I'm having a hard time learning tensors. I got upto the Christoffel symbols. This thread is intended to be a place of discussion (more or less mathematical) concerning tensors. Anyone who is having difficulty learning tensors is invited. Everyone else is invited too. Interested?

Good thread. I wish lethe contributes more in this thread.

Lethe!? I thought he was never coming back?

I've always wanted to know what tensors are used for. Personally, I have never had to use them before. Perhaps a good place to start is with a definition and maybe some basic examples. We can work our way up from there.

I'll direct you to a previous post of mine on the topic:

http://www.physicsforums.com/showpost.php?p=263615&postcount=6

- Warren

Very excellent description chroot. Thanks!

I'll ask a specific problem I'm having. After I derived the Geodesic equations in terms of Christoffel symbols (CS), I wanted to try this out using an example. I tried calculating them in the case of a sphere embedded in 3D space. It seemed like a lengthy procedure to me, involving lots of summations (which is to expected when dealing with tensors). Then I tried an example given in the book I was using. I don't remember the specific example now, but they seemed to have calculated the values of each of the CS in a single step. Either that, or they skipped steps. I was wondering if there was some shortcut which I had missed. Would someone please give an example (preferably using a common figure like the sphere) and show the procedure for calculating the CS? It would do to derive only {11,1}, or any other one demonstrating the advantage of the procedure.

I was wondering if there was some shortcut which I had missed. Would someone please give an example (preferably using a common figure like the sphere) and show the procedure for calculating the CS? It would do to derive only {11,1}, or any other one demonstrating the advantage of the procedure.I'm not entirely convinced by this, but let's try, abbreviated pretty well to the point of opacity.

Consider an arbitrary (0,2) tensor A_qr defined at a point p at coordinate x^pon a manifold M, with M embedded in ambient space defined with coordinates {y^a}. Now form the derivatives required for abitrary translation of A_qr(x^p) relative to ambient coordinates.
∂A_qr/∂x^p = ∂/∂x^p[(∂y^a/∂x^q)(∂y^a/∂x^r)] = (∂²y^a/∂x^p∂x^q)(∂y^a/∂x^r) + (∂²y^a/∂x^r∂x^p)(∂y^a/∂x^q)

Holy Mary, I'm going to call that A_qr,p = y_a,pqy_a,r + y_a,rpy_a,q

Now cyclically permute the indices, getting

A_rp,q = y_a,qry_a,p + y_a,pqy_a,r

and

A_pq,r = y_a,rpy_a,q + y_a,qry_a,

Add the first two and subtract the third, giving

1/2(A_qr,p + A_rp,q - A_pq,r)

substitute the metric tensor for A and you have defined {pq,r}, Christoffel I

I've always wanted to know what tensors are used for. Personally, I have never had to use them before. Perhaps a good place to start is with a definition and maybe some basic examples

How about starting with a less precise, verbal description, just to get an idea of these things.

Tensors are physically useful quantities which have components with respect to some set of axes - i.e some viewpoint. The components change if the viewpoint is changed but some of the properties of tensors are 'invariant', and don't depend on our point of view. It would be ridiculous for a concrete bridge to fall down if we simply changed our viewpoint, and the maximum (principal) stress component, which determines whether the concrete will break, is indeed an invariant of the stress, which is a tensor. Stress is a 2nd order tensor because it has 3^2 = 9 components in 3-D. The 9 components can be represented as a 3x3 matrix.

A 'vector' (in the elementary sense) is a simpler type of tensor. Force is thus a tensor - it's components depend on our choice of axes but the magnitude of the force is an invariant and is independent of the axes. Indeed, the transformation laws linking the components in different sets of axes are there specifically to preserve the invariants. A vector is called a 1st order tensor because it has 3^1 = 3 components in 3-D. The 3 components can be represented as a 3x1 matrix (strictly, 1x3 but let's leave that at the moment).

