# In spacetime, how do we interpret the “4th dimension”?

Relativistic Spacetime is a "merger" (or whatever the technical term is) of 3 spatial dimensions and 1 time dimension into a 4-dimensional Lorentzian manifold.

However, I am confused as to how I should interpret the "time dimension", for the following reason:

The "time" that a particle takes over the course of a trajectory through spacetime, is simply the "length" (according to the Lorentzian metric tensor) of that path in spacetime. The result is of course that the time a particle spends, depends on its trajectory, even if two trajectories land at the same point in spacetime.

This results for example in the twin scenario, where one twin stays put, and the other travels away with close to the speed of light and back. The second twin will be younger.

I understand that time is therefore relative, and there is not an absolute time function from the lorentzian manifold to $R$. We cannot speak of "the time" at a particular point in spacetime.

While I understand this, it leaves me with a puzzle. To make it clear: assume the simple "flat" spacetime $(t, x, y, z) \in R^4$ of special relativity, where the metric tensor $g_{ij}=\eta_{ij}=\delta^i_j \cdot \{1 \text{ if} i=1, -1 \text{ otherwise }\}$. Assume a time orientation $T^i_{(x)}=(1,0,0,0)$.

What confuses me about this is: while time is not an absolute thing, but differs between particles and depends on their trajectory through spacetime, there is nevertheless an absolute time dimension, given by the first coordinate in $R^4$ (in the special relativity case. Ofc in the general case we can't simply map it to $R^4$ but nevertheless the same point holds).

We don't have to look at the value of the "time coordinate" of each point in spacetime, because even if we don't, I am confused about what the time dimension is/means, given that we've defined time already as something different, namely the "lorentzian length" of curves through spacetime.

That is my main question, but we could even go further and define an absolute time function by mapping every point in spacetime simply to its first coordinate. What would be the meaning of that "absolute time function"? The fact that this is possible seems to contradict the proposition that there is no absolute time in relativity.

I hope my question is clear, but I find my confusion hard to explain.

You might want to have a look at What is time, does it flow, and if so what defines its direction? though I suspect this is a bit too basic for you and you already understand most of what it covers.

Anyhow, one of the hard things to grasp about relativity is that there is no one absolute coordinate system. Any coordinate system is as good as any other coordinate system when it comes to describing the physics. Any coordinate system will have one time-like dimension, but different coordinate systems will not necessarily agree of what the time-like coordinate is.

This is responsible for the time dilation that you describe. Both the moving observer and the twin on Earth use coordinate systems in which they are at rest. However their two time dimensions are not the same. Each observes the other's time dimension to be a mixture of time and space, so they disagree about the displacement along the time axis.

• Dear John, thanks for this answer. If I may ask, regarding the last paragraph, is there a simple way of seeing why each observer sees the other's time dimension as a mixture of time and space? I admit I don't understand what this mixture means either. Many thanks for any clarifications or hints. – user929304 Jun 14 '17 at 8:42

Your coordinate system $t,\,x,\,y,\,z$ has a time coordinate; the reason time is relative is because a Lorentz transformation gives a new coordinate system $t',\,x',\,y',\,z'$ for which the change in $t$ between two spacetime events is in general not equal to the change in $t'$. See here for a simple example.

In particular, simultaneity is relative. Your chosen coordinate system allows you to define the events that occur "at" a given time as a locus of the form $t=t_0$. However, that locus of events is not of the form $t'=t_0'$.

Relativistic Spacetime is a "merger" (or whatever the technical term is) of 3 spatial dimensions and 1 time dimension [...]

Spacetime (at least in "sufficiently sensible, simple cases"; such as globally hyperbolic spacetime), i.e. a set $\mathcal S$ of events, can be represented as a (suitably general) product of (the "structure" of) spacelike hypersurfaces and (the "structure" of) (the image of) a curve; formally $$\mathcal S := \Sigma \times \mathcal I.$$

In other words, for such a spacetime we can consider its foliation, or slicing into an ordered set (one-dimensional set) of disjoint, not necessarily exactly equal, three-dimensional spacelike hypersurfaces.

The order of these leaves, or slices, is thereby consistent with (or indeed derived from) the (individual, timelike) order in which each individual participant took part in events.

So far on the gist of your main question.
Now some additional remarks:

The "time" that a particle takes over the course of a trajectory through spacetime,

... the duration $\tau$ of the particle ("identifiable material point", "participant") from having indicated its participation in one event, until having indicated its participation in another, subsequent event ...

is simply the "length" [...] of that path in spacetime.

Sure, the path of a specific participant $A$ (i.e. the ordered set of events in which it took part), between a specific initial event $\varepsilon_{A J}$, and a specific final event $\varepsilon_{A Q}$, can "simply" be assigned some specific (suitably generalized) "length" value: ...

$\tau A[ \, \_J, \_Q \, ] :=$
$\text{inf}\left[ \, \left\{ \, \sum_{k = 1}^n \ell[ \, \varepsilon_{A P_{(k - 1)}}, \varepsilon_{A P_{(k)}} \, ] \, {\large{|}} \, n \in \mathbb N, \varepsilon_{A J} \equiv \varepsilon_{A P_{(0)}} \ll \varepsilon_{A P_{(1)}} \ll ... \ll \varepsilon_{A P_{(n)}} \equiv \varepsilon_{A Q} \, \right\} \, \right]$

... provided that suitable distance values $\ell$ (so-called Lorentzian distances) are given (as a corresponding pseudo-metric space).

