Notices
Results 1 to 38 of 38
Like Tree2Likes
  • 1 Post By Markus Hanke
  • 1 Post By wallaby

Thread: Tensors

  1. #1 Tensors 
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    hey guys, recently when i was hearing lectures on gravity by Witten and Susskind, i head the term Tensors alot. Now i tried to look up these tensors, and i cant seem to drag out a simple understanding( well obviously i cant expect it to be simple).. e.g Ricci tensor, Einstein tensor etc...

    Does anyone know what a tensor is???
    thank you for all helpful or susceptible replies.


    Reply With Quote  
     

  2.  
     

  3. #2  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    Well, a tensor is simply an element in the vector space of all multilinear maps that send elements in its dual space to the real numbers.

    Any wiser? Probably not - you would need to know what is meant by a vector space, by a multilinear map, by a dual vector space and also by a double-dual space.

    I suspect you are trying to run before you walk, but I could easily be wrong.

    Say what, I probably have somewhere on my hard drive a little "tutorial" I made for another forum. If I can find it, I might post it. But - and moderator please note - this would technically count as cross-posting. Please advise, by PM if needed.


    Reply With Quote  
     

  4. #3  
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    A tensor is a general term for quantities with a dimension. A single number is a scalar (0 dimensions), a vector is a quantity with a direction (1 dimension) and can be seen as a tensor of order one. A matrix is usually a 2 dimensional entity, i.e. a tensor of order 2 and so forth.

    More on this here: http://en.wikipedia.org/wiki/Tensor

    All I know is that there is an Einstein notation for calculating with tensors. This is used, when you add up individual elements of such a tensor.
    Reply With Quote  
     

  5. #4 Re: Tensors 
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by Heinsbergrelatz
    hey guys, recently when i was hearing lectures on gravity by Witten and Susskind, i head the term Tensors alot. Now i tried to look up these tensors, and i cant seem to drag out a simple understanding( well obviously i cant expect it to be simple).. e.g Ricci tensor, Einstein tensor etc...

    Does anyone know what a tensor is???
    thank you for all helpful or susceptible replies.
    Yes, lots of people know what tensor are.

    But there is not a simple explanation.

    And you ought to be aware that what a physicist means by a tensor is what a mathematician calls a tensor field.

    You can find treatments of tensor fields in texts on differential geometry. But be aware that differential geometry is relatively sophisticated mathematics, well beyond calculus for instance.

    The particular tensors that you have listed are variations on the theme of curvature tensors, and there is no simple explanation.
    Reply With Quote  
     

  6. #5  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    i see.. as expected no general definition or explanation could simply be provided for this, but i appreciate the glimpse of definition given from Dishmaster and Guitarist. and you are right Dr.Rocket, i was reading stuffs on tensors and things really start to get out of hand from the whole defining a tensor field in space....

    this is a question out of this topic, but is there a real professor, as in professor's who give lectures in University who uses this forum. not as in the Forum professor, but real teachers and professors from schools, and uni.?????
    :? :?
    Reply With Quote  
     

  7. #6  
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by Heinsbergrelatz
    i see.. as expected no general definition or explanation could simply be provided for this, but i appreciate the glimpse of definition given from Dishmaster and Guitarist. and you are right Dr.Rocket, i was reading stuffs on tensors and things really start to get out of hand from the whole defining a tensor field in space....

    this is a question out of this topic, but is there a real professor, as in professor's who give lectures in University who uses this forum. not as in the Forum professor, but real teachers and professors from schools, and uni.?????
    :? :?
    Yes.

    I did that many years ago.

    salsaonline is doing it now.
    Reply With Quote  
     

  8. #7  
    Forum Bachelors Degree
    Join Date
    Mar 2009
    Posts
    421
    Tensors are not that complicated once you get the basic idea behind them.

    The basic example to think of is a vectorfield. Let's think, for example, of wind moving along the surface of the earth. At every point on the surface, we can think of a little arrow that points in the direction that the wind is moving. And as the points vary along the surface, the vectors vary smoothly.

    This is the first defining property of a tensor: A smooth family of things like "vectors" attached to different points in space.

    The second property of a tensor is how it "changes" when we change our coordinate system.

    For example, imagine that we look at the wind velocities in a very small region of the surface of the earth. Specifically, we want to study the wind velocity at the intersection of Westwood Blvd. and Le Conte in Los Angeles. We pick Le Conte as our x-axis and Westwood as our y-axis. So near this intersection, we can use these coordinates to completely define all of our wind velocity vectors.

