Skip to main content
Post Reopened by Qmechanic
edited title; edited tags
Link
Qmechanic
  • 212.9k
  • 48
  • 589
  • 2.3k

Representing $\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}|^3}$ in polar coordinates

deleted 263 characters in body; edited tags
Source Link
QED
  • 323
  • 1
  • 11

Is there a wayIn his book introduction to representelectrodynamics, Griffiths uses derives the vector field rr|rr|32 where r is a constant vector and r isidentity ˆrr2=4πδ3(r) Using the position vectorformula for divergence in polar coordinates in terms of the basis vectors \hat{\mathbf{r}},\hat{\mathbf{\theta}}\boldsymbol{\hat{\phi}}?

The reason I'm asking this is for deriving the identity \nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}\right)=4\pi\delta^3(\mathbf{r-r'}).

I know how to show. He then states that "more generally" \nabla \cdot \frac{\mathbf{\hat{r_s}}}{r_s^2} = 4\pi\delta^3(\mathbf{r_s}) Where \nabla\cdot\frac{\hat{\mathbf{r}}}{r^2}\mathbf{r_s} is \delta^3(\mathbf{r}) using polar coordinates, and I loosely understand how changing the argument to \mathbf{r-r'} should "shift" the divergence becauseseperation vector, \frac{\partial }{\partial x}\left(f(x-a)\right)=f_x(x-a)\mathbf{r}-\mathbf{r'} and I can show this\mathbf{r'} is the casea constant. I'm trying to find a way to derive this identity using cartesianpolar coordinates, but I was wondering how to use polar coordinatesam struggling to solve this. My main issue is findingfind a way to relate the valuesrepresent \frac{\mathbf{r-r'}}{{|r-r'|^3}} in terms of the vector field to the basis vectors, since in the case wherepolar coordinates \mathbf{r'}=\mathbf{0} they always point(\mathbf{\hat{r}}, \hat{\boldsymbol{\phi}},\hat{\boldsymbol{\theta}}) as v_r\mathbf{\hat{r}}+v_\theta\boldsymbol{\hat{\theta}}+v_\phi\boldsymbol{\hat{\phi}} for some scalar functions v_r, v_{\theta}, v_{\phi} to use in the radial direction, butformula for divergence in general I don't see how to derive a relationship between the vector field and the polar basis vectorscoordinates.

Is there a way to represent the vector field \frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}} where \mathbf{r'} is a constant vector and \mathbf{r} is the position vector in polar coordinates in terms of the basis vectors \hat{\mathbf{r}},\hat{\mathbf{\theta}}\boldsymbol{\hat{\phi}}?

The reason I'm asking this is for deriving the identity \nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}\right)=4\pi\delta^3(\mathbf{r-r'}).

I know how to show that \nabla\cdot\frac{\hat{\mathbf{r}}}{r^2} is \delta^3(\mathbf{r}) using polar coordinates, and I loosely understand how changing the argument to \mathbf{r-r'} should "shift" the divergence because \frac{\partial }{\partial x}\left(f(x-a)\right)=f_x(x-a) and I can show this is the case using cartesian coordinates, but I was wondering how to use polar coordinates to solve this. My main issue is finding a way to relate the values of the vector field to the basis vectors, since in the case where \mathbf{r'}=\mathbf{0} they always point in the radial direction, but in general I don't see how to derive a relationship between the vector field and the polar basis vectors.

In his book introduction to electrodynamics, Griffiths uses derives the identity \nabla \cdot \frac{\mathbf{\hat{r}}}{r^2} = 4\pi\delta^3(\mathbf{r}) Using the formula for divergence in polar coordinates. He then states that "more generally" \nabla \cdot \frac{\mathbf{\hat{r_s}}}{r_s^2} = 4\pi\delta^3(\mathbf{r_s}) Where \mathbf{r_s} is the seperation vector, \mathbf{r}-\mathbf{r'} and \mathbf{r'} is a constant. I'm trying to find a way to derive this identity using polar coordinates, but am struggling to find a way to represent \frac{\mathbf{r-r'}}{{|r-r'|^3}} in terms of the basis vectors in polar coordinates (\mathbf{\hat{r}}, \hat{\boldsymbol{\phi}},\hat{\boldsymbol{\theta}}) as v_r\mathbf{\hat{r}}+v_\theta\boldsymbol{\hat{\theta}}+v_\phi\boldsymbol{\hat{\phi}} for some scalar functions v_r, v_{\theta}, v_{\phi} to use in the formula for divergence in polar coordinates.

