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Prove the transformation is scalar

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    1.) Prove that the infinitesimal volume element d3x is a scalar
    2.) Let Aijk be a totally antisymmetric tensor. Prove that it transforms as a scalar.

    2. Relevant equations


    3. The attempt at a solution
    1.) Rkh = ∂x'h/∂xk

    By coordinate transformation, x'h = Rkh xk

    dx'h = (∂x'h/∂xk) (∂xk/∂x'j) (∂x'i/∂xj) dxj

    dx'h = δih Rji dxj

    dx'h = Rjh dxj

    This shows that the differential doesn't affect the transformation hence by performing the differential three times it would not affect the transformation, that is, it is a scalar. Can anyone verify if this is correct?

    2.)
    A'mnl = ( T'mnl - T'lnm )
    = RimRjnRklTmnl - RklRjnRimTlnm
    = RimRjnRkl(Tmnl - Tlnm)

    What should I do next here?
     
  2. jcsd
  3. Jan 27, 2016 #2

    ChrisVer

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    1. I am not sure what you are trying to prove with that? In fact I am not even sure that the infinitesimal volume is indeed a scalar...? It should be transformed with the Jacobian determinant of the transformation (take for example the [itex]dV_{spherical}[/itex] and the [itex]dV_{cartesian}[/itex]. Obviously going from (x,y,z) to (x',y',z')=(r,θ,φ) doesn't keep dV the same. In cartesian it's dx dy dz (fine), in spherical it's not dr dθ dφ.

    2. What is T?
     
  4. Jan 27, 2016 #3
    It is an orthogonal transformation so I'm trying to prove that it is a scalar under that transformation. T is an antisymmetric tensor.
     
  5. Jan 28, 2016 #4

    ChrisVer

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    You ended up in:
    [itex]dx'^h = R^{jh}dx^j[/itex]
    right?
    Now you should check what is:
    [itex]d^3x' \equiv dx'^1 dx'^2 dx'^3[/itex]
    why? because by the form you ended at, I cannot see how you deduce that it's a scalar, but instead that dxi are the components of a vector.

    2. Well if you have:
    [itex]\epsilon'^{ijk} = R^{im} R^{js} R^{kt} \epsilon^{mst}[/itex]
    can you show that if [itex]i=j[/itex] is [itex]\epsilon^{ijk}=0[/itex]? Similar for the rest cases?
     
  6. Jan 28, 2016 #5
    For the first part I think the argument will be the same as part two because,

    [itex]d^3x' \equiv (dx'^1)^i (dx'^2)^j (dx'^3)^k = R^{im}R^{js}R^{kt}(dx^1)^m (dx^2)^s (dx^3)^t[/itex]

    So I need to do something to loose those R's (which is the case in part two)

    For part two, by setting i=j then take the sum,

    [itex]Σ\epsilon'^{iik} = ΣR^{im} R^{is} R^{kt} \epsilon^{mst}[/itex]

    [itex]Σ\epsilon'^{iik} = δ^{ms} R^{kt} \epsilon^{mst}[/itex]

    [itex]Σ\epsilon'^{iik} = R^{kt} \epsilon^{mmt}[/itex]

    By definition, [itex]\epsilon^{mmt}=0[/itex] if m=s so, [itex]Σ\epsilon'^{iik} = 0[/itex]

    Is this what you want me to show? Is it correct that by definition [itex]\epsilon^{mmt}=0[/itex] if m=s?
     
  7. Jan 28, 2016 #6

    ChrisVer

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    I don't think that "summing" can help... you are trying to prove that for example [itex]\epsilon'^{11a}=\epsilon'^{22a}=\epsilon'^{33a} =0[/itex] and not [itex]\epsilon'^{11a}+\epsilon'^{22a}+\epsilon'^{33a} =0[/itex]...
    just set the index [itex]\epsilon'^{ijk}[/itex] i=j (don't sum)... and see what you get...

    and yes, by definition that's how it is (that's why it's a totally-antisymmetric tensor)...
    if [itex]\epsilon^{mnr}[/itex] is a totally antisymmetric tensor then the following is true:
    [itex]\epsilon^{mnr}=-\epsilon^{nmr}[/itex] (along with other permutations).
    so if [itex]m=n[/itex] let's say it's m=n=1 then:
    [itex]\epsilon^{11r}=-\epsilon^{11r}[/itex] (you ended up with a=-a, so a must be 0)
     
  8. Jan 28, 2016 #7

    ChrisVer

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    No indices over dx^1,2,3 and dx'^1,2,3 ...
    in other words just replace the dx'h= Rhidxi you obtained but for h=1,2 and 3:
    [itex]d^3 x' = dx'^1 dx'^2 dx'^3 = R^{1i}R^{2j} R^{3k} \epsilon^{ijk} dx^1 dx^2 dx^3[/itex]
    from linear algebra, what is:
    [itex]\epsilon_{ijk}R^{1i}R^{2j} R^{3k}=?[/itex]
     
    Last edited: Jan 28, 2016
  9. Jan 28, 2016 #8
    It should be [itex]\epsilon_{123}[/itex], but by your post above what does [itex]\epsilon^{ijk}=0[/itex] have to do with this?
     
  10. Jan 28, 2016 #9

    ChrisVer

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    nothing, it concerns question 1.
     
  11. Jan 28, 2016 #10
    Oh wait, how come [itex]\epsilon^{ijk}[/itex] appeared from the transformation

    [itex]d^3 x' = dx'^1 dx'^2 dx'^3 = R^{1i}R^{2j} R^{3k} \epsilon^{ijk} dx^1 dx^2 dx^3[/itex]

    and I thought the transformation should involve indices? Why use numbers and introduce epsilon?
     
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