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Why tensor analysis ?

  1. Apr 5, 2006 #1
    why tensor analysis ???

    Hi

    Im studying tensor analysis this course but i dont know why we study it ? is it helpfull ? what is the history 4 it ? what tensor mean ???

    :confused: :confused: :confused:


    i hope i can find answer !
     
  2. jcsd
  3. Apr 5, 2006 #2

    robphy

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    What course are you referring to?
    (A related question you might ask yourself: Why vector analysis?)
     
  4. Apr 5, 2006 #3
    It's pretty much the language of relativity. Vectors and matrices are special forms of tensors and they are hugely useful, since vectors are how we describe physical things like force and acceleration and matrices are the mainstay of linear algebra, which covers just about everything in some way or another.
     
  5. Apr 5, 2006 #4

    robphy

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    There's also rigid-body dynamics, fluid mechanics, electrodynamics, (because the real world has anisotropies)...
    ... of course, there's also [multi]linear algebra and differential geometry.
     
  6. Apr 6, 2006 #5

    dextercioby

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    Is this a course for mathematics students ? It would then be weird, cause i'm sure a rigorous course in differential geometry would account very well any "tensor analysis course".

    Daniel.
     
  7. Apr 6, 2006 #6
    ...solid mechanics, stress analysis, thermodynamics...

    Almost all of engineering, in fact.
     
  8. Apr 7, 2006 #7
    thanks everybody

    i think the picture is better now :)

    and this course is optional course for mathematics students in my university
     
  9. Apr 9, 2006 #8
    In GR the setting is very much different from your regular flat Euclidean plane. In fact, we may want to change the setting (by possibly changing the coordinate system, or the metric.) to suit our various needs.

    What good would a formula be if it changed completely under one of these changes? For example, some differential equations can easily be solved in one variable but almost impossible in another.

    We dont want this, we want all our formulas to be the same, no matter what coordinate system we are in. Luckily for us, we can write all our equations using tensors, then under any change they remain the same.

    So basically, tensors are mathematical objects which are invariant under a change of coordinates, therefore very useful when what you are doing is a lot of coordinate changing.
     
  10. Apr 9, 2006 #9
    The components of tensors also transform in particularly simple ways under co-ordinate transformations, which also helps.
     
  11. Apr 14, 2006 #10

    mathwonk

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    you might as well ask, "why multiplication? or to quote bill cosby, "why is there air?"
     
  12. Apr 14, 2006 #11

    HallsofIvy

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    To fill basketballs, of course! (oh, my God, are we showing our age!)


    The crucial property of matrices is that they transform "homogeneously". That is, the components of a tensor in one coordinate system are a sum of products of numbers times the components in any other coordinate system.

    In particular, if we state an equation in terms of tensors- that is, A= B, then A- B= 0 so in any other coordinate system we have A'- B'= 0 or A'= B'
    That is, if an equation, written in terms of tensors, is true in one coordinate system, then it is true in any coordinate system. Since physical laws are neccessarily independent of the coordinate system (which is "man made"), tensors are the natural form in which to write equation representing natural laws.
     
  13. Apr 19, 2006 #12
    Since it sounds like sara_math feels her question is answered, may I ask a question?

    The components of the tensor change when you change coordinate systems. So what is meant by the object is invarient?

    Is there another way of looking at it?

    Hmmm...
    I\'m actually very confused on this at the moment.
    In the Special and General Relativity section of this forum, someone found this paper:
    D.R. Gagnon et al.
    Guided-wave measurement of the one-way speed of light
    Physical Review 38A(4), 1767 (1988)

    The authors take Maxwell\'s equations in tensor form, transform into some non-inertial frame (ie non-lorentz transformation), show what Maxwell\'s equations look like in this coordinate system (by looking at the components of the tensors), and then somehow claim they experimentally prove this formulation is incorrect. (And then claim they can therefore declare one-way velocities of light a coordinate system independent value... because they have \"experimentally disproved\" non-lorentz transformations.)

