tensors

[RichD]

I've been trying to learn tensors, without success. The notation
sucks, for one thing. But II don't see the purpose. Einstein
needed
it for his theory, apparently. But it's also used for solids; stress
and
strain. Why can't you use vector algebra?

I'm not looking for an explanation of what/how, but the why and
whatfor.
Is there any intuition, or is it just formal, abstract manipulation?

--
RIch

[dilettante]

"RichD" <r_dela...@yahoo.com> wrote in message
news:921d6879-8361-439f...@rg9g2000pbc.googlegroups.com...

If you have access to university library, you may be able to find a quaint
little book, "The Einstein Theory of Relativity" by Lillian Lieber, which
has a very down to earth explanation of tensors and the kind of thing they
are useful for. At a more technical level, Spivak's "A Comprehension
Introduction to Differential Geometry" has some material you may find
helpful, though it can be rough going. At your local bookstore you'll
probably find "Relativity Demystified" by David McMahon. This looks like the
most promising introduction to me in terms of gettng both the mathematical
nuts and bolts, and some explanation of the physical significance as you go.

[Lord Androcles, Zeroth Earl of Medway]

"RichD" wrote in message
news:921d6879-8361-439f...@rg9g2000pbc.googlegroups.com...

======================================================
Your question is akin to asking "why use multiplication when repeated
addition works just fine?"
a) 3+3+3+3+3+3 = 18.
b) 6*3 = 18.

a) is easy to compute but clumsy -- ideal for a computer
b) requires you learned your multiplication tables by rote when you were 6
years old.

There is a description and history here:
http://en.wikipedia.org/wiki/Tensor

Try this problem:
A theatre has booked a famous entertainer for Saturday night.
All patrons want to see the show and some will pay more for a
front row seat, others have little money but will pay for a seat
"up in the gods", the upper gallery. There are 10,000 seats, 50
wide by 100 deep in the stalls and 100 wide, 50 deep in the
horseshoe gallery.
Front gallery seats in the arms of the horseshoe are closer to
the entertainment and more desirable. The nearer the seat is
to the stage the more you can charge per seat.
As manager your job is to price the seats for the maximum profit.
The richest will pay $1,000 for one seat. The poorest will pay $1,
so you are guaranteed a take of $10,000.
If you charge $1000 per seat you have the potential take of
$10 million, but only one patron will come to the show and your take
will be $1000, your profit will be negative as you still have to pay
overheads and the entertainer, who wants $100,000 for his
performance.
If you charge $1 per seat everybody will come, your take will be
$10,000 and your profit still negative, but that's 10 times better
than charging $1000 per seat with 9,999 empty seats.

What is your seat pricing strategy?
At first you may compute distance (seat to centre stage) as
d = sqrt( x^2 + y^2 +z^2) and say price = 1/d * k
You may be surprised to learn that you maximise profit when you
have some empty seats.
As the price comes down the number of seats taken goes up,
but there comes a point where lowering the price any further
reduces the profit.
The real question is, should I hire you to manage my theatre
while I'm off sailing in my yacht in the Mediterranean?
Do you have "knows tensors and how to apply them" on your
resume?
And that is the why and whatfor, I don't give a hoot for
Einstein's theory, it doesn't buy me anything.

-- Lord Androcles, Zeroth Earl of Medway

[dilettante]

"dilettante" <n...@nonono.no> wrote in message
news:k23abg$o5k$1...@dont-email.me...

>
> "RichD" <r_dela...@yahoo.com> wrote in message
> news:921d6879-8361-439f...@rg9g2000pbc.googlegroups.com...
>> I've been trying to learn tensors, without success. The notation
>> sucks, for one thing. But II don't see the purpose. Einstein
>> needed
>> it for his theory, apparently. But it's also used for solids; stress
>> and
>> strain. Why can't you use vector algebra?
>>
>> I'm not looking for an explanation of what/how, but the why and
>> whatfor.
>> Is there any intuition, or is it just formal, abstract manipulation?
>>
>> --
>> RIch
>
> If you have access to university library, you may be able to find a quaint
> little book, "The Einstein Theory of Relativity" by Lillian Lieber, which
> has a very down to earth explanation of tensors and the kind of thing they
> are useful for. At a more technical level, Spivak's "A Comprehension
> Introduction to Differential Geometry"

Of course, I meant "Comprehensive", not "Comprehension", in that title.

[Tom Roberts]

On 9/3/12 9/3/12 - 3:50 PM, RichD wrote:
> I've been trying to learn tensors, without success. The notation
> sucks, for one thing. But II don't see the purpose. Einstein
> needed
> it for his theory, apparently. But it's also used for solids; stress
> and
> strain. Why can't you use vector algebra?

You cannot just use vector algebra, because:
A) it is unable to distinguish the different ranks of tensors (a dyad
represents rank-2, but higher ranks are not covered; in GR, tensors
with ranks up to 4 are essential)
B) some aspects of vector algebra are specific to three dimensions
(e.g. the cross product, the curl); relativity needs four dimensions
C) it ignores the important distinction between vectors and co-vectors
D) it ignores the essential difference between co- and contra-variant
quantities (related to (C) but more general)
E) It introduces a confusing and very limited notion of "pseudo vector"
(see (B) above)

> I'm not looking for an explanation of what/how, but the why and
> whatfor.
> Is there any intuition, or is it just formal, abstract manipulation?

Any intuition is directly connected to geometry.

The best understanding for me came from the early chapters of Misner, Thorne,
and Wheeler, _Gravitation_, and also Baez and Muniain,
_Gauge_Fields,_Knots,_and_Gravity_.

A rank-N tensor is a multi-linear function of N vectors and co-vectors onto the
Reals. The co- and contra-variance of tensors depends on their behavior under a
mapping of the underlying manifold: a covariant tensor naturally behaves simply
under a push-forward of the map, while a contravariant tensor naturally behaves
simply under the pull-back of the map.


Tom Roberts

[Lord Androcles, Zeroth Earl of Medway]

"Tom Roberts" wrote in message news:WfSdnXo85Zv...@giganews.com...

-- "Tom Roberts" wrote in message
news:9LqdnZVOesi...@giganews.com...
It is not me who is wrong.
Tom Roberts
===========================================================
It must be Einstein, then.
Bwhahahahaha!
"Tom Roberts" wrote in message news:t5CdnQfQvY4...@giganews.com...
The problem is with your poor reading skills, not with the clocks. Einstein
NEVER said "they are slowed", because he knew full well that they are not
(see above).
"Thence we conclude that a balance-clock at the equator must go more slowly,
by a very small amount, than a precisely similar clock situated at one of
the poles under otherwise identical conditions." -- Einstein
--Lord Androcles, Zeroth Earl of Medway

[Koobee Wublee]

Ordinary matrices can be 0, 1, or 2 dimensions --- rank 0, rank 1, and
rank 2. Matrices with higher dimensions are covered by tensors.
Think tensors as matrices with any dimensions, and you won’t go wrong.

After the Riemann tensor (rank 4 = 4 x 4 x 4 x 4) is reduced for no
apparent reasons to the Ricci tensor (rank 2), for all practical
considerations, GR only deals with 2-dimensional matrices (4 x 4) ---
rank-2 tensor. For all practical applications, GR deals with
symmetric matrices, and this simplifies the mathematics greatly.

A vector is still a scalar. When you take the derivative of a vector,
you have to deal with its unit vectors, and that can get very
cumbersome. Tensors represent the derivative of a vector without
using the unit vectors, and that actually becomes simpler.

There is actually no such thing as tensor calculus. It is all about
doing calculus on matrices which still follows exactly how Leibniz had
laid out centuries ago. For example,

** dF(x, y, z) = @F/@x dx + @F/@y dy + @F/@z dz

Where

** F(x, y, z) = function of x, y, z
** @/@x = partial derivative with respect to x

The modern notation of tensor calculus is actually quite neat. It
eliminates the repeatable tasks of writing down the summation
operators. That can save a lot of time when you tried to take the
derivative of anything. You can manually add in the summation stuff
back in if you like. Thus, the same example above becomes

** dF(q0, q1, q2..., qN) = sum(n = 0 to N)[@F/@q_n dq_n]

Or

** dF = @F/@q_n dq_n

At the end of the day, there is nothing special about tensor
calculus. So, don’t be intimidated by these words. <shrug>

[Pete Weber]

On Tue, 04 Sep 2012 18:09:53 +0100, Lord Androcles, Zeroth Earl of Medway
wrote:

[snip crap]

> Androcles, Zeroth Earl of Medway

why don't you better use standard usenet quoting like
everybody else, what you do looks stoopid

[Uwe Hayek]

If you want to understand General Relativity, do *not* go for the tensors.

