Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors
of rank 0


then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.


I really, really don't see why you think you are fooling me with this
spew.
----------------------------------------------------------------------------------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?

On Apr 18, 2:59 am, " wrote:
Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors

of rank 0

then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.

I really, really don't see why you think you are fooling me with this
spew.
---------------------------------------------------------------------------*-------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?

link: http://groups.google.com/group/sci.p...134ac437846e5a

On Apr 18, 3:06 am, " wrote:
On Apr 18, 2:59 am, " wrote:





Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors

of rank 0

then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.

I really, really don't see why you think you are fooling me with this
spew.
---------------------------------------------------------------------------**-------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?

link:http://groups.google.com/group/sci.p...0134ac4378...- Hide quoted text -

- Show quoted text -

************************************************** *****

Eric the Diligent further informs us:
"Even if you ask Google really, really nicely to remove the post from
Google's archives, folks will still have seen how stupid you are."

Really!!!

Sir Eric then continues
" Plus there are a ****load of forum--USENET interfaces that keep
your drek
forever and ever. "

Oh joy! This posting here will be Eternaly and Forever@!@@

On Apr 17, 10:59 pm, " wrote:
Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors

of rank 0

then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.

I really, really don't see why you think you are fooling me with this
spew.
----------------------------------------------------------------------------------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?

Temperature is a function of velocity [This follows from
thermodynamics, which you never learned]. It can also be represented
as a function of energy. Neither 3-velocity or energy are frame
independent quantities - temperature will be different in different
coordinates.

A simpler example - the magnitude of your 3-velocity. In one frame, it
can have magnitude 1. In a co-moving frame, it has magnitude 0.

A tensor is invariant under a coordinate transformation, and a rank 0
tensor is the _SAME_ in all coordinate systems. These scalars are not
rank 0 tensors.

Am I wasting my time in typing this? We'll find out.

On Apr 18, 9:35 pm, Eric Gisse wrote:
On Apr 17, 10:59 pm, " wrote:



Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors

of rank 0

then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.

I really, really don't see why you think you are fooling me with this
spew.
----------------------------------------------------------------------------------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?

Temperature is a function of velocity [This follows from
thermodynamics, which you never learned]. It can also be represented
as a function of energy. Neither 3-velocity or energy are frame
independent quantities - temperature will be different in different
coordinates.

A simpler example - the magnitude of your 3-velocity. In one frame, it
can have magnitude 1. In a co-moving frame, it has magnitude 0.

A tensor is invariant under a coordinate transformation, and a rank 0
tensor is the _SAME_ in all coordinate systems. These scalars are not
rank 0 tensors.

Am I wasting my time in typing this? We'll find out.

i zuzpegd you confuse scalars with scalars field

any scalar is / can be function of somthin else

than all scalars are rank 0 tensors no mater what

On Apr 18, 12:28 pm, HB wrote:
On Apr 18, 9:35 pm, Eric Gisse wrote:



On Apr 17, 10:59 pm, " wrote:

Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors

of rank 0

then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.

I really, really don't see why you think you are fooling me with this
spew.
----------------------------------------------------------------------------------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?

Temperature is a function of velocity [This follows from
thermodynamics, which you never learned]. It can also be represented
as a function of energy. Neither 3-velocity or energy are frame
independent quantities - temperature will be different in different
coordinates.

A simpler example - the magnitude of your 3-velocity. In one frame, it
can have magnitude 1. In a co-moving frame, it has magnitude 0.

A tensor is invariant under a coordinate transformation, and a rank 0
tensor is the _SAME_ in all coordinate systems. These scalars are not
rank 0 tensors.

Am I wasting my time in typing this? We'll find out.

i zuzpegd you confuse scalars with scalars field

Dyslexic troll shows a little intelligence...

C'mon. Pick a pseudonym and post with adult English, and folks will
happily answer your questions.