A 'scalar' is an even simpler, one might say degenerate, tensor. Temperature is thus a tensor. Temperature has only one 'component' even in 3-D and could be represented by a 1x1 matrix. A scalar is called a Zero'th order tensor because it has 3^0 = 1 component in 3-D.

So there's a whole hierarchy of tensors that are used routinely, perhaps unknowingly, by schoolchildren, never mind scientists and engineers.

RDT2: Nice post. I was thinking of doing something much more abstract and technical, but I leave the stage to you.

Oh. And welcome to the madhouse!

I have Wald's "General Relativity", and was very pissed off when he defined the Christoffel symbols as tensors. this probably has confused a lot of people

I'm having a hard time learning tensors. I got upto the Christoffel symbols. This thread is intended to be a place of discussion (more or less mathematical) concerning tensors. Anyone who is having difficulty learning tensors is invited. Everyone else is invited too. Interested? Tensors are a rather simple mathematical definition. I am surprised there is this much discussion going on. Perhaps the difficulting is in your efforts to understand what a tensor represents and how this is applied in a theoretical world?

To be as clear as possible, I am going to take a step out of the theoretical world that those in this thread may be in to explain tensors. I am going to take a step into the world of Computational Fluid Dynamics in the hopes of explaining a real world tensor. You need not fret because knowledge in CFD is not needed - it just happens to be the place that came off the top of my head for real world application.

A tensor is merely the mathematical result of taking the outer product of two equally lengthed vectors. That is, in CFD, the velocity component of fluid is 3-dimensional, so the velocity is expressed in terms of a vector of length three. Let V be a velocity vector defined as: V = u i + v j + w k where i, j, and k are orthogonal unit vectors.

Now, the outer product is just another way of saying normal matrix multiplication, that is, if we take V*V, we get the tensor:

\left(\begin{array}{cc}u\\v\\w\end{array}\right) \left(\begin{array}{cc}u&v&w\end{array}\right) = \left(\begin{array}{cc}u^{2}&uv&uw\\vu&v^{2}&vw\\wu&wv&w^{2}\end{array}\right)

Edit: Ah ha! apparently I cannot use latex, I guess I forgot what forum I was at - I'll try to edit this later.

I have Wald's "General Relativity", and was very pissed off when he defined the Christoffel symbols as tensors. this probably has confused a lot of peopleDid he really? Then he's wrong, of course. The Christoffel symbols have tensor components, but are not themselves tensors.

OK, what is a tensor? I'm going to give a Micky Mouse account to start with.

All definitions I know of revolve around the rather unhelpful notion that "tensors are mathematical objects that obey certain transformation rules". Let's not try and improve on that, so first let's see what's meant by transformation.

Given a set of coordinates, the act of in some way moving an object clearly changes that objects coordinate components. The act of moving the object is called a transformation, and we know we can translate, reflect or rotate our object, or any combination of them. Clearly the exact same effect can be acheived by moving the coordinates rather than the object (no, this is not relativity!) and this is called coordinate transformation.

The important thing to realise is that in the physical sciences these objects have a real physical existence (like RDT2's bridge), and we are not at liberty to change their real properies just by mucking around with our coordinates.

Now a scalar, being just a number, is invariant under coordinate transformations, and we may write x' = x, where the prime denotes the transformed entity. But a vector in 3-dimensional Euclidean space, for example, requires three quantities to fully specify it, and it follows that any coordinate transformation applied to a vector also requires three quanities to specify it. Let's write v'_i = ∑a_ijv_j where i,j = 1,...3. Note that I'm rather sloppily saying vector transformation when strictly I should be saying transformation of vector components, as it's the coordinates that are really transforming.

But there lot'sof physically important objects that don't behave like either of these under coordinate transformation. So let's invent a set of objects (still in 3-space for our example) that transform according to 3ⁿ quantities in the following way

T'_ij = ∑_k∑_la_jka_ilT_kl.

