1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof the trace of tensor

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Could someone guide me to proof following equation.

    For any A is a tensor; a, b, c are vectors. Proof that:

    Tr (A) a.(b x c) = Aa.(b x c) + a.(Ab x c) + a. (b x Ac)

    with (.) is dot product, and (x) is cross product of vector

    2. Relevant equations



    3. The attempt at a solution

    My proof is below:

    (1): Aa.(b x c) = Ai1(ei [itex]\otimes e1[/itex] aiei. (bjej x ckek) = Ai1δi1aiei.(bjej x ckek) = A11aiei.(bjej x ckek)

    similar to two remain part

    (2): a.(Ab x c) = A22aiei. (bjej x ckek)
    (3): a. (b x Ac) = A33aiei.(bjej x ckek)

    from (1), (2), and (3) I get the result.

    But there may have something wrong. Could anybody check for me?

    Thanks
     
  2. jcsd
  3. Mar 22, 2012 #2

    hunt_mat

    User Avatar
    Homework Helper

    Write what you know:
    [tex]
    \mathbf{A}\mathbf{a}=A_{ij}a_{j}\quad (\mathbf{b}\times\mathbf{c})_{i}=\varepsilon_{ijk}b_{j}c_{k}
    [/tex]
    And so:
    [tex]
    \mathbf{A}\mathbf{a}\cdot (\mathbf{b}\times\mathbf{c})=A_{il}a_{l} \varepsilon_{ijk} b_{j}c_{k}=A_{il}\varepsilon_{ijk}a_{l}b_{j}c_{k}
    [/tex]
    Do these for the other two and then add. The trick is then to relabel the dummy indicies to get the required LHS.
     
  4. Mar 22, 2012 #3
    Thanks to Hunt mat

    Could I derive Aa.(b x c) as following:

    Aa.(b x c) = Amn(em [itex]\otimes[/itex] en) aiei.(bjej x ckek)
    = Amnδniaiem.(bjej x ckek)
    = Amiaiem.εijkbjckei
    = Amiδmiεijkaibjck
    = Ammεijkaibjck

    Is it similar to your guidance?

    Thanks
     
  5. Mar 22, 2012 #4

    hunt_mat

    User Avatar
    Homework Helper

    Can you use LaTeX please, I can't read all that.
     
  6. Mar 22, 2012 #5
    Hi Hunt mat
    Here I just type using Latex. Could you check it for me?

    Aa.(b x c) = Amn(em [itex]\otimes[/itex] en)aiei . (bjej [itex]\times[/itex] ckek) = Amn(en.ei)aiem) . [itex]\epsilon[/itex]ijkbjckei = Amiem.ei[itex]\epsilon[/itex]ijkaibjck = Amm[itex]\epsilon[/itex]ijkaibjck

    Thanks
     
  7. Mar 22, 2012 #6

    hunt_mat

    User Avatar
    Homework Helper

    I don't think this is right as you have no contraction between the A and the a.
     
  8. Mar 22, 2012 #7

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's still kind of hard to follow your notation. If the same index appears more than twice in the same product of tensors, you're doing something wrong. The ones that are being summed over should appear exactly twice.

    I don't think you need to mention the basis vectors all the time. For example, the left-hand side is ##(\operatorname{Tr} A)(a\cdot(b\times c))=A_{ii}a_j(b\times c)_j=A_{ii}a_j\varepsilon_{jkl}b_k c_l##. No need to mention basis vectors here.

    You may want to check out the LaTeX guide for the forum.
     
  9. Mar 22, 2012 #8
    Thanks to Hunt mat, and Fredrik

    I think for any tensor A and vector a, b, c. We should write in their component as:
    [tex]
    \mathbf{A}\mathbf{a}=A_{mn}a_{i}\quad (\mathbf{b}\times\mathbf{c})_{i}=\varepsilon_{ijk}b_{j}c_{k}
    [/tex]
    And so:
    [tex]
    \mathbf{A}\mathbf{a}\cdot (\mathbf{b}\times\mathbf{c})=A_{mn}a_{i} \varepsilon_{ijk} b_{j}c_{k}=A_{mn}\varepsilon_{ijk}a_{i}b_{j}c_{k}
    [/tex]

    Because indicies of the tensor A and the vector a should be independent.

