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I'm trying to understand how the divergence formula in curvilinear coordinates is derived, but unfortunately my textbook doesn't go into much detail. Here is what they show:

I was wondering if someone could answer a couple of questions for me?

Firstly, is the vector field assumed to be constant?

Secondly, how exactly do they get to the line (8.9)? I don't really understand it. Is it done using surface integrals or something simpler?

If someone can answer these questions for me I would really appreciate it!

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    $\begingroup$ The vector field is assumed to be differentiable, which means that in a small enough region we may assume it is linear. If it were constant, then we could expect that the flux in one face was equal to the flux out of the opposite face. $\endgroup$
    – Arthur
    Jul 26, 2017 at 8:03
  • $\begingroup$ Okay that makes sense, thanks! Would you mind explaining to me how they get line (8.9)? $\endgroup$
    – user190290
    Jul 26, 2017 at 8:15
  • $\begingroup$ I'll see what I can do. Just for the record, are you asking about how they set up the left-hand side, how they got from the left-hand side to the right-hand side, or both? By the way, the first line on page 300 seems to have a typo: it should be $h_1h_2h_3$ instead of $h_2h_2h_3$. $\endgroup$
    – Arthur
    Jul 26, 2017 at 8:21
  • $\begingroup$ Thank you for spotting the typo! Very much appreciated :) Yeah all of it I'm afraid :( if you can help in any way that would be great but if you can't that's completely fine too :) $\endgroup$
    – user190290
    Jul 26, 2017 at 8:23

1 Answer 1

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The flux through the left face of the "curvi-cube" is approximately $(A_1h_2h_3dx^2dx^3)|_{(x^1,x^2,x^3)}$ (assuming $A_1$ doesn't change much along the surface spanned by $h_2dx^2, h_3dx^3$, or more specifically, that how $A_1$ changes over the surface doesn't change much from the left to the right face, so we may as well pick a corner to evaluate it), while the flux through the right face is $(A_1h_2h_3dx^2dx^3)|_{(x^1 + dx^1,x^2,x^3)}$. Thus the net flux out if the cube through those two faces is $$ (A_1h_2h_3dx^2dx^3)|_{(x^1 + dx^1,x^2,x^3)} - (A_1h_2h_3dx^2dx^3)|_{(x^1,x^2,x^3)} $$ Since $dx^2$ and $dx^3$ doesn't change between the two faces, we may factor them out. This gives the left-hand side.

As for the right-hand side, we multiply by $1 = \frac{dx^1}{dx^1}$, and note that $\frac{1}{dx^1}$ goes together with the bracket to form the definition of the partial derivative of $A_1h_2h_3$ with respect to $x^1$. The remaining factor $dx^1dx^2dx^3$ stays outside the derivative.

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  • $\begingroup$ Thank you so much for taking the time to help me! I'm just trying to wrap my head around it. I was wondering where on the above diagram you would place the points $(x^1,x^2,x^3)$ and $(x^1+dx^1,x^2,x^3)$? $\endgroup$
    – user190290
    Jul 26, 2017 at 8:49
  • $\begingroup$ Intuitively, on the left and right end respectively of the lowermost edge on the close side of the cube. Although that's just because I think of the coordinate system as oriented the conventional way (positive $x^1$ direction to the right, $x^2$ into the page and $x^3$ upwards) and I think of the $dx^i$ as positive. It could be any of the four left-to-right edges, oriented whichever way you choose. $\endgroup$
    – Arthur
    Jul 26, 2017 at 8:53
  • $\begingroup$ Okay got it, thanks! I might be looking into it too much but does that mean $dx^1 =h_1x^1$? $\endgroup$
    – user190290
    Jul 26, 2017 at 8:59
  • $\begingroup$ Also are we assuming that $A_1, A_2$ and $A_3$ are in the same direction as $x^1, x^2$ and $x^3$ respectively? $\endgroup$
    – user190290
    Jul 26, 2017 at 9:09
  • $\begingroup$ Hmm... I am not sure what to make of the $h_i$, actually. The book says that the cube has dimensions $h_idx^i$, but then they talk about the point $(x^1+dx^1, x^2, x^3)$ instead of $(x^1 + h_1dx^1, x^2, x^3)$. I find that strange. And yes, the vector field $\mathbf A$ has components $A_1, A_2, A_3$ in the directions of $x^1, x^2, x^3$. That's what the components of a vector field are, after all. $\endgroup$
    – Arthur
    Jul 26, 2017 at 9:13

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