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A question regarding General Stokes' Equation

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]X^{i}[/itex] be a vector field in Minkowski space [itex]R^{4}_{1}[/itex]. We define the integral of this vector over a 3-dimensional hypersurface as the integral of the 3-form [itex]X^{i}dS_{i}[/itex]. where
    [itex]dS_{i}=\frac{1}{6}\sqrt{|g|}ε_{jkli} dx_{j}\lambda dx_{k}\lambda dx_{l}[/itex](don't know how to type exterior derivative)

    Prove that:
    [itex]\int_{∂V} X^{i}dS_{i} = \int_{V} \frac{\partial X^{i}}{\partial x^{i}}dV[/itex]

    From "Modern Geometry, Methods and Applications, Vol I", 26.5 Exercise 2

    2. Relevant equations
    Stokes' Equation


    3. The attempt at a solution
    I thought it was pretty simple so I just simply apply the Stokes' Equation which give me something like this:
    [itex]\int_{V} \frac{\partial X^{i}}{\partial x^{i}}\frac{1}{6}\sqrt{|g|}ε_{jkli} ε_{ijkl} dx_{1}\lambda dx_{2}\lambda dx_{3}\lambda dx_{4}[/itex]

    while we can calculate the value of ε_{jkli} ε_{ijkl} by substituting indexes and considering its relation to the determinant of δ, the coefficient 1/6 is something I cannot get rid of. And it seems the more general form of this equation can be written as:
    [itex]\int_{∂V} X^{i}dS_{i} = \int_{V} ∇_{i} X^{i}dV[/itex] where the coefficient becomes [itex]\frac {1}{(n-1)!}[/itex]. But now I cannot make connections with [itex]\frac {1}{(4-1)!}[/itex]
     
    Last edited: Oct 6, 2012
  2. jcsd
  3. Oct 10, 2012 #2
    I think I got something, although not sure it is correct:
    I made a mistake there that it should be [itex]ε_{imnh} ε_{ijkl}[/itex]. The indexes should not be identical. Then taking advantage of the identity that [itex]ε_{abcd} ε_{efgh} = ε_{abcd} ε_{efgh}δ_{11}δ_{22}δ_{33}δ_{44} = ε_{abcd}δ_{1e}δ_{2f}δ_{3g}δ_{4h}[/itex] which corresponds to a determinant. And in the case of [itex]ε_{imnh} ε_{ijkl}[/itex] we must perform a contraction of the tensor, thereby leaving a (4-1) order determinant which has (4-1)! components of δ with different indexes. However these indexes are identical and can be switched, so this fill explains the coefficient of [itex]\frac{1}{6}[/itex] and can also be extended to the case of n-form
     
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