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Hello all my tensor questions

  1. Jun 1, 2007 #1
    Must a physical law be able to expressed by tensor equations (forms)? If there is a quantity like finite rotation, which cannot be expressed by any tensor, then there is no corresponding tensor equations for it? What is the limitation of tensor analysis then?
  2. jcsd
  3. Jun 1, 2007 #2
    tensor component

    There seems confusion between tensor and tensor components often. Let's admit both of them can express the total entity of a tensor well and correctly.But how do we know tensor components before knowing it as a tensor. This problem degenerate to a simpler question " how do we know vector components before knowing it as a vector?"

    In tensor analysis, a general definition of tensor is something like a group of functions (or components) under different coordinate systems which transform according to some specified transformation rules. But how do we know the components are just tensor components? How do we know the vector components in the new coordinate systems are just tensor components and NOT PHYSICAL COMPONENTS (in general coordinate systems, physical components do not transform like tensor components do)?

    Therefore, how do know a quantity is a tensor according to the subject in tensor analysis?
  4. Jun 1, 2007 #3
    tensor component and tensor

    If we know a group of functions transform as a tensor's compoents do, under two different coordinate systems. Then in tensor analysis, we say the group of functions make a tensor.

    But we only checked in two coordinate systems, the group of functions transform like a tensor, we really also need to confirm that in all other different coordinate systems, the group of functions transform like a tensor.

    Is there a good answer to this? More or less, "group" theory may help with this. Could somebody confirm and give a clear explanation related to this?

    Even if the group theory has sorted out the above question (let's say so first), then another question arises. If a quantity can be described by a Cartesian tensor (defined in the Cartesian coordinate systems), must it also be a general tensor (defined in the general coordinate systems)? As is well known, a Cartesian tensor (partial derivative of a vector, say) is not generally a tensor in general coordinate system.
  5. Jun 1, 2007 #4
    You have to have a definition of the quantity in two arbitrary coordinate systems. Then you check if the components trasform like components of a tensor.

    There are quantities that are not tensors like partial derivatives of a vector field with respect to a give coordinate system and the Christoffel symbols in a given coordinate system. They are combined to form a tensorial quantity, the covariant derivative, that has the tensorial transformation law.
  6. Jun 1, 2007 #5
    scalar and invariant

    In my textbooks on soild mechanics, scalar and invariant seem to mean the same thing.

    However, I am not sure about whether there is difference between these two. My worry is as follow,

    We can say a vector is an invariant since the total entity of a vector does not change under change of coordinate systems? but it is not a scalar.
  7. Jun 1, 2007 #6

    Chris Hillman

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    Hi, uiulic,

    I think you might be reading the wrong books. One thing you need to understand is that a tensor is an algebraic object, a generalization of "linear operator". Indeed, a tensor is nothing more or less than a multilinear map taking a bunch of vectors and/or covectors to the field--- usually the field of real numbers--- of the vector space where these vectors and covectors live). As such, they have many properties similar to the properties of linear mappings taking pairs of vectors to the reals. If you don't fill in all the "slots", you get multilinear maps producing vectors and/or covectors instead of real numbers.

    Once you understand tensors as multilinear maps, and understand the basic features of tensor algebra (including decompositions, behavior under change of vector space basis, and invariants of tensors), then you can study tensor fields, which are constructed much the way that vector fields are constructed. We construct fiber bundles called tensor bundles over our manifold M, such that global sections define tensor fields on M. Many algebraic properties of tensors carry over to the realm of tensor fields with little change (including decompositions, behavior under change of frame field, and invariants), but now we have various notions of differentiation to play with, including Lie derivatives and (if a connection has been defined on M) covariant derivatives, and (if our tensor field is antisymmetric and thus an exterior form), exterior derivative.

    Try Frankel, The Geometry of Physics.
  8. Jun 1, 2007 #7
    An invariant is something that is not transformed under a transformation of the background coordinates. A vector is transformed, so is not invariant. A scalar is not transformed.
  9. Jun 1, 2007 #8
    linear vector function

    As in https://www.physicsforums.com/showthread.php?t=171462, a linear vector function is sometimes called a second-order tensor, at least accepted by continuum mechanics people.

    The argument and output of this function must be both vector in order that the function connecting them is a second-order tensor. The question is that

    this argument vector and the output vector are in the same vector space?

    Maybe the answer of this question will tell the difference between the tensors I use and you use.
  10. Jun 1, 2007 #9
    When we say that something transforms like a tensor, we generally test it under a general transformation, thus not limiting ourselves to a single pair of coordinate systems.
  11. Jun 1, 2007 #10
    if not linear transformation

    Tensor is connected with linear transformation. It seems as follows,

    Multilinear mapping between a vector space and its dual space (some people seem give dual space a different name).

