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Need mental picture

  1. May 10, 2007 #1
    Hello Gals,

    I know what a scalar is.
    I know what a vector is.
    I know what a linear transformation is.

    But what in the name of sweet aunt petunia is a rank 3 tensor?

    Love,
    Plx Mny
     
  2. jcsd
  3. May 10, 2007 #2

    Demystifier

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    I believe that it is misleading to try to visualize higher-rank tensors. Think of vector as collections of 3 or 4 numbers, not as arrows. Then the algebraic generalization to matrices, rank-3 tensors, etc, is trivial.
     
  4. May 10, 2007 #3
    It is often recommended that you don't try to visualize a 3rd rank tensor, or any tensor for that matter, but to think of it merely as a multi-linear function which maps vectors and 1-forms to to the set of real numbers. The higher the rank the tougher the visualization becomes. In fact I never try to visualize a tensor myself.

    Pete
     
  5. May 10, 2007 #4
    Hmm. I take it you've never read David Hume.

    Anyways, I thought a little about this on my lunch break. I came up with "a
    linear combination of linear combinations". Doesn't seem like a concept worth worrying about.
     
  6. May 10, 2007 #5
    There are always people who anyone never read. So what does David Hume have to do with this and why didn't you explain it when you posted his name? That's an irritating habit for you to fall into. :yuck:

    Pete
     
  7. May 10, 2007 #6

    pervect

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    A rank 3 tensor inputs three generalized vectors (i.e. either a vector or their dual vector), and spits out a scalar.

    One can also think of it as inputting 2 generalized vectors (or a rank 2 tensor), and outputting a vector, or inputting 1 generalized vector, and outputing 2 vectors (or a rank 2 tensor).
     
  8. May 11, 2007 #7
    I was wondering this same thing. I can't even visualize linear transformations. I keep wondering: does not being able to visualize it make it impossible to do things like GR if you're a "geometric thinker" like me as opposed to a "formula thinker" like... pretty much everyone else?

    I'm really bad. My eyes spin in circles whenever I see a sum in sigma notation, and I have to write it out with the ellipsis before I understand what it's saying!
     
  9. May 11, 2007 #8
    Well I am just learning this stuff and it's pretty clear that there is a lot of obfuscation going on. So I am taking very small bites.

    I can definitely visualize a linear transformation. Maps a vector to another vector. Period.
     
  10. May 11, 2007 #9
    That holds true for any kind of transformation and not specifically to a linear one. A linear transformation is of the form

    Y = aX + B

    I.e. all linear transformations have this form.

    Pete
     
  11. May 11, 2007 #10
    I know what a rank 10 tensor is
    I know what a rank 11 tensor is...
     
  12. May 12, 2007 #11
    As pmb_phy mentioned, that's just a "transformation". That I can visualize. It's the "linear" part that gets me.
     
  13. May 12, 2007 #12
    That's not right. A linear transformation is not allowed to offset the vector (your B), and it can do a lot more than just scale the vector by a scalar "a". It can skew it, rotate it, and other things.
     
  14. May 12, 2007 #13
    Many tensors can be visualized as ellipsoids.

    e.g. the inertia tensor, or the polarizability tensor.
     
  15. May 12, 2007 #14
    Thank you for that correction Xezlec. You're correct of course.

    Pete
     
  16. May 12, 2007 #15
    That's interesting. That got me thinking and exploring Wikipedia. I guess the best way to visualize a general linear transformation is to visualize the three basis vectors of a 3D coordinate system and think of just changing/moving any/all of them in any way, with their tails remaining stuck together at the origin. Then I can picture the effect on any object living in the space "attached" to those vectors.

    Then, a rank-3 tensor is like picturing those three arbitrarily changed/moved vectors in a different coordinate system, and applying a different linear transformation to each one.
     
  17. May 30, 2009 #16

    m=(((1,2),(2,4)),((2,3),(5,6))) is a rank 3 tensor with dimension 2, a vector of matrices.
    A tensor is a nested list.

    An example Eigenmath http://eigenmath.net/ script with a rank 3 tensor is:

    --Maxwell equations in tensor form.
    --See the book Gravitation p. 81.
    --
    -- F + F + F = 0
    -- ab,c bc,a ca,b
    --
    -- ab a
    -- F = 4 pi J
    -- ,b
    --
    --For this demo, use circular polarized light.
    --
    EX = sin(t+z)
    EY = cos(t+z)
    EZ = 0
    BX = cos(t+z)
    BY = -sin(t+z)
    BZ = 0
    FDD = (( 0, -EX, -EY, -EZ),
    ( EX, 0, BZ, -BY),
    ( EY, -BZ, 0, BX),
    ( EZ, BY, -BX, 0)) --See p. 74. Here, DD means "down down" indices.
    X = (t,x,y,z) --Coordinate system
    FDDD = d(FDD,X) --Gradient of F
    T1 = transpose(transpose(FDDD,2,3),1,2) --Transpose bca to abc
    T2 = transpose(transpose(FDDD,1,2),2,3) --Transpose cab to abc
    check(FDDD + T1 + T2 = 0)
    guu = ((-1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1))
    FDDU = contract(outer(FDDD,guu),3,4) --Easier to make FDDU than FUUD.
    check(contract(FDDU,2,3) = 0) --For light J is zero.
    "OK"

    The gradient of a rank 2 tensor (matrix) in a coordinate system (vector), is a rank 3 tensor. (GAMUDD) in example below.
    Another example is the gradient of the metric in general relativity which is the connection.
    An example of a 4th rank tensor is the Riemann curvature of spacetime. RUDDD in http://eigenmath.net/examples/bondi-metric.txt .
     
    Last edited: May 30, 2009
  18. Jun 3, 2009 #17
    Now some one try to visualize contravariant tensors! I tried to twenty years ago and then decided to do grad school in engineering. At least with fluids you don't get beyond three dimensions!
     
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