Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/26/2003 1:04 PM US Mountain Standard Time
Message-id:

(Gauge) wrote in message
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(WaiteDavid137) wrote in message
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Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/25/03 3:59 AM US Mountain Standard Time
Message-id:

Tom Roberts wrote in message
...
On 9/20/2003 8:18 AM, Pmb wrote:
"Randy Poe" wrote in message
om...
Tom Roberts wrote in message

But if one means "element fo the Lorentz group", then there is
indeed a transformation between spherical coordinates in two different
inertial frames;

Since when? Lorentz transformations include boosts, rotations and
relfections.

The Lorentz transformation between two spherical coordinates in this
discussion
IS just a boost.

You're not paying attention again waite. A transformation from a frame
S with Cartesian coordinates to the same frame S with Spherical
coordinates is NOT a Lorentz transformation

Seems I didn't complete that. A transformation which maps
(r,theta,phi) to (r',theta',phi') is not a Lorentz transformation.


You completed your thought and it was wrong. Now you completed a seperate
thought and it is wrong to. The Lorentz transformation between two spherical
coordinate systems is

ct' = gamma[ct - beta*r*cos(theta)sin(phi)]
r'cos(theta')sin(phi') = gamma[rcos(theta)sin(phi) - beta*ct]
r'sin(theta')sin(phi') = rsin(theta)sin(phi)
r'cos(theta') = rcos(theta)

This satisfies the Lorentz group's property of the interval's invariance as it
yields:

dct'^2 - dr'^2 - r'^2[dtheta'^2 + sin^2(theta')dphi'^2] = dct^2 - dr^2 -
r'^2[dtheta^2 + sin^2(theta)dphi^2]

(WaiteDavid137) wrote in message ...
Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/26/2003 1:04 PM US Mountain Standard Time
Message-id:

(Gauge) wrote in message
.com...
(WaiteDavid137) wrote in message
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Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/25/03 3:59 AM US Mountain Standard Time
Message-id:

Tom Roberts wrote in message
...
On 9/20/2003 8:18 AM, Pmb wrote:
"Randy Poe" wrote in message
om...
Tom Roberts wrote in message

But if one means "element fo the Lorentz group", then there is
indeed a transformation between spherical coordinates in two different
inertial frames;

Since when? Lorentz transformations include boosts, rotations and
relfections.

The Lorentz transformation between two spherical coordinates in this
discussion
IS just a boost.

You're not paying attention again waite. A transformation from a frame
S with Cartesian coordinates to the same frame S with Spherical
coordinates is NOT a Lorentz transformation

Seems I didn't complete that. A transformation which maps
(r,theta,phi) to (r',theta',phi') is not a Lorentz transformation.


You completed your thought and it was wrong.

Bzzzt! Wrong again waite.

Now you completed a seperate
thought and it is wrong to. The Lorentz transformation between two spherical
coordinate systems is

ct' = gamma[ct - beta*r*cos(theta)sin(phi)]
r'cos(theta')sin(phi') = gamma[rcos(theta)sin(phi) - beta*ct]
r'sin(theta')sin(phi') = rsin(theta)sin(phi)
r'cos(theta') = rcos(theta)

Wrong. That is a transformation between S and S' expressed in
Sphercical coordinates. It is NOT a Lorentz transformation.

Whassa matta waite? No speaka de Englaise? You keep making this same
old lame mistake. You START WITH/ASSUME an invalid definition then you
go about stating examples of that invalid definition. Your problem is
that you simply don't know what the "Lorentz Transformation" means.
I've given you resourse after resource from that literature which
DEFINES terms. Thorne's new text is a great resource, especially on
this point. Too bad you refuse to read it or state why you claim
Thorne is wrong

You're becomming more boring than usual waite.

Tell me waite - you've had several days now to contact Thorne. What
was his response to your bogus claims as to why his new text is wrong?
Or are you too ashamed to post his reply?


