I'm a bit confused. Please help me

The geometric distinction between timelike and spacelike distances is mirrored in the devices used to measure them. A clock is a device that measures timelike distances; a ruler is a device for measuring spacelike ones. Two nearby points or a timelike world line are timelike separated, $d s^2<0$. To measure the distance along a particle's world line, it is convenient to introduce $$ d \tau^2 \equiv-d s^2 / c^2 . $$

Preceding this paragraph, the author defined that $(ds)^2$ as the squared distance between points in spacetime. $d s^2=-(c d t)^2+d x^2+d y^2+d z^2$.

I've two questions :

  • Why do we say clocks measure timelike distances? Why don't we say clocks measure timelike time? And why insist on 'Timelike'? What about clocks measuring spacelike distances/times?

  • If we say that $(ds)^2$ is the squared distance between points in spacetime then the distance is imaginary for timelike events?

  • $\begingroup$ In physical terms clocks measure the flow of energy from a local energy reservoir (spring, battery etc.) to infinity. They are, if you like, the ideal implementation of an irreversible system. It doesn't get any more "irreversible" than a clock (that's why time can't "go backwards"). The surprising thing is that in the physical vacuum this flow of energy from any given point to infinity seems to have geometric properties very similar to those of space itself. We can treat it like a quasi-distance. Where does this come from? Ultimately it comes from the relativity principle. $\endgroup$ Sep 29, 2022 at 8:00

1 Answer 1


Before answering directly your questions, I think a little summary could become useful.

THE ROLE OF $\mathbf{(ds)^2}$ IN RELATIVITY. Using the metric with signature (-+++), the one your book is using, we can identify three different cases, which can be further visualized using light cones (I strongly suggest for this Section 5.3 of Rindler's book "Relativity: Special, General and Cosmological" or any other introductive book you find yourself comfortable with):

  1. $\mathbf{(ds)^2<0}$, so that the events are said to be timelike separated. As a consequence they can be connected with a signal whose speed doesn't overcome the speed of light $c$. They can therefore influence each other and such events are in the causal past or causal future of one another.
  2. $\mathbf{(ds)^2=0}$, so that the events are said to be lightlike or null separated. In this case only light can connect those two events.
  3. $\mathbf{(ds)^2>0}$, so that the events are said to be spacelike separated and not even light can connect the signals. They cannot therefore influence each other and such events are in the causal present of one another.

This is where the terms spacelike and timelike come from.

Returning to your set of questions, the concept of distance in Relativity has to be interpreted as a space-temporal one, when time and space coordinates are inevitably interlaced and cannot be further separated. For this reason clocks are used to measure distances as well as rulers, with the difference that we can use clocks just to measure timelike distances since we cannot use signals that overcome the speed of light. Furthermore, clocks do not measure "timelike time" because time is a tricky concept in Relativity and it's better to think in terms of proper time, which is a concept strongly related to the body which is moving, since observers would not agree on a specific value of time.

Concerning your last question, $(ds)^2$ is known as a relativistic invariant, so that its value it's exactly the same for different observers. Generally, you don't take the square root of it in order to find distances, since you would lose a lot of information for your specific problem. For this reason, I don't think it's useful to talk about "imaginary distance", since you're probably used to Euclidian geometry, but Relativity works in hyperbolic geometry, which can become quite hard sometimes to translate it into common sense. For timelike events you can simply state that $(ds)^2<0$, so that $$ dx^2+dy^2+dz^2<c^2 dt^2$$ Then, taking the square root of it you can simply state that the Euclidean distance $$dr=\sqrt{dx^2+dy^2+dz^2}$$ is smaller than the product of $dt$ times the speed of light. I'm not sure you can go any deeper on that.


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