In spacetime, I understand that we multiply time by the speed of light to deal with homogeneous distances over the four axis, space and time.

But what does $t$ refers to precisely? Where is $t$ measured?

Is $t$ a property of the observer linked to the reference system, in other words, the time as measured by a clock linked to the observer?

Should this time $t$ be considered universal time?

  • 2
    $\begingroup$ It's the time in a conceptual lattice of clocks & rods which is at rest in the frame. See physics.stackexchange.com/q/514964/123208 There's no universal time in relativity. $\endgroup$
    – PM 2Ring
    Commented Aug 16, 2022 at 13:35

2 Answers 2


The $t$ corresponds to the wristwatch time (the proper time) of the inertial observer drawing the spsacetime diagram where this observer is at rest.

In special and general relativity, that $t$ is not a universal time since (as @PM2Ring said in a comment).

Structurally speaking, $t$ is first a property of the worldline of the inertial observer at rest in her spacetime diagram. Some further construction (e.g. radar measurements or some other procedure) is needed to use this $t$ to help label and assign coordinates to events not this worldline... to figuratively "spread time over space".


I was going to add a comment in response to @kiriloff , but then felt it was better to update my initial response.

Given any worldline, $\tau$ is the proper time along that worldline, which is analogous to the arc-length of a curve. As $\tau$ advances along that worldline, the corresponding event on the worldline varies its $(t,x,y,z)$ coordinates, where the coordinates are those set up by the inertial observer (call her Alice) at rest in the spacetime diagram.

So, for Alice's worldline [which is vertical through the origin in her spacetime diagram], as her $\tau_{Alice}$ advances, the corresponding event on Alice's worldline varies its $t$-component (with $t=\tau_{Alice}$) but its $(x,y,z)$ coordinates remain $(0,0,0)$ since Alice is at rest at origin in her own spacetime diagram.

Here is the radar-measurement procedure to assign coordinates to events in spacetime, especially to events not on her worldline. For a distant event $Q$,

  • Alice arranges to send a light signal to $Q$. She notes the time $\tau_{send}$ on her wristwatch.

  • Alice waits for the reflected echo light signal from $Q$. She notes the time $\tau_{receive}$ on her wristwatch.

  • Alice defines \begin{align} t_Q &=\frac{\tau_{receive}+\tau_{send}}{2}\\ x_Q &=c\frac{\tau_{receive}-\tau_{send}}{2}\\ \end{align}

  • (When $Q$ is on her worldline, then $\tau_{receive}=\tau_{send}$. So, $x_Q=0$ and $t_Q=\tau_{send}=\tau_{receive}$.)

So, by using the light-signals, this is how Alice assigns a time coordinate to a distant event $Q$---essentially assigning "the wristwatch reading $\tau$" of the event on her worldline that Alice says is simultaneous with the distant event $Q$.


Note that $\tau_{receive}=(t_Q +x_Q/c)$ and $\tau_{send}=(t_Q -x_Q/c)$ so that $$\tau_{receive}\tau_{send}=(t_Q +x_Q/c)(t_Q -x_Q/c)=t_Q^2-(x_Q/c)^2.$$

Every inertial observer through the origin event will generally assign different coordinates to event $Q$. But they will find (by experiment or by using the Bondi k-calculus, for example) that the product of their radar times $\tau_{receive}\tau_{send}$ will be equal (i.e. an invariant... the invariant squared-interval). In terms of their rectangular coordinates, this product equals $t_Q^2-(x_Q/c)^2$.


Here's a spacetime diagram of the radar measurement,
in particular, how the Blue inertial observer uses Blue's wristwatch and radar to assign the time-coordinate $t=2$ to events not on Blue's worldline. $$\tau_{receive}=7 \quad \tau_{send}=-3 \qquad\mbox{so }\tau_{receive}\tau_{send}=-21 $$

$$t_Q=\frac{7+(-3)}{2}=2\quad x_Q=\frac{7-(-3)}{2}=5 \qquad\mbox{so } t_Q^2-x_Q^2=(2)^2-(5)^2=-21$$


Here is that radar experiment viewed from a different inertial frame where Blue has, say, $v_{Blue\ wrt\ Red}=(3/5)c$ (so, Doppler $k=2$).


By the way, although not drawn in, Red can also measure $Q$ with radar $$\tau_{red,receive}=14 \qquad \tau_{red,send}=-3/2 \qquad \mbox{so }\tau_{red,receive}\tau_{red,send}=-21 $$

$$t_{rQ}=\frac{14+(-3/2)}{2}=\frac{25}{4}=6.25\quad x_{rQ}=\frac{14-(-3/2)}{2}=\frac{31}{4}=7.75$$ $$\qquad\mbox{so } t_{rQ}^2-x_{rQ}^2=(6.25)^2-(7.75)^2=-21$$

  • $\begingroup$ I think the other answer from @Dale is quite different, could you comment? He says proper time is tau as defined per the mentioned difference. $\endgroup$
    – kiriloff
    Commented Aug 16, 2022 at 20:18
  • $\begingroup$ @kiriloff I updated my answer. $\endgroup$
    – robphy
    Commented Aug 16, 2022 at 21:27

But what does t refers to precisely?

The $t$ you are talking about is coordinate time. It is simply one of four coordinates used to label events in spacetime. Not all coordinate systems even have a $t$ coordinate, so it is not something essential or fundamental.

Where is t measured?

Coordinate time is not measured. What is measured is called proper time and it is often denoted with $\tau$ in the relativity literature. In an inertial frame the relationship between $t$ and $\tau$ is $$c^2 d\tau^2= c^2 dt^2 -dx^2 -dy^2 -dz^2$$

It is this quantity that clocks measure.

Is t a property of the observer linked to the reference system, in other words, the time as measured by a clock linked to the observer?

Should this time t be considered universal time?

The coordinate time is just a convention adopted for a specific coordinate system. Because it is frame-dependent it is not considered universal.

  • $\begingroup$ How can I understand intuitively that clocks measure c2dτ2=c2dt2−dx2−dy2−dz2 in which some some space differences occur? $\endgroup$
    – kiriloff
    Commented Aug 16, 2022 at 20:13
  • $\begingroup$ Intuitively, a clock measures the “length” of its worldline in spacetime, where “length” is not given by the Pythagorean theorem but the above formula instead $\endgroup$
    – Dale
    Commented Aug 16, 2022 at 21:13

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