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Homework Help: Help with proof

  1. Jan 20, 2004 #1
    How would I solve E1jk without the summation? I know how to solve it using the summation symbol but don't know howto do it without it.


    Also, I need help proving that |torque|^2 = |r x F|^2= r^2F^2sin@(thetarF ). r dot F = rF cos (thetarF . Would I have to use (r x F) dot (r x F)?
     
  2. jcsd
  3. Jan 20, 2004 #2
    The Qs are not clear to me
    Does E represents Electric field

    And What do u want to prove for Torque
     
  4. Jan 21, 2004 #3
    here is an attachment with the questions

    I need help with the third one in number 1 and numbers 2(proof) and 3. For number 3, how would I expand?
     

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  5. Jan 21, 2004 #4
    I still dont know what does epsilon delta represents may u have some notations

    But as far as Q3 goes
    The angle between [tex]\vec r x \vec F[/tex] is zero
    Hence [tex](\vec r x \vec F).(\vec r x \vec F) = |(\vec r x \vec F)^2|[/tex]
     
    Last edited: Jan 21, 2004
  6. Jan 21, 2004 #5

    HallsofIvy

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    As himashu121 pointed out, in order to find ε1jk, you have to know what εijk means! I suspect I do know what it means since it is just a matter of looking up a definition, it would be much better for you to do that.

    Right out the formula for εijk, and substitute i= 1- in fact, write out all the components and then just copy down those that have i=1.
     
  7. Jan 21, 2004 #6
    I still want to know epsilon and delta i believe these are vector components

    Though for Part2:

    Write a=wx(wxr) = (w.r)w-(w.w)r where all are vectors and x is a cross product.

    If r is perpendicular than w.r=0
     
  8. Jan 22, 2004 #7

    HallsofIvy

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    Since jlmac2001 hasn't responded: In tensor analysis, δij is the tensor represented by the unit matrix: 1 if i=j, 0 otherwise.

    εijk is the "alternating" tensor. It is defined to be: 1 if ijk is an even permutation of 123, -1 if ijk is an odd permutation of 123, 0 otherwise (i.e. if any one of the indices is repeated).

    εsub]1jk[/sub] is therefore:
    ε111= 0
    ε112= 0
    ε113= 0
    ε121= 0
    ε122= 0
    ε123= 1
    ε131= 0
    ε132= -1
    ε133= 0

    Written as a matrix this would be:
    [0 0 0]
    [0 0 1]
    [0 -1 0]
     
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