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Tensors question.

  1. Jul 5, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    Given that Aij is a contravariant tensor of rank 2, is the following a contravariant tensor of rank 3: Aijxi/xk?


    3. The attempt at a solution
    Using the chain rule, I have found xi/xk to be a contravariant tensor of rank 1:
    [itex]\bar{x}[/itex]i/[itex]\bar{x}[/itex]k = [itex]\bar{x}[/itex]i/xl * xl/[itex]\bar{x}[/itex]k
    Is that correct? Does the above product indeed yield a contravariant tensor of rank 3?
     
  2. jcsd
  3. Jul 5, 2013 #2

    Fredrik

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    That doesn't look correct. How and why did you use the chain rule?

    I think you should just try to determine what ##A^{ij}x_i/x_k## transforms to.
     
  4. Jul 5, 2013 #3
    May this be a better attempt?
    Following your advice, I got:
    ∂[itex]\bar{q}[/itex]i/∂ql * ∂[itex]\bar{q}[/itex]j/∂qp * ∂qt/∂[itex]\bar{q}[/itex]s * ∂[itex]\bar{q}[/itex]a/∂qr
    Hence, a composite tensor of rank 4? :S
     
  5. Jul 5, 2013 #4

    Fredrik

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    I don't understand what you're doing. Why is there no A, no k, and no division in that result? And why is there an i in it?

    By the way, are you sure the problem statement is correct? You can tell immediately from the fact that there are only two free indices that it can't be a rank-3 tensor.
     
  6. Jul 5, 2013 #5
    There was no division as I multiplied by the inverse. May I not do so?
    The 'i' stems from the contravariant component of A.
    'a' should have been 'k', sorry for that.
    Should it have been:
    [∂[itex]\bar{q}[/itex]i/∂ql * ∂[itex]\bar{q}[/itex]j/∂qp * ∂qt/∂[itex]\bar{q}[/itex]s * ∂[itex]\bar{q}[/itex]k/∂qr]Alpxtxr)?
     
  7. Jul 5, 2013 #6
    If this is wrong, I am then not really sure how to proceed.
    I am not sure how to transform the original expression and was merely trying to infer based on what I have been hitherto taught.
     
  8. Jul 5, 2013 #7

    Fredrik

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    Since repeated indices are summed over, i is a dummy variable in the original expression. ##A^{ij}x_i/x_k## can also be written as ##A^{mj}x_m/x_k##. This expression may or may not denote the ##{}^j_k## component of a tensor. So it's natural to check what the ##{}^j_k## component is in another coordinate system. That's why I don't expect to find a free ##i## in the final result.

    You seem to be handling j and the first i correctly. I don't understand the rest of what you're doing. In particular I don't understand the comment about why there's no division. The transformation of the denominator is going to look something like this:
    $$\frac{1}{\bar x_k} =\frac{1}{M_k^px_p}.$$ Here M is the matrix defined by writing the transformation of the basis vectors as ##\bar e_i=M_i^j e_j##. I don't see a way to rewrite this so that there's no division.
     
  9. Jul 5, 2013 #8
    Could I write the division thus:
    1/[itex]\bar{x}[/itex]k = 1/(∂qs/∂[itex]\bar{x}[/itex]k)?
     
  10. Jul 5, 2013 #9

    Fredrik

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    What's q? And what's s? You could write this:
    $$\frac{1}{\bar x_k}=\frac{1}{\displaystyle\frac{\partial x^p}{\partial\bar x^k}x_p}.$$ By the way, I think a notation like this makes it easier to do the calculation:
    \begin{align}
    M^p_k &=\frac{\partial x^p}{\partial\bar x^k}\\
    (M^{-1})^p_k & =\frac{\partial\bar x^p}{\partial x^k}
    \end{align}
     
  11. Jul 5, 2013 #10
    So I would get:
    ∂[itex]\bar{q}[/itex]i/∂ql * ∂[itex]\bar{q}[/itex]j/∂qp * Alp*$$\frac{1}{\displaystyle\frac{\partial x^p}{\partial\bar x^k}x_p}.$$
    Right?
     
  12. Jul 5, 2013 #11

    Fredrik

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    You're getting closer, but what happened to the ##x_i## that you had in the original expression? Also, I would recommend to not use the same dummy index for two different summations in the same expression. (There are four p's in that expression).
     
  13. Jul 5, 2013 #12
    My bad :). So I would get:
    (∂[itex]\bar{q}[/itex]i/∂ql)(∂[itex]\bar{q}[/itex]j/∂qp)(Alp)(∂qn/∂[itex]\bar{q}[/itex]i)(xn)($$\frac{1}{\frac{\partial x^b}{\partial\bar x^k}x_b}$$)
    Right?
     
  14. Jul 5, 2013 #13

    Fredrik

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    What is q? Shouldn't it be x? If yes, then you're finally getting the result I'm getting. It can be simplified a bit because of the two i's.

    I'm going to bed, so I probably won't post again tonight.

    Did you see my question in post #4 (my second post)? I doubt that the problem statement is OK, since we don't actually have to do any of these calculations to see that we're not dealing with a tensor of rank 3. All we have to do is to count the number of free indices and see that it's not 3.
     
  15. Jul 6, 2013 #14
    Yes, it should have been x. But why is the use of the q's in the beginning legit (in the transformation of A)? That is, why was I "allowed" to use q's for transforming A itself whereas I could only use x's to transform x? Is it because q's denote the vectors spanning A?
    Isn't that a composite tensor of rank 4 then?
     
  16. Jul 6, 2013 #15

    Fredrik

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    I thought that q was just an alternative notation, i.e. sometimes you denote the coordinate by q and sometimes by x. In that case, you should pick one of the symbols and stick to it. In this case, I assume that ##x_k## denotes the ##k##th coordinate of a point in space. If that's the case, then the problem has already chosen the notation for us: You have to use x.

    Since there are only two free indices, this is either a tensor of rank 2 or not a tensor at all. That's why I suspect that you got the problem statement wrong. It's immediately obvious that it can't be a tensor of rank 3. No calculations necessary.

    You should try to simplify your result.
     
    Last edited: Jul 6, 2013
  17. Jul 6, 2013 #16
    Would you please care to explain why there are only two free indices? In other words, which are the two?
     
  18. Jul 6, 2013 #17

    Fredrik

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    j and k. See the first sentence in post #7.
     
  19. Jul 6, 2013 #18
    Okay, I understand. BUT -
    1) What do l,p,n and b denote (in the transformation)? Are they dummy as well?
    2) By what means at this point may it be determined then whether this is at all a tensor?
     
  20. Jul 6, 2013 #19

    Fredrik

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    Every index that appears twice is summed over, and is therefore a dummy index.

    Simplify the expression. Then compare your formula for ##\bar A^{ij}\bar x_i/\bar x_k## with the tensor transformation law for rank-2 tensors. If it looks exactly the same, then we're dealing with a tensor. If it doesn't, then we almost certainly aren't. It's a bit tricky to make absolutely sure. You would have to evaluate your result and the result you get from the tensor transformation law in a specific coordinate system and see if they're the same. It's possible that you have to repeat this procedure with several different coordinate systems. The first coordinate system you try may tell you something like "if this is a tensor, then the 11 component is 0". If you keep trying with other coordinate systems, you may find that this is a tensor only if all the components are 0.
     
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