Tensor and scalar substitution

On Apr 12, 5:54 pm, Gc wrote:
On 12 huhti, 08:26, Koobee Wublee wrote:

a tensor is nothing but a multi-dimensional matrix. shrug

There are important differences. In physics tensor, like a vector,
often is a physical being. Matrix represents a linear transformation
with respect to a particular base and that`s it. With tensors you need
to consider how it changes with respect to change of base(yes, in the
classical pictures it stays the same even if it changes)

So, in engineering, for example, a tensor deals with stress level in a
structure can be treated as if it a matrix, but in GR’s case where the
tensor deals with the curvature of spacetime has magical properties.
It sounds like a myth to yours truly. Who can you fool? Only self-
styled physicists. shrug

This myth came from the misunderstanding of what a metric represents
as the myth can be brought into light in the following post.

http://groups.google.com/group/sci.p...61e49769?hl=en

shrug

I can only do it in tripolar coordinates. note that
the "three-dimensional Navier-Stokes eqaution" has
to be parameterized with time, or "t," but
one doesn't habitually refer to that as a phase-space (or
Minkowski's neologism of "spacetime," which seems
to derive from the general usage of "configuration spaces,"
as with teh foregoing Hamiltonians and Lagrangians.

it's nice to hide behind a whole bunch
of matrix transformations, but, then,
what is to be expected of the aging of some astronaut,
if he were able to make a round trip approaching teh speed
-- not the velocity of corpuscles --
of lightwaves (not "planar waves," which are just
one of those approximations, wheras curvature
is the most impoprtant thing, since Gauss treeted of it
-- and measured the local curvature of Universe,
with his God-am theodolite, for the goment of France.

oops; forgot to shrug,
which is a great exercise if done correctly, and
repeatedly, and strensuosly, like Atlas and
Ayn Remington Rand -- the patron saint
ot typewriting!

On Apr 12, 3:42 am, "J.B. Wood" wrote:
On 04/12/2011 06:12 AM, Han de Bruijn wrote:



Piezoelectric, yes. But aerodynamics?

Han de Bruijn

Hello, and since aerodynamics is a branch of fluid dynamics, tensors
have application there as well. Piezoelectric materials belong to a
class of materials whose electrical properties are classified as
anisotropic and are typically characterized by tensors. (Ferrites are
also anisotropic). Basically any materical whose properties can't be
adequately described by a number (scalar) or a vector (magnitude +
direction) can in many instances be characterized by an n-rank
(n-valence) tensor. In many physical (mechanical, electrical) cases n=3
or 4. In the realm of mathematical physics n can be greater. Sincerely,
J. B. Wood e-mail:

Even simpler is what I often do in aerodynamics.
I begin with Pitch, Yaw ,Roll, (Spherical CS),
transform that to Cartesian dx, dy, dz, then integrate
those to X,Y,Z and then place in again into
spheroidal and Long-Lat (Mercator) CS.

An orbiting satellite appears as a sine wave in Mercator,
usually bounded by latitude, until it re-orientates and
fires a thruster.
Regards
Ken S. Tucker

On Mon, 11 Apr 2011 23:28:49 -0700 (PDT), Eric Gisse
wrote:

On Apr 11, 10:26*pm, Koobee Wublee wrote:
On Apr 11, 10:23*pm, Eric Gisse wrote:

On Apr 11, 10:20*pm, chemguy wrote:

A tensor may be represented as a ratio of stress and a scalar may be represented as a ratio of power. A stress-power equation may represent the Einstein Field Equation.
Please see; *https://sites.google.com/site/tensorsandscalars/

No. Please learn what a tensor is.

Yes, a tensor is nothing but a multi-dimensional matrix. *shrug

Tensors have transformational properties, matrices do not. Open a book
on the subject.
More specifically tensors transform like their vectors do. This also
applies to the metric matrix gij which cannot be transformed and is
therefore not a tensor.
John Paulos

On Apr 18, 10:16 pm, Tom Roberts wrote:

One must first define the quantities that appear in physical theories.

Yes, indeed. One must decide on what is invariant and what is not.
shrug

Today we do so mathematically.

Well, not quite. Self-styled physicists do so matheMagically.
shrug

Tensors are an important part of all modern physical
theories.

Whatever. shrug

Vectors are just a specific type of tensor (rank-1).

Vectors are no matrices. Vectors indeed represent the geometry which
is observer independent. shrug

It is rather difficult to work with vectors. However, it is much
easier to work with the self dot product of the same vector, ds^2.
shrug

What you are describing is that tensors describe the geometry. Since
the geometry is invariant, tensors must be invariant. Well, that is
fine. However, the application is totally wrong in GR. What you have
called the metric is not what you have described as a tensor. For
example, given the most studied metric below,

** [1, 0, 0]
** [0, 1, 0]
** [0, 0, 1]

There is no indication that this metric is describing flat space if
the coordinate system is not specified. As for a sub-example, only
the linearly rectangular coordinate system describes flat space. Any
non-linear coordinate system connecting with this metric above
describes curved space. shrug

So, according to a segment of spacetime below,

** ds^2 = g_ij dq^i dq^j

ds^2 is the geometry which you what you call tensor, and it has only
one element. You can describe this element as sqrt(ds^2) or d[s] (a
vector). Remember that a vector cannot be universally represented by
a matrix. PD and yours truly had agreed on this one before. The
bottom line is that g_ij is the metric, and you have no right to call
it a tensor as well. shrug

With this subtle understanding, GR can be checkmated in a few simple
moves all within the logic learnt by junior high school kids. Only
idiots would not see what is coming and continue to gulp down the
fermented diarrhea of Einstein the nitwit, the plagiarist, and the
liar. shrug

On Apr 19, 7:49 am, Tom Roberts wrote:

That may be so for other transformations, but for transformations related to
changes of coordinates, there are no effects on vectors and tensors. And in
physics, any measurement is likewise completely independent of coordinates. It
is not happenstance that these different quantities share the property of
coordinate independence -- that's why tensors are so important in modern models
of the world, and why tensors are used to model measurements in the world.

Your description of what tensor is is the geometry itself. shrug

It is totally useless in GR. shrug

The reason is that what you have called the metric is not a tensor in
your definition and description. shrug

What you have called a metric is a square matrix with no information
as to what specific coordinate system. That is the metric can only
possibility describing the geometry if the choice of coordinate system
is specifically given. Otherwise, a given metric is meaningless.
shrug

Given the equation describing a segment of space or spacetime below,

** ds^2 = g_ij dq^i dq^j

What you have called the tensor is represented by ds^2. shrug

What you have called the metric is represented by g_ij as the elements
to the matrix that you have called the metric. shrug

The self-styled physicists need to get their acts together and de-
mystify themselves. shrug

This should be a very basic logical concept in the realm of no more
advanced than junior high school or, perhaps, higher grade school
level. shrug

isn't it generally assumed that the metric is "locally approx. flat,"
thus the metic is treated to be so, infinitessimally?... I mean,
it may be wrong to do so, as with "plane waves," but
it doesn't violate teh twins' God-am "paradox" so-called e.g.;
does it?

oh; you never say, what actually happens,
other than your hyper**** drive.