    If this mapping is not linear, then the transformation will not be called a tensor. Then there is something else that tensor cannot describe. This will relate to my previous questions.

    Why linear mapping is so important? Mathematicians care about this linear mapping too? I think at least non-linear mapping is a general one and should be the interest more of mathematicians.
  12. Jun 1, 2007 #11
    I have asked so many naive questions on tensor, because I am really a new student on this and my work is not much related to tensor at all. Therefore, I, as an engineering student, hope you can answer the small (in your eyes) questions with patience. I appreciate all your attention. Many other engineering students may also benefict here.
  13. Jun 1, 2007 #12

    Chris Hillman

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    Invariant vs. scalar?

    Ditto MaWM, but I'd add this:

    The word invariant typically refers (in mathematics and applied mathematics) to a quantity (often a real or complex number) associated with a geometric object such as a linear operator or tensor, which does not change under specified coordinate transformations. (In elasticity, which I think you mentioned, BTW, you will encounter objects which are only "tensorial" wrt affine transformations, not general coordinate transformations. Elsewhere in physics you encounter objects which are "vectorial" wrt proper transformations, but not improper or determinant reversing ones.)

    Something to be aware of is the distinction between tensors and tensor fields.

    Let me first say something about invariants of tensors and linear operators. Typical examples of invariants in linear algebra include: the characteristic polynomial of a linear operator and its coefficients including trace and determinant, the minimal polynomial of a linear operator, the eigenvalues and eigen spaces of a linear operator. As you can see, some of these are numbers (those can be called "scalar invariants"), some are polynomials, some are vector subspaces! What they have in common is that they characterize the operator in ways which are independent of choice of basis. In contrast, "the" matrix of a linear operator is highly basis dependent! But one way to understand the invariance of trace, say, is that applying the "change of basis" transformation to a matrix doesn't change its trace. Tensors are just multilinear mappings and they too have invariants in this sense. In particular, the components of a tensor are highly basis dependent, and you know the change of basis transformation. But you can form various "traces" by "contraction" of the tensor and products of tensors, and these will be invariants. For example, [itex]T^{abcd}[/itex] has scalar invariants including [itex]{T^{mn}}_{mn}, \; {T^{mn}}_{jk} \, {T^{jk}}_{mn}[/itex]. Invariant theory is closely related to representation theory, btw--- both are very important in physics. A good way to see this is to simultaneously study invariants and representations of finite groups--- two lovely and closely intertwined theories!

    You form tensor fields by bundling tensors, and this automatically leads to notions of invariants for tensor fields. For example, if [itex]T^{abcd}[/itex] is now regarded as a tensor field on M, then [itex]{T^{mn}}_{mn}, \; {T^{mn}}_{jk} \, {T^{jk}}_{mn}[/itex] and so on become functions on M. If various such scalar invariants (functions!) have algebraic relationships such as satisfying certain identities, this property will not only be invariant but can be used to confirm in coordinate free fashion that [itex]T^{abcd}[/itex] belongs to some class of tensor fields. Pursuing this idea for exterior forms (which are completely antisymmetric tensors) led Cartan to a classification procedure for manifolds using frame fields. This has been implemented for Lorentzian manifolds and leads to a procedure which computes invariant relationships satisfied by certain adapted frames, which in favorable cases completely characterize a manifold up to local isometry. (In practice, unfortunately, this Karlhede algorithm can be hard to carry out.)
    Last edited: Jun 1, 2007
  14. Jun 2, 2007 #13


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    As requested, I have merged the threads with your questions.
  15. Jun 2, 2007 #14
    Thanks very much, Hurkyl and Chris.
  16. Jun 2, 2007 #15

    Chris Hillman

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    Hope that wasn't a lot of work!

    Thanks, Hurkyl!
  17. Jun 2, 2007 #16

    Chris Hillman

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    Uiulic's many many questions...

    OK, now that Hurkyl has consolidated the thread, I'll try to address uiulic's other questions.

    Well, there's no physical law stating that! And there's not a mathematical theorem stating that either, unless you add lotsa conditions. What you might see is an informal (or even formal) principle to the effect that in classical field theories (relativistic or otherwise), the fundamental physical laws are best stated in terms of tensors. This is because

    1. physical laws typically are PDEs with boundary conditions,

    2. tensor equations hold true in any coordinate system, and in setting up a boundary value problem it generally helps to use coordinates adapted to the boundary!