    But happens if we decide to use a different coordinate system? For example, at the Westwood/Le Conte intersection, they've painted diagonal crosswalks (cool, huh?). So maybe I want to use the direction of these diagonal crosswalks as my coordinate axes. What happens to the way I express my wind vectors when I do this change of coordinates? Simple: they get multiplied by the rotation matrix that changes one coordinate system into the other.

    And this second property is really what makes a tensor a tensor. That is, when we rotate our local coordinate system, tensors transform in a manner that's more or less analogous to a rotation.

    One can contrast tensors with spinors, which are a bit like tensors in that they are families of "vector-like" objects attached to a shape. But the local values of a spinor don't transform like a rotation when we rotate coordinates. (For the mathematically sophisticated, tensors arise from representations of the rotation group, whereas spinors arise from representations of the double cover of the rotation group.)

    Hope this helps.
    Reply With Quote  
     

  9. #8  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    The basic example to think of is a vectorfield. Let's think, for example, of wind moving along the surface of the earth. At every point on the surface, we can think of a little arrow that points in the direction that the wind is moving. And as the points vary along the surface, the vectors vary smoothly.

    This is the first defining property of a tensor: A smooth family of things like "vectors" attached to different points in space.
    thank you for the help, oo and by the way..
    why does this first characteristic of tensors remind me of the Divergence theorem?
    Reply With Quote  
     

  10. #9  
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    Quote Originally Posted by Heinsbergrelatz
    this is a question out of this topic, but is there a real professor, as in professor's who give lectures in University who uses this forum. not as in the Forum professor, but real teachers and professors from schools, and uni.?????
    :? :?
    I am not a professor, but I have given a few lectures at undergraduate level at a university and I have led exam classes.
    Reply With Quote  
     

  11. #10  
    Forum Ph.D. Heinsbergrelatz's Avatar
    Join Date
    Aug 2009
    Location
    Singapore
    Posts
    994
    I am not a professor, but I have given a few lectures at undergraduate level at a university and I have led exam classes.
    ah i see, so wow, apart from students like me, actual experienced people actually use this forum.. ah thank you for the reply.
    Reply With Quote  
     

  12. #11  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    "If you can't explain something simply, you do not understand it well enough." - Einstien

    Although his some of his books are filled with tensor field, I have never seen an explanation of what a tensor is by Einstein. Perhaps even he has some difficulty with the idea. Seeing arrows at points, adding dimensions and understanding how all those arrows interact is quite difficult for me to visualize.
    Reply With Quote  
     

  13. #12  
    Moderator Moderator Markus Hanke's Avatar
    Join Date
    Nov 2011
    Location
    Ireland
    Posts
    7,302
    Perhaps you could take a look at this site, and the documents posted there :

    visualrelativity.com

    That's probably as good as it gets so far as geometrical interpretations of tensors are concerned
    ReMakeIt likes this.
    Reply With Quote  
     

  14. #13  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    Perhaps you could take a look at this site, and the documents posted there :

    visualrelativity.com

    That's probably as good as it gets so far as geometrical interpretations of tensors are concerned
    Of those links, I liked this one http://www.phy.syr.edu/~salgado/pape...AAPT-01Sum.pdf

    Thanks
    Reply With Quote  
     

  15. #14  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    Reply With Quote  
     

  16. #15  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    Reply With Quote  
     

  17. #16  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    Rather than just exchanging the results of internet searches, how about a little tutorial?

    First thing to say is that most physics texts (and even some math texts) content themselves with woolly statements like "A tensor is a mathematical that behaves as {so-and-so} under a coordinate transformation......". This of course tells you nothing about what a tensor actually IS.

    Let's try and rectify this. Start here, remembering that what follows applies to both flat and non-flat manifolds, and to both Lorentzian and non-Lorentzian manifolds, flat or otherwise

    First I have to assume we all know what is meant by a vector space (if you don't, you're probably not ready for tensors yet!). However, I remind you of a coupla relevant facts:

    A vector space is a set of mathematical objects endowed with a closed operation which is in every way analogous to arithmetic addition. Closed means that for every then

    There is also an associated scalar field such that for any then . For what follows, I will take the generic field , the real numbers.