Read Clifford Will's "Was Einstein Right".
http://www.amazon.com/Was-Einstein-Right-2nd-Relativity/dp/0465090869/

By his applying the relativity and other deep physical principles in his
thought experiments, Einstein gained an insight how the universe works,
and it took him more than 3 years to cast it into tensors, with the help
of Grossman and Hilbert. The latter even tried to steal his tensor
formulation, without knowing how the universe works.

Uwe Hayek.

[Koobee Wublee]

On Sep 4, 12:18 pm, Uwe Hayek wrote:

> If you want to understand General Relativity, do *not* go for the tensors.

You don’t know what you are talking about. <shrug>

> Read Clifford Will's "Was Einstein Right".
> http://www.amazon.com/Was-Einstein-Right-2nd-Relativity/dp/0465090869/

Who cares about what a nitwit, a plagiarist, and a liar said? <shrug>

> By his applying the relativity and other deep physical principles in his
> thought experiments, Einstein gained an insight how the universe works,

Nonsense. GR is all in the mathematics. GR does not define the
universe. <shrug>

> and it took him more than 3 years to cast it into tensors, with the help
> of Grossman and Hilbert.

You really don’t know you tensors, eh? <shrug>

> The latter even tried to steal his tensor
> formulation, without knowing how the universe works.

You are totally out of your mind.

The field equations centers around the Ricci tensor (rank 2 of 4 x 4)
which was created by Levi-Civita after reducing the Riemann tensor
(rank 4 of 4 x 4 x 4 x 4).

What created the Riemann tensor?

The geodesic equations were derived through minimizing the action of
the shortest distance. The geodesic equations are nothing but Euler-
Lagrange equations that yield the Christoffer Symbols of the 2nd
kind. Defining the covariant operation within the geodesic equations
that yield a null four-velocity, that was what Ricci did. Taking the
double covariant derivative yields the Riemann tensor.

The whole GR is man-made mathematical artifact. <shrug>

Tensor was not invented by Grossmann, Hilbert, Ricci, Levi-Civita, or
Einstein the nitwit, the plagiarist, and the liar. Its application
was first tossed around by Voigt --- the same person who first
discovered the Voigt transform that satisfies the null results of the
MMX.

Unlike the Lorentz transform which satisfies the principle of
relativity, the Voigt transform says THE AETHER MUST EXIST. To this
day, all experiments that have verified SR also verifies the Voigt
transform where it is an antithesis to SR. <shrug>

[Tom Roberts]

On 9/4/12 9/4/12 - 2:16 PM, Uwe Hayek wrote:
> If you want to understand General Relativity, do *not* go for the tensors.
>
> Read Clifford Will's "Was Einstein Right".
> http://www.amazon.com/Was-Einstein-Right-2nd-Relativity/dp/0465090869/

If you want to understand General Relativity at the conceptual, non-mathematical
level, this is a better book -- Will discusses experiments, while Geroch
discusses concepts:

Geroch, _General_Relativity_from_A_to_B_.


If you want to really understand GR, tensors are indispensable, as is geometry.
MTW [referenced earlier] was very important to me.


Tom Roberts

[Lord Androcles, Zeroth Earl of Medway]

"Tom Roberts" <tjrobe...@sbcglobal.net> wrote in message news:LIednZKqrtA...@giganews.com...

The tail wagging the dog is all you understand, proven liar Roberts.

