Deriving invariance of interval I'm reading a book called A First Course In General Relativity, and when I got to the invariance of the interval, the book didn't derive it. Also, when the book  derived time dilation and Lorentz contraction, it said that the invariance of the interval was -1 and 1. But it didn't derive it, so I was wondering if anyone can help show me how to derive these. 
Also, I still don't understand how they derive the secondary observer's time and position world lines relative to the primary observer. Finally, I don't understand why the invariance of the interval of a primary observer is the same as the invariance of the interval of the secondary observer. Can you please help answer these questions?
Thanks!
 A: I'm assuming you mean Bernard Schutz's book A first course in general relativity, and you're looking at section 1.6 Invariance of the interval.
Schutz's book is an excellent introduction to relativity for readers who are mainly just curious and not intending to do research in GR, but Schutz does skip over things he thinks are obvious. In this case he defines the interval as:
$$ (\Delta s)^2 = -(\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 \tag{1} $$
and he considers the specific case of light rays for which $\Delta s = 0$. If we substitute $\Delta s = 0$ and rearrange then we get:
$$ \frac{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}{(\Delta t)^2} = 1 \tag{2} $$
But the fraction on the left hand side is just the speed of light squared. Pythagoras' theorem tells us that:
$$ (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 = d^2 $$
where $d$ is the distance between the start and end points in space. In this context $d$ is the distance the light has moved in a time $\Delta t$. Putting this into equation (2) we get:
$$ \frac{d^2}{(\Delta t)^2} = v^2 = 1 $$
so:
$$ v = \pm 1 $$
where $v$ is the speed of light. We've done this calculation for some specific observer using their coordinates $(t,x,y,z)$, but Schutz's point is that if we start with the assumption that the speed of light is the same for all observers then our initial equation (1) must be true for all observers.
A few quick comments: relativists tend to use units in which $c=1$, which is why the speed of light turns out to be $1$. It's $\pm 1$ because the light ray could be moving in either direction. Finally, there's a more detailed explanation of how time dilation and length contraction are related to the invariance of the interval in How do I derive the Lorentz contraction from the invariant interval?.
A: The general properties of spacetime intervals arise from their general definition, which refers to geometric relations (distances, durations) between members of inertial systems (of course without referring to any incidental coordinate assignments):
timelike intervals: If the same one member $A$ of an inertial system had taken part (successively) in two distinct events $\varepsilon_{A~P}$ (i.e. the event of $A$ and a suitable participant $P$ having been coincident, having met each other in passing) and $\varepsilon_{A~Q}$ (i.e. the event of $A$ and a suitable participant $Q$ having been coincident, having met each other in passing)
then the (value of the) spacetime interval between these two events, $s^2[~\varepsilon_{A~P}, \varepsilon_{A~Q}~]$, is defined as proportional to the square of participant $A$'s duration between having indicated the coincidence with $P$ and having indicated the coincidence with $Q$: $$ s^2[~\varepsilon_{A~P}, \varepsilon_{A~Q}~] := -c^2~\tau A[~{}_P, {}_Q~],$$
 where $-c^2$ is a conventional non-zero proportionality factor.
 (If $\varepsilon_{A~P}$ and $\varepsilon_{A~Q}$ are actually one and the same event then $A$'s corresponding duration is of course Null, and so is therefore the spacetime interval between one event and the identical same event.)
Considering the geometric relations between participant $A$ and members of another inertial frame (to which $A$ does not belong), say $F$ and $G$ (who were and remained at rest wrt. each other), such that