    Now define the Cartesian product of sets such that every element of is of the form ; this is called an "ordered pair", since . Applying the vector space axioms we have that, for , then .

    We also will have that, for all then

    With me so far?

    Now it is a fact that, to any vector space I can associate a "sister" space called the dual space, also a vector space , with the property that, so that whenever then for any .

    As my battery is getting dangerously low, let me pose the rhetorical question: Is there a mapping, based on the dual space, that sends ? You may be surprised at the answer!

    Meantime, I had better quit.

    Ask questions if you want
    Reply With Quote  
     

  18. #17  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    If your intention was to show that you understand tensors, than I believe you, however that was no help at all.
    Reply With Quote  
     

  19. #18  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    Quote Originally Posted by ReMakeIt View Post
    If your intention was to show that you understand tensors, than I believe you, however that was no help at all.
    I think that was just the intro to the tutorial Guitarist proposed. I'm sure the relation to tensors will become more apparent once more instalments are released. Hopefully the wait isn't too long.
    Markus Hanke likes this.
    Reply With Quote  
     

  20. #19  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    I hope Guitarist does bring more to this thread, but this:

    t, so that whenever then for any

    is unforgivable
    Reply With Quote  
     

  21. #20  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    So where wuz I? Ah yes, I posed the rhetorical question

    if and if is also a vector space, is there a mapping that sends the Cartesian product of vector spaces: to the Real numbers. The answer is "Yes".

    One defines the tensor or outer or direct product as follows:

    Given ReMakeIt's favourite linear mapping and given that the product exists, then there is a bilinear mapping with the following property:

    For any and given that then where (an ordered pair, remember). Don't be dazzled by this - recalling that both and are real numbers, this as no more than simple arithmetic multiplication. Or if you must, you can think of this as a sort of braiding the tensor product with the Cartesian product

    Now the gadget above - that is - is called a TENSOR, in fact we call it a type (0,2) tensor, and the tensor product is the vector space of all such tensors. In short, this space nothing more (nor less) that the vector space of bilinear functions on the Cartesian product - strictly speaking these "functions" should be called functionals, but you don't really need to know the difference at present

    I am sure you all think I am off my head - surely this the wrong notation for a tensor! Well there is a convention that will bring my notation into register with what one usually reads in physics texts.

    But in order not to give the impression this convention is entirely arbitrary, we need to do some work, so later for that

    PS Wallers - did we not go through this before; that is before Latex was installed? Or have I mis-remembered?
    Last edited by Guitarist; February 1st, 2012 at 12:31 PM.
    Reply With Quote  
     

  22. #21  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    We started talking about tensors, but i was way out of my depth. (i probably still am to be honest)
    Reply With Quote  
     

  23. #22  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    Let's see if I can rectify that.

    So I promised to fix up my notation. First consider the garden-variety vector space . From this space we can always choose a set of vectors such that, for any we may write that where i = 1,2,....,n.

    Let's call this the "expansion of a vector on some basis". The objects are numbers from the scalar field over which our vector space in defined. In the case at hand this is the real number field. They are called "the components" of our vector

    A requirement for the existence of a basis, whose choice is otherwise arbitrary, is that no member of the set can be similarly expanded in terms of other members of the same set together with any element of this field.

    Now turn to the dual space . By a useful notational convention one writes, for that where the set are now the set of basis vectors for the dual space, under exactly the same conditions as before.

    Notice how I write the indices different for vectors and their duals.This is a meaningless but useful convention

    Now turn to our type (0,2). tensor, which I called . Expanding this on bases I will have that whenever .

    With me? probably not. Try this.

    Even though the basis vectors of a vector space and its dual are in some sense privileged, they are nonetheless vectors in their respective spaces whose components are obviously all 1. Now the basis for the dual space is always defined with respect to the basis of the "normal" space by whenever i = j and 0 otherwise.

    So = 1 whenever i= j and k = h

    Rewinding the definition of the bilinear map I gave, we have in this circumstance that

    So simplifying the summation limits we have that .

    This is the standard notation for a type (0,2) tensor written in component form. There is a good reason for this apparent laziness, but later for that (if indeed you want it)
    Reply With Quote  
     

  24. #23  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    Does the above post have something to do with a manifold?
    I guess what I see is that you have 2 basis in 1 real n-space and are doing a transformation from 1 to the other with the matrix A and it's analog phi.
    Reply With Quote  
     

  25. #24  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    I have pretty much no idea what are asking here. Yes it is true that any vector space of dimension n over the reals is trivially a manifold (Tell me why!) and yes it is true that the space of all rank n tensors is a vectors space. Make the connection. But no, I was not explicitly doing any topology

    But is a single vector space and so is each with a single basis.So I am a bit confused as to your question.