 

 

-- "Tom Roberts" wrote in message news:9LqdnZVOesi...@giganews.com...
It is not me who is wrong.
Tom Roberts
===========================================================
It must be Einstein, then.
Bwhahahahaha!
"Tom Roberts" wrote in message news:t5CdnQfQvY4...@giganews.com...
The problem is with your poor reading skills, not with the clocks. Einstein
NEVER said "they are slowed", because he knew full well that they are not
(see above).
"Thence we conclude that a balance-clock at the equator must go more slowly,
by a very small amount, than a precisely similar clock situated at one of
the poles under otherwise identical conditions." -- Einstein

--Lord Androcles, Zeroth Earl of Medway

 

 

[John Polasek]

This is obviously a complex topic but I can point out a few things to
you.
I assume you're familiar with vectors and matrices.

A matrix M can represent a second rank tensor which generally operates
on a vector X to produce another vector Y. (M, X, Y just to be
specific).
Y = MX or Y = X'M are both true if X' is a row vector and X is a
column vector. Tensors are all about subscripts:
y_i = m_ij*x_j (multiply like indices, here j, and sum)

You are probably only acquainted with column vectors, and are unaware
that there is such a thing as a row vector or what it means but you
can think of X (column) as a covariant and X' as a contravariant
tensor of the first rank.
We can make two products with X and X':
X'X = x^2, a scalar tensor: P = x_i*x_j delta ij = 0 rank
(delta makes i = j forces contraction to 0 rank).
XX' = a 3x3 outer product Pij = xi*xj
The connection between the matrix and tensor is that for each one
there are rules how to select the subscripts of the matrix and vector
and how to multiply the pairs of terms together and how to sum them to
produce a new term.
Tensors are an extremely rare form of matrix. The second rank tensor
contains partial differential coefficients linking the pertinent
vectors, but these vectors all have to be the same sort.
The restriction on a tensor is that it must transform like its
vectors. the physical interpretation of this statement is that the
object represented in physics by a tensor should be able to be rotated
to any position and still be valid.
Einstein's metric tensor does not meet this test: time is not at all
like XYZ.
I've never found a book that was useful for learning tensor algebra,
except maybe Jeffries. MTL's gravitation is pretty useless and verbose
and leaves the interesting stuff as an exercise for the student.
when I wrote a paper for the Journal of the optical Society of America
regarding mirrors and prisms as tensors, I made up my own tensors for
mirrors it's astonishingly easy. You start with a primitive tensors
which is stunningly simple and then simply transform it by rotation to
the Prism that you'd really like to have.
To give you an idea, a mirror with its vector normal along the X. axis
has its primitive form as simply the identity matrix with element 11
changed to -1. Call it P.

Now rotate the mirror matrix by any angle you want with the rotation
matrix R:
P' = R'PR
The original primitive is put into a matrix sandwich with R and R' on
each side, and you get a new mirror.
R Is of course the same transformation you use to rotate the vectors
themselves, but in this case rotate the object which is a mirror and a
second rank tensor. The tensor equation looks like this:
pij = delta ij - 2NiNj
where N is the mirror normal vector, and NiNj the outer product, 2d
rank.

John Polasek

[Steve Watson]

On Sep 4, 7:52 pm, Koobee Wublee <koobee.wub...@gmail.com> wrote:
> On Sep 3, 1:50 pm, RichD <r_delaney2...@yahoo.com> wrote:
>
> > I've been trying to learn tensors, without success. The notation
> > sucks, for one thing. But II don't see the purpose. Einstein
> > needed
> > it for his theory, apparently. But it's also used for solids; stress
> > and
> > strain. Why can't you use vector algebra?
>
> > I'm not looking for an explanation of what/how, but the why and
> > whatfor.
> > Is there any intuition, or is it just formal, abstract manipulation?
>
> Ordinary matrices can be 0, 1, or 2 dimensions --- rank 0, rank 1, and
> rank 2. Matrices with higher dimensions are covered by tensors.
> Think tensors as matrices with any dimensions, and you won’t go wrong.

with all do respect for your relativity, you do a
mistake here, the rank of a matrix does not relates
to its dimension

by definition matrix are 2d, you cant change that!

a rank to a matrix can be any, it relates to the number
of independent rows or columns, i hope i do no mistake

mister Dirk may back me up

good bye

[Steve Watson]

i think anybody may safely skip A to B and
get the B to C directly

[Shmuel Metz]

In <26qe48lss6saom7h5...@4ax.com>, on 09/05/2012

at 12:01 PM, John Polasek <jack...@hotmail.com> said:

>I assume you're familiar with vectors and matrices.

