(WaiteDavid137) wrote in message ...
Subject: Rest mass or inertial mass?
From: (Radi Khrapko)
Date: 8/13/2003 3:24 AM US Mountain Standard Time
Message-id:

Joe Fischer wrote in message
...
Radi Khrapko wrote:
:Joe Fischer wrote:
: Why, did they quit making balance beams and
: standard weights to compare an unknown mass to?
:
: The point is, does mass change with velocity?

Why should it, does it pick up bugs on the
windshield and those become part of the mass?

: Your operation must be applied to a moving body.

Please forget such nonsense and use the
same instruments used in any physics lab course.

Joe Fischer

Radi to Joe 071632 (#199)
Dear Joe!
You agree that the definition {mass is a Poincare invariant} is nonsense.
You cannot define mass.
No one of my opponents can define mass!
It is shame!
Note, you must point out an operation to determine mass of a moving body


You are in denial.
Radi to Wait 131240 (194)
What is denial?
I have read in a vocabulary:
{My husband is still in denial about my rape}.
And what is mass?

On 8/14/2003 12:28 PM, Gauge wrote:
(Bilge) wrote in message ...
Gauge:
(Bilge) wrote in message
ue-al.net...
WaiteDavid137:
proper mass is not a Poincare invariant. It's a Lorentz invariant.
[...]

I don't know where this sudden spate of attempting to distinguish
"Poincare' invariants" from "Lorentz invariants" started, but it is
nonsense.

When one applies a coordinate transform between two frames, one must
determine mathematically the effect of the transform on the components
of tensors. One _CANNOT_ simply apply the same transform to tensor
components that one applied to the coordinates. One can do that from
first principles, or one can do it via the usual transformation rules of
tensors (which of course came from first principles):

Given a Poincare' transform from an unprimed to a primed
frame (the {L} and {b} are constants):
x^i' = L^i'_j x^j + b^i'
the attempt to write
P^i' = L^i'_j P^j + b^i'
has an obvious problem: the units of P are different from
the units of b.

The correct transform for the components of any 4-vector V
(including the momentum P) is:
V^i' = L^i'_j V^j
because L^i'_j = dx^i'/dx^j; this immediately generalizes
to the contravariant components of other-rank tensors.
Analysis shows that the units are ALWAYS correct, including
cases where different coordinates have different units.


Tom Roberts

"Tom Roberts" wrote

[pmb wrote]
Sorry tom but I don't follow. Please post an example of a poincare
transformation from one inertial frame to another such that "mass" is
invariant.

As has been discussed endlessly around here, mass is an invariant,
meaning it is not affected by ANY coordinate transformation whatsoever.

You missed the point. I *know* that 'rest mass' is invariant. But that has a
very specific meaning. It means that (1) its a scalar - tensor of rank zero
(2) it's the magnitude of the particle's 4-momentum which is, of course, a
4-vector

In special relativity a quantity which remains unchanged by a Lorentz
transformation is called a "Lorentz scalar" or "Lorentz invariant."

If a quantity is to be called a "Poincare invariant" then that suggests that
the quantity remain unchanged by *all* possible Poincare transformations.

Recall the reason that I asked - bilge made the claim that rest mass is a
*Poincaire* invariant.

As it stands I don't see how that can be. It be clear if an example was
given where a general Poincare transformation was used and the result left
the magnitude of the 4-momentum invariant.

The point was always that rest mass is an inherent property of any particle
and using the comment "Poincare invariant" is not required. However since it
was brought up and I can't find the definition anywhere and there are no
examples of a Poincare invariant in any text that I have. wchogg indicated
that we were on the same wave length. But I didn't uinderstand what he
wrote.


That is, at base, a GR perspective, in which neither Lorentz nor
Poincare' transforms have any special stature (their special stature in
SR is, from a GR perspective, due to the fact that they are global
isometries of the Minkowski manifold). You are presumably interested in
SR:

Yes.

The translation of a Poincare' transform will not affect the components
of any 4-vector.

That is untrue. Simply try it. As you know - a poincare transformation, of
the 4-vector A, can be expressed as A' = LA + d where L = Lorentz
transformation and d = spacetime translation.

