# In spacetime what is the time $t$ on the $ct$ axis?

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?

• 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. Commented Aug 16, 2022 at 13:35

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".

update:

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 $$Q$$.

Update2:

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 $$t_Q^2-(x_Q/c)^2$$.

Update3:

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 $$\tau_{red,receive}=14 \qquad \tau_{red,send}=-3/2 \qquad \mbox{so }\tau_{red,receive}\tau_{red,send}=-21$$

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

• 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. Commented Aug 16, 2022 at 20:18
• @kiriloff I updated my answer. 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.

• How can I understand intuitively that clocks measure c2dτ2=c2dt2−dx2−dy2−dz2 in which some some space differences occur? Commented Aug 16, 2022 at 20:13
• 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
– Dale
Commented Aug 16, 2022 at 21:13