    I want to mention the basic vectors here to make it clearly to my understanding. I am not good in calculus then easy to confuse. I will rewrite my opinion as:

    [tex]
    \mathbf{A}\mathbf{a}\cdot (\mathbf{b}\times\mathbf{c})=A_{mn}(e_{m}\otimes e_{n})a_{i}e_{i}\cdot (\mathbf{b}\times\mathbf{c}) = A_{mi}a_{i}e_{m}\cdot (\mathbf{b}\times\mathbf{c})
    [/tex]

    Follow my derive, I don't know how to continue the operation to get the result. Please help me to understand.

    Thanks
     
    Last edited: Mar 22, 2012
  10. Mar 22, 2012 #9

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Are you sure that Aa isn't supposed to be interpreted as the product of an n×n matrix and an n×1 matrix? That would make more sense to me, because what's the dot product of a tensor with 3 indices and a tensor with 1 index?
     
  11. Mar 22, 2012 #10
    To Fredrik,

    Yeah, the Aa is supposed to be interpreted as the dot product of an n x n and nx 1 but that is special case. I would like to make a proof in general case then indices of tenor A and a should be independent.

    Thanks
     
  12. Mar 23, 2012 #11

    hunt_mat

    User Avatar
    Homework Helper

    I think that if A completely independent that the identity not actually true.
     
  13. Mar 23, 2012 #12
    Hi Hunt mat,

    You can find that identity on the book "Introduction to Continuum mechanics for engineers, revised edition 2007 by Ray M.Bowen". The equatuion is placed on page 266, Appendix A, equation No. A.5.37, but without proof.

    That why I can not understand.
     
  14. Mar 23, 2012 #13

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I found a first edition. It says that the trace of an endomorphism of V is defined by
    $$(\operatorname{tr} \mathbf{A})\mathbf{u} \cdot(\mathbf{v}\times\mathbf{w}) = (\mathbf{Au})\cdot(\mathbf{v}\times\mathbf{w}) +\cdots$$ This equality has number (A.5.37) in the first edition.

    If it's definition, it it wouldn't make sense to try to prove it.

    Edit: OK, maybe I should have read a few more lines. The next thing he does is to prove that ##\operatorname{tr}\mathbf A=A_{ii}##. Is that what you want to do?
     
  15. Mar 24, 2012 #14
    I understand trA = Aii. But I want to proof this definition. Because the definition from most other book is: trA = Aii
     
  16. Mar 24, 2012 #15

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Definitions can't be proved. I guess what you want to do is to prove that the two definitions are equivalent, i.e. that they assign the same meaning to the notation "tr A". Didn't the author do that immediately after that formula?
     
  17. Mar 26, 2012 #16
    I think the proof can be made for general equation when we use definition TrA = Aii. The author prove the equation using 3 basic vector only. Could the author's definition be true for any tensor A, vetors a, b, and c? That is what I want to know.
    Thanks
     
  18. Mar 26, 2012 #17

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It doesn't make sense to ask if a definition is true. A definition assigns a meaning to a term or a notation. This definition assigns a meaning to the notation tr A. The author proves that if tr A is defined by (A.5.37), then tr A=Aii. I don't understand your concern about using the fact that he used the basis vectors instead of arbitrary vectors in the proof he did. The definition is a "for all" statement. If the equality holds for all u,v,w, then it holds for i1, i2
    , i3.

    However, it makes sense to be concerned about whether the definition makes sense at all. What he should have proved (preferably before the definition), but didn't, is that there's a unique real number r such that ##r\mathbf{u} \cdot(\mathbf{v}\times\mathbf{w}) = (\mathbf{Au})\cdot(\mathbf{v}\times\mathbf{w}) +\cdots## for all ##\mathbf{u},\mathbf{v},\mathbf{w}\in\mathbb R^3##. Then it makes sense to define tr A to be that number.

