What physical conditions are packed into insisting the invariance of the spacetime separation interval among inertial observers?

Question

I am trying to construct the Lorentz transformation from first principles. There are two observers $S$ and $S'$ moving relative to one another. We assume there to be a base event at $(x,t) = (x', t') = (0,0)$ that both observers agree on: their clocks read the same time and they measure the event at the origin of their lattice of measuring sticks. Then, any other pair of places and times to either observer describes the elapsed time and displacement from the base event. In other words, $\Delta x = x$ and $\Delta t = t$, and we don't need to think about deltas with the understanding that all coordinates compare against a common, agreed-upon base event.

This motivates a linear transformation between coordinate values of observers for the same event as $x' = Ax + Bt$ and $t' = Cx + Dt$. If we insist that the frame $S$ always sees the origin of $S'$ moving at $v$ and $S'$ sees the origin of $S$ moving at $-v$, then the two conditions 1) $(0,t')$ whenever $(vt,t)$ and 2) $(-vt',t')$ whenever $(0,t)$ yield $x' = A(x - vt)$ and $t' = Cx + At$.

To find $A$ and $C$, we must have two more physical conditions that relate events to each observer. The short answer is that if we force $I$ to be the same for a given event compared to the base event in both $S$ and $S'$, where $I$ is constructed as $I = c^2t^2-x^2$ so that the condition $c^2t^2-x^2 = c^2t'^2-x'^2$ must be true, then we recover the Lorentz transformation by equating terms on the polynomial.

But I am struggling to understand the implications of invariance of the constructed $I$ and what kind of physical statements are being made. I know that one implication is that an event a distance and time away from $S$ and $S'$ that occurs at $X$ seconds in elapsed time and $X$ light-seconds in displacement for $S$ must occur at $X'$ seconds in elapsed time and $X'$ light-seconds in displacement for $S'$, i.e., the speed of light is constant for both observers. But if we insist only on the constancy of speed of light to all observers, this only gives us one condition to determine $A$ and $C$ in the transform; we need one more. This is somehow packed into $I$ invariance.

The answer to this question touches on it in the way I would like to understand How to motivate the importance of the spacetime interval

It seems that the answerer is insisting on two conditions: 1) All observers calculate and agree upon the proper time between the base event and any other event and 2) Invariance of $I$ is a statement about symmetry for a mid-way observer who sees $S$ moving at $-v/2$ and $S'$ moving at $v/2$. What does he mean by this? What is the physical statement that is being insisted upon?

This question also touches on it but the answers are confusing to me. What's the physical content in the invariance of spacetime interval in GR?

Thanks.

Answer 1

The word “invariance” is actually hiding a substantial amount of the heavy lifting in terms of the physical content. When we talk about something being invariant, it is always with respect to some specified class of transformations.

In this case the form of the spacetime metric is invariant under translations in time and space, rotations in space, and boosts in three directions. That is physically homogeneity in spacetime, isotropy in space, and the principle of relativity.

Those requires physical symmetries identify either the Galilean transformation or the Lorentz transformation. The specific form of the spacetime interval selects the Lorentz transformation and adds the invariance of $c$ to the list.

I would also include the clock postulate and all of the physics principles you can derive from the Lorentz transform such as time dilation and relativistic Doppler.