Lorentz transformation, problem with derivation I have a question about the Lorentz transformation: 
In the derivation it's said that two systems S and S' should at $t=0$ and $x=0$, overlap. We get the following transformation rules: 
$t'=\gamma_0(t-v_0x/c^2)$
$x'=\gamma_0(x-v_0t)$
$y'=y$
$z'=z$
My question is: What happens if at $t=0$ they dont overlap? Can I just add a constant to both the time and coordinate $x$ transformation equation, to account for the misalignment at $t=0 $?
$t'=\gamma_0(t-v_0x/c^2)  \color{Red}{+ T}$
$x'=\gamma_0(x-v_0t)  \color{Red}{+ X}$
 A: There are three types of transformations that preserve the spacetime interval. 


*

*Boosts: these are transformations like the one you gave. They transform between systems moving relative to each other, which have the same origin. If we write spacetime coordinates as column vectors like
$$ X = \left( \begin{matrix}
t \\
x \\
y \\
z \end{matrix} \right), $$
then Lorentz boosts are of the form $X \rightarrow X' = \Lambda X$ where $\Lambda$ is a $4\times4$ matrix obeying $\eta = \Lambda^T \eta \Lambda$ where $\eta$ is the metric. For example, the Lorentz boost in your question is given by the matrix
$$ \Lambda = \left( \begin{matrix}
\gamma & -\gamma v_0 & 0 & 0 \\
-\gamma v_0 & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{matrix} \right). $$
(Here we have set $c=1$.)

*Rotations: these are transformations between systems that are oriented differently, which also have overlapping origins but no relative motion. There is no difference between rotations in special relativity and rotations in ordinary mechanics. They are of the form $X' = RX$ where
$$ R = \left( \begin{matrix}
1 & 0 \\
0 & O \\
\end{matrix} \right) $$
and $O$ is a $3\times3$ matrix satisfying $O^TO = I$ and $\det(O) = 1.$ $O$ is just a normal spatial rotation matrix. Notice that $R$ only acts on the spatial coordinates and leaves $t$ unchanged.

*Translations: these are transformations between systems with different origins, but no relative motion or differing orientation. They are of the form $X' = X + C$ where $C$ is just a constant column vector.
The most general type of transformation that preserves the spacetime interval can combine boosts, rotations, and translations. For example, if you want to transform between two systems with different origins that are moving with relative velocity, then you just combine a boost with a translation:
$$ X \rightarrow X' = \Lambda X + C. $$
This is indeed exactly what you conjectured, and your final equations are correct.
As for terminology, the first two transformations (boosts and rotations) are called Lorentz transformations. The set of all Lorentz transformations is called the Lorentz group and is denoted by $SO(1,3)$. The set containing both Lorentz transformations and translations is called the Poincare group.
A: To give a less "groupy" answer: always think first about the 3D-analogue to what you're doing in 4D. So in 3D we have these translations and rotations which are linear transforms preserving $x^2 + y^2 + z^2;$ in 4D we in addition have these "boosts" and all three are linear transforms preserving $w^2 - x^2 - y^2 - z^2$ where $w = c t.$ So just "downgrade" the boost to a rotation and ask yourself what you'd do in 3D.
So in 3D we know these really easy rotation matrices $R$ to rotate a vector about the origin. What do you do when you want to rotate your points about a point $\vec r_0$ that is not the origin? You form something complicated, $\vec r' = \vec r_0 + R(\vec r - \vec r_0),$ where you first translate your coordinates to the origin, then rotate, then translate them back so that for $\vec r = \vec r_0$ we have also $\vec r' = \vec r_0.$ And this procedure will work equally well for Lorentz boosts, $r_2^\mu = r_0^\mu + L^\mu_{~~\nu} (r_1^\nu - r_0^\nu).$
However: also reflect that the rotation about the point that is not the origin is really disorienting if you're at the origin, now all the things that you're talking about locally are rotated to some strange point $\vec p = (I - R) ~\vec r_0,$ very difficult to use.
So in practice, what we do in 3D and 4D is to choose some origin which is helpful to us: in 3D it is usually a point on an object which we're keeping track of; in 4D it is usually some instantaneous event which both observers can agree happened. And then even though both observers find this a little clumsy for the other points, we get a really nice translation between them that makes the math super-easy.
