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This isn't really isn't strictly a homework question; there wouldn't be solutions provided. I know the answer -- if anything it follows from other electrostatic results. I would like to make sure what I have written down as steps are legitimate.

I will also skip ahead some steps because typing everything out would be very time consuming, especially if I have no idea if the codes will turn out right.

Prove that

[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = 0 [/tex]

[tex]\nabla \times {\bf r} = 0 [/tex]

I begin with

[tex]\nabla \times {\bf r} = 0 [/tex]

Using Levi Civita notations,

[tex]\nabla \times {\bf r} = \epsilon_{ijk} \partial_{j} x_{k}=\epsilon_{ijk} \delta_{jk}[/tex]

Where

[tex] \partial_{j} = \frac{d}{dx_{j}}[/tex]

Here, I reason that for it to have a value, j=k in the kronecker delta, but if that is the case, then the levi civita tensor would be zero, hence, curl of a 1/r is zero.

Next,

[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = \epsilon_{ijk} \partial_{j} (x_{k} (x_{b} x_{b})^{-3/2}) [/tex]

[tex]=\epsilon_{ijk}( (x_{b} x_{b})^{-5/2} \partial_{j}(x_{k}) - 3x_{k}x_{j}(x_{k} (x_{b} x_{b})^{-5/2}[/tex]

[tex]=\frac{1}{r^{3}}\nabla \times {\bf {r}} - \frac{3}{r^5} {\bf {r}} \times {\bf r} = 0-3(0) = 0[/tex]

I will also skip ahead some steps because typing everything out would be very time consuming, especially if I have no idea if the codes will turn out right.

## Homework Statement

Prove that

[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = 0 [/tex]

[tex]\nabla \times {\bf r} = 0 [/tex]

## Homework Equations

## The Attempt at a Solution

I begin with

[tex]\nabla \times {\bf r} = 0 [/tex]

Using Levi Civita notations,

[tex]\nabla \times {\bf r} = \epsilon_{ijk} \partial_{j} x_{k}=\epsilon_{ijk} \delta_{jk}[/tex]

Where

[tex] \partial_{j} = \frac{d}{dx_{j}}[/tex]

Here, I reason that for it to have a value, j=k in the kronecker delta, but if that is the case, then the levi civita tensor would be zero, hence, curl of a 1/r is zero.

Next,

[tex]\nabla \times \frac{\bf \hat r}{r^{2}} = \epsilon_{ijk} \partial_{j} (x_{k} (x_{b} x_{b})^{-3/2}) [/tex]

[tex]=\epsilon_{ijk}( (x_{b} x_{b})^{-5/2} \partial_{j}(x_{k}) - 3x_{k}x_{j}(x_{k} (x_{b} x_{b})^{-5/2}[/tex]

[tex]=\frac{1}{r^{3}}\nabla \times {\bf {r}} - \frac{3}{r^5} {\bf {r}} \times {\bf r} = 0-3(0) = 0[/tex]

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