I am working on a project on which I need to calculate the geomagnetic field in different coordinates. When I use the conventional form of the dipole field in spherical coordinates: $$\vec{B}_{r,\phi}=\frac{B_0 R^3}{r^3}(-2 \cos \phi\, \hat{e}_r- \sin \phi\, \hat{e}_\phi),$$ with $B_0$ as the field intensity in the equator, $R$ the radius of the Earth, $r$ is an arbitrary distance from the center from the center, $\phi$ is the polar angle. The divergence of this field is zero. $$\nabla \cdot \vec{B}_{r,\phi}=0.$$ When I transform it to cartesian coordinates $\hat{I}$,$\hat{J}$,$\hat{K}$ $$\vec{B}_{X,Y,Z}=\frac{B_0 R^3}{(X^2+Y^2+Z^2)^{5/2}}[-3 X Z\, \hat{I} -3 YZ\, \hat{J}+(X^2+Y^2-2Z^2)\hat{K}],$$ the divergence of this field is also zero $$\nabla \cdot \vec{B}_{X,Y,Z}=0$$ The transformation on which I am having trouble is happening when I convert the $\vec{B}_{X,Y,Z}$ to the cyllindrical orbital coordinates through and ordinary rotation matrix 3-1-3, wehre the first rotation is about the $\hat{K}$-axis, the longitude of ascending node $\Omega$, the second around the new $\hat{X}^\prime$-axis, the inclination $i$, and the final rotation around the $\hat{K}^\prime$-axis, is the sum of the periapsos argument $\omega$ and the true anomaly $f$, so the total angle of this final rotation is $\omega+f$, and the variables $X$, $Y$ and $Z$ are given by: $$X=r (\sin (f) (\cos (i) \cos (\omega ) \sin (\Omega )+\sin (\omega ) \cos (\Omega ))+\cos (f) (\cos (i) \sin (\omega ) \sin (\Omega )-\cos (\omega ) \cos (\Omega )));$$ $$Y=-r(\cos (i) \cos (\Omega ) \sin (f+\omega )+\sin (\Omega ) \cos (f+\omega ));$$ $$Z=-r \sin (i) \sin (f+\omega ).$$ When I do these transformations, I obtain a magnetic field in this form: $$\vec{B}_{r,f}=\frac{B_0 R^3}{r^3}[-2 \sin i \sin (\omega+f)\, \hat{r} + \sin i \cos(\omega+f)\, \hat{f} + \cos i\, \hat{k}],$$ since is a cyllindrical frame, I choosed to name the the basis as $\hat{r}$, $\hat{f}$ and $\hat{k}$, wehre the variable $r$ has the same interpretation of distance and the variable $f$, the true anomaly is the angle analogous to an angle of cyllindrical coordinates.

However when I take the divergence of $\vec{B}_{r,f}$, it is not equals zero, actually: $$\nabla \cdot \vec{B}_{r,f}=\frac{3 B_0 R^3 \sin i \sin (\omega+f)}{r^4}.$$ I know that Maxwell's equations states that the divergence of any magnetic field is zero so I am just wondering why this is happening, I am pretty sure that the transformations are correct, but I can't actually understand why the divergence of this field is non-zero.

Obs: I also made the transformation using the electromagnetic field tensor formalism for the rotation and it gives the same result as the commom vector transformation by a matrix of rotation.

  • 6
    $\begingroup$ Since $\nabla \cdot B = 0$ is written in terms of covariant objects like the divergence and a vector field $B$, if that equation holds in one coordinate system, then it must hold in every coordinate system. Therefore, your transformation from Cartesian to cylindrical coordinates cannot be correct. $\endgroup$
    – Andrew
    Mar 26 at 3:08
  • 2
    $\begingroup$ I suspect the issue lies not in the transformation into cylindrical coordinates, but in taking the divergence incorrectly in curvilinear coordinates. For instance, it's no longer true that divergence is $\sum_i\frac{\partial}{\partial x_i}$ in more general coordinates $\endgroup$
    – DanDan0101
    Mar 26 at 3:24
  • $\begingroup$ @Andrew, It's not a simple transformation to cylindrical coordinates, it actually is a rotation through the three angles $\Omega$, $i$, $\omega$, before the transformation to cylindrical coordinates, and since the unitary transformation from Cartesian coordinates to cylindrical coordinates is a rotation around the $\hat{z}$-axis, it happens that this final rotation adds up to the $\omega$ rotation. $\endgroup$ Mar 26 at 4:17
  • $\begingroup$ @DanDan0101, I'm using the software Wolfram Mathematica, specifically the function Div[$\vec{B}_{r,f}$,{r,f,z}, "Cylindrical"] so it's not a problem of generalization through different coordinates. $\endgroup$ Mar 26 at 4:20
  • $\begingroup$ @EricD'Antona The point is the same. A rotation is a coordinate transformation as well. If the divergence is zero in one frame, it must be zero in all frames. However, Kurt G.'s answer indicates that the divergence of your field in Cartesian coordinates may not be zero, which if correct indicates that there is an error in the transformation from spherical to Cartesian coordinates. $\endgroup$
    – Andrew
    Mar 26 at 15:42

