Yes, you're right..
But if I use the Biot-Savart law I have:$$B = \displaystyle\frac{1}{c^2}v\times{}E\Longrightarrow{}B c^2= v\times{}E\Longrightarrow{}-B c^2=\frac{dr}{dt}\times{}\triangledown{}V$$How would this be to any coordinate system? Could I pass the velocity and the gradient...
Thanks for the answer, but it's not exactly what I was finding for ..
I know from Maxwell's third equation that:$$\triangledown{} \times{} \vec{E}= -\frac{{\partial B}}{{\partial t}}$$Therefore$$(\triangledown{} \times{} \vec{E}) {\partial t}= -{\partial \vec{B}} \Longrightarrow{}...
Hello,
For example, an electric field vector, such as the gradient of the potential, passes in the following way to any coordinate system:$$E = -\triangledown{}V = - \frac{{\partial V}}{{\partial x^i}}g^{ij}e_j$$
But what about a vector of a magnetic field? How would it be expressed in any...