That's a tensor. This one here is called a rank two tensor, because here transformation depends on 3ⁿ quantities with n = 2.

By this definition it should be clear, as RDT2 said, a vector is a rank 1 tensor and a scalar is rank zero.

This is already too long and very probably patronising - sorry aboout that. It gets better later on!

Am I boring you? Let's see....

The eagle-eyed among you have noticed something strange aboout the tensor transformation I gave earlier, namely

T'_ij = ∑_k∑_la_jka_ilT_kl.

But before going into that, take a look at the equation. It has, as part of its instructions "sum over the indices k and l". Notice that these indices appear twice in the RHS. So let's introduce a notational shorthand which says..... if an index appears more that once in an equation like this, we will take summation as implicit, and the summation sign may as well be omitted.

Right. The transformation above is orthogonal (a rotation). Now orthogonal transformations have some rather special properties. The origin remains fixed, as does one axis for each rotation. And, importantly, three successive rotations in 3-space give different results depending on the order in which they are carried out. In other words, orthogonal transformations are non-commutative. We might therefore expect that tensors defined relative to orthogonal transformations to have correspondingly special properties, and as they do we will say

i) they are not typical of tensors in general, and
ii) have a special name. You will find them called Cartesian tensors, affine tensors and orthogonal tensors.

So let's talk about them a little, realising that much of what we say may not carry over to the more general case.

Consider an arbitray vector in 3-space, and rotate the coordinates around the x axis. It is easily seen that,a rotation of the axes "clockwise" is the same as rotating the vector anti-clockwise with the axes fixed. In other words, the vector transforms oppositely to the coordinates. If our vector is a tensor, let's call it a contravariant tensor.

Easy enough. Now let's get silly, for the sake of illustration. Suppose we were to somehow shrink all our axes, the vector, in its new coordinates would be "longer", as would all other vectors. That's fine, as the inner product of two similarly "enlarged" vectors would be the same in both coordinate systems.

But there is an entity, whose existence is required for rather technical reasons, called a dual vector. Let's not worry about him over much, but only to say he is defined as the linear functional f_v such that
f_v(w) = v·w, the latter being the inner product of the vectors v and w.

In order for this equality to hold in our shrunk coordinates (and it must - remember that scalars are invariant), if v is "bigger" the f_v must be "smaller". i.e. it transforms in the same way as the coordinates. And if it's a tensor, we'll call it covariant.

And before somebody pounces, I know it's silly, approximate and controversial, at least with regard to vectors. (Not only that, in point of fact, there is no distinction between co- and contravariant tensors in linear orthogonal transformations. More on that later, perhaps)

Well, Quarkhead, I already knew that (about the Christoffel Symbols). This, is the method that I referred to as a long one. I know it's this way in theory, but can the actual calculation be made easier in particular examples?

Can anyone show (work out) the steps used to calculate the CS taking a particular numerical example (like that of the surface of a sphere)? I want to see whether any shortcuts exist? A direct application of the equation (as in the derivation given by Quarkhead) seemed too long to me (because the book apparently carried this out in one step, and did not explain). It is the CS of the second kind that makes real trouble.

Aer, I'm not having any difficulty with the concept. Especially since I'm also learning GR.

Can anyone show (work out) the steps used to calculate the CS taking a particular numerical example (like that of the surface of a sphere)? Numerical? As in actual numbers? I can barely count my change!

Seriously, I don't really think numbers are going to help here. I want to see whether any shortcuts exist? A direct application of the equation (as in the derivation given by Quarkhead) seemed too long to me (because the book apparently carried this out in one step, and did not explain).I gave the shortest cut I could come up with, and I know it was pretty obscure. I don't think it can be done in one step (but, like you, I am a relative neophyte)
Longer cuts are always better, in the first instance. It is the CS of the second kind that makes real trouble. Why? Having derived Christoffel I, simply do

Γ^j_hk = ∑g^jlΓ_hkl

Anybody want me to continue my monologue? It's interesting stuff, but if you don't care.....