Now consider the 4-momentum P = (E/c p_x, p_y, p_z) and d = (a, b, c, d).

The term "mass is invariant" to me means this == P*P = (m_o*c^2)^2

i.e. its the magnitude of the particle's 4-momemtum. Now consider the
Poincare transformation

P' = LP + d = (E'/c p'_x, p'_y, p'_z) + (a, b, c, d)

P'*P' does not equal P*P


Remember that the basis of vectors wrt a given set of
coordinates is

{d/dx^i} (partial differentiation)

The definition of Lorentz 4-vector is "anything that transforms in the same
way as the coordinates under a Lorentz transformation"

A Lorentz invariant is single number that remains unchanged by a Lorentz
transformation.

For a quantity to be "Scooby Doo invariant" then it must remain unchanged
under a "Scooby Doo transformation"

Therefore for a quantity to be a "Poincare invariant" then it must remain
unchanged under a "Poincare transformation"

What you've posted is a definition of a "general vector." Recall that we've
had this discussion before.

However the definition that I've given above may not be what is meant by
"Poincare invariant" in the literature.

Do you have a sold referance in which the term "Poincare invariant/tensor"
is defined and/or used?


IOW: Remember that position is NOT a 4-vector, ..

We've already had this discussion tom. Position is not a *general* 4-vector.
It *is* a Lorentz 4-vector. And hence my question.

If one is to assume a general coordinate transformation then they transform
as you've indicated. That is not generally the case since rules for defining
tensors can vary slighty and thus change the meaning

bilge is a arrogant prick so he didn't pick up on this and therefore decided
that he need to flame me.

However I will assume that you understand what I'm saying and will answer
the question I've asked in respons - i.e. do you have a solid referance?

What does IOW mean? (forgive me but I forget these things)

Pmb

"Radi Khrapko" wrote in message
om...
(Bilge) wrote in message
...
Radi Khrapko:

There are many Poincare invariants.

The only poincare invariants in minkowski space are mass and spin.

Radi to Bilge 130719 (200)
Angular momentum?

To say {mass is a poincare invariant} means to say nothing.

To you, perhaps not.
{mass is a Poincare invariant} does not define mass.
{Bilge is human being} does not define Bilge.

Forgive me if I'm telling you what you alreadyu know Radi but bilge is
following the leader as usual. And in doing so he's made a huge error here
by trying to use that as a definition. The comment "mass is an invariant" is
supposed to mean that its the magnitude of the 4-momentum, i.e. its the M in
(let c=1)

E^2 - p^2 = M^2

However that relation is derived from the relation E = gamma*. But even so -
what's p? And what's E? E is defined as

E = Kinetic Energy + Rest Energy

Momentum is defined as

P = mv = gamma*M*v

If we're supposed to measure E through E = gamma*M then his logic is
circular. If we're supposed to measure E through E = K + E_o then we don't
know how to measure it unless we know what M is. Suppose that I tried to
measure E by ramming the particle into a calorimeter. I measure the heat
rise and I get a value. But that's only the kinetic energy. We don't know
how E is broken up so we still don't know what E is. If we have the mass and
the kinetic energy THEN we know E.

So these arguements on defining mass as the magnitude of the 4-momentum are
given by people like bilge and waite who have not thought things through.

Pmb

On Thu, 14 Aug 2003 20:00:58 +0000, Pmb wrote:

What does IOW mean? (forgive me but I forget these things)

"In other words".

Jeff

"Tom Roberts" wrote in

I dispute your terminology, and your attempt to bifurcate the set of
4-vectors.

If that's your position then please define the term "angular momentum"
within the context of special relativity. Use MTW if you like - Section 5.11

Pmb

Pmb wrote:
The topic is "poincare invariant" - the qualifier seems to imply something
different that it should be qualified with Poincare.

It does not matter what the topic is, when one applies a coordinate
transform, one must COMPUTE how the components of vectors are affected
by the change in coordinate basis. For Poincare' transforms the
translation does not contribute. shrug


There is a very important point regarding this with Lorentz tensors and thus
Lorentz invariants. That the *definiton* of "Lorentz tensor" is "that which
transforms as the position vector does."

