- #1
mingshey
- 5
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Jackson("Classical Electrodynamics", Ch.6)
uses the theorem of curl of curl to separate current density into transverse and parallel,
[tex]\vec J = \vec{J_p}+\vec{J_t}[/tex] to say,
[tex]\begin{align*}\vec{J}(\vec{x}) &= \int\vec{J}(\vec{x'})\delta(\vec{x}-\vec{x'})d^{3}x'\\
&= -{1\over{4\pi}}\int\vec{J}(\vec{x'})\nabla^2 \left({1\over|\vec{x}-\vec{x'}|}\right)d^{3}x'
\end{align*}[/tex]
Since the del is about [tex]x[/tex] and independent of the integral variable,
[tex]\begin{align*}{}&=-{1\over{4\pi}}\nabla^2\int{\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'
\end{align*}[/tex]
And using the theorem
[tex]\nabla\times(\nabla\times\vec{A})=\nabla(\nabla\cdot\vec{A})-\nabla^2\vec{A}[/tex]
[tex]\begin{align*}\vec{J}(\vec{x}) &=
{1\over{4\pi}}\nabla\times\nabla\times\int{\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'-{1\over{4\pi}}\nabla\left(\nabla\cdot\int{\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'\right)\end{align*}[/tex]
But here Jackson take some hidden procedure to get from the second term of ther right side
[tex]-{1\over{4\pi}}\nabla\left(\int{\nabla'\cdot\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'\right)={1\over{4\pi}}\nabla\left(\int{\partial\rho(\vec{x'})/\partial t
\over|\vec{x}-\vec{x'}|}d^{3}x'\right)[/tex]
to use the continuity theorem to get the term about a time derivative of charge density at [tex]x'[/tex].
And I cannot see how is the differential about [tex]x[/tex] changed into a differential about [tex]x'[/tex] and got inside the integral, and is only applied to the current density, but not the denominator.
Can somebody explain it for me? Thank you.
uses the theorem of curl of curl to separate current density into transverse and parallel,
[tex]\vec J = \vec{J_p}+\vec{J_t}[/tex] to say,
[tex]\begin{align*}\vec{J}(\vec{x}) &= \int\vec{J}(\vec{x'})\delta(\vec{x}-\vec{x'})d^{3}x'\\
&= -{1\over{4\pi}}\int\vec{J}(\vec{x'})\nabla^2 \left({1\over|\vec{x}-\vec{x'}|}\right)d^{3}x'
\end{align*}[/tex]
Since the del is about [tex]x[/tex] and independent of the integral variable,
[tex]\begin{align*}{}&=-{1\over{4\pi}}\nabla^2\int{\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'
\end{align*}[/tex]
And using the theorem
[tex]\nabla\times(\nabla\times\vec{A})=\nabla(\nabla\cdot\vec{A})-\nabla^2\vec{A}[/tex]
[tex]\begin{align*}\vec{J}(\vec{x}) &=
{1\over{4\pi}}\nabla\times\nabla\times\int{\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'-{1\over{4\pi}}\nabla\left(\nabla\cdot\int{\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'\right)\end{align*}[/tex]
But here Jackson take some hidden procedure to get from the second term of ther right side
[tex]-{1\over{4\pi}}\nabla\left(\int{\nabla'\cdot\vec{J}(\vec{x'})
\over|\vec{x}-\vec{x'}|}d^{3}x'\right)={1\over{4\pi}}\nabla\left(\int{\partial\rho(\vec{x'})/\partial t
\over|\vec{x}-\vec{x'}|}d^{3}x'\right)[/tex]
to use the continuity theorem to get the term about a time derivative of charge density at [tex]x'[/tex].
And I cannot see how is the differential about [tex]x[/tex] changed into a differential about [tex]x'[/tex] and got inside the integral, and is only applied to the current density, but not the denominator.
Can somebody explain it for me? Thank you.
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