Especially since I'm also learning GR.Here's a very good source.
http://people.hofstra.edu/faculty/Stefan_Waner/diff_geom/tc.html

Choose the PDF option. It's a really good lecture. (Keep you busy for a while!)

Hey, hey! Don't think of it as a monologue. I'm getting here whenever I can. The reason I wanted to use numerical examples is because I wanted to see the formula work out in practice. And the reason the CS II make trouble is that there's a lot more summation involved. Have you ever tried it out? Of course, the derivations are pretty simple, and I've used them elsewhere, as in the covariant derivative. Okay. I guess there is no easy way. Let's continue with other things.

And the reason the CS II make trouble is that there's a lot more summation involved. Naturally, but if you collapse the extra terms to the metric, it's a piece of cake!Have you ever tried it out? Using numbers? No. I don't do numbers. ..as in the covariant derivative. Now that is a slippery customer! I'm pretty sure I've got it nailed, but I'm not sure.

What do you mean by collapsing the extra terms to the metric? I think this is what I've been looking for.
When I said numbers, I meant some actual example, like the sphere or something.

Don't bother posting any derivation of the covariant derivative. I was up last night deriving the equation of a covariant derivative for a covariant tensor of order 'n'. I'll post it if you want.

What do you mean by collapsing the extra terms to the metric? I think this is what I've been looking for. g^mkΓ_nsk = Γ^m_nsis what I meant (by index contraction and multiplication). I gave the derivation of the metric, at your request, a little earlier.

Don't bother posting any derivation of the covariant derivative. I was up last night deriving the equation of a covariant derivative for a covariant tensor of order 'n'. I'll post it if you want.OK, I won't bother, especially since I've had a couple of beers.

But yes, please post your derivation, and I'll get back.

This is such fun!

<B>Covariant Differentiation</B><Br>I'm not sure you're going to read this through. All you can do is work it out yourself and use this as a reference. You won't be able to imagine how I typed this up. And it'll only be harder to read. You're going to have one hell of time. Enjoy. Since we're dealing with a rank 'n' tensor, invlolving many products of partial derivatives, I've used the Π notation to represent products (I did that when I derived it the first time on paper, too). If we continue these exchanges for a while, we might even come up with a 'Multiplication Convention', whereby the Π is dropped for repeated sub-subscripts and sub-superscripts!<Font face= "georgia" style= "font-size: 16px;"><Br>
I've abbreviated the product (<B>∂</B>x<SUP>α<SUB id=s>1</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>1</SUB></SUP>)(<B>∂</B>x<SUP>α<SUB id=s>2</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>2</SUB></SUP>)...(<B>∂</B>x<SUP>α<SUB id=s>n</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>n</SUB></SUP>) to <B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<BR>The rule for transformation of a covariant tensor of rank 'n' from unprimed system (x<SUP>i</SUP>) to primed system (y<SUP>i</SUP>) is:<Br><Br>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB> = A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB><B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<Br><Br>Differentiate w.r.t y<SUP>j</SUP>, and expand using the product rule for differentiation, and we obtain:<Br><Br>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) = (<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>)<B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>) + A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>(<B>∂</B><SUP>2</SUP>x<SUP>α<SUB id=s>1</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>1</SUB></SUP><B>∂</B>y<SUP>j</SUP>)<B>Π</B><SUB id=s>θ=2</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>) + A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>(<B>∂</B>x<SUP>α<SUB id=s>1</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>1</SUB></SUP>)(<B>∂</B><SUP>2</SUP>x<SUP>α<SUB id=s>2</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>2</SUB></SUP><B>∂</B>y<SUP>j</SUP>)<B>Π</B><SUB id=s>θ=3</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>) + ... + A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB><B>Π</B><SUB id=s>θ=1</SUB><SUP>n-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B><SUP>2</SUP>x<SUP>α<SUB id=s>n</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>n</SUB></SUP><B>∂</B>y<SUP>j</SUP>)<Br><Br>We can further shorten this using a Σ and write it as:<BR><BR>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) = (<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>)<B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>) + <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB><B>Π</B><SUB id=s>θ=1</SUB><SUP>φ-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B><SUP>2</SUP>x<SUP>α<SUB id=s>φ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>φ</SUB></SUP><B>∂</B>y<SUP>j</SUP>)<B>Π</B><SUB id=s>θ=φ+1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)
<Br>We can get the following expression for the second order partial derivative from the transformation law of the Christoffel symbols:
<Br>(<B>∂</B><SUP>2</SUP>x<SUP>γ</SUP>/<B>∂</B>y<SUP>i</SUP><B>∂</B>y<SUP>j</SUP>)
= <B>Γ<I>'</I></B> <SUP>k</SUP><SUB>ij</SUB>(<B>∂</B>x<SUP>γ</SUP>/<B>∂</B>y<SUP>k</SUP>) - <B>Γ</B> <SUP>γ</SUP><SUB>αβ</SUB>(<B>∂</B>x<SUP>α</SUP>/<B>∂</B>y<SUP>i</SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>)
<Br>Substitute this in place of (<B>∂</B><SUP>2</SUP>x<SUP>α<SUB id=s>φ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>n</SUB></SUP><B>∂</B>y<SUP>j</SUP>) and we have:
<Br>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) = (<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>)<B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>) + <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>[<B>Π</B><SUB id=s>θ=1</SUB><SUP>φ-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<B>Π</B><SUB id=s>θ=φ+1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)][ <B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB>(<B>∂</B>x<SUP>α<SUB id=s>φ</SUB></SUP>/<B>∂</B>y<SUP>k</SUP>) - <B>Γ</B> <SUP>α<SUB id=s>φ</SUB> </SUP><SUB>μν</SUB>(<B>∂</B>x<SUP>μ</SUP>/<B>∂</B>y<SUP>i<SUB id=s>φ</SUB></SUP>)(<B>∂</B>x<SUP>ν</SUP>/<B>∂</B>y<SUP>j</SUP>) ]<BR>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) = (<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>)<B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>) + <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>[<B>Π</B><SUB id=s>θ=1</SUB><SUP>φ-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<B>Π</B><SUB id=s>θ=φ+1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)][ (<B>∂</B>x<SUP>α<SUB id=s>φ</SUB></SUP>/<B>∂</B>y<SUP>k</SUP>)<B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB>] - <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>[<B>Π</B><SUB id=s>θ=1</SUB><SUP>φ-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<B>Π</B><SUB id=s>θ=φ+1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)][ (<B>∂</B>x<SUP>μ</SUP>/<B>∂</B>y<SUP>i<SUB id=s>φ</SUB></SUP>)(<B>∂</B>x<SUP>ν</SUP>/<B>∂</B>y<SUP>j</SUP>) ]<B>Γ</B> <SUP>α<SUB id=s>φ</SUB> </SUP><SUB>μν</SUB>