    Alternatively, and equivalently, he could have proved that if tr A is defined by tr A=Aii, then ##(\operatorname{tr} \mathbf{A})\mathbf{u} \cdot(\mathbf{v}\times\mathbf{w}) = (\mathbf{Au})\cdot(\mathbf{v}\times\mathbf{w}) +\cdots## for all ##\mathbf{u},\mathbf{v},\mathbf{w}\in\mathbb R^3##. It appears that what he did prove tells us that if tr A=Aii, then ##(\operatorname{tr} \mathbf{A})\mathbf{i}_1 \cdot(\mathbf{i}_2\times\mathbf{i}_3) = (\mathbf{A i}_1)\cdot(\mathbf{i}_2\times\mathbf{i}_3) +\cdots##

    I'm saying "appears" because I'm having some trouble understanding his definitions. Obviously, it's at least partially because I haven't read them all. Is V=ℝ3? First he says that A is an endomorphism of V. (At least I think that's what he says). Then he says that ##\mathbf{A}=A_{kj}\mathbf{i}_k\otimes\mathbf{i}_j##, something I would interpret as a function from V×V into ℝ, not from V into V. But I suppose he may have defined ##\mathbf{i}_k\otimes\mathbf{i}_j## as the map that takes ##\mathbf u## to ##(\mathbf{i}_k\mathbf{u})\mathbf{i}_j##, where ##\mathbf{i}_k## is interpreted as a cotangent vector and ##\mathbf{i}_j## as a tangent vector. This would mean that ##(\mathbf{i}_k\otimes\mathbf{i}_j)(\mathbf u)=u_3\mathbf{i}_j##, and that ##\mathbf{Ai}_1=A_{kj}(\mathbf{i}_k \mathbf{i}_1)\mathbf{i}_j=A_{kj}\delta_{k1}\mathbf{i}_j=A_{1j}\mathbf{i}_j##. Uh, that's not the result he's getting. However, if I instead guess that ##\mathbf{i}_k\otimes\mathbf{i}_j## is the map that takes ##\mathbf u## to ##(\mathbf{i}_j\mathbf u)\mathbf{i}_k##, I get the result he does. Is that really how he defines ##\otimes##?

    If you want us to help you figure this out, you need to first figure out what the definitions are, and then show us an attempt to use them to prove what you need to prove. I'm assuming that what you need to prove is that if tr A=Aii, then tr A satisfies (A.5.37).

    By the way, I don't see how to prove that (A.5.37) follows from the simpler definition of the trace. I'm actually having a hard time believing that it can be true.
     
  19. Mar 27, 2012 #18
    Thanks to your help,

    The author define that
    ##(\operatorname{tr} \mathbf{A})\mathbf{u} \cdot(\mathbf{v}\times\mathbf{w}) = (\mathbf{Au})\cdot(\mathbf{v}\times\mathbf{w}) +\cdots## for all ##\mathbf{u},\mathbf{v},\mathbf{w}\in\mathbb R^3##.

    And he proved the TrA = Aii using the properties of unit basic vectors. I understand what he proved. And I think we can follow the method that he used to proof the prove the his definition.

    I think can use below operation
    [tex]
    \mathbf{a}\cdot (\mathbf{b}\times\mathbf{c})=\varepsilon_{ijk}a_{i}b_{j}c_{k}
    [/tex]

    1. If i,j,k = 1,2,3 and using the author's proof, I can get the right - hand side as:
    $$(\mathbf{Aa})\cdot(\mathbf{b}\times\mathbf{c}) + (\mathbf{a})\cdot(\mathbf{Ab}\times\mathbf{c}) + (\mathbf{a})\cdot(\mathbf{b}\times\mathbf{Ac}) = (a_{1}b_{2}c_{3})(\operatorname{tr} \mathbf{A}).$$

    2. I can continue with the other notation of i,j,k following the permutation symbol. Then I take summation all situation of i,j,k; I can get the result.

    What do you think about this way to prove?

    Thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof the trace of tensor
  1. Trace proof (Replies: 1)

  2. Tensor proof (Replies: 3)

  3. Trace of matrix proof (Replies: 14)

Loading...