    Well, never say never. A rotation by [itex]\pi/3[/itex] is just a linear operator and thus a special case of a tensor. The theory of linear representations of finite groups, which I mentioned as being closely intertwined with the theory of invariants of finite groups, is much concerned with such "discrete symmetries"!

    This reads like a philosophal question, but I'll guess your real puzzlement concerns how the mathematical definitions work. For that I refer you to any good modern textbook on manifold theory.

    You need to read a modern textbook, I think. The definition is that a tensor is a multilinear map. And you said "tensor" when you meant "tensor field". A tensor field is constructed from tensors by taking a section of the appropriate tensor bundle.

    Again, you can't understand your confusion until you have mastered the levels of structure. Beginning with tensors (multilinear maps), which are generalizations of linear operators/bilinear forms and which obey very similar algebraic laws, have invariants, and so on, versus tensor fields. Then you need to understand coordinate bases versus frame fields. You can't do that unless you have mastered linear algebra ("abstract vector spaces" and linear operators, vector space bases, bilinear forms).

    No, that's badly garbled.

    Huh? This is clearly wrong, but I can't make sufficient sense of it to understand the nature of the confusion.

    I don't think anyone can "confirm" your guess because your questions mostly don't make sense to us yet. It is true that the theory of finite groups and also the theory of Lie groups and Lie algebras is intimately related to representation theory and invariant theory, but apart from the issue of "tensorial under group G", I don't think group theory is what you want here. When you say "group of components", you are using the term in a way which is quite different from the technical meaning of "group" in mathematics.

    Gee, I wish you hadn't jumped to that conclusion, because its most likely wrong.

    No, the implication goes the other way. And I have been trying to tell you that in the term "cartesian tensor", the word "Cartesian" should not suggest cartesian coordinate charts on euclidean space but rather affine transformations on euclidean space viz. general diffeomorphisms. A cartesian tensor can be represented in other coordinate charts.

    That's not a good example.

    As far as I can, you were the only one who used that odd-sounding phrase (is German your native tongue, by any chance?), and I already asked once that you define it because it is not a standard term in the English language literature. The obvious guess is that you mean a linear function which in index gymnastics notation could be written [itex]V^a \mapsto {T^a}_b \, V^b[/itex], aka a linear operator. Indeed, a linear operator can be viewed as second rank tensor whose two slots accept a vector and covector respectively. Notice that these are vectors, covectors, and tensors, not vector fields and so on.

    The definitions I am using are perfectly standard. If you don't understand this, the most likely cause is that you are studying books published before 1940 (or perhaps, more recent engineering books which avoid matrices, much less tensors). If you have a Dover book, note that the relevant date is the original publication date, which is typically many decades before the date of the Dover reprint.

    Who told you that one tests tensorial property with respect to a pair of coordinate charts? Thats wrong. First, tensors are algebraic objects, so vector space bases are relevant, but coordinate charts makes no sense in the context of the vector space structure.

    Right, tensors are multilinear mappings, so very special functions. Tensor fields are obtained by bundling tensors, so again very special objects.

    Because linearity greatly simplifies all kinds of things!

    It would be an exxajeration, but not such a huge one, to say that progress in mathematics has most often come when someone succeeds in linearizing some problem. In particular, the huge success of Lie theory derives from the fact that Lie algebras are linear objects, and thus much easier to work with that Lie groups.

    I like to define mathematics as the art of reasoning about simple phenomena without getting confused. Mathematicians spend a lot of time trying to figure out how to extract from complex situations some simple theoretical setup. Then they often spend further time thinking about how to treat that simpler setup using linearized equations.
    Last edited: Jun 2, 2007
  18. Jun 3, 2007 #17
    Would it be acceptable for me to also put up a question (this can be split off if felt more appropriate)?

    An unsymmetric tensor (call it nabla a, where a is a vector) contained within a system partial differential equation is found to have complex eigenvalues.

    In the original tensor pde form, this tensor is found as :

    a . (nabla a) (where a = a vector, . = dot product, nabla a = tensor product)

    Since the solution field is considered to be only Real, & the eigenvalues of (nabla a) can become complex, how does the system 'unwind' itself from this situation, to produce an only Real solution?
  19. Jun 3, 2007 #18
    after Chris's wonderful replies

    Dear Chris,

    It means much more than the answers to my questions themselves.

    My past confusion on tensor is mainly because of my mixing tensor and tensor field, which is also why I had so many problems.