Post Undeleted by Qmechanic
Post Deleted by QED
Post Closed as "Needs details or clarity" by Qmechanic
added 23 characters in body; edited tags; edited title
Source Link
Qmechanic
  • 212.9k
  • 48
  • 589
  • 2.3k

Representing $\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}$ in polar coordinates

Is there a way to represent the vector field \frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{3}}\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}} where \mathbf{r'} is a constant vector and \mathbf{r} is the position vector in polar coordinates in terms of the basis vectors \hat{\mathbf{r}},\hat{\mathbf{\theta}}\boldsymbol{\hat{\phi}}?

The reason I'm asking this is for deriving the identity \nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{3}}\right)=4\pi\delta^3(\mathbf{r-r'})\nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}\right)=4\pi\delta^3(\mathbf{r-r'}).

I know how to show that \nabla\cdot\frac{\hat{\mathbf{r}}}{r^2} is \delta(\mathbf{r})\delta^3(\mathbf{r}) using polar coordinates, and I loosely understand how changing the argument to \mathbf{r-r'} should "shift" the divergence because \frac{\partial }{\partial x}\left(f(x-a)\right)=f_x(x-a) and I can show this is the case using cartesian coordinates, but I was wondering how to use polar coordinates to solve this. My main issue is finding a way to relate the values of the vector field to the basis vectors, since in the case where \mathbf{r'}=\mathbf{0} they always point in the radial direction, but in general I don't see how to derive a relationship between the vector field and the polar basis vectors.

Representing \frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{3}} in polar coordinates

Is there a way to represent the vector field \frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{3}} where \mathbf{r'} is a constant vector and \mathbf{r} is the position vector in polar coordinates in terms of the basis vectors \hat{\mathbf{r}},\hat{\mathbf{\theta}}\boldsymbol{\hat{\phi}}?

The reason I'm asking this is for deriving the identity \nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{3}}\right)=4\pi\delta^3(\mathbf{r-r'})

I know how to show that \nabla\cdot\frac{\hat{\mathbf{r}}}{r^2} is \delta(\mathbf{r}) using polar coordinates, and I loosely understand how changing the argument to \mathbf{r-r'} should "shift" the divergence because \frac{\partial }{\partial x}\left(f(x-a)\right)=f_x(x-a) and I can show this is the case using cartesian coordinates, but I was wondering how to use polar coordinates to solve this. My main issue is finding a way to relate the values of the vector field to the basis vectors, since in the case where \mathbf{r'}=\mathbf{0} they always point in the radial direction, but in general I don't see how to derive a relationship between the vector field and the polar basis vectors.

Representing $\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}$ in polar coordinates

Is there a way to represent the vector field \frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}} where \mathbf{r'} is a constant vector and \mathbf{r} is the position vector in polar coordinates in terms of the basis vectors \hat{\mathbf{r}},\hat{\mathbf{\theta}}\boldsymbol{\hat{\phi}}?

The reason I'm asking this is for deriving the identity \nabla \cdot \left(\frac{\mathbf{r-r'}}{|\mathbf{r-r'}|^{\frac{3}{2}}}\right)=4\pi\delta^3(\mathbf{r-r'}).

I know how to show that \nabla\cdot\frac{\hat{\mathbf{r}}}{r^2} is \delta^3(\mathbf{r}) using polar coordinates, and I loosely understand how changing the argument to \mathbf{r-r'} should "shift" the divergence because \frac{\partial }{\partial x}\left(f(x-a)\right)=f_x(x-a) and I can show this is the case using cartesian coordinates, but I was wondering how to use polar coordinates to solve this. My main issue is finding a way to relate the values of the vector field to the basis vectors, since in the case where \mathbf{r'}=\mathbf{0} they always point in the radial direction, but in general I don't see how to derive a relationship between the vector field and the polar basis vectors.

added 4 characters in body
Source Link
QED
  • 323
  • 1
  • 11
Loading
Source Link
QED
  • 323
  • 1
  • 11
Loading