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Index/Einstein notation: from/to matrix form

  1. Feb 7, 2013 #1
    So I've just started working with the index/einstein notation for matrices and vectors the other day. I've been doing a few exercises from a booklet I have, but I am still a bit confused. I am pretty sure my confusion is rather stupid though, so I apologize in advance.

    1. The problem statement, all variables and given/known data
    So one question that has me puzzled is writing the following in matrix form:
    Dβ[itex]\nu[/itex] = A[itex]\mu\nu[/itex]B[itex]\alpha\mu[/itex]C[itex]\alpha\beta[/itex]

    The second question, I suppose is similar, and it's writing the following as matrix multiplication:
    Dαβ = Aα[itex]\mu[/itex]B[itex]\mu\nu[/itex]C[itex]\beta\gamma[/itex]

    3. The attempt at a solution

    Alright, so my main confusion comes from the fact that, the way I've been thought is that in order to multiply matrices they need to be the same size. So an (MxN) matrix can only be multiplied with an (RxP) one, if N = R. Now in the first, and in the second one, this is (seems?) not the case, as for example a mu by nu matrix A is multiplied with an alpha by mu matrix B. I suppose that this is not actually the case though, and that I am misunderstanding. My question is, what am I doing wrong here? Could you help me in the right direction? The second equation is giving me same issue really, as I don't know how to represent those matrices in a correct way, as to me they seem to be of the wrong dimensions.

    Kind regards
     
  2. jcsd
  3. Feb 7, 2013 #2
    All the matrices you are multiplying are square matrices. The subscripts are not the dimensions of the matrices, they are the row-column addresses of a specific element in each matrix. When you use this type of notation, you are summing over repeated indices. When you write AαμBμσ, what your are really calculating the product of the two matrices A and B. Note that you are summing over the products of the row elements of A and the column elements of B.

    Chet
     
  4. Feb 7, 2013 #3
    Mhm alright yes, that makes sense. In general they are all N by N matrices I suppose, as you let your sum (which is omitted) range to N.
    However, I'm afraid I am still confused.. All the examples that I saw so far have the form AαβBβγ, so the second index of the first matrix would match the first index of the second matrix. Here, that is different. How is that reflected in the answer?
    Would AμνBαμ simply be Eμμ? I suppose not, but I don't really get why..

    I apologize, as this is a very basic topic, I just don't see exactly how it works just yet.

    Is the answer to the first question simply
    2eanlzd.jpg

    No right, there has to be more to it, as this just leaves out the indices..
     
    Last edited: Feb 7, 2013
  5. Feb 7, 2013 #4
    AμνBαμ would be BA.

    AμνBαμ=[itex]\sum_{\mu=1}^{\mu=N}B_{\alpha\mu}A_{\mu\nu}[/itex]
     
  6. Feb 7, 2013 #5

    vela

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    More specifically, it would be ##(BA)_{\alpha\nu}##.

    Remember that ##A_{\mu\nu}## and ##B_{\alpha\mu}## are just numbers, so you can reorder them: ##A_{\mu\nu}B_{\alpha\mu} = B_{\alpha\mu}A_{\mu\nu}##.
     
  7. Feb 7, 2013 #6
    Thank you both, that makes a lot more sense. Though that leaves me a little confused.. The rearrangement works with B and A, from the aforementioned equations, but the indices of C do not match up with either B, A, or their product, do they? So how does that work? Again, I apologize for something that is probably so obvious..
     
  8. Feb 7, 2013 #7

    vela

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    Your original expressions are messed up, which is probably why you're confused. The indices should match up the way you probably think they should.
     
  9. Feb 7, 2013 #8
    As in, they are incorrect? If that is the case I am going to punch myself for spending so much time on it instead of just asking my professor tomorrow. Thanks a lot though, at least I am more at ease with the general concept now, and I can move on to the actual topics associated with it!
     
  10. Feb 7, 2013 #9

    vela

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    Well, the second one, Dαβ = AαμBμνCβγ, is. (The first one is fine — I misread it earlier.) On the lefthand side, you have ##\alpha## and ##\beta## as free indices. On the righthand side, you have ##\alpha##, ##\beta##, ##\gamma##, and ##\nu##. It probably should have read ##D_{\alpha\beta} = A_{\alpha\mu}B_{\mu\nu}C_{\beta\nu}##.
     
  11. Feb 8, 2013 #10
    You are entirely correct, that gamma should have been a nu, strange mistake on my part.
    Giving it some more thought, the solution to the first one is probably D = CTBA, and the second one would be ABCT, or is that not a legal operation?
     
  12. Feb 8, 2013 #11

    vela

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    Yes, those are correct.
     
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