Classical Electromagnetism and Special Relativity - Transformation of The Fields Between Inertial Frames - The E and B Fields

The E and B Fields

This equation, also called the Joules-Bernoulli equation, considers two inertial frames. As notation, the field variables in one frame are unprimed, and in a frame moving relative to the unprimed frame at velocity v, the fields are denoted with primes. In addition, the fields parallel to the velocity v are denoted by while the fields perpendicular to v are denoted as . In these two frames moving at relative velocity v, the E-fields and B-fields are related by:

\begin{align} & \mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}\\ & \mathbf {{B}_{\parallel}}' = \mathbf {{B}_{\parallel}}\\ & \mathbf {{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) \\ & \mathbf {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right) \end{align}

where

is called the Lorentz factor and c is the speed of light in free space. The inverse transformations are the same except v → −v.

An equivalent, alternative expression is:

\begin{align} & \mathbf{E}' = \gamma \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right ) - \left ({\gamma-1} \right ) ( \mathbf{E} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}} \\ & \mathbf{B}' = \gamma \left( \mathbf{B} - \frac {\mathbf{v} \times \mathbf{E}}{c^2} \right ) - \left ({\gamma-1} \right ) ( \mathbf{B} \cdot \mathbf{\hat{v}} ) \mathbf{\hat{v}}\\ \end{align}

where vĖ‚ is the velocity unit vector.

If one of the fields is zero in one frame of reference, that doesn't necessarily mean it is zero in all other frames of reference. This can be seen by, for instance, making the unprimed electric field zero in the transformation to the primed electric field. In this case, depending on the orientation of the magnetic field, the primed system could see an electric field, even though there is none in the unprimed system.

This does not mean two completely different sets of events are seen in the two frames, but that the same sequence of events is described in two different ways (see Moving magnet and conductor problem below).

If a particle of charge q moves with velocity u with respect to frame S, then the Lorentz force in frame S is:

In frame S', the Lorentz force is:

If S and S' have aligned axes then:

\begin{align} & u_x'=\frac{u_x+v}{1 + (v \ u_x)/c^2}\\ & u_y'=\frac{u_y/\gamma}{1 + (v \ u_x)/c^2}\\ & u_z'=\frac{u_z/\gamma}{1 + (v \ u_x)/c^2} \end{align}

A derivation for the transformation of the Lorentz force for the particular case u = 0 is given here. A more general one can be seen here.

Component by component, for relative motion along the x-axis, this works out to be the following:

\begin{align} & E'_x = E_x & \qquad & B'_x = B_x \\ & E'_y = \gamma \left( E_y - v B_z \right) & & B'_y = \gamma \left( B_y + \frac{v}{c^2} E_z \right) \\ & E'_z = \gamma \left( E_z + v B_y \right) & & B'_z = \gamma \left( B_z - \frac{v}{c^2} E_y \right). \\ \end{align}

The transformations in this form can be made more compact by introducing the electromagnetic tensor (defined below), which is a covariant tensor.

Read more about this topic:  Classical Electromagnetism And Special Relativity, Transformation of The Fields Between Inertial Frames

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