<Br>The second term in right hand side of the above equation,<BR><B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>[<B>Π</B><SUB id=s>θ=1</SUB><SUP>φ-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<B>Π</B><SUB id=s>θ=φ+1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)][ (<B>∂</B>x<SUP>α<SUB id=s>φ</SUB></SUP>/<B>∂</B>y<SUP>k</SUP>)<B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB>]<BR>can be written in the following form by expanding the Π,<BR><B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>[(<B>∂</B>x<SUP>α<SUB id=s>1</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>1</SUB></SUP>)(<B>∂</B>x<SUP>α<SUB id=s>2</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>2</SUB></SUP>)...<I>(<B>∂</B>x<SUP>α<SUB id=s>φ</SUB></SUP>/<B>∂</B>y<SUP>k</SUP>)</I>...(<B>∂</B>x<SUP>α<SUB id=s>n</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>n</SUB></SUP>)]<B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB><BR>According to the transformation rule, this is nothing other than,<BR><B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...k...i<SUB id=s>n</SUB></SUB><B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB>
<Br>In the third term,<BR><B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>[<B>Π</B><SUB id=s>θ=1</SUB><SUP>φ-1</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)<B>Π</B><SUB id=s>θ=φ+1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)][ (<B>∂</B>x<SUP>μ</SUP>/<B>∂</B>y<SUP>i<SUB id=s>φ</SUB></SUP>)(<B>∂</B>x<SUP>ν</SUP>/<B>∂</B>y<SUP>j</SUP>) ]<B>Γ</B> <SUP>α<SUB id=s>φ</SUB> </SUP><SUB>μν</SUB><BR>α<SUB>φ</SUB>, μ, and ν are dummy indices. I'll replace α<SUB>φ</SUB> by γ, and then, μ by α<SUB>φ</SUB>, and ν by β.<BR><B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...γ...α<SUB id=s>n</SUB></SUB><B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>)<B>Γ</B> <SUP>γ</SUP><SUB>α<SUB>φ</SUB>β</SUB>
<Br>So now our equation has taken the form,<BR>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) = (<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>)<B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>) + <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...k...i<SUB id=s>n</SUB></SUB><B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB> - <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...γ...α<SUB id=s>n</SUB></SUB><B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>)<B>Γ</B> <SUP>γ</SUP><SUB>α<SUB>φ</SUB>β</SUB>
<Br>Transfer the second term in the R.H.S, to the L.H.S<BR>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) - <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...k...i<SUB id=s>n</SUB></SUB><B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB> = (<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>)<B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>) - <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...γ...α<SUB id=s>n</SUB></SUB><B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>)<B>Γ</B> <SUP>γ</SUP><SUB>α<SUB>φ</SUB>β</SUB>
<Br>And the final result is:
<Br>(<B>∂</B>A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB>/<B>∂</B>y<SUP>j</SUP>) - <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A'<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...k...i<SUB id=s>n</SUB></SUB><B>Γ<I>'</I></B> <SUP>k</SUP><SUB>i<SUB id=s>φ</SUB> j</SUB> = <B>(</B>(<B>∂</B>A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...α<SUB id=s>n</SUB></SUB>/<B>∂</B>x<SUP>β</SUP>) - <B>Σ</B><SUB id=s>φ=1</SUB><SUP>n</SUP> A<SUB>α<SUB id=s>1</SUB>α<SUB id=s>2</SUB>...γ...α<SUB id=s>n</SUB></SUB><B>Γ</B> <SUP>γ</SUP><SUB>α<SUB>φ</SUB>β</SUB><B>)</B><B>Π</B><SUB id=s>θ=1</SUB><SUP>n</SUP> (<B>∂</B>x<SUP>α<SUB id=s>θ</SUB></SUP>/<B>∂</B>y<SUP>i<SUB id=s>θ</SUB></SUP>)(<B>∂</B>x<SUP>β</SUP>/<B>∂</B>y<SUP>j</SUP>)
<Br>This is the transformation rule for a covariant tensor of rank 'n+1'. Hence, we can conclude that the quantity on the left is a tensor. The covariant derivative is defined as:
<Br>A<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB>, j</SUB> = (<B>∂</B>A<SUB>i<SUB id=s>1</SUB></SUB><SUB>i<SUB id=s>2</SUB></SUB>...<SUB>i<SUB id=s>n</SUB></SUB>/<B>∂</B>j) - A<SUB><STRONG>k</STRONG>i<SUB id=s>2</SUB>...i<SUB id=s>n</SUB></SUB><B>Γ</B> <SUP>k</SUP><SUB>i<SUB id=s>1</SUB> x<SUP>j</SUP></SUB> - A<SUB>i<SUB id=s>1</SUB><STRONG>k</STRONG>...i<SUB id=s>n</SUB></SUB><B>Γ</B> <SUP>k</SUP><SUB>i<SUB id=s>2</SUB> j</SUB> - ... - A<SUB>i<SUB id=s>1</SUB>i<SUB id=s>2</SUB>...<STRONG>k</STRONG></SUB><B>Γ</B> <SUP>k</SUP><SUB>i<SUB id=s>n</SUB> j</SUB>
</Font><Style> sup {font-size: 14px; font-family: times new roman;} sub {font-size: 14px; font-family: times new roman;} sub #s {font-size: 11px;} sup #s {font-size: 11px;}</Style>

