# How to derive the spacetime interval from the Lorentz transformation?

I've seen plenty of derivations of the Lorentz transformation from the spacetime interval.

Can this process be reversed, to derive the spacetime interval from the Lorentz transformation and the two postulates that the principle of relativity applies and the speed of light is absolute?

• What precisely are you assuming about the "Lorentz transformations" that you want to derive the spacetime interval from? That the interval is invariant under Lorentz transformation is a straightforward computation, what exactly do you want to derive? Aug 16, 2017 at 9:15
• @ACuriousMind If we apply the Lorentz transformation to the spacetime interval, we can see that it is invariant, but what if we didn't know the form of the spacetime interval so we could test it - how could we make the Lorentz transformation "spit out" the invariance of the spacetime interval? Aug 16, 2017 at 9:19

I don't know which derivation of the form invariant spacetime interval $\Rightarrow$ Lorentz transformation you have in mind, but the easiest derivations all make inferences which are "if and only if" inferences, i.e. the reasoning and inference chain can be run in both directions.

For example, writing the Lorentz transformation as a general $4\times4$ real element matrix $\Lambda$ acting on $4\times1$ real element columns representing 4-vectors $X\in \mathbb{R}^{1+3}$, then the assertion of the spacetime interval's invariance is:

$$X^T\,\Lambda^T\,\eta\,\Lambda\,X\, = X^T\,\eta\,X;\quad\forall X\in\mathbb{R}^{1+3}\tag{1}$$

where, naturally, $\eta=\mathrm{diag}(1,\,-1,\,-1,\,-1)$. Now choose two in general different $X,\,Y\in\mathbb{R}^{1+3}$ and write down (1) for the sum $X+Y$: that is $(X+Y)^T\,\Lambda^T\,\eta\,\Lambda\,(X+Y)\, = (X+Y)^T\,\eta\,(X+Y)$, expand this little beast out and then apply (1) again to show that (1) implies:

$$X^T\,\Lambda^T\,\eta\,\Lambda\,Y = X^T\,\eta\,Y;\quad\forall X,\,Y\in\mathbb{R}^{1+3}\tag{2}$$

Now choose the sixteen different combinations of the usual basis vectors for $X$ and $Y$ and you thus show (witnessing that $\eta$ is nonsingular, i.e. defines a nondegenerate billinear form):

$$\Lambda^T\,\eta\,\Lambda = \eta\quad\tag{3}$$

(3) trivially implies (1), so (3) and the assertion of spacetime interval are logically equivalent. They imply, and are implied by, one another.

Now I, and I believe many people, would think of (3) as the definition of the Lorentz transformation: its use in a set builder notation gives us a complete characterization of the Lorentz group. But you may be wanting to work with other characterizations of the Lorentz group, or the special Lorentz group (proper, orthochronous ones) perhaps. You can trivially check that a boost in the $x$ direction fulfills (3), thus, by our logical equivalence, leaves the spacetime interval invariant. Likewise, do the same for a rotation $R$, for which (3) is equivalent to $R^T\,R=\mathrm{id}$. Thus, if you define the proper orthochronous Lorentz group as the smallest group that contains the $x$ boost and the rotations, i.e. as the group of all finite products of the form $U_{1}\,R_{1}\,U_2\,R_2\,\cdots$ where the $U_i$ and $R_i$ are all $x$-boosts and rotations, respectively and apply (3) to such a chain inductively, you can show that this group conserves the spacetime interval. Note that, of course, a boost in any direction can be written in the form $R\,U\,R^T$, where $U$ is an $x$-boost and $R$ a rotation.

• Pfft, "naturally" $\eta={\rm diag}(-1,1,1,1)$ because some of us a "positive" people :D. Aug 16, 2017 at 10:15

Ok, so we take it given that we know that the transformations among the frames that we are allowed to traverse are the Lorentz transformation. The most concise and contentful definition of the Lorentz transformations would be the following: $$\Lambda ^{T}\eta\Lambda=\eta$$ Or, in the index notation, $$\Lambda^{\alpha'}_{\mu}\Lambda^{\beta'}_{\nu}\eta_{\alpha'\beta'}=\eta_{\mu\nu}$$ Now, make sure you appreciate that we are treating the matrix $\eta$ (whether written in the simple matrix form or in the index notation) simply as a matrix. We don't pretend to know apriori that it acts as the metric. But rather, we will now motivate as to why it would be wise to use it as the metric and (thus) to define the interval as $\eta_{\mu\nu}dx^\mu dx^\nu$.

The first step is to notice that the matrix $\eta$ (simply defined to be the diagonal matrix with entries $1,-1,-1,-1$) is really a tensor under the Lorentz transformations. This can be directly read off from the very definition of the Lorentz transformations when the definition is expressed in the index notation as expressed above.

Now that we have recognized that $\eta$ is really a $(0,2)$ tensor, using the definition of a $(0,2)$ tensor, we should appreciate that it is the very map that takes a pair of vectors to a real number (or, in other words, a scalar--a frame-invariant quantity). Thus, the most natural way to construct a frame-invariant quantity out of a displacement vector $\vec{dx}$ would be to feed this vector into both the slots of the tensor $\eta$ and to define the outcome as the spacetime interval. That is to say, to define $$I=\eta(\vec{dx},\vec{dx})=\eta_{\mu\nu}dx^{\mu}dx^{\nu}=dt^2-dx^2-dy^2-dz^2$$ as the spacetime interval.

Of course, if one insists, one can argue that this is all just a motivation to define the spacetime interval to be $dt^2-dx^2-dy^2-dz^2$ and nothing more. And that is true. Because, after all, we are defining something and there is no word of God about how we should define something! It is just that if we see a very useful property in a thing then we give it some name. The same is true for spacetime interval. The observation that $dt^2-dx^2-dy^2-dz^2$ is invariant is an important one and thus, we call it the spacetime interval. That is the bottom line. What I have tried to demonstrate here is how one can very easily stumble upon this observation if one already has in mind that she is looking for some invariant quantity.