In Special Relativity, the spacetime interval $$\mbox{d}s^2 = \mbox{d}t^2 - \mbox{d}x^2 - \mbox{d}y^2 - \mbox{d}z^2 \tag{$\star$}$$ between two events is well known to be invariant under Lorentz transformations, i.e. identical for inertial observers.

Once one assumes the speed of light to be constant for all inertial observers, it is easy to see that $(\star)$ is indeed invariant if the events are lightlike separated. If I recall correctly, it was possible (assuming homogeneity and isotropy of space) to then also derive that $(\star)$ must be invariant for arbitrary events (i.e. also ones which are timelike or spacelike separated) but I don't recall the details.

Can anyone help me out with this?

  • $\begingroup$ IIRC, there is a (questionable?) derivation on Landau&Lifshitz. Something about second-order differentials being proportional for some reason. I don't remember the details. $\endgroup$ – AccidentalFourierTransform Sep 2 '18 at 20:47
  • $\begingroup$ Yeah I know... I read that on a wiki page but I couldn't find it anymore. In what book exactly is it? One would argue that the differentials differ by a constant which can only depend on the magnitude of the velocity (by isotropy of space) and then do some kind of transitivity argument with the cosine law. $\endgroup$ – Thomas Bakx Sep 2 '18 at 21:18
  • $\begingroup$ Here it is, under 'invariance of interval': en.wikipedia.org/wiki/…. I don't really understand why the argument needs to be made using infinitesimals specifically, because we already know the 'lightlike' equality to be true in the general (non-infinitesimal) case. $\endgroup$ – Thomas Bakx Sep 2 '18 at 21:45
  • $\begingroup$ A treatment in this style is Palash Pal, "Nothing but relativity," arxiv.org/abs/physics/0302045 $\endgroup$ – user4552 Sep 3 '18 at 0:12
  • $\begingroup$ @BenCrowell The treatment by Pal is indeed based on homogeneity and isotropy requirements but it derives both the invariance of the speed of light and the invariance of the metric at one stroke. It doesn't explicitly show how the invariance of metric that is not guaranteed when we postulate only the invariance of the speed of light becomes guaranteed when we further postulate the homogeneity and isotropy requirements. Correct me if I am not seeing the forest for the trees. $\endgroup$ – Dvij D.C. Sep 3 '18 at 1:07

As AccidentalFourierTransform pointed out the answer is given in "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition. Copy-paste :

$\boldsymbol{\S}\: \textbf{2. Intervals}$

In what follows we shall frequently.......


As already shown, if $\:ds=0\:$ in one inertial system, then $\:ds'=0\:$ in any other system. On the other hand, $\:ds\:$ and $\:ds'\:$ are infinitesimals of the same order. From these two conditions it follows that $\:ds^2\:$ and $\:ds'^{\,2}\:$ must be proportional to each other: \begin{equation*} ds^2=a ds'^{\,2} \end{equation*} where the coefficient $\:a\:$ can depend only on the absolute value of the relative velocity of the two inertial systems. It cannot depend on the coordinates or the time, since then different points in space and different moments in time would not be equivalent, which would be in contradiction to the homogeneity of space and time. Similarly, it cannot depend on the direction of the relative velocity, since that would contradict the isotropy of space.

Let us consider three reference systems $\:K,K_1,K_2\:$ and let $\:\boldsymbol{V}_{1}\:$ and $\:\boldsymbol{V}_{2}\:$ be the velocities of systems $\:K_1\:$ and $\:K_2\:$ relative to $\:K$. We then have : \begin{equation*} ds^2=a\left(\boldsymbol{V}_{1}\right) ds_{1}^{2}\,, \quad ds^2=a\left(\boldsymbol{V}_{2}\right)ds_{2}^{2} \end{equation*} Similarly we can write \begin{equation*} ds_{1}^{2}=a\left(\boldsymbol{V}_{12}\right)ds_{2}^{2} \,, \end{equation*} where $\:\boldsymbol{V}_{12}\:$ is the absolute value of the velocity of $\:K_2\:$ relative to $\:K_1$. Comparing these relations with one another, we find that we must have \begin{equation} \dfrac{a\left(\boldsymbol{V}_{2}\right)}{a\left(\boldsymbol{V}_{1}\right)}= a\left(\boldsymbol{V}_{12}\right). \tag{2.5}\label{eq2.5} \end{equation} But $\:\boldsymbol{V}_{12}\:$ depends not only on the absolute values of the vectors $\:\boldsymbol{V}_{1}\:$ and $\:\boldsymbol{V}_{2}\:$, but also on the angle between them. However, this angle does not appear on the left side of formula \eqref{eq2.5}. It is therefore clear that this formula can be correct only if the function $\:a\left(\boldsymbol{V}\right)\:$ reduces to a constant, which is equal to unity according to this same formula. Thus, \begin{equation} ds^2= ds'^{\,2}\,, \tag{2.6}\label{eq2.6} \end{equation} and from the equality of the infinitesimal intervals there follows the equality of finite intervals: $\:s=s'$.

Thus we arrive at a very important result: the interval between two events is the same in all inertial systems of reference, i.e. it is invariant under transformation from one inertial system to any other. This invariance is the mathematical expression of the constancy of the velocity of light.

  • $\begingroup$ Thanks for sharing the section. I suppose that once we know that $a$ only depends on the velocity, the rest is pretty clear. The spacetime interval as defined is invariant under rotations in 3-space, so if one observer rotates his coordinates, this should not change the expression for $a$. I don't understand how homogeneity of space and time imply dependence on $v$ only, though. Any thoughts? $\endgroup$ – Thomas Bakx Sep 2 '18 at 22:42
  • $\begingroup$ @Thomas Bakx : Read carefully the paragraph below the first equation. $\endgroup$ – Frobenius Sep 2 '18 at 22:47
  • $\begingroup$ I did, but I think I am missing the point. What is it about $\frac{ds^2}{ds'^2}$ that makes this work? It seems like I could just insert any other function and argue that the quotient 'unprimed' / 'primed' cannot depend on space and time because of homogeneity... I hope this clarifies my problem. If I understand correctly, homogeneity of space and time asserts (more or less) that any invariant physical quantity can only depend on the differences between coordinates. This makes the proposed $ds^2$ seem at least reasonable. $\endgroup$ – Thomas Bakx Sep 2 '18 at 22:53
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    $\begingroup$ @ThomasBakx What is it about $\dfrac{ds^2}{ds'^2}$ that makes this work? I would say nothing. It would work for absolutely any two quantities that are required to vanish simultaneously. But clearly, it is only this specifically constructed quantities that would be required to vanish simultaneously as long as we are postulating only once invariant speed. $\endgroup$ – Dvij D.C. Sep 3 '18 at 0:51
  • $\begingroup$ As for why $a$ would not depend on the coordinates: We have $ds^2=a ds'^2$. This means, we have $\eta_{\alpha\beta}dx^\alpha dx^\beta=a\eta_{\mu '\nu '}dx'^\mu dx'^\nu$. Now, unless $a$ is a constant, it is not ensured that every $\dfrac{\partial^2x'^\mu}{\partial x^\alpha \partial x^\beta}$ vanishes. But we do want all of them to vanish for homogeneity. Thus, proven. $\endgroup$ – Dvij D.C. Sep 3 '18 at 0:56

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