1 Answer 1


I confirm that the divergence of $$\tag{1} \vec{B}_{X,Y,Z}=\frac{B_0 R^3}{(X^2+Y^2+Z^2)^{5/2}}[-3X Z\, \hat{I} -3YZ\, \hat{J}+(X^2+Y^2-2Z^2)\hat{K}], $$ which I prefer to write as $$\tag{2} B(x,y,z)=\frac{1}{(x^2+y^2+z^2)^{5/2}}\begin{pmatrix}-3xz\\-3yz\\x^2+y^2-2z^2\end{pmatrix} $$ is zero: \begin{align} \nabla\cdot B&=\frac{-3z-3z-4z}{(x^2+y^2+z^2)^{5/2}}-\frac{5}{2}\frac{-6x^2z-6y^2z+2z(x^2+y^2-2z^2)}{(x^2+y^2+z^2)^{7/2}}\\[2mm] &=\frac{-10z(x^2+y^2+z^2)+15x^2z+15y^2z-5x^2z-5y^2z+10z^3}{(x^2+y^2+z^2)^{7/2}}=0\,.\tag{3} \end{align} The conversion of the components of any vector field from Cartesian coordinates $(x,y,z)$ into cylindrical coordinates $(\rho,\varphi,z)$ is done as follows: \begin{align}\tag{4} \begin{pmatrix}B_\rho\\B_\varphi\\B_z\end{pmatrix}=\begin{pmatrix}\cos\varphi & \sin\varphi& 0\\-\sin\varphi &\cos\varphi&0\\0&0&1\end{pmatrix} \begin{pmatrix}B_x\\B_y\\B_z\end{pmatrix}\,. \end{align} With (2) we get \begin{align} \begin{pmatrix}B_\rho\\B_\varphi\\B_z\end{pmatrix}&= \frac{1}{(x^2+y^2+z^2)^{5/2}}\begin{pmatrix}-3xz\cos\varphi-3yz\sin\varphi\\3xz\sin\varphi-3yz\cos\varphi\\x^2+y^2-2z^2 \end{pmatrix}\\ &=\frac{1}{(\rho^2+z^2)^{5/2}}\begin{pmatrix}-3\rho z\cos^2\varphi-3\rho z\sin^2\varphi\\ 3\rho z\cos\varphi\sin\varphi-3\rho z\cos\varphi\sin\varphi\\ \rho^2-2z^2 \end{pmatrix}\\ &=\frac{1}{(\rho^2+z^2)^{5/2}}\begin{pmatrix}-3\rho z\\0\\\rho^2-2z^2\end{pmatrix}\,.\tag{5} \end{align} The divergence in cylindrical coordinates is $$ \nabla\cdot B=\frac{\partial_\rho(\rho B_\rho)}{\rho}+\frac{\partial_\varphi B_\varphi}{\rho}+\partial_z B_z\,.\tag{6} $$ Using (5) this is zero according to Wolfram Alpha as it must.

  • $\begingroup$ Howevwr, for my problem I need to apply those rotatione I cited $\Omega$, $i$ and $\omega$ in order to derive a formula for the geomagnetic field in the orbital frame, also known as perifocal frame. Then, in this perifocal frame transform into cylindrical coordinates, where $f$ is the true anomaly. I used a conventional rotation matrix of 3-1-3 type to make this first transformation $R(\Omega, i, \omega)$, and since the unitary transformation of cartesian to cylindrical is identical to the it adds up to the previous equation as $R(\Omega,i,\omega+f)$. $\endgroup$ Mar 26 at 20:08
  • $\begingroup$ @EricD'Antona Had to correct (5) which does not change the fact that divergence is zero. Regarding your nomenclature: I have honestly no clue what an orbital or perifocal frame is and what your variables $\Omega,i,\omega,f$ specify. We have now checked $B$ in three different coordinate systems which are well-known (spherical, Cartesian, cylindrical). You may want to take that for granted and give more background on the non-standard stuff you are using. $\endgroup$
    – Kurt G.
    Mar 26 at 21:34
  • $\begingroup$ I am assuming your conversion to your cylindrical coordinates also works for $i=0$ and that your $f$ is my $\varphi$. Your equation $$\vec{B}_{r,f}=\frac{B_0 R^3}{r^3}[-2 \sin i \sin (\omega+f)\, \hat{r} + \sin i \cos(\omega+f)\, \hat{f} + \cos i\, \hat{k}]$$ becomes then $$\vec{B}_{r,f}=\frac{B_0 R^3}{r^3}[0\, \hat{r} + 0\, \hat{f} + \hat{k}]$$ which is different from (5). This is the heart of the problem. Your conversion looks wrong. $\endgroup$
    – Kurt G.
    Mar 28 at 17:30
  • $\begingroup$ G, actually it is right, because if one sets $i=0$, then the inclination relative to the equator is zero, it implies that $z=0$, if one sets $z=0$ in your equation (5), then we'll find both the same result. $\endgroup$ Mar 29 at 3:38
  • $\begingroup$ For the life of me I don't see how $\cos i$ is zero when $i=0.$ You haven't aswered my question what $\Omega, i,\omega,f$ specify. $\endgroup$
    – Kurt G.
    Mar 29 at 5:59

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