    I agree, it seems like total nonsense. But it was published in a respected peer review journal .. and hasn\'t been retracted or refuted. Strange.

    It seems like the authors took the well respected special relativity and totally twisted it to mean we can\'t have non-inertial frames or something.

    I\'ll just continue watching the discussion and see how it turns out. But I am correct in thinking this is total nonsense, right? Tensor equations work in non-inertial frames as well, right?
     
    Last edited: Apr 19, 2006
  14. Apr 19, 2006 #13
    A tensor is not defined by it's components. A tensor is defined as a linear functional on a vector space and is independant of the coordinate system, i.e. that it is a multi-linear map [tex]\mathbf{T}:V\times \cdots\times V\times V^*\times\cdots\times V^*\rightarrow \mathbb{R}[/tex], i.e. a map that takes N members of a vector space and M members of the dual of this vector space into the reals and is linear in each of its arguments, [tex]\mathbf{T}(a\vec{V}+b\vec{W})=a\mathbf{T}(\vec{V})+b\mathbf{T}(\vec{W})[/tex].

    When one says a tensor is invariant one is not referring to the numbers that are its components one is referring to the invariance of the value of this mapping. This mapping is invariant under coordinate transformations as these transformations are such that they preserve inner product, i.e. contractions between covariant and contravariant indices. For example, the inner product is a tensor (the metric) the value of which upon two vector fields is a coordinate invariant quantity, though the components of the metric itself differ according to the coordinate system.
     
  15. Apr 19, 2006 #14

    Hurkyl

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    The idea is essentially the same as this:

    Suppose you have a line in the Euclidean plane.

    Well, in order to study it, let's say you choose a point in the plane to be the origin, and draw in some coordinate axes. Now that you've done all of this, you can write the equation for the line.

    Well, maybe you changed your mind -- you actually wanted some other point to be the origin, and the coordinate axes to be something else as well. Now that you've made all those changes you can write the equation for the line.

    The two equations will obviously be different -- but we're always talking about the same line.
     
  16. Apr 19, 2006 #15

    HallsofIvy

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    The components of a tensor change, under change of coordinates, "homogeneously"- that is, the new components are sums of numbers (typically derivatives) times the old components. Specifically, that means that if the components are all 0 in one coordinate system then they are all 0 in every coordinate system.

    Suppose A= B is a correct equation in some coordinate system (where A and B are tensors). That's saying that A- B= 0. Transforming into a new coordinate system (A- B)'= 0 (' indicating the components of the tensor in the new coordinate system). But (A-B)'= A'- B'= 0 so A'= B' in the new coordinate system. That is, if an equation, in tensors, is true in one coordinate system then it is true in every coordinate system.

    Since coordinate system are "man-made", not natural things, any physical law must be independent of the coordinate system- thus tensor equations are the natural way to represent them.
     
  17. Apr 20, 2006 #16
    Okay, I think I see. But let me try repeating this in my own words to make sure.

    Perturbation,
    You are saying that a tensor is actually a map on a vector space -> reals.
    We may change our basis for representation, but the vector is still the same ... and similarly the map is still the same (the tensor is still the same). Correct?

    HallsofIvy,
    Unfortunately, I am not understanding your explanation about components of a tensor changing \"homogeneously\". For instance, what if the change in coordinates only involves a translation of the origin. That is not a homogenous change, right? (for instance the moment of inertia tensor for a collection of particles would not change homogeneously here, right?) Maybe I am still misunderstanding.


    Also, I am having trouble with this:
    A tensor is a linear functional on a vector space. This helps me understand its invarience under coordinate transformations. However, if I perform a non-linear coordinate transformation, the fact that the tensor is a _linear_ functional on the vector space confuses me about what it looks like in this new coordinate system ... it seems to suggest that vectors can no longer be thought of as displacements from one coordinate point to another. So what is a vector now? Maybe I have always been thinking of vectors wrong?


