Notion of Present

Question

Can't I sync all watches in spacetime and call this time slice the present? In Carlo Rovelli's book he tried to explain that the notion of the present is local only, which I could not follow.

Answer 1

This is a nuanced topic, and one that is the subject of a lot of philosophical debate, much of which appears ill-informed.

The statement that 'the notion of the present is local' is analogous to the notion of horizontal being local. Within a given distance it makes sense on Earth to talk about the horizontal, but what is horizontal in an East-West direction for me in England will seem tilted by 15 degrees from the perspective of someone in New York. It is not possible to define a plane which is 'horizontal' for everyone.

Of course, the answer to the conundrum is that there is a (roughly) spherical surface of the Earth, and 'horizontal' at any given point on the surface means a tangent to the surface at that point. The disagreements only arise if you project the tangents too far.

Wherever you meet someone on the surface of the Earth you will both agree that you are together on the the same physical level. However, if one of you uses a coordinate system in which the Z direction is normal to the Earth in London, and another uses a coordinate system in which the Z direction is normal to the Earth in New York, you will attribute conflicting values of height to the point at which you are standing.

The same is true of the present. At any point in the universe, regardless of their respective frames of reference, two observers who meet will always agree that they are together in 'the present'. However, depending on their frames of reference they will have different values for the date and time of their meeting. One might say it is noon on Wednesday, and the other might says it is 10.37 on Friday, but they are just using different time coordinates to label the same event.

You cannot project a plane through a point on your time access and claim it represents the present everywhere any more than you can project a plane through your Z axis and claim it represents the horizontal everywhere.

Answer 2

In special relativity, you can pick an inertial reference frame, and then in that frame you can do essentially what you describe: place clocks all over space (not spacetime) and synchronize them all. The synchronization can be done by various equivalent methods, such as transporting clocks slowly or Einstein synchronization.

In general relativity, this doesn't work anymore for a general spacetime. It works only in a static spacetime, which is one that is not changing over time and is not rotating. In a non-static spacetime, Einstein synchronization is not transitive, so synchronizing clock A with clock B and B with C does not mean that A is synchronized with C.

Answer 3

The "present" in general relativity is a thorny topic. It's already no bouquet of roses in special relativity, but there is at least a fairly simple theorem for it.

Two events are causally related if there exists a causal curve (timelike or null) linking them. If no such curve exists, we say that these two events are synchronous. A synchronization is a foliation of our spacetime by achronal spacelike hypersurfaces (ie : we have a time function $\mathfrak{t} : M \to \mathbb{R}$ such that $\mathfrak{t}^{-1}(t) = \Sigma_t$, $\Sigma_t$ a spacelike hypersurface), for which we say that if two events evaluate to the same value under this time function, then they are synchronous.

Given some inertial observer $\gamma$, there exists, for Minkowski space, an objective way of performing a synchronization of our spacetime using only physical processes. If, for every $p_\tau \in \gamma$, such that $\gamma(\tau) = p_\tau$, our observer emits light rays in all directions (each of them labelled by time of emission and angles $\theta, \phi$), then by a theorem of Weyl, you can find a synchronization for every point. If the signal $(\tau_1, \theta, \phi)$ is reflected by another point and returns to our observer $\gamma_0$ at $\tau_2$, then we say that this point (let's call it $p_{\tau_1\tau_2\theta\phi}$) is synchronous with a point of $\gamma_0$ somewhere between $p_{\tau_1}$ and $p_{\tau_2}$. The exact point we get is given by the Reichenbach synchronization :

\begin{equation} \tau = \tau_1 + \varepsilon(\tau_2 - \tau_1) \end{equation}

with $\varepsilon \in (0,1)$. Those all roughly correspond to every possible spacelike straight lines crossing $\gamma_0$. In particular, the standard synchronization we usually use is the Einstein synchronization, $\varepsilon = 1/2$. This corresponds to the notion that, in this coordinate system, light is measured as $c$ (other Reichenbach conventions, light has on average the speed of light but its "coordinate speed" is faster or slower depending on how it travels). Given this, we can assign to every point of Minkowski space an angle, time, and a distance from the Synge formula :

$$\| p \| = \sqrt{(t - t_1)(t - t_2)}$$

Which basically just means that it is at a distance far enough that this distance divided by the time for the trip is the speed of light (with $c = 1$ here).

