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Antisymmetrized tensor product

  1. Apr 8, 2007 #1
    Could someone explain to me what this is and explain the formula to me? I don't think I understand the formula.

    I don't think I quite understand why that's the antisymmetrized tensor product. Maybe its because i don't want o think about it too much.
    Last edited: Apr 8, 2007
  2. jcsd
  3. Apr 8, 2007 #2

    [tex]F_{ab}=\partial_a A_b - \partial_b A_a[/tex]

    See it is antisymmetric in its 2 indices: therefore

    [tex]F_{ab} = -F_{ba}[/tex]

    which further implies that

    [tex]F_{aa} = 0[/tex]
  4. Apr 8, 2007 #3


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    These are the properties of an antisymetric tensor. I think Terilien is talking about the antisymmetrized tensor product, which is something like

    [tex]W_{ab}=T_{[a}S_{b]} = \frac{1}{2!}(T_aS_b-T_bS_a)[/tex]

    More generally,it creates an antisymetric tensor of valence (a+c,b+d) out of two tensors of valence (a,b) and (c,d).

    This definition is pretty straightfoward and unambiguous (to me at least). So what don't you understand about this?

    We have two tensors [itex]\mathbf{T}[/itex] and [itex]\mathbf{S}[/itex] whose components are [itex]T_a[/itex] and [itex]S_b[/itex] respectively, and we define a new tensor [itex]\mathbf{W}[/itex] by constructing its components from the components of [itex]\mathbf{T}[/itex] and [itex]\mathbf{S}[/itex]. And it is straightfoward to verify that [itex]\mathbf{W}[/itex] defined in this way is indeed a tensor (i.e. that the set of numbers [itex]W_{ab}[/itex] transform like the components of a tensor)
    Last edited: Apr 8, 2007
  5. Apr 8, 2007 #4
    Yes but how exactly do we get the general formula (p +q)!/p!q!A_[u,1...u,pB_u,p+1......u,p+q]. for some reason I'm scared to figure this out on my own. I usally do, but I'm too anxious. Please help. I think i understand the P!Q! but I don't quite get the the term (p+q)!. As in i don't know why its in the formula.

    so essentially I don't understand the coefficients that well.
    Last edited: Apr 8, 2007
  6. Apr 8, 2007 #5


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    hint: nCr = n!/r!(n-r)!
    it is all about finding all possible pairs of indices. because total anti-symmetric means [tex]A_{ij}=-A_{ji}\; \forall \, i, j[/tex]
  7. Apr 8, 2007 #6


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    Your question was posed quite badly. What you want to know is what motivated people to define the antisymetrized tensor product that way, and particularly, what's with the coefficients.
  8. Apr 8, 2007 #7
    Ok well could someone tell me? I'm assuming it has to do with integration. frankly i'd like to leanr about differential forms in such a way that I know how they arise.
  9. Apr 17, 2007 #8

    your question about differential forms seems to have two parts:

    1) why do we need antisymmetric tensors as diff. forms? in other words, why is it that "dxdy" in the integral of f(x,y)dxdy needs to be interpreted as an antisymmetric product of dx and dy, rather than an ordinary product?

    2) how to define the antisymmetric product of two antisymmetric tensors in the general case.

    Regarding 1), I am not sure if a clear explanation of this can be found in an existing textbook. You may look at the lecture notes here, chapter 1 - maybe it will help. Regarding 2), the antisymmetric factors are largely cosmetic and some people define them differently. But the idea is that an antisymmetric n-form is a function of a completely antisymmetric combination of n different vectors. So it is natural to introduce a factor of 1/n! into the formula. When we compute the product of two antisymmetric tensors of valence m and n, they will come with factors 1/m! and 1/n! and then there will be an overall factor (m+n)! from the resulting tensor. Hope this helps. Try an example with m=1 and n=2 to see how it works.
  10. Apr 17, 2007 #9
    One is already pretty clear. It always has been.
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