I may soon post a photograph of a written version. The only problem is that my handwriting isn't very good.

This almost impossible to follow, due your truly ghastly notation.

First, why dive straight in with a rank n tensor, 1 or 2 would suffice to extract the principle.

Second, there's really no need for subscripted subscripts, you should find a way round it (most people manage to)

Third, some motivation would be useful, as would some commentary. Otherwise it looks like you're just shuffling symbols for the sake of it.

Third, it looks, on a quick glance, that you have too many terms in your last equation. I'm fairly sure you only need two, although not completely.

If you look back, you'll see that I mentioned posting the derivation for a rank 'n' tensor. This is why there are sub-indices, and this is why there are many terms in the last equation.
What changes do you want me to make to my notation?
And I'll post the derivation for a rank 1 tensor as soon as I can. I thought you had already learnt worked that out.

Too bad, Quarkhead is a busy person, and that no one else knows enough math.

Too bad, Quarkhead is a busy person, and that no one else knows enough math.Yeah, I am (or was that irony?). As I am still struggling with this subject, I have a lot of thinking to do before responding. But I will, I promise.

But, not to confuse threads, look at space and time, I have a preliminary for you there.

Please reply to my post above that one.

Please reply to my post above that one.Now, now, don't get tetchy! I am mega-busy right now, and I hope neither you nor any other member will be affronted when I say that posting here is pretty low on my life-priority list.

But to answer your question, your notation in your posts is your own affair, I can only say what I would do, which is this.

I would define a rank n tensor, using the notation you used. I would then do the calculus using a rank 2 tensor, type (1,1), (0,2) or (2,0) and point out that extension to higher ranks is a (relatively) straightforward affair.

I am in the process of preparing, piecemeal, a rather more gentle walk through the calculus, in the hope of carrying others with us.

OK, I'm going to take this a lot more slowly than Rosnet, mainly because I'm still feeling my way in this area.

We agreed what a tensor is, and introduced, in a very rough and ready way, the concept of co- and contra-variant tensors. Recall that we had restricted ourselves to affine tensors, those that are subject to linear orthogonal transformations. Let's continue to do that for now, and see what we can do with these beasts.

We can add them, provided they are of the same rank. The result is another tensor of the same rank e.g.

A_ij + B_mn = C_pq

We can multiply by a scalar, or strictly speaking multiply each component by a scalar, the result being a tensor of the same rank.

We can multiply two or more tensors on the same or different ranks, the result being a new tensor whose rank is the sum of the original ranks, e.g.

A_iB_jk = C_pqr. When we get to work in curvilinear coordinates, this last operation becomes rather more sophisticated.

Now, it should be apparent that these operations can only be performed "locally", from which we go on to insist that a tensor is always defined at a point P, and that the operations above can only be performed on tensors defined at P. We will say that the set of tensors at P is a vector space over P, with dimension equal to the number r of different components of the tensors comprising it i.e. dim = n^r.

So far, so simple. There is, however, another operation of great importance in tensor calculus, known as contraction. Take any tensor of rank 2 or greater (you'll see why) say A_ijk, and set any two indices equal to each other (that's allowed, as they are what are called "dummy" indices - they are only matrix row/column labels) and sum over indices, so in the above, set i = j and we get

A_jjk = B_k. Thus contraction of a tensor of rank r results in a tensor of rank r - 2. You should easily see that contraction of a rank 2 tensor results in a scalar, sometimes referred to as the "trace" of that tensor.

