Week Two – The Summation Convention

Here are the slides for the second week.[gview file=”http://oafak.com/wp-content/uploads/2019/08/Week-Two.pdf”]
I note that there have been no comments or questions from Systems Engineering students. There will be a test two weeks time on the first four weeks. Those who do not practice the Q&As as instructed will have themselves to blame.

18 comments on “Week Two – The Summation Convention

  1. Ayomide emmanuel says:

    Good day sir,
    Please I don’t understand the component form of vector product in the textbook sir pages 31 and 32.

    • oafak says:

      “I don’t understand…” is not a question. It is a comment that is not actionable. While I desire comments and questions, this particular one cannot be helped. You can tell me where you followed the notes up to and at what particular step you have an issue. If there are ten steps in an issue, “I don’t understand can mean anything from “I do not even know what the issue is all about”, or “I do not follow step 8”! The ball is still in your court to become more specific in what in the component form you have difficulty with.

      • Ayomide emmanuel says:

        Thank you sir. I have understood it now. I rushed through it instead of reading it calmly sir. The slide helped me.

        • oafak says:

          Very happy to know your problem is solved. Please ask your quetions after you have read what you already have. I am always available to assist you.

  2. ben says:

    Good day.
    from question 1.16 of the textbook “Chapter-1.-Review-of-Vectors-1”
    (a × b)⋅(b × c)x(c × a) = (a × b) ⋅ [(b × c)⋅a]c − (a × b)⋅[(b × c)⋅c]a
    = (a × b) ⋅ [(b × c) ⋅ a]c
    = (a × b ⋅ c)((b × c) ⋅ a)
    = (a × b ⋅ c)(a ⋅ b × c )
    = (a × b ⋅ c) 2
    from line 4 of this comment, (a × b)⋅[(b × c)⋅c]a = 0
    please can i get a further break down of this

    • oafak says:

      The triple product \bold{(b\times c)\cdot c} is a scalar quantity; call it \alpha. (You should be able to see, after you have read to the end of this comment that \alpha=0). Your expression is therefore, \alpha \bold{(a\times b)\cdot a}. This scalar triple product vanishes for at least two reasons:
      1. It is the volume of a parallelepiped with sides \bold{a,a,b} which vanishes because of linear dependency of the vector edges.
      2 The vector \bold{(a\times b)} is perpendicular to \bold{a}. Taking its dot product with a brings in the cosine of 90 degrees. You know what that means.
      Watch out for such simple applications of elementary knowledge.

  3. eniola says:

    Good day sir.
    Products of Alternating tensors.page 30

    when we are expanding the matrix from page 30.

    11 + 22 + 33i
    i was trying to break it down myself using the Kronecker Delta the answer is dri.
    but when i broke it down i had dri + dri + dri so am thinking there should be a 3 before the final answer dri.

    • eniola says:

      Trying to copy and paste formulas and symbols from the textbook doesnt show on this website maybe something can be worked out on the site sir.
      the 11 +22 + 33i
      is dr1d1i + dr2d2i + dr3d3i.

  4. Ayomide emmanuel says:

    Sir,in page38, I don’t understand how equation61 came into play.

    • eniola says:

      which part there is it where the Kronecker Delta acted as a substitution symbol?
      we have 6 parts there which one exactly or all?

      • oafak says:

        Eniola, for the Kronecker delta to ack as substitution symbol, it MUST share an index with another symbol. In that case remove the Kronecker delta and replace the unshared index of the other symbol. It is easier done than explained. Look at the examples in class and in the book.

    • oafak says:

      Ayomide, Look on the previous page, you see the expression for aij. as for bij, you will need to take the dot product of equation 59 with xi alpha to get it. Ask me in class if its still unclear

  5. Ajibola Areo says:

    Good day sir, from practice question 1.20

        \[ (\mathbf{w} \times \mathbf{u}) \times (\mathbf{w} \times \mathbf{v}) = [(\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v}]\mathbf{w} - [(\mathbf{w} \times \mathbf{u}) \cdot \mathbf{w}]\mathbf{v})  \]

    please sir can you explain? what happened to “- [(\mathbf{w} \times \mathbf{u}) \cdot \mathbf{w}]\mathbf{v})
    that the next line became

        \[ =[(\mathbf{w} \times \mathbf{u}) \cdot \mathbf{v}]\mathbf{w} \]

    Ajibola Areo
    Systems Engineering

    • oafak says:

      For the term \left[\bold{w,u,w}\right] observe that you are taking the triple product of three vectors that are linearly dependent. Looked at in another way, you are looking at the volume of a parallelepiped where two sides are the same vector! Such a parallelepiped will be on a plane – giving you a zero volume!
      If you want to look at it as the dot product of \bold{w} \times \bold{u} and \bold{w}, remember that \bold{w} \times \bold{u} is perpendicular to \bold{w} and that the scalar product of two perpendicular vectors is zero.

  6. Olalekan Ayomide David says:

    Sir, When I was reading some notes under the Levi Civita symbol. I came upon the word Abelian.
    Sir, what does it mean for a set of operators to be Abelian

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