1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook