
fission in www.geocities.com/physics_world/mass_energy.htm?

Or are you going to avoid answering - again.

Seems that's your trade mark - when you finally figure out that your
wrong you avoid the subject all together.


Mr. Pmb

"Gauge" wrote

Are you going to directly state what you claim is wrong regarding the
fission in www.geocities.com/physics_world/mass_energy.htm?

Or are you going to avoid answering - again.

The page has been updated. The new page is at
http://www.geocities.com/physics_wor...ass_energy.htm

Pmb

"Gauge" wrote

Another note on the conservation of mass. This assumes that the potention
energy of the particles is zero. That is to say that the total energy of all
particles is the sum of their kinetic energies and their rest energies. In
that case mass is conserved.

Proof is given in

"The Classical and Relativistic Concepts of Mass," Erik Eriksen and Kjell
Voyenli, Foundations of Physics, Vol. 6, No. 1, 1976. Pages 115
to 124

as well as in

"Concepts of Mass in Contemporary Physics and Philosophy," Max Jammer,
Princeton University Press, (2000), page 58 (but this is just Jammer
explaining Eriksen's proof) Jammer and Eriksen also explain why it's wrong
to identify rthe rest mass of a particle with its classical mass.

Pmb

Subject: Rest mass or inertial mass?
From: "Pmb"
Date: 8/11/03 3:28 AM US Mountain Standard Time
Message-id:


"Gauge" wrote

Another note on the conservation of mass. This assumes that the potention
energy of the particles is zero. That is to say that the total energy of all
particles is the sum of their kinetic energies and their rest energies. In
that case mass is conserved.

Proof is given in

"The Classical and Relativistic Concepts of Mass," Erik Eriksen and Kjell
Voyenli, Foundations of Physics, Vol. 6, No. 1, 1976. Pages 115
to 124

as well as in

"Concepts of Mass in Contemporary Physics and Philosophy," Max Jammer,
Princeton University Press, (2000), page 58 (but this is just Jammer
explaining Eriksen's proof) Jammer and Eriksen also explain why it's wrong
to identify rthe rest mass of a particle with its classical mass.

Pmb




You are obviously confused about what was written or how it relates to what we
are talking about. Center of momentum frame energy is always conserved whether
there is a potential or not. In the case that there is one, the Lorentz quage
potential contributes to the mass of a particle-
m^2*c^2 = g_mu_nu(P^mu - qA^mu)(P^nu - qA^nu)
( see problem 3.1.8 at http://www.geocities.com/zcphysicsms/chap3.htm )
, but the mass is still the rest frame energy and is still conserved. If you
argue that it is not then you are arguing against conservation of energy!