    Some people give the definition of tensor as some linear operator, while others define it as a set of quantities transforming in accordance with some specific laws under change of coordinate systems. To my current understanding, the latter definition is related to tensor field because the set of quantities are often (directly or indirectly) picked as functions of the field variables (the arguments) since we are more interested in the evolution (change) of the set of quantities in that referenced field (actually a kind of subspace related to our physical space as often in fluid mechanics, or a physical body as often in solid mechanics, which we are concerned about).

    Engineering people (including me) fear the words like bundle, isomorphism, and vector space etc, although these are good for understanding the meaning (geometrical, analytical, and physical etc) of tensor in a modern generalized sense. You can even say engineers fear the words tensor. Your explanations of the subject can release our pain on tensor to a large extent. What tensor can offer us? Your answers imply that tensor is introduced more because of convenience (not only for compactness of the equations) for analysizing problems. After you, a physical law is not necessarily expressed by tensor equations, which I now accept. But a physical law (equation) should not change when our reference frame (system) changes, e.g. when the coordinates change (some people use the words a homogeneous and isotropic reference frame related), or else the equation is only fortuitous and accidental; and allow me to call this “invariance”. However, if a physical quantity can be endowed a tensorial character (sometimes by our effort), then the “invariance” can be easily confirmed. If not (because of inability or impossibility), the physical law is still there(and inability is only because of limited experience in a sense).

    You also give us a good answer to “ what is mathematics”, which also implies how mathematics as a tool can be applied to physics and engineering. Of course, tensor a tool too.The physical world does not change with mathematics, but on the contrary mathematics helps us understand the world better.

    If you disagree or have more comments, please tell us.

    [I am not German, but I know my standard of English is poor although I have been studying in England for three years.]

    Thank you very much!
  20. Jun 3, 2007 #19

    Chris Hillman

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    You'd have to give more information. If that pde occurs in linear elasticity, I bet you forgot to symmetrize the gradient of your vector!
  21. Jun 3, 2007 #20

    Chris Hillman

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    Promoting better knowledge of usefule math among engineers

    Yes, I know. This isn't the first time I've tried to promote greater use of manifold theory, exterior calculus, and differential geometry within engineering. Many engineering professors seem to prefer to stick with the methods they know. For example, many engineering textbooks on strengths of materials avoid using tensors when they discuss the theory of elasticity, which (not surprisingly) I feel is not doing their students any favors. But the best engineering schools, e.g. Cornell, have been trying for decades to teach topics such as exterior calculus to their students.

    We (mathematicians) aim to please :wink:
  22. Jun 3, 2007 #21
    Thanks for your input Chris.

    Let's add a time-variant term up front & set the right-side to some arbitrary function f(x,y,z,t). {All derivatives are partials. a=a(x,y,z,t)}

    da/dt + a.(nabla a)=f(x,y,z,t)

    This is a fairly common pde form in physical applications. The tensor is most definitely unsymmetric in this case.
    Last edited: Jun 3, 2007
  23. Jun 5, 2007 #22

    Chris Hillman

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    I still don't see what's puzzling you. I take it you are working with a Cartesian chart? If so, it probably helps to recognize that
    [tex]\vec{X} \cdot \nabla \vec{X} = \nabla_{\vec{X}} \vec{X}}[/tex]
    is a covariant derivative, in fact a path curvature. Does the situation you have in mind involve convective differentiation?
    Last edited: Jun 5, 2007
  24. Jun 6, 2007 #23
    If the principal (eigen-) values of the [tex]\nabla \vec{X}[/tex] tensor, in

    [tex]\vec{X} \cdot \nabla \vec{X} = \nabla_{\vec{X}} \vec{X}}[/tex]

    become complex, and [tex]\vec{X}[/tex] has to be real in a cartesian frame, what can we make of this? Where do the complex parts go to in order for only a real solution to remain - remembering that the principal values are generally max-intermediate-min for the tensor values?

    Think of the stress tensor, with max-intermediate-min stresses shown in the principal (eigenvalue) directions (eigenvectors). What happens with a non-symmetric tensor is the possibility of complex principal values. If [tex]\vec{X}[/tex] is now dotted with the [tex]\nabla \vec{X}[/tex], a few complex terms should be present at principal conditions, no?
    Last edited: Jun 6, 2007
  25. Jun 6, 2007 #24

    Chris Hillman

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    I still don't see what's troubling you, but assuming you already know all about real versus complex in the context of eigenthings for linear operators, the textbook by Barret O'Neill, Semi-riemannian geometry, should help. Eigenthings work a bit differently in Lorentzian and Riemannian manifolds.
  26. Jun 6, 2007 #25
    ^ Thanks for the reference, Chris. I love the word 'eigenthings'... :rofl:
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