In 4-dimensional spacetime, when we study the spacetime interval, why did Einstein put a negative sign in it?

$$x_1=x$$ $$x_2=y$$ $$x_3=z$$ $$x_4=ct$$ $$ds^2=dx^2+dy^2+dz^2-(cdt)^2$$

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    $\begingroup$ Comment to the post (v1): What do you mean? There seems to be no $i$'s in Einstein's original 1905 papers. $\endgroup$
    – Qmechanic
    Commented Oct 21, 2017 at 15:53
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    $\begingroup$ Or are you asking about why there is a sign in the Minkowski metric? $\endgroup$
    – Qmechanic
    Commented Oct 21, 2017 at 15:55
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    $\begingroup$ It's Minkowski who came up with the idea. $\endgroup$
    – Nemo
    Commented Oct 21, 2017 at 16:08
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    $\begingroup$ Related: physics.stackexchange.com/q/107443/2451 , physics.stackexchange.com/q/168532/2451and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 21, 2017 at 16:26
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    $\begingroup$ Possible duplicate of Minkowski Metric Signature $\endgroup$ Commented Oct 25, 2017 at 13:56

3 Answers 3


By spacetime interval, we want to define a quantity associated with a pair of events that remain invariant among all the inertial frames. In order to ensure that this happens, we can use the knowledge of Lorentz transformation that relates the coordinates of an event as observed by different inertial observers. The Lorentz transformations, in the simplest form, look like this:

$$ \begin{aligned} x'&=\dfrac{x-vt}{\sqrt{1-v^2}}\\ t'&=\dfrac{t-vx}{\sqrt{1-v^2}} \end{aligned} $$

Now, one can verify that if we devise our quantity as $dx^2-dt^2$ then it remains the same in the primed frame as well, i.e., $dx'^2-dt'^2 = dx^2 - dt^2$. Further, using some more mathematics, it can be shown that $dx^2-dt^2$ is the only such quantity (except for some obvious quantities constructed out of $dx^2-dt^2$) that remains invariant among all the inertial frames.

Thus, we define the spacetime interval as $dx^2-dt^2$. Historically, Einstein only reached up to Lorentz transformations. It was Minkowski who realized that there is a very nice geometric interpretation of these results and introduced the concept of spacetime interval. Notice that in the usual Euclidean geometry, the transformations look like this:

$$ \begin{aligned} x'&=\phantom{-}x\cos\theta+y\sin\theta\\ y'&=-x\sin\theta+y\cos\theta \end{aligned} $$ Here, the only quantity that remains invariant among all the frames is $(dx)^2+(dy)^2$ and that is the reason we identify it as the space interval.

It is not that someone came up with an out-of-the-box idea of introducing a minus sign instead of the plus sign. It is the properties of the transformations between the frames that decide what quantity is justified to be identified with the spacetime interval.

  • $\begingroup$ is there a $y$ missing in the last equation? I'm not sure about that... $\endgroup$ Commented Oct 21, 2017 at 17:58
  • $\begingroup$ @ErikPillon Yes, of course. Thanks for pointing it out. I have corrected my answer. $\endgroup$
    – user87745
    Commented Oct 21, 2017 at 18:59

It turns out that this thing of $ict$ is not done anymore nowadays. The issue is that Galilean transformations together with rotations preserved the distance in $\mathbb{R}^3$

$$|\mathbf{x}_2-\mathbf{x}_1|^2=\Delta x^2+\Delta y^2+ \Delta z^2$$

while Lorentz transformations together with the rotations preserved the interval between events in $\mathbb{R}^4$

$$\Delta s^2=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2.$$

It turns out Einstein's original paper dated of 1905 didn't present a much geometric approach.

Now, looking at the two things above it is somewhat clear that if you define $$(x^0,x^1,x^2,x^3)=(ict,x,y,z)$$

Then the interval $\Delta s^2$ is merely the usual "distance in $\mathbb{R}^4$" with the traditional inner product. Namely, you consider the set of all tuples $(x^0,x^1,x^2,x^3)$ endowed with the interval function

$$d(x_1,x_2)^2=\sum_{\mu=0}^4 (\Delta x^\mu)^2,\quad \Delta x^\mu = x_2^\mu-x_1^\mu.$$

The $i$ appears to make the minus sign appear in the interval as expected.

Today this isn't done anymore. One simply defines directly that the "metric tensor" of flat spacetime $\mathbb{R}^4$ (which is just a sort of inner product) is given by

$$\eta(v,w)= \eta_{\mu\nu} v^\mu w^\nu,\quad (\eta_{\mu\nu})=\begin{pmatrix}-1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}.$$

Notice that with this definition, spacetime is really $\mathbb{R}^4$ (no strange looking imaginary coordinate) and that given events $x,y\in \mathbb{R}^4$ we recover

$$\Delta s^2 = \eta(y-x,y-x).$$

As you recover the distance from the dot product in $\mathbb{R}^3$.

Now, why there is a minus sign by the way? The issue is that there was one apparent inconsistency between classical electrodynamics and galilean invariance. Galilean invariance required that the laws of mechanics stayed in the same form when written with respect to reference frames moving with respect to each other at a constant speed in a straight line.

The laws of electrodynamics did not behave like that. It was thus believed that galilean invariance didn't apply to electrodynamics and that there was a prefered reference frame (that of the aether) to formulate those laws. The frame of the aether though couldn't be detected.

It was shown by some physicists, Einstein among them in the 1905 paper, that by introducing two quite reasonable postulates that essentialy lead to relaxing the hypothesis of absolute time, one would find new transformation rules between these same galilean frames and these rules would keep Maxwell's electrodynamics invairant.

This quantity $\Delta s^2$ appears quite a lot, and by these new Lorentz transformations it is kept invariant. It was later discovered in the geometric formulation, that everything could be formulated around the invariance of this quantity and that the sought transformations themselves could be simply characterized as those that preserve the metric tensor $\eta$ and this lead to a geometric formulation of SR which directly lead to GR later on.


Because it worked.

Today we don't do that. Instead we recognize that associating a hyperbolic geometry to space-time is conceptually more satisfying, and mathematically more elegant and easier to use.

By the way, if you are learning relativity from a book that uses $ict$, you are learning an outdated method that no one uses today. Find another resource.


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