Why f is no longer a function of x and y ?

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SUMMARY

The discussion centers on the mathematical concept of functions of multiple variables, specifically addressing why a function \( f(x,y) \) can appear to lose its dependency on \( x \) and \( y \) when restricted to a defined rectangular region. Participants clarify that while \( f \) remains a function of \( x \) and \( y \), the derivatives may be zero due to the nature of dummy variables in integration, as illustrated by the Biot-Savart Law and the example of Coulomb's force. The distinction between primed and unprimed variables is emphasized, highlighting that \( J \) is a function of \( x' \) rather than \( x \).

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  • #31
PeroK said:
The same difference as differentiating with respect to ##y_1## and ##y_2## in the above examples.
That solves my whole problem. Thank you so much.
 
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  • #32
Adesh said:
So, what’s the difference between differentiating with respect to ##x## and ##x’## ?
In that response I was trying to illustrate to you the way we think of axes, but:
Define ##\vec a=f(x)\vec e_1=(f(x), 0)## and ##\vec c=g(x')\vec e_1=(g(x'),0)##. You know how to do differentiation on vectors as you are reading Griffith's book, so what do you think?
Adesh said:
That solves my whole problem. Thank you so much.
I think the main point that you need to remember is to think of the axes as the set of real numbers. The function you have of both ##x## and ##x'## who are on the same "axis" can be seen as ##\vec a = f(x, x')\vec e##.
 
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  • #33
On #4
\nabla\times J=[\frac{\partial J_y(x',y',z')}{\partial z}-\frac{\partial J_z(x',y',z')}{\partial y}]\hat x + ...
is zero.
 
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