    OK, there are circumstances where it is permissible to think of scalars (the as I wrote) as linear transformations (what are these circumstances?) but this is not one of them..

    I found your question hard to follow, so it is possible that it is me that is confused.
    Reply With Quote  
     

  26. #25  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    I found your question hard to follow, so it is possible that it is me that is confused.
    I would think I am likely more confused

    I am asking if V has a basis and V* has a basis do they span the same space? Or do the Terms V x V and V* (outerproduct) V* span the same space?
    Reply With Quote  
     

  27. #26  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    I'm not sure if i understand, so i thought i'd try a bit of context. Am i correct in saying that the tensor product of two tangent vectors, to some point on a 3d surface, results in the metric tensor?
    Reply With Quote  
     

  28. #27  
    Forum Freshman MrStupid's Avatar
    Join Date
    Feb 2012
    Posts
    9
    I have a sort of cumulative definition of a tensor that I have come up with after some experience in thinking about them and reading about general relativity.

    A tensor is a multidimensional array of numbers. The number of dimensions is called the rank. Scalars, vectors, and matrices are tensors of rank 0, 1, and 2 respectively. The Riemann curvature tensor is an example of a higher rank tensor. It has four dimensions and completely characterizes curvature, i.e., how a manifold (curved space) differs from a Euclidean space in an infinitesimal neighborhood of a point.

    Tensors are coordinate free representations of some physical operation or condition, for example, the curvature of spacetime in GR.

    You may wonder at this point, as I did, how an array of numbers can be a coordinate free representation. Don't the numbers have to reference some coordinate system? In a very abstract sense, no they don't, they can be symbolic, but I like the concrete, so I think of a tensor as an object (in software terms) that carries its coordinate system around with it. This is an important feature, because it permits software procedures to combine tensors with different coordinate systems in a way that hides that complexity from the user. Thus, a tensor represented with a specific coordinate system can represent the class of all equivalent tensors, and in that sense it is coordinate free. You can choose to inspect the contents of the tensor in whatever coordinate system you prefer.

    When I see tensors used in physics they are to be interpreted as multilinear functions and they are often written with superscripts and subscripts indicating how vectors along the corresponding dimensions of the tensor are to be used. For example, the pseudometric tensor g in GR is may be written as gab, gab, or gab. The subscripts and superscripts are often used with so-called Einstein notation to indicate summation. For example, gabxaxb indicates a summation is to be performed for vectors along the dimensions of the repeated indices. In this case the result would be a scalar that is a weighted average of the xaxb products. Conventions on subscripts vary, which is another source of confusion.
    Reply With Quote  
     

  29. #28  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    Hey, hold on chaps. Don't try to run before you can walk!

    It is true that the metric tensor is a type (0,2) tensor, but not all type (0,2) tensors are metric tensors. Very briefly, the metric tensor on a finite real vector space induces an inner product which, again, roughly speaking, gives information about length, angle and distance.in an Euclidean vector space.

    The metric tensor on a manifold that may or may not be globally Euclidean is a generalization of the concepts of length and distance.

    Anyway, let's talk now about type (2,0) tensors.

    So the vector space is a perfectly genuine vector space, and, as such, is entitled to its own dual space. This is usually written as and is called the "double dual" to . So, while it is evident that the spaces and are isomorphic, they are not naturally so - this isomorphism depends on a choice of basis for each.

    It is also evident, for the same reason, that and are isomorphic, and since isomorphism is transitive, then and are isomorphic. Moreover this isomorphism does NOT depend upon a choice of basis; one calls this a "natural isomorphism (the proof is fiddly but not hard - you can take my word for it, or work a proof for yourselves).

    Under this circumstance is customary to take, for any and some that . This implies that we may regard as a mapping on , so that and thus .