Anyway ... I'm not confusing scalars and scalar fields. Temperature
might not be the best example in that case, because you can call it a
scalar field. Try magnitude of 3-velocity - no scalar field there.


any scalar is / can be function of somthin else

True.


than all scalars are rank 0 tensors no mater what

Incorrect.

A scalar is a quantity without direction.

A rank 0 tensor is a scalar with the additional property that it is
the _same number_ under a coordinate transformation. Go read about how
tensors transform and what it means to be a tensor.

Consider a rank 2 tensor. Every rank 2 tensor can be represented as a
matrix, but not every matrix is a rank 2 tensor. Only matrices with
certain properties can represent a rank 2 tensor.

I have a reasonable expectation that you are able to do this because
you, whether you like it or not, show a degree of intelligence in your
dyslexic trolling.

On Apr 18, 8:59 am, " wrote:
Here is Eric Gisse Divine words and wisdow:

Get a clue. You have no idea what tensors are.
I understand these subjects. You do not.

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

I wrote: mass, temperature, and other scalar quantities are tensors

of rank 0

then Sir Eric the Wise wrote:

I can construct the scalar quantity "mass" using specific tensors.
That makes it a rank 0 tensor. I cannot do the same thing with
temperature.

I really, really don't see why you think you are fooling me with this
spew.
---------------------------------------------------------------------------*-------------------

Is temperature a rank_0 tensor as I wrote, or is Eric the Wise correct?


From _Physical Properties as Tensors_ [1]:

[BLOCKQUOTE
The pyroelectric tensor, (essentially a vector) represents the
relation between a first-rank tensor (the vector of electric
polarization) and a zero-rank tensor (the temperature).
]

See also [2]. However, Gisse is right on that non-relativistic
temperature is not a relativistic invariant (a trivial thought
however).


[1] http://www.iucr.org/iucr-top/comm/ct.../18/node2.html

[2] http://www.geol.umd.edu/pages/facili...mdr/press.html

On Apr 18, 9:35 pm, Eric Gisse wrote:

Temperature is a function of velocity [This follows from
thermodynamics, which you never learned].

I do not know if you are replying learned
thermodynamics, from your writings, you do not.

[i] You confound kinetic temperature with general concept of
temperature.

Kinetic temperatures are a subset of temperature. Kinetic temperature
is directly related to 'velocity', how follows from a _kinetic_ model.
But temperature also exists beyond kinetic models...

[ii] But still when limiting the discussion to kinetic models you are
being very imprecise (as usual).

[Kinetic] temperature depends on v^2, which is a scalar can be derived
from the module of the velocity, often called speed.

[Speed /= velocity] as explained in any elementary course on
mechanics.

It is more precise and clever to define kinetic temperature as Wiki
does in the section _Temperature in Gases_ [1]:

[BLOCKQUOTE
For a monatomic ideal gas the temperature is related to the
translational motion or average speed of the atoms.
]

It can also be represented
as a function of energy. Neither 3-velocity or energy are frame
independent quantities - temperature will be different in different
coordinates.

Right if you are so clever to check a relativistic LT for non-
relativistic temperature; wrong if you are more clever still and do
for the adequate generalization, the four-temperature.


[1] http://en.wikipedia.org/wiki/Tempera...ature_in_gases

On Apr 21, 5:35 am, "Juan R."
wrote:
On Apr 18, 9:35 pm, Eric Gisse wrote:

Temperature is a function of velocity [This follows from
thermodynamics, which you never learned].

I do not know if you are replying learned
thermodynamics, from your writings, you do not.

A bold assertion.
[i]

You confound kinetic temperature with general concept of
temperature.

No, I don't.

I don't write down everything I know, especially when replying to
someone who isn't going to understand it.


Kinetic temperatures are a subset of temperature. Kinetic temperature
is directly related to 'velocity', how follows from a _kinetic_ model.
But temperature also exists beyond kinetic models...

Duh.


[ii] But still when limiting the discussion to kinetic models you are
being very imprecise (as usual).

[Kinetic] temperature depends on v^2, which is a scalar can be derived
from the module of the velocity, often called speed.

Duh. x2


[Speed /= velocity] as explained in any elementary course on
mechanics.

Duh x3.

However, it remains that what I said is correct. Temperature is a
function of velocity - T = 1/2 m * v dot v.

Why do you waste your time on such pointless pedantry?


It is more precise and clever to define kinetic temperature as Wiki
does in the section _Temperature in Gases_ [1]:

[BLOCKQUOTE
For a monatomic ideal gas the temperature is related to the
translational motion or average speed of the atoms.
]

Wiki is neither precise or clever on this account, and it has not told
me anything I did not already know. Temperature can also be a function
of pressure, volume, electric and/or magnetic fields - even
gravitational acceleration. These cases are interesting but are
completely irrelevant - bringing them up will only confuse someone who
is already confused enough.


It can also be represented
as a function of energy. Neither 3-velocity or energy are frame
independent quantities - temperature will be different in different
coordinates.

Right if you are so clever to check a relativistic LT for non-
relativistic temperature; wrong if you are more clever still and do
for the adequate generalization, the four-temperature.

Uhhhhh....no

Temperature does _not_ become a four-vector in relativity. I haven't
the faintest goddamn idea where you get that notion from - temperature
remains a scalar, but the additional condition is imposed that the
relevant thermodynamic equations are based on temperature taken in the
_rest frame_ of the junk in question.

My thermodynamics knowledge comes from Reif's Statistical and Thermal
Physics - an actual text on the subject rather than random Wiki
reading.


[1] http://en.wikipedia.org/wiki/Tempera...ature_in_gases

On Apr 22, 12:46 am, Eric Gisse wrote:
Temperature is a function of velocity [This follows from
thermodynamics, which you never learned].

I do not know if you are replying learned
thermodynamics, from your writings, you do not.

A bold assertion.

No more bold your first one, but mine is based in your wrong postings
and half-true assertions.

However, it remains that what I said is correct. Temperature is a
function of velocity - T = 1/2 m * v dot v.

Why do you waste your time on such pointless pedantry?

It is not pedantry. It is called precision. There is some difference
between scalar products [x * x] and vectorial ones [v * v].

Therefore is not suprising that is in _Temperature in Gases_ [I cited
before]:

[BLOCKQUOTE
For a monatomic ideal gas the temperature is related to the
translational motion or average speed of the atoms.
]

uses speed rather than velocity.

It can also be represented
as a function of energy. Neither 3-velocity or energy are frame
independent quantities - temperature will be different in different
coordinates.

Right if you are so clever to check a relativistic LT for non-
relativistic temperature; wrong if you are more clever still and do
for the adequate generalization, the four-temperature.

Uhhhhh....no

Temperature does _not_ become a four-vector in relativity. I haven't
the faintest goddamn idea where you get that notion from - temperature
remains a scalar, but the additional condition is imposed that the
relevant thermodynamic equations are based on temperature taken in the
_rest frame_ of the junk in question.

As usual you got some basic stuff, ignore more advanced treatments and
then you believe i invented. But, by 21th time (you are really slow to
learn), just because you did not study stuff and never read literature
on a topic does not mean i invented.

In relativity, temperature scalar can be substituted by corresponding
four generalization (as velocity is by a four-velocity for instance).

See 2nd edition of Moller, The theory of relativity. He uses inverse
temperature by commodity [theta = 1/T]. I.e. instead [S = U/T] writes
[S = U*theta]

Since you claim no copy of that textbook, I have found an example
online [2] He uses notation [beta = 1/T] and introduces the four
temperature (its four inverse) in equation (5).

Another online resource _contradicting_ you is [3] also introducting a
four-temperature concept, no matter how many times you write that is
not possible.

[1] http://en.wikipedia.org/wiki/Tempera...ature_in_gases

[2] http://arxiv.org/pdf/physics/0505004

[3] http://www.rish.kyoto-u.ac.jp/isss7/...PDF/P-3-18.PDF