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Solving with index notation

  1. May 8, 2005 #1
    I've been stuck with this problem since a while.. thought I'd ask here;

    [tex]\nabla \times \dfrac{\vec{A} \times \vec{r}}{2}[/tex]
    solving normally isn't any problem, but I have to do it with index notation, where A is an arbitrary vector field and r is the position vector)

    This is how far I can come:
    (leaving out the vector-lines above A and r)
    [tex]\frac{1}{4}(\nabla \times (A \times r)) = \frac{1}{4}\epsilon_{ijk}\cdot \partial_{j}(A \times r)_k = \frac{1}{4}\epsilon_{ijk}\cdot \partial_{j}(\epsilon_{klm} \cdot A_l \cdot r_m) = \frac{1}{4}\epsilon_{ijk} \epsilon_{klm}\cdot \partial_{j}(A_l \cdot r_m) = \frac{1}{4}\epsilon_{ijk} \epsilon_{klm}(r_m \cdot \partial_j A_l + A_l \cdot \partial_j r_m)[/tex]

    But then what...? I'm not even sure I'm allowed to bring in that [tex]\epsilon_{klm}[/tex]. I'm very new to this notation, and don't know much more than einstein's summation convention.
     
  2. jcsd
  3. May 8, 2005 #2

    xanthym

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    SOLUTION HINTS:

    You did very well so far!!
    When indexing a vector result, remember to indicate the LEFT side component with the FREE index ("i" in this case). Thus, your result should have been (note the "i" subscript added on the LEFT side):
    (Note: To save time, the constant fraction factor is not shown.)

    [tex] 1: \ \ \ \ (\nabla \, \times \, (\vec{A} \, \times \, \vec{r}))_{\displaystyle i} \ \ = \ \ \epsilon_{ijk} \, \epsilon_{klm} \, (r_m \cdot \partial_j A_l \ + \ A_l \cdot \partial_j r_m) \ \ = \ \ \epsilon_{kij} \, \epsilon_{klm} \, (r_m \cdot \partial_j A_l \ + \ A_l \cdot \partial_j r_m)[/tex]

    What to do next?? Almost always at this point, the following identity is invoked:

    [tex] 2: \ \ \ \ \epsilon_{\displaystyle kij} \, \epsilon_{\displaystyle klm} \ \, = \ \, (\delta_{\displaystyle il}\,\delta_{\displaystyle jm} \, - \, \delta_{\displaystyle im}\,\delta_{\displaystyle jl}) [/tex]

    Placing the above identity into Eq 1, we get:

    [tex] 3: \ \ \ \ (\nabla \, \times \, (\vec{A} \, \times \, \vec{r}))_{\displaystyle i} \ \ = \ \ (\delta_{il}\,\delta_{jm} \, - \, \, \delta_{im}\,\delta_{jl}) \cdot (r_m \cdot \partial_j A_l \ + \ A_l \cdot \partial_j r_m) [/tex]

    [tex] 4: \ \ \ \ \ \Longrightarrow \ \ \ (\nabla \, \times \, (\vec{A} \, \times \, \vec{r}))_{\displaystyle i} \ \ = \ \ (\delta_{il}\,\delta_{jm})\cdot(r_m \cdot \partial_j A_l) \ \, + \ \, (\delta_{il}\,\delta_{jm})\cdot(A_l \cdot \partial_j r_m) \ \, - \ \, (\delta_{im}\,\delta_{jl})\cdot(r_m \cdot \partial_j A_l) \ \, - \ \, (\delta_{im}\,\delta_{jl})\cdot(A_l \cdot \partial_j r_m) [/tex]

    FIRST term on the right of Eq 4 is simplified below. Use this example to further simplify the entire expression and derive the final result.

    [tex] 5: \ \ \, \ (\delta_{\displaystyle il}\,\delta_{\displaystyle jm})\cdot(r_{\displaystyle m} \cdot \partial_{\displaystyle j} A_{\displaystyle l}) \ \ = \ \ (r_{\displaystyle j} \cdot \partial_{\displaystyle j} A_{\displaystyle i}) \ \ \ \ \ \ \ \ \ \ \ \ \color{red} \textsf{(Example 1st term on right Eq 4)} [/tex]

    (Remember to apply the constant fraction factor (1/2) when finished.)

    ~~
     
    Last edited: May 8, 2005
  4. May 8, 2005 #3

    dextercioby

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    Where does that [itex] \frac{1}{4} [/itex] come from ?To my mind,there was supposed to be only [itex]\frac{1}{2} [/itex].

    Daniel.
     
  5. May 8, 2005 #4

    xanthym

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    Good observation. "Sunshine" should make that correction for the final result (see Msg #2).


    ~~
     
    Last edited: May 8, 2005
  6. May 8, 2005 #5
    Thanks xanthym... extremely good explanation! I wasn't aware of identity 2.

    As for the 1/2 that became 1/4, I was thinking of the equation as [tex]\nabla \times (\dfrac{\vec{A}}{2} \times \dfrac{\vec{r}}{2}) [/tex] whick isn't the case. Thank you for noticing. :)
     
  7. May 9, 2005 #6
    stuck with the same problem: I know that

    [tex]\partial_{\displaystyle i} A_{\displaystyle i}[/tex] means [tex]\nabla \cdot A[/tex] but is [tex]\partial_{\displaystyle j} A_{\displaystyle i}[/tex] equivalent to [tex]\nabla A[/tex]?
     
  8. May 9, 2005 #7

    dextercioby

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    Nope,that's a second rank tensor in cartesian coordinates.You can see it has 9 components,i.e.the components of the GRADIENT of the vector field [itex] \vec{A} [/itex].

    Daniel.
     
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