Mr. Pmb - waite's relativity teacher

(WaiteDavid137) wrote [bogus stuff]

This satisfies the Lorentz group's property of the interval's invariance as it
yields:

dct'^2 - dr'^2 - r'^2[dtheta'^2 + sin^2(theta')dphi'^2] = dct^2 - dr^2 -
r'^2[dtheta^2 + sin^2(theta)dphi^2]

Tsk tsk tsk. A property of the Lorentz transformation is to leave ds^2
invarian. A property of all allowable coordinate transformations is
that it leaves ds^2 invariant as well. It is meaningless to claim that
the Lorentz group has anything to do with speherical coordinates since
the Lorentz group is composed only of those transformations which map
Minkowski coordinates to Minkowski coordinates. However the Lorentz
transformations are not the only transformations in special relativity
- guess you didn't know that either huh waite?

However if you can prove that this definition as I've taught it to you
is wrong then provide proof. And NO "Cause I say so." is not proof.
That's the waite-parrot routine and not a valid arguement.

Mr. Pmb - waite's relativity teacher

(WaiteDavid137) wrote [bogus stuff]

This satisfies the Lorentz group's property of the interval's invariance as it
yields:

dct'^2 - dr'^2 - r'^2[dtheta'^2 + sin^2(theta')dphi'^2] = dct^2 - dr^2 -
r'^2[dtheta^2 + sin^2(theta)dphi^2]

Tsk tsk tsk. A property of the Lorentz transformation is to leave ds^2
invariant. A property of all allowable coordinate transformations is
that it leaves ds^2 invariant as well. It is meaningless to claim that
the Lorentz group has anything to do with speherical coordinates since
the Lorentz group is composed only of those transformations which map
Minkowski coordinates to Minkowski coordinates. However the Lorentz
transformations are not the only transformations in special relativity
- guess you didn't know that either huh waite?

However if you can prove that this definition as I've taught it to you
is wrong then provide proof. And NO "Cause I say so." is not proof.
That's the waite-parrot routine and not a valid arguement.

Mr. Pmb - waite's relativity teacher

Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/27/03 7:33 AM US Mountain Standard Time
Message-id:

(WaiteDavid137) wrote in message
...
Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/26/2003 1:04 PM US Mountain Standard Time
Message-id:

(Gauge) wrote in message
.com...
(WaiteDavid137) wrote in message
...
Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/25/03 3:59 AM US Mountain Standard Time
Message-id:

Tom Roberts wrote in message
...
On 9/20/2003 8:18 AM, Pmb wrote:
"Randy Poe" wrote in message
om...
Tom Roberts wrote in message

But if one means "element fo the Lorentz group", then there is
indeed a transformation between spherical coordinates in two
different
inertial frames;

Since when? Lorentz transformations include boosts, rotations and
relfections.

The Lorentz transformation between two spherical coordinates in this
discussion
IS just a boost.

You're not paying attention again waite. A transformation from a frame
S with Cartesian coordinates to the same frame S with Spherical
coordinates is NOT a Lorentz transformation

Seems I didn't complete that. A transformation which maps
(r,theta,phi) to (r',theta',phi') is not a Lorentz transformation.


You completed your thought and it was wrong.


Now you completed a seperate
thought and it is wrong to. The Lorentz transformation between two
spherical
coordinate systems is

ct' = gamma[ct - beta*r*cos(theta)sin(phi)]
r'cos(theta')sin(phi') = gamma[rcos(theta)sin(phi) - beta*ct]
r'sin(theta')sin(phi') = rsin(theta)sin(phi)
r'cos(theta') = rcos(theta)

Wrong. That is a transformation between S and S' expressed in
Sphercical coordinates. It is NOT a Lorentz transformation.

I just proved you wrong.

Tell me waite - you've had several days now to contact Thorne. What
was his response to your bogus claims as to why his new text is wrong?

You are a liar as you just now proved to everyone again.

Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/27/03 7:39 AM US Mountain Standard Time
Message-id:

(WaiteDavid137) wrote [bogus stuff]

This satisfies the Lorentz group's property of the interval's invariance as
it
yields:

dct'^2 - dr'^2 - r'^2[dtheta'^2 + sin^2(theta')dphi'^2] = dct^2 - dr^2 -
r'^2[dtheta^2 + sin^2(theta)dphi^2]

And NO "Cause I say so." is not proof.

I already mathematically proved you wrong. You just refused to read it all.

(WaiteDavid137) wrote in message ...
Subject: The meaning of the term "Four-Tensor"
From: (Gauge)
Date: 9/21/03 5:34 AM US Mountain Standard Time
Message-id:

(WaiteDavid137) wrote in message
...
Subject: The meaning of the term "Four-Tensor"
From: "Pmb"
Date: 9/20/2003 10:14 AM US Mountain Standard Time
Message-id:

then we call the vector
the "position 4-vector."

Position is not a 4-vector. X does not transform as dX

When someone simply says "tensor" it should be the more general definition
according to which the components transform tensorially in the same way as
the differential dX.

Finally got you to change your position I see. About time.

I've said that from day one.

No, You changed your position bewteen the first and second paragraphs quoted
from you in this very post.

Nonsense. I never changed my mind about anything in any post in any
thread with one possible exception - over the years I've stated that I
had no preferance of mass as rest mass or mass as relativistic mass. I
clearly stated several times that I'd reserve my opinion until I've
studied this in every single aspect that I could think of and in which
many others have explained in the peer reviewed literature. At this
time I do not believe that "mass = rest mass" is a valid way to think
of mass.

THAT is the only thing I've changed my mind on.

Mr. Pmb - waite's relativity teacher

"Tom Roberts" wrote in message
...
Gauge wrote:
Tom Roberts wrote in message
...
Hmmm. This of course depends upon what one means by "Lorentz transform".
If one means the specific equations used in SR then I agree that no such
transform relates spherical coordinates in two different inertial
frames. But if one means "element of the Lorentz group", then there is
indeed a transformation between spherical coordinates in two different
inertial frames;

Since when? Lorentz transformations include boosts, rotations and
relfections. Since when do they include Cartesian to Spherical
transformations?

Please READ WHAT I WROTE.

When you stop being a prick and stop yelling I might just do that - but I
don't see the point in this case since you don't seem to know what a Lorentz
group is.


You are confusing "element of the Lorentz group" with "Lorentz
transformation".

Oh come on now tom. I thought you knew better than this. People expect more
from you than this. *By definition* the "Lorentz Group" is the set of all
Lorentz transformations.

I can't even recall how many times that you've said that a "tensor is
a function on a manifold."

Goodness, I hope I never said that -- it's quite clearly wrong.

I stand (sit actually) corrected. You said that a tensor is a function.

That is indeed a definition of "tensor" (yes, I simplified the
definition and omitted covectors -- IIRC I said so immediately following
the passage you quoted).

A "function on a manifold" is a _FIELD_, not a tensor. But of course
there can be tensor fields (and in GR most of the tensors of interest
are tensor fields).


Are you now changing your mind?

Certainly not. But you need to learn how to read more carefully.

Don't be so condescending roberts. It doesn't become you. Or do you find it
neccesary to act like this?

You REALLY need to learn the difference between "misunderstanding"/"slip"
and needing to "read more carefully" roberts.




Actually a tensor is a function of vectors and 1-forms.

That's what the passage of mine you quoted said. You REALLY need to
learn how to read more carefully.

I saw no mention of one-forms. YOU? need to (1) pay more attention and (2)
stop acting like davy waite.

I tend to stop reading your posts when I see start being a prick.


And yet, in my earlier discussion the direction of my arm is a vector
(aka a rank-1 tensor), but it's sure not obvious how it is a
function.... If tensors are to be defined as functions of rank-1
tensors, that recursive definition must "bottom out" somewhere, and
vectors is where that happens.

Why? Vectors and 1-forms are not defined in terms of tensors. Not from what
I've seen.

Just READ WHAT I WROTE

and this is where you were well into being a prick.


Pmb

(WaiteDavid137) wrote [****]

That parrot is still here? Go away scooter and come back when you grow up

Mr. Pmb - waite's relativity teacher

(WaiteDavid137) wrote [childish nonsense]

Grow up scooter. When you change your mind and want to act like an
adult then you might have a chance at posting a logical arguement

Mr. Pmb - waite's relativity teacher