I have a scalar magnetic field in a volume expressed by the formula

$$B(x,y)=B_0 + \frac{\partial B}{\partial x}(x-x_0) + \frac{\partial B}{\partial y}(y-y_0)$$

which approximates the non-homogeneity of the magnetic field with linear gradients. This is defined in a specific frame of reference in one plane perpendicular to the $z$-axis. Multiple planes will provide the magnetic field in the volume.

I need to rotate this formula around the $z$-axis with an angle $\phi$. Can this be done and how will the derivatives transform in this case? How will this be done with higher order derivatives (secondary question)?


The solution I found is to rotate the gradient. If the gradient is defined by

$$\vec{\nabla}B=\frac{\partial B}{\partial x} \hat{x}+\frac{\partial B}{\partial y} \hat{y}+\frac{\partial B}{\partial z} \hat{z}$$

And we use the rotation matrix on that

$$R\cdot \vec{\nabla}B=\left( \begin{array}{ccc} \cos (\phi ) & -\sin (\phi ) & 0 \\ \sin (\phi ) & \cos (\phi ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\left( \begin{array}{c} \frac{\partial B}{\partial x} \\ \frac{\partial B}{\partial y} \\ \frac{\partial B}{\partial z} \\ \end{array} \right)=\left(\frac{\partial B}{\partial x}\cos\phi-\frac{\partial B}{\partial y}\sin\phi\right)\hat{x}+\left(\frac{\partial B}{\partial x}\sin\phi+\frac{\partial B}{\partial y}\cos\phi\right)\hat{y}$$

And now we substitute those gradients in the main expression of the Taylor expansion and rotate the offsets $x-x_0$ and $y-y_0$ around the point $(x_0,y_0)$:

$$B(x,y)=B_0 + \left(\frac{\partial B}{\partial x}\cos\phi-\frac{\partial B}{\partial y}\sin\phi\right)\left(x_0+(x-x_0)\cos \phi-(y-y_0)\sin \phi\right) + \left(\frac{\partial B}{\partial x}\sin\phi+\frac{\partial B}{\partial y}\cos\phi\right)\left(y_0 + (x-x_0)\sin \phi+(y-y_0)\cos \phi\right)$$


The magnetic field itself is not a scalar, it is a pseudo-vector. Therefore your formula seems to assume that the magnetic field is a scalar. Is that just a single component or is it the absolute value of the field?

If it is $$ \vec B(x,y)=\vec B_0 + \frac{\partial \vec B}{\partial x}(x-x_0) + \frac{\partial \vec B}{\partial y}(y-y_0), $$ then you could apply a rotation matrix.

I have the impression that you could just add a term with $$\frac{\partial \vec B}{\partial z}(z-z_0)$$ to it and have an approximation in the whole of 3D.

If you just want to transform the derivatives, they transform inversely to the vector elements. So let $A$ be the rotation matrix for the angle $\phi$, then you need to transform the derivatives with $A^{\mathrm T}$, the transpose of that transformation.

Background: Partial derivatives are covectors and transform like so if you have a transformation that goes $\phi\colon x \mapsto y$ with $x = \phi(y)$: $$\frac{\partial}{\partial y^\mu} = \frac{\partial x^\alpha}{\partial y^\mu} \frac{\partial}{\partial x^\alpha} $$

  • 1
    $\begingroup$ Be careful. $\vec B$ is not a three-vector, it is a three-pseudovector (or, more formally, the Hodge dual of a 2-form in 3D). This does not play a role when considering rotations, but it should be borne in mind. $\endgroup$ – ACuriousMind Jul 15 '14 at 11:01

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.