First, you confuse the tensor with its components; that may be part of
your problem.

Second your definition of "Lorentz transor" does not generalize to the
Poincare' transforms, because 4-element objects whose components
transform "as the position vector does" are not vectors. You "lucked
out" in that this simple definition happens to work for Lorentz transforms.

Third, your usage of "position vector" is an oxymoron. Position is NOT a
vector, no matter how much you wish it to be one. Only DIFFERENCES
between two positions are vectors (and then only in a flat manifold; in
general they must be infinitesimally-close).


The proper way to do this is to not attempt to distinguish "Lorentz
tensor" from "tensor" at all, but to recognize that for the restricted
case of a flat manifold and transformations belonging to its isometry
group, then all the complicated terms[#] in the equations go to 0. One
then obtains the usual equations of SR (rather than the more complicated
ones of GR). Yes, many/most physicist authors do not do it properly, and
some of them take the shortcut you decribe. shrug

[#] terms involving the connection or some curvature tensor.


Not neccesarily - again - please take note that this is **entirely** a
question on terminology and terminology only!

Yes. You repeatedly attempt to create your own teminology that is
superficially similar to standard terminology, and then insist that it
makes sense. shrug


As I said before, check your units -- your attempted addition does not
make sense.

What's the problem? P is a 4-vector. Each component has units of momentum.
The quantity (a, b, c, d) is the same thing except that I forgot the c and I
used that as a letter - damn. Here

P' = LP + d = (E'/c p'_x, p'_y, p'_z) + (A/c, B, C, D)

In MKS units, E'/c,p'_x,p'_y,p'_z all have units kilogram*meters/sec,
but A/c,B,C,D all have units meters. It makes no sense to try to add
them together. This is just basic Physics 101....


Tom Roberts

A few comments on this point and I'm done - unless you can get of this
kick of bothering me with your dislike for definitions and your
constant blaming me for them. Can you do that?

Tom Roberts wrote

I have no idea what your point is ...

Of course you do. You show that you do below

-- in MTW 5.11 angular momentum is not
a 4-vector,

Who said that it was?

... and this is not directly related to your attempted ..

You're wrong. It's not an "attempt" on my part - its a standard
definition of "Lorentz 4-vector". If you don't like it then write to
J.D. Jackon, Hans C. Ohanian, Lovelock etc. and ask them all to change
their definitions.


bifurcation. Note also that in their definition, x^u never appears
alone,

That's incorrect. It does appear alone. See Goldstein's text, second
and third ed (I haven't checked version one) where he defines
"relativistic angular momentum" and "center of mass"

Goldstein - second ed.
angular momentum - Eq. (7-128) page 318
center of mass - Eq. (7-320) page 320

Goldstein - third ed.
angular momentum - E.q (7.120) page 310
center of mass - Eq. (7-134) on page 312

it only appears in the DIFFERENCE between two positions.

And the coordinate 4-vector is a 4-vector because the origin is chosen
as a referance event from which all events are described. So it's the
difference between the events X and 0

In MTW that's overly obvious. The angular momentum tensor is a third
rank tensor. It's a third rank tensor becuase x^u is a 4-vector in
special relativity. And since the event a^u also is a 4-vector then
the difference is a 4-vector. a^u is the referance event. Let P be an
event in flat spacetime. Sometimes such an event is identified by its
vectorial separation x_p from some arbitrarily chosen event in
spacetime known as the "origin" .

Now why do you think they used a difference of two positions
rather than just position?

Same reason it's done in classical mechanics. Angular momentum, like
angular velocity is dependant on the origin. There is no unique origin
for angular momentum. However if you want to then you can specify the
arbitrary point as the origin.

Note that it must hold for *all* a^u. One can always choose the origin
as the referance event as Goldstein does. However the do generalize
later in the text to arbitrary events.

Simply show explicitly that what MTW call angular momentum is a third
rank tensor and you'll see this.

Do you know why the angular momentum tensor is a third rank tensor?

By the way. MTW is unique in this respect. Other texts I have simply
choose the origin as the arbitrary point.

