If I have a four vector of the form:

$$ \left( \begin{array}{ccc} T\\ \vec{X}\end{array} \right) $$ where $T$ is the analogous time component (i.e. energy, angular frequency, scalar potential, charge density) and $\vec{X}$ is the analogous space component (i.e. momentum, wave number, vector potential, current density).

Then would finding the respective 4-vectors in a different frame be a simple matter of multiplying any four vector by the Boost matrix (for the case where the relative velocity is in the x-direction): $$ \left( \begin{array}{ccc} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) $$ ?

In other words, is the Boost matrix the Lorentz transformation for all objects classified as 4-vectors? Or are there certain 4-vectors whose Lorentz transform is represented by a different matrix?

What about for the case of electric $\vec{E}$ and magnetic fields $\vec{B}$? There are 3 components for each mentioned field, meaning there are 6 quantities that transform in a change of reference frame. I realize there are equations for them, but they don't appear to be related to the Boost matrix.

  • 2
    $\begingroup$ Yes: thats the definition of four-vector (anything that transforms in the same way $(t,\boldsymbol{x})$ does). The electromagnetic field is not a four-vector, but rather a second rank tensor: its transformation properties are a bit more involved. $\endgroup$ Jan 17, 2016 at 9:50

1 Answer 1


Yes, all 4-vectors transform as you state under a Lorentz transform.

For the case of $\vec E$ and $\vec B$, they are indeed not 4-vectors. There are two ways of transforming the $\vec E$ and $\vec B$ fields to different coordinate frames. You can define the $A$ 4-vector in terms of the potential functions $\phi$ and $\vec A$, letting \begin{equation} A = \begin{pmatrix}\frac{\phi}{c}\\\vec A\end{pmatrix}. \end{equation} Then $A$ transforms as a 4-vector and you can get the fields back from the standard relations $\vec E = -\nabla \phi - \frac{d\dot A}{dt}$, $\vec B = \nabla \times \vec A$.

Or you can define a certain 4x4, rank-2 tensor $F$, which is written directly in terms of $\vec E$ and $\vec B$, where \begin{equation} F^{\mu \nu} = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}. \end{equation} Then you can transform this tensor, using standard rules for transforming tensors between coordinate systems, and pick out the components of $\vec E$, $\vec B$ in the transformed system. $F$ is used much more in Lagrangian treatments of electrodynamics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.