Now let's consider the case of curved spacetime. Immediately things go back, for a very wide variety of reasons.

• A light signal sent at local time $\tau_1$ can come back before $\tau_1$ due to bad causal behaviour.
• A future-pointing light signal may come back as a past-pointing light signal (it will come back in such a way that it seems to actually be emitted again by the original observer!)
• Two points may be separated in such a way that the return signal will never come back, due to horizons.
• If our manifold has a cut locus, two different null geodesics can end on the same point.

There are many reasons why things can go bad. Not all of them make it impossible to perform such a thing, but it is most certainly a much less trivial affair than before. I'll go over a few techniques, simplifications and mitigating factors.

First, some ground rules : we'll pick the most physically reasonable type of spacetime with respect to causality. In other words, a globally hyperbolic spacetime. This way, we know that a synchronization exists at least in principle. A globally hyperbolic spacetime is characterized by the existence of Cauchy surfaces, ie achronal spacelike hypersurfaces such that every causal curve intersects each such surfaces exactly once. This is equivalent to the previously mentioned time function $\mathfrak{t}$, and in this case we have $\mathfrak{t}^{-1}(t) = \Sigma_t$ being a Cauchy surface.

The synchronization of two points is then simply that $p$ and $q$ are synchronous if $\mathfrak{t}(p) = \mathfrak{t}(q)$. This notion is somewhat arbitrary of course : just like in special relativity, there are many different foliations of the spacetime we could do with different synchronous surfaces. This isn't much of a problem but a much bigger problem we have is that we only have an abstract definition, and not any measurable quantities here.

First let's see how the exchange of light signals work here : take our main observer $\gamma_0$ again here, with $p_1, p_2 \in \gamma_0$. A null geodesic $\ell_1$ is sent from $p_1$ to $q$, the point we wish to analyze, and $q$ sends back another one $\ell_2$ to $p_2$. If we switch the time-orientation of $\ell_1$ (so that we have a past-oriented null geodesic from $q$ to $p_1$), this means that both $p_1$ and $p_2$ are in the light cone of $q$. To be more specific, $p_1 \in J^-(q)$ and $p_2 \in J^+(q)$. Unless $q \in \gamma_0$, this means further that there is a section of $\gamma_0$ between $p_1$ and $p_2$ that is not causally related to $q$ : there exists a point between $p_1$ and $p_2$ which is simultaneous to $q$.

This is good news but not great news either. If $p_1$ and $p_2$ were to be proven to be on the boundary of $J(q)$, then we could pick any synchronization we wish, but proving that this is so may not be possible. Let's for now consider a few simpler cases, such as local synchronizations and static ones.

As is well known, in any spacetime there exists a local neighbourhood called the normal neighbourhood in which the exponential map is a homeomorphism : for any point $p$, we have a normal neighbourhood $N_p$ such that, for any point $q \in N_p$, then there exists a vector $v$ in $p$'s tangent space such that

$$q = \exp_p(v)$$

The exponential map basically states that we have a family of geodesics $\gamma_{p,v}$ starting at $p$ ($\gamma_{p,v}(0) = p$) and with initial tangent $v$ ($\dot{\gamma}(0) = v$) such that $\exp_p(v) = \gamma_{p,v}(1)$. In such a neighbourhood, two points are connected by a unique geodesic. Overall, we have the very practical fact that this neighbourhood is the image of a subset of the tangent space (ie, Minkowski space). So let's consider a causally convex (ie any causal curve between two points of $N_p$ is entirely within $N_p$) normal neighbourhood around our observer $\gamma_0$. From the exponential map we can simply directly apply the Reichenbach synchronization in the tangent space and map it to $N_p$, giving us locally a synchronization.