But, and it is a big but, contraction in this sense necessarily depends upon the orthogonality of the transformation properties of affine tensors. This is hugely important,and deserves a whole post of its own.

I'll try it if there's any interest, otherwise, Ill shut up. Any interest?

I only thought about contractions as an algebraic shortcut to make new tensors from existing ones, such as the Ricci tensor from the Riemann-Christoffel tensor. But, you apparently see some other significance to this. Can you clarify?

Oh, go on, you can do it! Let Aⁱ and B_i be rank 1 tensors, what will you call them? What is AⁱB_i equal to? Ring any bells? It should do! Contraction of higher rank tensors has something of the same flavour.

Sure, the inner product of two vectors is significant. But what good does it do if we sum up something like the squares of the diagonal elements of the metric tensor, say (Contaction of rank 2 tensor). And that too, in cases where the metric tensor, (or any other tensor), isn't diagonalized?

Sure, the inner product of two vectors is significant. But what good does it do if we sum up something like the squares of the diagonal elements of the metric tensor, say (Contaction of rank 2 tensor). And that too, in cases where the metric tensor, (or any other tensor), isn't diagonalized?Not sure I understand what you're getting at. Contraction of a rank 2 tensor is a scalar, as you saw. We can call it an inner product if we like, but it's better to call it the trace. Now the trace of any matrix is the sum of its diagonal entries, whether it's a diagonal matrix or not.

Moreover, the matrix representation of the metric tensor has entries which are the coefficients of the dxⁱdx^j. On the diagonal, these reduce to dxⁱdx^j, with i = j, i.e. contraction.

Maybe it would help if I gave an alternative, but perfectly acceptable defintion of tensor contraction: Contraction involves multiplying a tensor by the Kroenecker delta (I assume you know what that is?) and summinig over repeatd indices. So that, prior to summation, the matrix is diagonal.

But anyway, you're not taking squares of the diagonal elements, as these are coefficients on the squared derivatives. And the trace of a matrix is every bit as meaningful a characteristic as is the inner product of two vectors. In fact they are equivalent.

That was a bit garbled, wasn't it. Let me try again tomorrow.

The inner product is not a contraction of any tensor. Of course, you could make a second rank tensor as the outer (?) product of two vectors, and <I>then</I> contract this tensor, in order to obtain the inner product of the vectors. I don't see any meaning in this, though.

C<Sup>i</Sup><Sub>j</Sub> = A<Sup>i</Sup>B<Sub>j</Sub>

C<Sup>i</Sup><Sub>i</Sub>=A<Sup>i</Sup>B<Sub>i</Sub>

First, any tensor can be decomposed into it's constituent parts. Just as, any time we see a 6 we can think of it as 2 x 3. So that Aⁱ_j can be written BⁱC_j. Thus, if we choose to equate i with j, mutiplication is implicit in the contraction operation.

Morever, if you accept my alternative definition of contraction (multiplication by the Kroenecker delta) and recognise that the Kroenecker is a rank 2 tensor, you see again that mutiplication is implicit in contraction.

As we agreed, the product of two rank 1 tensors with matching indices is a scalar, which can be thought of as the inner product or, better, the trace.

So what's your problem here? I sincerely hope you're not trying to score points of some sort!

No, I wasn't.
I don't see any use in the <I>concept</I> of contraction, except in the case interpreting the inner product of vectors as a contraction of their outer product, in which case, it is quite unnecessary. Okay, leave the subject. There's lots more to worry about. I'll think about this a bit more.

I don't see any use in the <I>concept</I> of contraction, except in the case interpreting the inner product of vectors as a contraction of their outer product, in which case, it is quite unnecessary. But it's not! How else can are you going to introduce the notion of length and angle without the notion of an inner product, and it's higher derivatives? It's absolutely essential.