    And as for my other question:
    [regarding the Gagnon et al. publication] But I am correct in thinking this is total nonsense, right? Tensor equations work in non-inertial frames as well, right?

    The answer is: yes, that publication made some serious mistakes for the tensor equations are indeed still true in non-inertial frames. Correct?

    Sorry for all the questions. You guys are great though. Thank you very much.
     
    Last edited: Apr 20, 2006
  18. Apr 20, 2006 #17
    Yep. We have the components of some tensor [itex]R^a_b[/itex] and if I contract it with a covariant and a contravariant vector I'll be mapping my tensor into the reals [itex]V^b\omega_aR^a_b\in\mathbb{R}[/itex], corresponding to [itex]\mathbf{R}(\vec{V}; \tilde{\omega })[/itex]. Under a linear coordinate transformation [itex]\Lambda[/itex] we have

    [tex]\Lambda_b^{b'}\Lambda^a_{a'}V^b\omega_a\Lambda_a^{a'}\Lambda^b_{b'}R^a_b=V^b\omega_aR^a_b[/tex]

    As [tex]\Lambda_b^{b'}\Lambda^b_{b'}=1[/tex]. So the value of the tensor is conserved under a linear transformation of the coordinates.

    A non-linear coordinate transformation isn't a very nice transformation. Coordinate transformation are intended to preserve the properties of geometric objects, a non-linear transform won't do this. The point about a linear transformation is that it expresses the basis vectors of the new system as linear combinations of the basis vectors of the old system, which obviously makes sense geometrically because it's vector addition. Tensors are defined by their invariance under these coordinate transformations.


    Translation isn't a general coordinate transformation. A coordinate transformation merely changes how one measures position relative to a fixed origin (e.g. Cartesian into spherical [itex](x, y, z)\mapsto (r, \theta , \phi )[/itex]).

    Yes, otherwise tensors would be useless in general relativity, where lorentzian frames only exist locally, where the local physics can be extended to global physics. However, the covariant form of Maxwell's equation are defined for Minkowski space-time, where there is a notion of global inertial reference frames. As for mistakes I can't really comment because I've not read the paper, nor can I be bothered to read it.
     
    Last edited: Apr 20, 2006
  19. Apr 21, 2006 #18
    Thank you very much for you explanations. This is all making so much more sense when viewing Tensors as a mapping/functional.

    Hmm... but the whole thing that got me interested in this is that Einsteins field equations for GR are tensor equations. To look at an accelerated frame, don\'t I need to do a non-linear transformation? Or am I looking at this wrong again?

    Why isn\'t translation a general coordinate transformation? It is incredibly useful in physics as translation invarience leads to energy and momentum conservation. And rotation (kind of like a translation, since its just adding a constant to the angle coordinates) invarience leads to conservation of angular momentum.

    Maybe being too stuck in the physics is clouding my view of the math.

    Don\'t worry about it. Trust me, it is not worth the time to read it.

    The fact that we are human means that some incorrect papers are bound to be published. But this is the first one I have ran across personally, so it was kind of a shock. Oh well. Bound to happen eventually. At least the number of good papers I have read greatly dwarfs the number of incorrect ones.
     
  20. Apr 21, 2006 #19
    Yeah, ignore me there, you're right. I got confused between a linear coordinate transformation and a linear transformation on a vector space:eek:. The former being a linear map between coordinates, and the latter a linear combination of basis elements the coefficients given by the elements of the Jacobian, so it doesn't matter whether the coordinate transformation is linear or not for the latter. Replace where I've said "linear coordinate transformation" with "linear transformation". This does of course render my saying a translation isn't a linear transformation void, but everything else I've said, with the replacement I just made, I believe is correct.

    Gah, Dave, you're a fool. I'd cover my shame by editting my posts but it's too late to edit them :(
     
    Last edited: Apr 21, 2006
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