Tensors

Discussion in 'Physics & Math' started by Rosnet, Jun 28, 2005.

  1. Rosnet Philomorpher Registered Senior Member

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    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?
     
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  3. everneo Re-searcher Registered Senior Member

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    Good thread. I wish lethe contributes more in this thread.
     
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  5. oxymoron Registered Senior Member

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    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.
     
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  7. chroot Crackpot killer Registered Senior Member

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  8. superluminal I am MalcomR Valued Senior Member

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    Very excellent description chroot. Thanks!
     
  9. Rosnet Philomorpher Registered Senior Member

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    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.
     
  10. QuarkHead Remedial Math Student Valued Senior Member

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    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<sub>qr</sub> defined at a point p at coordinate x<sup>p</sup>on a manifold M, with M embedded in ambient space defined with coordinates {y<sup>a</sup>}. Now form the derivatives required for abitrary translation of A<sub>qr</sub>(x<sup>p</sup>) relative to ambient coordinates.
    &part;A<sub>qr</sub>/&part;x<sup>p</sup> = &part;/&part;x<sup>p</sup>[(&part;y<sup>a</sup>/&part;x<sup>q</sup>)(&part;y<sup>a</sup>/&part;x<sup>r</sup>)] = (&part;<sup>2</sup>y<sup>a</sup>/&part;x<sup>p</sup>&part;x<sup>q</sup>)(&part;y<sup>a</sup>/&part;x<sup>r</sup>) + (&part;<sup>2</sup>y<sup>a</sup>/&part;x<sup>r</sup>&part;x<sup>p</sup>)(&part;y<sup>a</sup>/&part;x<sup>q</sup>)

    Holy Mary, I'm going to call that A<sub>qr,p</sub> = y<sub>a,pq</sub>y<sub>a,r</sub> + y<sub>a,rp</sub>y<sub>a,q</sub>

    Now cyclically permute the indices, getting

    A<sub>rp,q</sub> = y<sub>a,qr</sub>y<sub>a,p</sub> + y<sub>a,pq</sub>y<sub>a,r</sub>

    and

    A<sub>pq,r</sub> = y<sub>a,rp</sub>y<sub>a,q</sub> + y<sub>a,qr</sub>y<sub>a,</sub>

    Add the first two and subtract the third, giving

    1/2(A<sub>qr,p</sub> + A<sub>rp,q</sub> - A<sub>pq,r</sub>)

    substitute the metric tensor for A and you have defined {pq,r}, Christoffel I
     
    Last edited: Jul 25, 2005
  11. RDT2 Registered Senior Member

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    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.
     
    Last edited: Jul 26, 2005
  12. QuarkHead Remedial Math Student Valued Senior Member

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    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!
     
  13. Lucas Registered Senior Member

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    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
     
  14. Aer Registered Senior Member

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    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.
     
  15. QuarkHead Remedial Math Student Valued Senior Member

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    Did he really? Then he's wrong, of course. The Christoffel symbols have tensor components, but are not themselves tensors.
     
  16. QuarkHead Remedial Math Student Valued Senior Member

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    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'<sub>i</sub> = &sum;a<sub>ij</sub>v<sub>j</sub> 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 3<sup>n</sup> quantities in the following way

    T'<sub>ij</sub> = &sum;<sub>k</sub>&sum;<sub>l</sub>a<sub>jk</sub>a<sub>il</sub>T<sub>kl</sub>.

    That's a tensor. This one here is called a rank two tensor, because here transformation depends on 3<sup>n</sup> 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!
     
  17. QuarkHead Remedial Math Student Valued Senior Member

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    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'<sub>ij</sub> = &sum;<sub>k</sub>&sum;<sub>l</sub>a<sub>jk</sub>a<sub>il</sub>T<sub>kl</sub>.

    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<sub>v</sub> such that
    f<sub>v</sub>(w) = v&middot;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<sub>v</sub> 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)
     
    Last edited: Jul 31, 2005
  18. Rosnet Philomorpher Registered Senior Member

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    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?
     
  19. Rosnet Philomorpher Registered Senior Member

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    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.
     
  20. QuarkHead Remedial Math Student Valued Senior Member

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    Numerical? As in actual numbers? I can barely count my change!

    Seriously, I don't really think numbers are going to help here.
    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.
    Why? Having derived Christoffel I, simply do

    &Gamma;<sup>j</sup><sub>hk</sub> = &sum;g<sup>jl</sup>&Gamma;<sub>hkl</sub>
     
    Last edited: Jul 29, 2005
  21. QuarkHead Remedial Math Student Valued Senior Member

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    Anybody want me to continue my monologue? It's interesting stuff, but if you don't care.....
     
  22. QuarkHead Remedial Math Student Valued Senior Member

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  23. Rosnet Philomorpher Registered Senior Member

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    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.
     

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