The issue here is fairly obvious : we typically don't know how far the exponential map extends. This could be the entire universe or a space smaller than the Planck scale for all we know, although it's commonly accepted that our spacetime is close enough to Minkowski space that we can trust such an approximation over fairly large distances.

Now let's consider something a bit more general. From before, we've seen that a synchronization always exists, and we could try to work backward from it, ie : take a time function $\mathfrak{t}$, a Cauchy surface $\Sigma_t$, an observer $\gamma_0$, and from this consider the light cone of the point $q$ on $\Sigma_t$ we're considering. Our two points of emission and reception are in $J^\pm(q)$, and we know that those points are not necessarily on the boundary of the light cone. Therefore, all we can really say about our synchronization is

$$t = t_1 + \varepsilon(x) (t_2 - t_1)$$

Our synchronization parameter could be any continuous function $\varepsilon : M \to (0,1)$.

Now to help out a bit, we'll have to consider something a bit excessive. Let's imagine now that our observer $\gamma_0$, at every instant, sends a light signal to every spacetime point, and every spacetime point reflects it back. This allows us the benefit of knowing that, for a sufficiently small neighbourhood of $\gamma_0$, we can use some manner of Einstein synchronization and work from there.

Now given a time function $\mathfrak{t}$ on $M$, this induces a foliation not only by Cauchy surfaces, but also by timelike geodesics. Let's consider our observer as one of those geodesics, and every point we synchronize as lying on one of them. Every light clock trip hitting $q$ is therefore parametrized only by its time of emission (or reflection at $q$, or reception, as those are all related) and by the position of the geodesic on the initial Cauchy surface, so that if $q \in \gamma_{y}$, where $\gamma_y(0) = y$, then we'll refer to it as the trip $\ell_{t_1, y} = \ell_1(p_1 \to q) \cup \ell_2(q \to p_2)$. There may be more than one such light trip (indeed an example using the Minkowski cylinder $\mathbb{R} \times S$ is easy to construct), but more importantly, there is a unique such shortest trip : the light trip obtained by connecting $q$ with $p_1$ and $p_2$ on the horismos $H(q)$ (the surface of the light cone) has a unique set of times $t_1$ and $t_2$ such that this is the shortest possible such trip.

So the information we have are :

• The time of emission $t_1$
• The time of reception $t_2$
• Angular data of the emission $\theta, \phi$
• A unique identifier of our object (this is important to make sure we found the correct shortest signal)

It can be shown that using the Einstein synchronization blindly with such data will not work. Even in Minkowski space with an accelerated observer, the coordinates generated by such a process will not be a bijection for our spacetime. What we can do though is to use all the data generated by all our observers to generate a local vector field around each observer. The time function will be such that $d\mathfrak{t}$ is normal to all the local basis defined by our synchronizations.

This is not a very realistic process, of course. We can't really fill the universe with infinitesimal observers, and having our spacetime filled with light beams coming from an infinity of sources, each encoding infinitely many data, is a bit of a stretch of the notion of "test field". But given a reasonable approximation of this process, we can define synchronizations up to some level of confidence.

Answer 4

The notion of 'the present' presupposes a notion of simultaniety and Einstein pretty much deconstructed this notion in his paper on special relativity when he promoted the constancy of the speed of light to a universal principle. This deconstruction is pushed even further when he developed his theory of General Relativity. And it is to this what Rovelli is referring to when he says the present can only be defined locally.

However, all is not lost for 'the present'. It's notorious that both QM and GR do not get along well. And Smolin in his book, Time Reborn, suggests that their different understanding of time is one of the major stumbling blocks. He suggests that QM shows that time is 'open'. And obviously the time just before time is open is the present.

It's worth recalling that Einstein had a famous debate with the then very famous French philosopher Bergson on the nature of time, which on all accounts the then even more famous physicist won. Bergson defined time through his notion of duration which he defines as:

Duration is essentially a continuation of what no longer exists into what does exist. This is real time, perceived and lived. Duration therefore implies consciousness and we place consciousness at the heart of things for the very reason that we credit them with a time that endures.