Why don't you search for a passage
where they call x^u a "4-vector" or a "Lorentz vector" or
somesuch? -- I suspect you won't find any....

I've done this several times in the past and you've ignored it. Why
would I expect you to pay attention this time?

But I'll give you the benifit of the doubt and assume that you missed
the constant posting of referances that I've given on this point when
I discussed this with you before.

From "Classical Electrodynamics - 2nd Ed.," J.D. Jackson


----------------------------------------------------
(page 535)

Comparison of the invariant length element (ds)^2 in (11.68) with the
similarly invariant scalar product(11.66) suggests that the covariant
coordinate 4-vector x_a can be obtained from the contravariant x^b by
contraction with g_ab, ...
----------------------------------------------------

On page 604 the angular momentum density is defined as

(12.109) M^abg = T^ab x^g - T^af x^b

See also Eq. (12.117)



From "Gravitation and Spacetime - 2nd Ed," Ohanian and Ruffini

----------------------------------------------------
(page 72)

"The quantities x^u and x_u are called, respectively, the
'contravariant' and 'covariant' components of the position vector."
----------------------------------------------------
(Page 76)

"A 'vector' (or tensor of rank one) is a four-component object that
transforms under Lorentz transformations in exactly the same way as
x^u; .. Accoring to this definition, x^u (and aslo dx^u) serves sa a
prototype for all vectors."
----------------------------------------------------
(page 307)

(gives definition of *general* tensor == Eq[8])
[Eq [1] x'^u = x'^u(x)]

"Note that x^u 'is not a vector' with respect to general coordinate
transformations; the transformation law for x^u is given by Eq[1], and
this does not have the form of Eq. [8]. In the exceptional case of a
'linear' (and homogeneous) transformation, Eq. [1] becomes x' = b^u_v
x^v [10] where b^u_v is a constant matrix. In this case x^u does have
the transformation law [8]since Eq 10] can be written as (general
tensor). This x^u is a vector with respect to such linear
transformations. This is why, in Chapter 2, we were able to treat the
rectangular coordinate x^u as a vector with respect to Lorentz
transformations"
----------------------------------------------------

In "Tensors, Differential Forms, and Variational Principles," David
Lovelock and Hanno Rund, page 53 - the authors define the term
"Lorentz tensor" as I've explained it to you.

Tom Roberts wrote [stuff]

I forgot an obvious one. Kip Thorne's and Roger Blanford's new text.

http://www.pma.caltech.edu/Courses/p...002/index.html

See "Chapter 1: Physics in Flat Spacetime: Geometric Viewpoint"

http://www.pma.caltech.edu/Courses/p...p01/0201.2.pdf

"1.9 Directional Derivatives, Gradients, Levi-Civita Tensor, Cross
Product and Curl" page 35
---------------------------------------------------
In this definition we have denoted points, e.g. P, by the vector x_p
that reaches from some arbitrary origin to the point.
---------------------------------------------------

It's still being written so you have the time to e-mail Kip and
explain to him why he's wrong. Please let me know his response. :-)

Pmb

"Gauge" wrote in message
om...
A few comments on this point and I'm done - unless you can get of this
kick of bothering me with your dislike for definitions and your
constant blaming me for them. Can you do that?

Tom Roberts wrote

I have no idea what your point is ...

Of course you do. You show that you do below

-- in MTW 5.11 angular momentum is not
a 4-vector,

Who said that it was?

... and this is not directly related to your attempted ..

You're wrong. It's not an "attempt" on my part - its a standard
definition of "Lorentz 4-vector". If you don't like it then write to
J.D. Jackon, Hans C. Ohanian, Lovelock etc. and ask them all to change
their definitions.


bifurcation. Note also that in their definition, x^u never appears
alone,

That's incorrect. It does appear alone. See Goldstein's text, second
and third ed (I haven't checked version one) where he defines
"relativistic angular momentum" and "center of mass"

Goldstein - second ed.
angular momentum - Eq. (7-128) page 318
center of mass - Eq. (7-320) page 320

Goldstein - third ed.
angular momentum - E.q (7.120) page 310
center of mass - Eq. (7-134) on page 312

it only appears in the DIFFERENCE between two positions.