    Given the same properties for the outer product, one call the gadget a type (2,0) tensor, and according to the argument and convention I gave earlier one writes this as
    Reply With Quote  
     

  30. #29  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    Presumably then the mixed type tensors, (1,1), are gadgets of the form, or . Where and such that or ?
    Reply With Quote  
     

  31. #30  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    BINGO!! Nice work wallers.
    Reply With Quote  
     

  32. #31  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    blah blah blah, i still dont follow... your notation goes beyond what I understand.
    Reply With Quote  
     

  33. #32  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    Quote Originally Posted by ReMakeIt View Post
    blah blah blah, i still dont follow... your notation goes beyond what I understand.
    Care to provide an example of notation that you don't understand?
    Reply With Quote  
     

  34. #33  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    Quote Originally Posted by ReMakeIt View Post
    I hope Guitarist does bring more to this thread, but this:

    t, so that whenever then for any

    is unforgivable
    this em yes em
    Reply With Quote  
     

  35. #34  
    Forum Sophomore ReMakeIt's Avatar
    Join Date
    Jan 2012
    Posts
    126
    I read that something like

    V*, Some permutation of V, and V are in R space, so that whenever psi is in V* then Psi of r is some multiple of something in R space for any v in the space of V.


    And i dont understand that
    Reply With Quote  
     

  36. #35  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    Quote Originally Posted by ReMakeIt View Post
    V*, Some permutation of V, and V are in R space, so that whenever psi is in V* then Psi of r is some multiple of something in R space for any v in the space of V.
    is the space of all linear mappings from the vector space (V) to the field, in this case the real numbers. Or more plainly, it is the vector space of objects that take vectors as input and give scalars as output. The later part of the statement talks specifics, defining as the function of vectors (v) that gives a scalar (a) as output.
    Reply With Quote  
     

  37. #36  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    Thanks wallaby for saving the trouble of explaining it in words. However, without meaning to sound unkind, I doubt that someone who needs perfectly standard mathematical notation explained in this way is ready for tensors.

    Whatever. Using this notation, we easily extend our existing tensors to those of higher rank. So that is the space of type (0,n) tensors, likewise for the space of type (n,0) tensors, and also, following wallaby for the space of type (n,m) tensors.

    Now it follows that the mapping is the space of type (1,0) tensor, which are vectors. But be careful - not all vectors are type (1,0) tensors! Likewise a (0,1) tensor is a dual vector, but again the inverse is false.

    It is a matter of convention that a type (0,0) tensor is a scalar. Actually its more that a convention - it has to do with how tensors in general transform under coordinate changes and a recognition that by definition scalars are invariant under coordinate transformation.

    Or to put it more accurately, any object that is invariant under arbitrary coordinate transformation is defined to be scalar.

    Now to make sense of this would require a brief foray into manifolds. But as it appears that the aspiring physicists here have little taste for the mathematics of their chosen subject, I shall not force it upon them
    Reply With Quote  
     

  38. #37  
    Moderator Moderator
    Join Date
    Jun 2005
    Posts
    1,620
    OK, so in spite of my little pout the other night, let's continue.

    Some interest was raised in the so-called metric tensor, so let's chat about that for a while. But I warn you all, the notation may be a little confusing, so I shall try to help you.....

    First I have to assume you know what is meant by an inner product on a finite-dimensional vector space - very roughly it gives information about the "length" of vectors and the "angle" between any two.

    So, in elementary treatments we find the inner product written as, for any that and physicists (bless them!) often use to mean the same thing. I confess I find this a nasty notation, but that's up to them.

    But we know better now, right? We know that the object in my first construction is an element the the vector space . So we have a mapping which we now recognize as a type (0,2) tensor ie. an element in

    Under the circumstance that this tensor defines the inner product, one calls this particular tensor a "metric tensor".

    The inner product becomes . Notationally, this is not so good: I gave the standard notation of such a tensor in component form as, say, . But convention has it it that the metric tensor and its components are written in lower case. Later for that

    For now let me alert you to a nice property of our tensor.We know that for any vector space there exists a dual space such that . In the special case that is an inner product space, for any there exists some unique , let's temporarily call it such that .

    This implies that, for any and some fixed we may write where the "dot" is an empty "slot".

    So, when we come to express all this in component form, something neat happens. In particular, when we transfer this whole set-up to a 4-manifold, you will be able to understand the reduced line element that Markus gave in one of the many "what is time" threads.
    As you will see if you are interested (but only if)
    Reply With Quote  
     

  39. #38  
    Forum Professor wallaby's Avatar
    Join Date
    Jul 2005
    Location
    Australia
    Posts
    1,521
    Until it becomes blatantly obvious that i don't understand something i'm happy to press on.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •