We can derive, how an arbitrary vector transforms under the rotation of coordinate system. For example, if a point in x-y plane is given by P(x,y). The same point in a rotated coordinate frame (conventionally we take - anticlockwise as a positive angle with respect to x axis i.e. angle $\alpha$). The new coordinates can be denoted as P(x',y').
If we want to go from one system to another, the relation between old and new coordinates axes is imminent. It can be easily verified these relations, x' = x cos$\alpha$ (projection of old x-axis on new x'-axis) + y sin$\alpha$ (projection of old y-axis with new x'-axis) and y' = x cos(90+$\alpha$) ( projection of old x-axis on new y'-axis) + y cos$\alpha$ ( projection of old y-axis on new y'-axis).
Simply they are written as, $$ x' = x cos\alpha + y sin\alpha \\ y' = -xsin\alpha + y cos\alpha $$
where x,y are measured in same units. In a similar way, Electric field and Magnetic field is the only thing you need to know, when you are dealing with Electrodynamics, which is analogues to our usual coordinate system.
Instead of any point, Any EM phenomena can be pointed in a plane as a point where we can put Electric field on the x-axis and Magnetic field on the y-axis.
Once we made this analogy, all the equations and condition we derive for coordinate axes can be transferred here with careful analysis. Now, we just need the above coordinate rotation property where we change variables (x,y) $\rightarrow$ (E,B)
E,B should be in the same units. So, we prefer Gaussian system. In SI we just need to use "cB" instead of B where "c" is the speed of light used for pure dimensional reasons.
In our new system, the rotation of coordinate system is given by, $$ E' = E cos\alpha + B sin\alpha \\ B' = -E sin\alpha + B cos\alpha $$ with the corresponding transformation of charge densities $$ \rho_e' = \rho_e cos\alpha + rho_m sin\alpha \\ \rho_m' = -\rho_e sin\alpha + \rho_m cos\alpha $$
This transformation specifically known as Duality transformation.
As in the previous case, Rotation of coordinate system doesn't change any physical fact about the location of the point, Maxwell's equations are invariant under this duality transformation. We can check it as follows,
Maxwell's equations before transformation,
$$\nabla\cdot \vec{E} = 4\pi{\rho_e} \\ \nabla\cdot\vec{B} = 4\pi\rho_m \\ \nabla\times \vec{E} = -\frac{4\pi}{c}\vec{J_m}- \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \frac{4\pi}{c}\vec{J_e}+\frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}}$$
After the transformation,
(1)
$$ \nabla\cdot\vec{E'} = (\nabla\cdot\vec{E}) cos\alpha + (\nabla\cdot\vec{B}) sin\alpha = 4\pi\rho_e cos\alpha + 4\pi\rho_m sin\alpha = 4\pi\rho'_e $$
(2)$$ \nabla\cdot\vec{B'} = (\nabla\cdot\vec{B}) cos\alpha - (\nabla\cdot\vec{E}) sin\alpha = 4\pi\rho_m cos\alpha - 4\pi\rho_e sin\alpha = 4\pi\rho'_m $$
(3) $$ \nabla \times \vec{E'} = (\nabla\times\vec{E}) cos\alpha + (\nabla\times\vec{B}) sin\alpha \\~\\= - \frac{4\pi}{c} \vec {J_m} cos\alpha + \frac{4\pi}{c} \vec{J_e} sin\alpha - \frac{1}{c} \frac{\partial{\vec{B}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}} sin\alpha \\~\\= - \frac{4\pi}{c} \vec{J_m'} - \frac{1}{c}\frac{\partial\vec{B'}}{\partial{t}} $$
(4)
$$\nabla \times \vec{B'} = (\nabla\times\vec{B}) cos\alpha - (\nabla\times\vec{E}) sin\alpha \\~\\= \frac{4\pi}{c} \vec {J_e} cos\alpha + \frac{4\pi}{c} \vec{J_m} sin\alpha + \frac{1}{c} \frac{\partial{\vec{E}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} sin\alpha \\~\\= \frac{4\pi}{c} \vec{J_e'} + \frac{1}{c}\frac{\partial\vec{E'}}{\partial{t}}$$
Also, the Lorentz force,
(5)
$$ F' = q_e' \left[ \vec{E'} + \frac{(\vec{v}\times\vec{B'})}{c}\right] + q_m' \left[\vec{B'} - \frac{(\vec{v}\times\vec{E'})}{c} \right] \\~\\ = (q_e cos\alpha + q_m sin\alpha) \left[\vec{E} cos\alpha + \vec{B} sin\alpha + \frac{(\vec{v} \times \vec{B}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{E}) sin\alpha}{c}\right] +\\~\\ (q_m cos\alpha - q_e sin\alpha ) \left[ \vec{B} cos\alpha - \vec{E} sin\alpha - \frac{(\vec{v}\times\vec{E}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{B}) sin\alpha}{c}\right] $$
After doing the arithmetic manipulations (8 terms will cancel out), the remaining 8 terms are, $$ F' = q_e\vec{E} cos^2\alpha + q_m\vec{B} sin^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c} cos^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} sin^2 \alpha +\\~\\ q_e \vec{E} sin^2\alpha + q_m \vec{B} cos^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c}sin^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} cos^2 \alpha \\~\\ = q_e \left[\vec{E} + \frac{(\vec{v}\times\vec{B})}{c}\right] + q_m \left[ \vec{B} - \frac{(\vec{v}\times\vec{E})}{c}\right] = F $$
Thus, we proved all the four Maxwell's equations with the Lorentz force is invariant under duality transformation.
We should note that, the angle $\alpha$ can vary arbitrarily. There is no any restriction to the values of $\alpha$ in the classical sense.
We will see the consequence in the next post.
If we want to go from one system to another, the relation between old and new coordinates axes is imminent. It can be easily verified these relations, x' = x cos$\alpha$ (projection of old x-axis on new x'-axis) + y sin$\alpha$ (projection of old y-axis with new x'-axis) and y' = x cos(90+$\alpha$) ( projection of old x-axis on new y'-axis) + y cos$\alpha$ ( projection of old y-axis on new y'-axis).
Simply they are written as, $$ x' = x cos\alpha + y sin\alpha \\ y' = -xsin\alpha + y cos\alpha $$
where x,y are measured in same units. In a similar way, Electric field and Magnetic field is the only thing you need to know, when you are dealing with Electrodynamics, which is analogues to our usual coordinate system.
Instead of any point, Any EM phenomena can be pointed in a plane as a point where we can put Electric field on the x-axis and Magnetic field on the y-axis.
Once we made this analogy, all the equations and condition we derive for coordinate axes can be transferred here with careful analysis. Now, we just need the above coordinate rotation property where we change variables (x,y) $\rightarrow$ (E,B)
E,B should be in the same units. So, we prefer Gaussian system. In SI we just need to use "cB" instead of B where "c" is the speed of light used for pure dimensional reasons.
In our new system, the rotation of coordinate system is given by, $$ E' = E cos\alpha + B sin\alpha \\ B' = -E sin\alpha + B cos\alpha $$ with the corresponding transformation of charge densities $$ \rho_e' = \rho_e cos\alpha + rho_m sin\alpha \\ \rho_m' = -\rho_e sin\alpha + \rho_m cos\alpha $$
This transformation specifically known as Duality transformation.
As in the previous case, Rotation of coordinate system doesn't change any physical fact about the location of the point, Maxwell's equations are invariant under this duality transformation. We can check it as follows,
Maxwell's equations before transformation,
$$\nabla\cdot \vec{E} = 4\pi{\rho_e} \\ \nabla\cdot\vec{B} = 4\pi\rho_m \\ \nabla\times \vec{E} = -\frac{4\pi}{c}\vec{J_m}- \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \frac{4\pi}{c}\vec{J_e}+\frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}}$$
After the transformation,
(1)
$$ \nabla\cdot\vec{E'} = (\nabla\cdot\vec{E}) cos\alpha + (\nabla\cdot\vec{B}) sin\alpha = 4\pi\rho_e cos\alpha + 4\pi\rho_m sin\alpha = 4\pi\rho'_e $$
(2)$$ \nabla\cdot\vec{B'} = (\nabla\cdot\vec{B}) cos\alpha - (\nabla\cdot\vec{E}) sin\alpha = 4\pi\rho_m cos\alpha - 4\pi\rho_e sin\alpha = 4\pi\rho'_m $$
(3) $$ \nabla \times \vec{E'} = (\nabla\times\vec{E}) cos\alpha + (\nabla\times\vec{B}) sin\alpha \\~\\= - \frac{4\pi}{c} \vec {J_m} cos\alpha + \frac{4\pi}{c} \vec{J_e} sin\alpha - \frac{1}{c} \frac{\partial{\vec{B}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}} sin\alpha \\~\\= - \frac{4\pi}{c} \vec{J_m'} - \frac{1}{c}\frac{\partial\vec{B'}}{\partial{t}} $$
(4)
$$\nabla \times \vec{B'} = (\nabla\times\vec{B}) cos\alpha - (\nabla\times\vec{E}) sin\alpha \\~\\= \frac{4\pi}{c} \vec {J_e} cos\alpha + \frac{4\pi}{c} \vec{J_m} sin\alpha + \frac{1}{c} \frac{\partial{\vec{E}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} sin\alpha \\~\\= \frac{4\pi}{c} \vec{J_e'} + \frac{1}{c}\frac{\partial\vec{E'}}{\partial{t}}$$
Also, the Lorentz force,
(5)
$$ F' = q_e' \left[ \vec{E'} + \frac{(\vec{v}\times\vec{B'})}{c}\right] + q_m' \left[\vec{B'} - \frac{(\vec{v}\times\vec{E'})}{c} \right] \\~\\ = (q_e cos\alpha + q_m sin\alpha) \left[\vec{E} cos\alpha + \vec{B} sin\alpha + \frac{(\vec{v} \times \vec{B}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{E}) sin\alpha}{c}\right] +\\~\\ (q_m cos\alpha - q_e sin\alpha ) \left[ \vec{B} cos\alpha - \vec{E} sin\alpha - \frac{(\vec{v}\times\vec{E}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{B}) sin\alpha}{c}\right] $$
After doing the arithmetic manipulations (8 terms will cancel out), the remaining 8 terms are, $$ F' = q_e\vec{E} cos^2\alpha + q_m\vec{B} sin^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c} cos^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} sin^2 \alpha +\\~\\ q_e \vec{E} sin^2\alpha + q_m \vec{B} cos^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c}sin^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} cos^2 \alpha \\~\\ = q_e \left[\vec{E} + \frac{(\vec{v}\times\vec{B})}{c}\right] + q_m \left[ \vec{B} - \frac{(\vec{v}\times\vec{E})}{c}\right] = F $$
Thus, we proved all the four Maxwell's equations with the Lorentz force is invariant under duality transformation.
We should note that, the angle $\alpha$ can vary arbitrarily. There is no any restriction to the values of $\alpha$ in the classical sense.
We will see the consequence in the next post.
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