And the coordinate 4-vector is a 4-vector because the origin is chosen
as a referance event from which all events are described. So it's the
difference between the events X and 0

In MTW that's overly obvious. The angular momentum tensor is a third
rank tensor. It's a third rank tensor becuase x^u is a 4-vector in
special relativity. And since the event a^u also is a 4-vector then
the difference is a 4-vector. a^u is the referance event. Let P be an
event in flat spacetime. Sometimes such an event is identified by its
vectorial separation x_p from some arbitrarily chosen event in
spacetime known as the "origin" .

Now why do you think they used a difference of two positions
rather than just position?

Same reason it's done in classical mechanics. Angular momentum, like
angular velocity is dependant on the origin. There is no unique origin
for angular momentum. However if you want to then you can specify the
arbitrary point as the origin.

Note that it must hold for *all* a^u. One can always choose the origin
as the referance event as Goldstein does. However the do generalize
later in the text to arbitrary events.

Simply show explicitly that what MTW call angular momentum is a third
rank tensor and you'll see this.

Do you know why the angular momentum tensor is a third rank tensor?

By the way. MTW is unique in this respect. Other texts I have simply
choose the origin as the arbitrary point.

Why don't you search for a passage
where they call x^u a "4-vector" or a "Lorentz vector" or
somesuch? -- I suspect you won't find any....

I've done this several times in the past and you've ignored it. Why
would I expect you to pay attention this time?

But I'll give you the benifit of the doubt and assume that you missed
the constant posting of referances that I've given on this point when
I discussed this with you before.

From "Classical Electrodynamics - 2nd Ed.," J.D. Jackson


----------------------------------------------------
(page 535)

Comparison of the invariant length element (ds)^2 in (11.68) with the
similarly invariant scalar product(11.66) suggests that the covariant
coordinate 4-vector x_a can be obtained from the contravariant x^b by
contraction with g_ab, ...
----------------------------------------------------

On page 604 the angular momentum density is defined as

(12.109) M^abg = T^ab x^g - T^af x^b

See also Eq. (12.117)



From "Gravitation and Spacetime - 2nd Ed," Ohanian and Ruffini

----------------------------------------------------
(page 72)

"The quantities x^u and x_u are called, respectively, the
'contravariant' and 'covariant' components of the position vector."
----------------------------------------------------
(Page 76)

"A 'vector' (or tensor of rank one) is a four-component object that
transforms under Lorentz transformations in exactly the same way as
x^u; .. Accoring to this definition, x^u (and aslo dx^u) serves sa a
prototype for all vectors."
----------------------------------------------------
(page 307)

(gives definition of *general* tensor == Eq[8])
[Eq [1] x'^u = x'^u(x)]

"Note that x^u 'is not a vector' with respect to general coordinate
transformations; the transformation law for x^u is given by Eq[1], and
this does not have the form of Eq. [8]. In the exceptional case of a
'linear' (and homogeneous) transformation, Eq. [1] becomes x' = b^u_v
x^v [10] where b^u_v is a constant matrix. In this case x^u does have
the transformation law [8]since Eq 10] can be written as (general
tensor). This x^u is a vector with respect to such linear
transformations. This is why, in Chapter 2, we were able to treat the
rectangular coordinate x^u as a vector with respect to Lorentz
transformations"
----------------------------------------------------

In "Tensors, Differential Forms, and Variational Principles," David
Lovelock and Hanno Rund, page 53 - the authors define the term
"Lorentz tensor" as I've explained it to you.

I forgot one. From "Electrodynamics and Classical Theory of Fields and
Particles, "Barut, Dover Pub., (1980)

Barut also defines 4-vectors as that which transforms as the coordinates do
under a Lorentz transformation and he also defines angular momentum (he
calls it J^uv) in the same was as Goldstein, i.e.

J^uv = x^u p^v - x^v p^u

Same thing for the definition of relativistic torque N^uv

N^uv = x^u K^v - x^v K^u

where K = 4-force (aka Minkowski force)

But I'm getting a new EM text in a few weeks - let's see what Shadowitz does

Pmb