FSVO familiar and FSVO vector. What you wrote seemed very 19th
century.

>A matrix M can represent a second rank tensor

In a specific basis.

>generally operates on a vector X to produce another vector Y.

No; it might be doubly covariant or doubly contravariant.

>Tensors are all about subscripts:

No. Tensors are all about multilinear maps and don't have subscripts.
It's the expression of a tensor in terms of a specific basis that has
subscripts.

>Tensors are an extremely rare form of matrix.

Rare? They're all over the place, and not a form of matrix.

>The second rank tensor contains partial differential coefficients
>linking the pertinent vectors,

Not even close. I suspect that you're thinking of the Jacobean.

>The restriction on a tensor is that it must transform like its
>vectors.

What is a mixed (co- and contra-variant) tensor, chopped liver?

>the physical interpretation of this statement is that the object
>represented in physics by a tensor should be able to be rotated to
>any position and still be valid.

Tensors aren't limited to the tangent bundle and in som,e applications
the term "rotated" is meaningless.

>Einstein's metric tensor does not meet this test:

So much the worse for the test.

>I've never found a book that was useful for learning tensor algebra,

Halmos's "Finite Dimensional Vector Spaces"; I found it an easy read
in HS. I don't know whether it is still in print.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[John Polasek]

On Wed, 05 Sep 2012 17:58:37 -0400, Shmuel (Seymour J.) Metz
<spam...@library.lspace.org.invalid> wrote:

>In <26qe48lss6saom7h5...@4ax.com>, on 09/05/2012
> at 12:01 PM, John Polasek <jack...@hotmail.com> said:
>
>>I assume you're familiar with vectors and matrices.
>
>FSVO familiar and FSVO vector. What you wrote seemed very 19th
>century.
>
>>A matrix M can represent a second rank tensor
>
>In a specific basis.

The basis must already be defined or you cannot write the tensor.

>>generally operates on a vector X to produce another vector Y.
>
>No; it might be doubly covariant or doubly contravariant.
>
>>Tensors are all about subscripts:
>
>No. Tensors are all about multilinear maps and don't have subscripts.
>It's the expression of a tensor in terms of a specific basis that has
>subscripts.

Please don't patronize me with this textbook recitation about
multilinear maps and their lack of subscripts. If they don't have
subscripts you can just admire them but if you want to do real
calculation you have to bring out the subscripts.

>>Tensors are an extremely rare form of matrix.
>
>Rare? They're all over the place, and not a form of matrix.

Tensors must meet a rigid set of requirements and almost no matrix
will do that. Specifically, the tensor must transform exactly as its
basis transforms.

>>The second rank tensor contains partial differential coefficients
>>linking the pertinent vectors,
>
>Not even close. I suspect that you're thinking of the Jacobean.

I guess you don't know what a tensor is for. More or less it's a
formalization of the chain rule.
You should be familiar with
@tau/@t = sqrt(-g00) (1)
Which is the corner differential coefficient in the metric tensor
written as a matrix. And so are they all.

>>The restriction on a tensor is that it must transform like its
>>vectors.
>
>What is a mixed (co- and contra-variant) tensor, chopped liver?
>
>>the physical interpretation of this statement is that the object
>>represented in physics by a tensor should be able to be rotated to
>>any position and still be valid.
>
>Tensors aren't limited to the tangent bundle and in som,e applications
>the term "rotated" is meaningless.
>
>>Einstein's metric tensor does not meet this test:
>
>So much the worse for the test.

I have already defined an x-axis mirror tensor, both in tensor
subscript language and as a square matrix. As the latter, it is the
identity matrix with m11 equal to -1.

I could get fancy, like you aesthetes, and proclaim that it has a
"signature" of -1 1 1. That makes it not only spiffy, but guarantees
that it is a mirror matrix.
And its principal function will be to turn an incoming righthanded
space into an outgoing lefthanded space.

Oh look, Einstein's metric tensor has the signature -1 1 1 1. then
quite clearly it is also a mirror tensor and therefore it would be
inverting space, which would be an intolerable flaw.

But, but, it can't be true, or else how did it get that way?

The answer can be found in MTL Gravitation in which they have a
section with the show-offy title "Farewell to ict".

Quite obviously the way to get rid of the mirror canard (and get
right) is to re-institute x00 = ict instead of ct or t.
The basis should be
ict x y z and cannot be ct x y z
But ICT isn't nice-we don't want an imaginary component of space.
In the scalar product xGx we must get -c2t2. "Why don't we just make
G00 negative?"
The answer is that you would then be violating a fundamental law of
nature, tampering with nature's coordinate system. The tensor is not
permitted to be a mirror-which specifically inverts the chirality of
space.
Take another look at equation 1 above. Doesn't that differential
coefficient sqrt(-g00) ring any bells with you? Perhaps some faint
feeling of revulsion? nay, even suspicion?
I think we have strong grounds to conclude that there is no such thing
as space-time or if it exists, it is a mighty wobbly one since it has
the metric tensor as its basis.

>>I've never found a book that was useful for learning tensor algebra,
>
>Halmos's "Finite Dimensional Vector Spaces"; I found it an easy read
>in HS. I don't know whether it is still in print.

I started going into all the basic details in order to help illuminate
the OP who is, as he said, confused by what he reads.
I find I have to explain this every two or three years.
John Polasek

[Ronald Benedik]

Since a matrix is full of scalars of a specific body and
the fact that the vector space of matrices of a specific kind
form a vector space a matrix is in fact a vector and so it
yields a contravariant tensor (0,1).

My 2 cents.....

[Lord Androcles, Zeroth Earl of Medway]

"Ronald Benedik" <Aon.91...@aon.at> wrote in message news:5048c9ea$0$1585$91ce...@newsreader04.highway.telekom.at...

Anyone that says “in fact” does so only to convince themselves

of their own lies. In fact, actually a matrix is not in fact a vector

actually, it has no fucking actual in fact direction. Keep your two

cents and don’t give up your day job.

-- Lord Androcles, Zeroth Earl of Medway

 

[Shmuel Metz]

In <h1eh48hrmv7fpcee7...@4ax.com>, on 09/06/2012

at 11:56 AM, John Polasek <jack...@hotmail.com> said:

>Please don't patronize me with this textbook recitation about
>multilinear maps and their lack of subscripts.

Please don't make mistakes that beg for patronization.

>If they don't have subscripts you can just admire them but if
>you want to do real calculation you have to bring out the
>subscripts.

You're confusing a concept with a calculation. I can't calculate the
volume of a bucket of water without using numbers[1], but that doesn't
make the bucket a number.

>Specifically, the tensor must transform exactly as its
>basis transforms.

Not even close. You're still confusing tensors with a specific
application of tensors.

>I guess you don't know what a tensor is for.

As long as you continue making guesses instead of thinking, I'll
continue to patronize you.

>More or less it's a formalization of the chain rule.

LOTF,LMAO!

>You should be familiar with
> @tau/@t = sqrt(-g00) (1)

Not if I use the other sign convention ;-)

>I have already defined an x-axis mirror tensor,

K3wl! That's an example of a tensor, and nothing more.

>Oh look, Einstein's metric tensor has the signature -1 1 1 1. then
>quite clearly it is also a mirror tensor and therefore it would be
>inverting space, which would be an intolerable flaw.

Write me when you're sober.

>The answer can be found in MTL Gravitation

I don't have that book; V qb unir ZGJ Tenivgngvba.

>Quite obviously the way to get rid of the mirror canard

Is to tell the truth.

>The answer is that you would then be violating a fundamental law of
>nature,

No, tonto, the anwer is that you are confused.

>tampering with nature's coordinate system.

There is no "nature's coordinate system."

>I think we have strong grounds to conclude that there is no such
>thing as space-time or if it exists, it is a mighty wobbly one since
>it has the metric tensor as its basis.

If we do, you haven't found it.

>I started going into all the basic details

No, you started expounding nonsense.

[1] Or geometric equivalent.