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Homework Help: Smarter way to solve a continuity equation?

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data
    The density in 3-D space of a certain kind of conserved substance is given by
    [tex]\[\rho (x,y,z, t) = At^{-\frac{3}{2}}e^{-\frac{r^2}{4kt}}\][/tex]

    where [tex]\mathbf r = x\mathbf i + y\mathbf j +z\mathbf k[/tex] and [tex] r = |\mathbf r|[/tex]. The corresponding flux vector is given by
    [tex]\mathbf J(\mathbf r, t) = Bt^{-\frac{5}{2}}e^{-\frac{r^2}{4kt}}\mathbf r[/tex]
    Here, A, B, k and positive constants.


    2. Relevant equations

    Show that [tex]$\rho, \mathbf J$[/tex] satisfy the conservation equation [tex]\frac{\partial \rho}{\partial t}[/tex][tex] + \nabla \cdot \mathbf J = 0[/tex] only if [tex]$ A = 2B$[/tex]

    3. The attempt at a solution
    So I've looked at this, found the derivative for the density function, had a fair play with the div function, i'm just wondering if there is a smarter way to solve this then actually deriving the partial derivative and the div function and re-arranging? I have a feeling there is something inherent, for example like the divergence theorum, that i can use?

    mind you in the time it took me to get the tex working i could have solved the thing, but i'm still curious
     
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 29, 2010 #2
    One smarter way would be by using (first understanding the derivation, of course) Eqs. (49,50) from the entry "http://mathworld.wolfram.com/SphericalCoordinates.html" [Broken]".
     
    Last edited by a moderator: May 5, 2017
  4. Sep 29, 2010 #3
    Hmm. To be honest I dont know where to start. For example, how to i manage the position vector? Do i replace x, y, z with [tex] r cos \theta sin \phi[/tex] etc, then do the divergence? It starts to look very messy ... [tex] t^{-\frac{5}{2}}e^{-\frac{r}{4kt}} r cos \theta sin \phi[/tex] for the x component for example
     
  5. Sep 30, 2010 #4
    [tex]r[/tex] in the exponential is not the position vector. It is its length - one of the coordinates of the spherical system. It will stay as such. [tex]\theta,\phi[/tex] simply do not appear in the formula - which simplifies the calculations.
     
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