1
$\begingroup$

A textbook I'm studying with described finding vector magnetic potential $\vec{\text{A}}$ from Biot-Savart law as below.

$\vec{\text{H}_2}=\int_{\text{vol}}\frac{\vec{\text{J}_1}\times\hat{\text{a}_{\text{R}12}}}{4\pi\text{R}_{12}}dv_1$
      $=\frac{1}{4\pi}\int_{\text{vol}}\vec{\text{J}_1}\times(-\nabla_2\frac{1}{\text{R}_{12}})dv_1$
      $=\frac{1}{4\pi}\int_{\text{vol}}[(\nabla_2\times\frac{\vec{\text{J}_1}}{{\text{R}_{12}}})-\frac{1}{\text{R}_{12}}(\nabla_2\times\vec{\text{J}_1})]dv_1$
      $=\frac{1}{4\pi}\int_{\text{vol}}(\nabla_2\times\frac{\vec{\text{J}_1}}{{\text{R}_{12}}})dv_1$             ($\because\nabla_2\times\vec{\text{J}_1}=0$)

  • $\vec{\text{H}_2}$: magnetic field intensity on point 2 ($x_2$, $y_2$, $z_2$)
  • $\vec{\text{J}_1}$: current density on point 1 ($x_1$, $y_1$, $z_1$)
  • $dv_1$: infinitesimal volume on point 1 (equal to $dx_1dy_1dz_1$)
  • $\text{R}_{12}$: distance between point 1 and 2 (equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$)
  • $\nabla_2$: del operator which takes derivation on $x_2$, $y_2$, $z_2$

It also described about verifying Ampere's law with $\vec{\text{A}}$ as below.

$\nabla\times\vec{\text{H}}=\frac{1}{\mu_0}\nabla\times\nabla\times\vec{\text{A}}=\frac{1}{\mu_0}[\nabla(\nabla\cdot\vec{\text{A}})-\nabla^2\vec{\text{A}}]$

$\rightarrow\nabla_2\cdot\vec{\text{A}_2}=\frac{\mu_0}{4\pi}\int_{\text{vol}}\nabla_2\cdot\frac{\vec{\text{J}_1}}{{\text{R}_{12}}}dv_1$
                    $=\frac{\mu_0}{4\pi}\int_{\text{vol}}[\vec{\text{J}_1}\cdot(\nabla_2\frac{1}{\text{R}_{12}})+\frac{1}{\text{R}_{12}}(\nabla_2\cdot\vec{\text{J}_1})]dv_1$
                    $=\frac{\mu_0}{4\pi}\int_{\text{vol}}\vec{\text{J}_1}\cdot(\nabla_2\frac{1}{\text{R}_{12}})dv_1$            ($\because\nabla_2\cdot\vec{\text{J}_1}=0$)
                    $=\frac{\mu_0}{4\pi}\int_{\text{vol}}\vec{\text{J}_1}\cdot(-\nabla_1\frac{1}{\text{R}_{12}})dv_1$            ($\because\nabla_1\frac{1}{\text{R}_{12}}=-\nabla_2\frac{1}{\text{R}_{12}}$)

  • $\nabla_1$: del operator which takes derivation on $x_1$, $y_1$, $z_1$

As you can see, there are some expressions that result zero just because the variables are different to each other ($x_1$, $y_1$, $z_1$ and $x_2$, $y_2$, $z_2$).

This made me wonder, why distinguish point 1's coordinate as ($x_1$, $y_1$, $z_1$) and point 2's as ($x_2$, $y_2$, $z_2$)? Are they not in same coordinate system?

If they are in same coordinate system, why can't point 1's coordinate variables take derivation on point 2's and vice versa?

And if they aren't, how can $\text{R}_{12}$ exist that connects point 1 and point 2 when they are in different coordinate system?

$\endgroup$
2
$\begingroup$

Those two points are in the same coordinate system, but different and independent points nonetheless. Taking the derivative $\nabla_2$ with respect to point 2 means "how much does the given expression change if I vary point 2". And point 1 does not change if you vary point 2, therefore the derivatives $\nabla_2 x_1, \nabla_2 y_1, \nabla_2 z_1$ are zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.