tensors

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

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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.

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Of course, I meant "Comprehensive", not "Comprehension", in that title.
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"RichD" wrote in message
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 =====================================================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

On 9/3/12 9/3/12 - 3:50 PM, RichD wrote:
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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)
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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

"Tom Roberts" wrote in message 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

On Tue, 04 Sep 2012 18:09:53 +0100, Lord Androcles, Zeroth Earl of Medway wrote:
[snip crap]
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why don't you better use standard usenet quoting like everybody else, what you do looks stoopid

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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>

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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

On 9/3/2012 10:50 PM, RichD wrote:
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If you want to understand General Relativity, do *not* go for the tensors.
Read Clifford Will's "Was Einstein Right". (Amazon.com product link shortened) />/
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.

On Sep 4, 12:18 pm, Uwe Hayek wrote:
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You don’t know what you are talking about. <shrug>
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Who cares about what a nitwit, a plagiarist, and a liar said? <shrug>
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Nonsense. GR is all in the mathematics. GR does not define the universe. <shrug>
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You really don’t know you tensors, eh? <shrug>
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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>

On 9/4/12 9/4/12 - 2:16 PM, Uwe Hayek wrote:
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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

reading skills, not with the clocks. Einstein<BR>NEVER said "they are slowed", because he knew full well that they are not <BR>(see above).<BR>"Thence we conclude that a balance-clock at the equator must go more slowly, <BR>by a very small amount, than a precisely similar clock situated at one of <BR>the poles under otherwise identical conditions." -- Einstein<BR>--Lord Androcles, Zeroth Earl of Medway</EM></DIV> <DIV><EM><FONT size=4></FONT></EM>&nbsp;</DIV> <DIV><EM><FONT size=4></FONT></EM>&nbsp;</DIV></DIV></DIV></BODY></HTML>
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i think anybody may safely skip A to B and get the B to C directly

On Mon, 3 Sep 2012 13:50:21 -0700 (PDT), RichD
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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    

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FSVO familiar and FSVO vector. What you wrote seemed very 19th century.
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In a specific basis.
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No; it might be doubly covariant or doubly contravariant.
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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.
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Rare? They're all over the place, and not a form of matrix.
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Not even close. I suspect that you're thinking of the Jacobean.
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What is a mixed (co- and contra-variant) tensor, chopped liver?
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Tensors aren't limited to the tangent bundle and in som,e applications the term "rotated" is meaningless.
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So much the worse for the test.
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Halmos's "Finite Dimensional Vector Spaces"; I found it an easy read in HS. I don't know whether it is still in print.

On Wed, 05 Sep 2012 17:58:37 -0400, Shmuel (Seymour J.) Metz
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The basis must already be defined or you cannot write the tensor.
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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.
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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.
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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.
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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.
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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

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Please don't make mistakes that beg for patronization.
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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.
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Not even close. You're still confusing tensors with a specific application of tensors.
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As long as you continue making guesses instead of thinking, I'll continue to patronize you.
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LOTF,LMAO!
Not if I use the other sign convention ;-)
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K3wl! That's an example of a tensor, and nothing more.
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Write me when you're sober.
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I don't have that book; V qb unir ZGJ Tenivgngvba.
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Is to tell the truth.
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No, tonto, the anwer is that you are confused.
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There is no "nature's coordinate system."
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If we do, you haven't found it.
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No, you started expounding nonsense.
[1] Or geometric equivalent.

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.....