Let's define the interval as $I=\eta_{\mu\nu}dx^\mu dx^\nu$ where $\eta_{\mu\nu}$ is defined to be a frame invariant diagonal matrix whose diagonal elements are $-1,+1,+1,+1$ . By definition, $\eta_{\mu\nu}$ is the same in all the frames. In other words, I am defining how I would construct the interval once and for all frames. A priori, I don't know if $I$ is frame independent or not. Notice that I have not assumed $\eta$ to be a tensor and thus, $I$ is not a scalar - at least not in a trivial way.

Now, the postulates of SR are:

  1. Laws of Physics should be the same in all the inertial frames.

  2. The speed of light is invariant among all the inertial frames.

I will use the natural units. Thus, my second postulate requires that if $I=0$ in one frame then it will be zero in all the frames. That is to say that $I$ will be frame-invariant if it is equal to zero. But the second postulate seems silent as to whether $I$ will be invariant or not if it doesn't vanish. We usually get over this problem by demanding that the transformation law between the inertial frames must be first order in coordinates. (i.e., if we demand $x'^\alpha= \Lambda^\alpha _\mu x^\mu +c^\alpha$ with $\Lambda$ and $c$ being independent of the coordinates then we can show that if $\eta_{\mu\nu}dx^\mu dx^\nu=0$ has got to imply $\eta_{\sigma\rho}dx'^\sigma dx'^\rho=0$ then we require the interval to be frame-invariant even when it is not zero.)

But without explicitly assuming the transformation law to be $x'^\alpha= \Lambda^\alpha _\mu x^\mu +c^\alpha$, can I prove that the interval has to be a scalar? I mean it should be okay if the postulates of SR themselves somehow imply such a restriction on the transformation law, but I intend not to supply it externally.

Edit 1

I would like to add that in a qualitative way, Rindler argues that the first postulate itself demands spacetime to be both homogeneous and isotropic. Then, it becomes obvious that the transformation law must be of the form $x'^\alpha = \Lambda^\alpha_\mu x^\mu + c^\alpha$. The argument goes something like this: Assume a frame is inertial. Then all frames obtained by translating the origin (either in space or time) or by rotating the spatial coordinate axes should also be inertial. Because they are all kinematically equivalent to the original frame. But the first postulate requires that all the inertial frames behave identically and thus, the spacetime better be homogeneous and isotropic. Now, the boosted inertial frames should also see the spacetime homogeneous and isotropic because according to the first postulate the physics should be invariant among frames. This means the transformation should be of the form stated above. I would probably like to do something like this but in a more rigorous (and preferably mathematically sound so that mathematicians don't abuse it as a conjecture) way.

Edit 2

There is an interesting physical argument that demands the transformation to be linear and thus demands the interval to be generically invariant if it has to be invariant when it is null. It's the argument that the statement that a particle is moving at a constant velocity should be frame-independent. That is to say that if the velocity of a particle is constant in one (inertial) frame then it must be constant in all other (inertial) frames. A little scribbling would immediately show that this requires the transformation to be linear, i.e. of the form $x'^\mu = \Lambda^\mu_\nu x^\nu +c^\mu$.

  • $\begingroup$ Note that the invariance is a little stronger than that. It's not just intervals, but angles: $C = \eta_{\mu\nu} \operatorname{d}x^\mu \operatorname{d}y^\nu.$ $\endgroup$ Jun 16, 2017 at 22:19
  • $\begingroup$ Sorry, I don't get it. What do you mean by angles here? Invariance is $\eta_{\mu\nu}dx^\mu dx^\nu$ being same in all the frames, right? $\endgroup$
    – user87745
    Jun 16, 2017 at 22:28
  • $\begingroup$ With an ordinary dot product you have $\vec{A}\cdot\vec{B} = |A| |B| \cos\theta$. In this case the interpretation isn't so simple because the metric has that minus sign, but the principle is the same. Just as rotations preserve lengths and angles (thus $\vec{A}\cdot\vec{B}$ is constant), Lorentz transformations preserve $A_\mu B^\mu$ for any two 4-vectors. $\endgroup$ Jun 16, 2017 at 22:39
  • $\begingroup$ Lorentz transformation will most certainly preserve the interval. But without assuming the transformations to be of the form $x'^\alpha = \Lambda^\alpha _\mu x^\mu+c^\alpha$, how do I get in the first place that the allowed transformations are Lorentz transformations? $\endgroup$
    – user87745
    Jun 16, 2017 at 22:41
  • $\begingroup$ Linearity is an assumption that goes into this process. $\endgroup$ Jun 16, 2017 at 22:43

