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This is no question on sign convention, and it is no question if ds or $ds^2$ shall be considered as the spacetime interval: I have taken my personal decision to opt for the signature (+,-,-,-) convention, and to consider ds as the spacetime interval (that means I extract the root of the squared interval $ds^2$).

With this personal decision, I follow Landau Lifschitz: "The classical theory of fields" (see equations 2.3 and 2.4). However, there is one problem: Equation 2.4 reads there:

$$ds^2 = cdt^2 - dx^2 - dy^2 - dz^2$$

that means that ds has the unit of a space distance. In contrast, Sexl Urbantke: "Relativity, Groups, Particles", considers proper time as the "physical interpretation of the spacetime interval ds", and accordingly, they state in chapter 2.6 "Proper time and time dilation"

$$ds = dt \sqrt{1-v^2} < dt$$

So the question is: Has the spacetime interval a time unit, a space unit or both, and how can this be derived from special relativity (or is it an assumption only?)

Personally, I agree that proper time is the "physical interpretation of the spacetime interval ds". How is it possible then to assign to the spacetime interval a space distance unit?

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    $\begingroup$ This is similar to asking how it is possible to assign the mass of the proton a value of 938 MeV, which is an energy, not a mass. Please read about natural units. Physicists are very flexible about units, because there is no actual physics in them! $\endgroup$
    – G. Smith
    Dec 15 '19 at 16:52
  • $\begingroup$ @G.Smith, No physical difference if we treat spacetime intervals as meters or as seconds? $\endgroup$
    – Moonraker
    Dec 15 '19 at 17:13
  • $\begingroup$ Not to physicists who set $c=1$, which is pretty much anybody doing physics beyond intro physics. To a relativist, the physics distinguishing space from time is not their units but the fact that their contributions to the spacetime interval have opposite signs. $\endgroup$
    – G. Smith
    Dec 15 '19 at 17:17
  • $\begingroup$ The key point here is that the value of dimensional constants like $c$ is an arbitrary human choice. We can make them have any numerical value we want by choosing our units. So we might as well just choose units in a way that makes them dimensionless and disappear. $\endgroup$
    – G. Smith
    Dec 15 '19 at 17:26
  • $\begingroup$ For example, it should be clear that we can say that $c$ is 1 light-second per second. The next step is just to say that we are measuring distance in seconds rather than light-seconds, or time in light-seconds rather than seconds, and make $c$ the dimensionless value 1. $\endgroup$
    – G. Smith
    Dec 15 '19 at 17:29
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The most common convention is to give the spacetime interval units of distance. However, it’s also common to choose units such that $c = 1$ such that time and distance have the same units.

In the second expression you quote, $c = 1$ has been chosen. This can be seen by the $\sqrt{1 - v^2}$ factor: this is dimensionally inconsistent in SI units. You could rewrite it in different units as

$$ds = c \, dt \sqrt{1 - \frac{v^2}{c^2}}.$$

Note that it’s also possible to define a Lorentz-invariant interval with units of time:

$$(d\tilde{s})^2 = dt^2 - \frac{1}{c^2}(dx^2 - dy^2 - dz^2)$$

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  • $\begingroup$ Thank you, DavidH +1. The question is arising why units of distance are chosen if proper time may be considered as the "physical interpretation of the spacetime interval". $\endgroup$
    – Moonraker
    Dec 15 '19 at 9:33
  • $\begingroup$ @Moonraker is the quote made in the context of time-like intervals? Because for space-like intervals, the physical interpretation would be proper distance, for null intervals it would be - well i don't know in this case: the spacetime interval is simply 0. $\endgroup$
    – Umaxo
    Dec 15 '19 at 14:30
  • $\begingroup$ @Moonraker and i think the units of distance were chosen, because in Newtonian mechanics space already has its geometry. You have vectors, distances - everything...In STR you are bringing time into this whole machinery to extend all the 3-vectors to 4-vectors (etc.), so it is natural to pick distance over time, since the space "was there first" $\endgroup$
    – Umaxo
    Dec 15 '19 at 14:46
  • $\begingroup$ Yes, @Umaxo, I am mainly interested in timelike intervals. - Thank you for your opinion on the "why" - perhaps you can formulate an answer? $\endgroup$
    – Moonraker
    Dec 15 '19 at 15:10
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Personally, I agree that proper time is the "physical interpretation of the spacetime interval ds". How is it possible then to assign to the spacetime interval a space distance unit?

The spacetime interval can be divided into 3 categories - timelike, spacelike and null. The interpretation of spacetime interval being proper time can be made only for timelike interval (which was probably the context of your textbook). For spacelike intervals, the interpretation would be proper distance. For null interval neither of interpretations make sense. The units of spacetime interval were chosen without relevance to these interpretations, since there is no natural way to choose one defining type of interval over the others.

Now, the interpretation works, because for given type of spacetime interval you can find lorentz frame in which the spacetime interval reduces to proper time/distance (up to $c$ coefficient for units consistency), so f.e. if spacetime interval is time-like, then it is a measure of proper time in certain lorentz frame.

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I think it's just related to the choice of units. In some subjects of physics, we choose $c=1$, and also we can set other fundamental constants to 1 such as $\hbar=1$ and $k_B=1$.

In this case $c=1$ then \begin{equation} ds^2=dt^2-dx^2-dy^2-dz^2 \end{equation}

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  • $\begingroup$ Pinkman, it is definitely no choice-of-unit or c=1 problem: The equation of Landau/Lifschitz includes expressly c. This is not the case for the equation of Sexl/Urbantke, but they state clearly that "proper time is the physical interpretation of the spacetime interval". $\endgroup$
    – Moonraker
    Dec 15 '19 at 9:07
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    $\begingroup$ In General Relativity, where we set $c=1$, if $ds^2>0$, you can find a Lorentz transformation that takes you to a manifest time-like interval, i.e $x^{\mu}$'$=(\tau,0)$. In this case $ds^2=d\tau^2$ and therefore space time interval in nothing but proper time. $\endgroup$
    – devCharaf
    Dec 15 '19 at 9:21

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