Is the matrix representation of the Lorentz transform the same for all 4-vectors?

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.

• 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. – AccidentalFourierTransform Jan 17 '16 at 9:50

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.