1 Answer 1


What's going on here is that there is an unstated assumption that the transformation has to be continuous. That is, it is described by some set of parameters that we can take to be infinitesimal and integrate up their effects in a predictable way. The mathematics behind rotations and translations have been well understood for a long time, and aren't unique to special relativity, so we can limit ourselves to discussing frames of reference that are related to each-other by boosts (reference frames moving with respect to one another). In particular, we can look at boosts in the $x$-direction only, since rotations applied before and after applying the boost can fill out the rest of the transformation.

So, we need some transformations of the form: \begin{align} t' &= f(x,t,v,c) \\ x' &= g(x,t,v,c), \end{align} to describe how the coordinates in the primed and un-primed frames are related.

Here's where linearity comes in - how does this system of equations transform under a change of units? Say, $x \rightarrow a x,\ t\rightarrow bt$. Since any inertial observer is capable of using the same unit system (even if they have to accelerate into each-other's frames to synchonize) the boost transformations need to obey: \begin{align} bt' & =f\left(ax, bt, \frac{a}{b} v, \frac{a}{b}c\right) \\ ax' & = g\left(ax, bt, \frac{a}{b} v, \frac{a}{b}c\right). \end{align} In more technical terms - boosts preserve the ratios of lengths of artifacts and relative timing of timekeeping devices. All inertial observers agree how many centimeters to a meter, how many seconds to a minute.

This isn't quite enough to get to linearity, though. For example, $\frac{x}{|x|} \sqrt{x^2 + \left(ct\right)^2}$ would satisfy the requirement for $f$. Eliminating those requires additional facts. The first such fact is that the primed frame is defined by the relation $x' = 0 \Leftrightarrow x = vt$, thus $g(vt, t, v, c) = 0$.

Further, the rule for boosting has to work when we do it more than once. If we have three frames, labeled $0$, $1$, and $2$, then we have to have: \begin{align} t_2 &= f(g(x_0, t_0, v_{10}, c), f(x_0, t_0, v_{10}, c), v_{21}, c) = f(x_0, t_0, v_{20}, c) \\ x_2 &= g(g(x_0, t_0, v_{10}, c), f(x_0, t_0, v_{10}, c), v_{21}, c) = g(x_0, t_0, v_{20}, c), \end{align} where $v_{ij}$ is the velocity of frame $i$ with respect to frame $j$. Note that combining above relations implies: \begin{align} v_{21} &= \lim_{\Delta t \rightarrow 0} \frac{g(v_{20}\Delta t, \Delta t, v_{10}, c)}{f(v_{20}\Delta t, \Delta t, v_{10}, c)}. \end{align} I don't know if the above is sufficient or if we need to also impose correspondence with the Galilean addition of velocities formula at low speeds. In the nonrelativistic limit ($|v_{20}| \ll c$ and $|v_{10}| \ll c$) it's: $$v_{21} = v_{20} - v_{10}.$$

We also have the zero velocity requirements: \begin{align} t &= f(x, t, 0, c) \\ x &= g(x, t, 0, c). \end{align}

I'm still not sure if all of the above is sufficient to prove linearity, if some other additional fact(s) are needed, or if linearity ultimately has to be assumed. Try combining them and let us know if you find a counter-example (a non-linear rule that fits all of the above).


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