*

*$F$ also took part in event $\varepsilon_{A~P} \equiv \varepsilon_{A~F~P}$, and

*$G$ also took part in event $\varepsilon_{A~Q} \equiv \varepsilon_{A~G~Q}$,


then (as has been proven separately, e.g. here)
$$\tau A[~{}_P, {}_Q~] = \sqrt{ (\tau G[~{}_{AQ}, {}_{\circledS F \circ AP}~])^2 - \left(\frac{1}{c}~d[~F, G~]\right)^2 } = \tau G[~{}_{AQ}, {}_{\circledS F \circ AP}~]~\sqrt{1 - (\beta_{FG}[~A~])^2}, \tag{1}$$
where $\tau G[~{}_{AQ}, {}_{\circledS F \circ AP}~]$ is the duration of participant $G$ from having indicated the coincidence with $A$ and $Q$ until $F$'s indication simultaneous to $F$'s indication of coincidence with $A$ and $P$,
$d[~F, G~]$ is the distance between $F$ and $G$, and
$\beta_{FG}[~A~]~c$ is the speed of $A$ wrt. the inertial system of which $F$ and $G$ are members.
On the other hand:
spacelike intervals: Considering two members of the same inertial frame (i.e. two participants who were and remained at rest wrt. each other), $A$ and $B$, where $A$ took part in one event, $\varepsilon_{A~J}$, and $B$ took part in another event $\varepsilon_{A~K}$, and if the corresponding indications of $A$ and of $B$, resp., at these events, namely indications $A_J$ and $B_K$ were simultaneous to each other, then 
then the (value of the) spacetime interval between these two events, $s^2[~\varepsilon_{A~J}, \varepsilon_{B~K}~]$, is defined as proportional to the square of the (chronometric) distance between $A$ and $B$:
$$ s^2[~\varepsilon_{A~J}, \varepsilon_{B~K}~] := \alpha~d[~A, B~],$$
where the common non-zero constant of proportionality is still to be determined, in relation to the above convention.
(Again, and consistently with the conclusion reached above already, if $\varepsilon_{A~J}$ and $\varepsilon_{B~K}$ are actually one and the same event, such that $A$ and $B$ are in fact names referring to one and the same participant then $A$'s corresponding distance is of course Null, and so is therefore the spacetime interval between one event and the identical same event.)
Considering the geometric relations between participants $A$ and $B$ and members of another inertial frame (to which $A$ and $B$ don't not belong), say $F$ and $G$ (who were and remained at rest wrt. each other), such that


*

*$F$ also took part in event $\varepsilon_{A~J} \equiv \varepsilon_{A~F~J}$, and

*$G$ also took part in event $\varepsilon_{B~K} \equiv \varepsilon_{B~G~K}$, where (for definiteness in the following) $G$'s indication of coincidence with $B$ and $K$ was before $G$'s indication simultaneous to $F$'s indication of coincidence with $A$ and $J$,


then (as has been proven separately)
$$d[~A, B~] = \sqrt{ (d[~F, G~])^2 - c^2~(\tau G[~{}_{BK}, {}_{\circledS P \circ AJ}~])^2 } = d[~F, G~]~\sqrt{1 - (\beta_{FG}[~B~])^2}. \tag{2}$$
The formal similarity between eqs. (1) and (2) suggests to set $\alpha := 1$ and to arrive at the formal, generalized expression for interval values in terms of geometric relations between members of a general inertial frame: 
$$s^2[~\text{event in which }F\text{ took part}, \text{event in which }G\text{ took part}~] := -c^2~(\tau G[~{\small{\text{indication of own participation}, \text{indication simultaneous to }F\text{'s indication of participation}}}~])^2 + (d[~F, G~])^2, \tag{3}$$ 
or reduced to a mere mnemonic:
$$s^2 := -c^2~\text{duration}^2 + \text{distance}^2.$$
Finally,
lightlike intervals: if neither of the two cases discussed above applies, but instead participants in one event observed the signalfront of (indications of participants having taken part in) the other event, then the corresponding interval value between these two events is defined as Null:
$$s^2[~\text{pair of lightlike related events}~] = 0.$$
This is also consistent with the formal expression~(3), with the chronometric distance between any two members of one inertial system, such as participants $F$ and $G$, being proportional to the ping duration of either one wrt. the other.