In turn, values of Lorentzian distance $\ell$ might "simply" be expressed in terms of the durations (path lengths) of any and all participants; for example the specific value

$$\ell[ \, \varepsilon_{A J}, \varepsilon_{A K} \, ] := \text{sup}\left[ \, \left\{ \, \tau P[ \, \_J, \_K \, ] \, {\large{|}} \, \varepsilon_{A J} \equiv \varepsilon_{P J} \equiv \varepsilon_{A J P}, \varepsilon_{A K} \equiv \varepsilon_{P K} \equiv \varepsilon_{A K P} \, \right\} \, \right].$$

Therefore neither of these two relations gives an actual clue about how to obtain any of the required values $\ell$ or $\tau$. But at least they allow to express and to appreciate that, indeed, the durations $\tau$ of distinct participants between a common initial event and a common final event are not necessarily equal to each other, if these participants had been separated from each other, visiting different events inbetween.

(according to the Lorentzian metric tensor)

Well, coordinates obviously may be sprinkled onto the events of set $\mathcal S$ under consideration; and concrete values $g_{\mu\nu}$ of the metric tensor may then be calculated as derivatives (if they even exist) of Lorentzian distances $\ell$ wrt. coordinate differences. But this doesn't tell you either how to obtain values $\ell$ in the first place, either.

The Lorentzian length of a timelike path is proper time along that path. Eg. time as measured by the observer which moves along that path in spacetime.

If you are given a Lorentz-frame $(t,x,y,z)$, this can be interpreted as what a certain inertial observer sees. This observer is at rest with respect to his or her inertial frame, so his trajectory is given by $dx^i=0$ ($i$ ranges from 1 to 3), so we have (in $c=1$) $ds^2=-dt^2$, so for him, coordinate time IS proper time, but only for him!

We do not, however have "absolute time", because the line element $ds^2=-dt^2+dx^2+dy^2+dz^2$ is left invariant by Lorentz-transformations. Lorentz-transformation is essentially the linear transformation between "pseudo-cartesian" inertial frames that have the same origin.

Which means, that given any Lorentz-transformation $\Lambda^\mu_{\ \nu}$, the coordinates $x^{\mu'}=\Lambda^{\mu'}_{\ \nu}x^\nu$ also form a valid inertial frame. If the Lorentz-transformation $\Lambda$ contains boosts, then this new inertial frame represents the frame of an observer that moves with respect to the original observer with some constant velocity. In this case the new time coordinate is $t'=\Lambda^0_{\ \nu}x^\nu=\Lambda^0_{\ 1}t+\Lambda^0_{\ 1}x+\Lambda^0_{\ 2}y+\Lambda^0_{\ 3}z$, and this time coordinate is just as good as the previous one.

Probably what confuses you is that when we model Minkowski spacetime as $\mathbb{R}^4$, we choose a preferred coordinate system, which is the standard one.

But this IS a choice on our part, and ANY other coordinate system related to this chosen one via a Lorentz transform is just as good a coordinate system.

A better model would be to let $V$ be a 4-dimensional real vector space, and let $\langle,\rangle$ be an indefinite inner product of signature $(-+++)$. In this case, we can choose a basis of $V$, $(e_0,e_1,e_2,e_3)$, for which $\langle e_\mu,e_\nu\rangle=\eta_{\mu\nu}$, but we have an infinitude of such bases, all related by Lorentz-transforms, and none of them are preferred over the other.

The trajectory-dependent time, the one that an observer followong a given trajecrory measures, is called the proper time. The time coordinate of $\mathbb{R}^4$ is not an absolute time, but just the proper time of trajectories with constant spatial coordinates.

Of course, what constant spacial coordinates are depends on the choice of reference frame. We can change coordinates in Minkowski space using Lorentz transformations and get the coordinates corresponding to a different inertial frame. In it, time might now be different, and its non-absolutness is clear.

Maybe a slightly more technical view can put things in a more general perspective.

If you look at the Minkowski space as a vector space, any space-time event corresponds to a unique Minkowski vector (space-time point) $x$, while an inertial reference frame corresponds to a Minkowski vector basis $\{{\hat e}_\mu\}_\mu$ orthogonal w.r.t. the flat metric, meaning ${\hat e}_\mu \cdot {\hat e}_\nu = g_{\mu\nu}$.

This implies that one of the basis 4-vectors is necessarily time-like, while the other 3 must be space-like, regardless of the particular metric signature convention. Moreover, it is not hard to see that frames that share the same time-like basis 4-vector can only differ by a space symmetry operation (rotation, reflection, inversion, or translation).

So we can safely say that an inertial reference frame is defined by a unique time-like 4-vector. Then the "time" of any event $x$ w.r.t that frame is always the event coordinate w.r.t to the frame's time-like basis 4-vector, however we choose to list it in the 4-basis. That is, for the frame $\{{\hat e}_\mu\}_\mu$, we have $x =x^\mu {\hat e}_\mu$, and if ${\hat e}_0$ is the time-like vector, then the event time is $x^0$. Note that because the metric is pseudo_Euclidian, formally this is not $x\cdot {\hat e}_0 = x^\mu g_{\mu 0} = x_0$, but $x^0 = g^{0\mu}x_\mu = g^{0\mu} x\cdot {\hat e}_\mu$.

The event's proper time is of course the event time relative to the time-like vector of its "rest frame", where its spatial coordinates are null. This is your "lorentzian length", since in this case the "magnitude" of 4-vector $x$, $x\cdot x = g_{\mu\nu}x^\mu x^\nu$, is given by its proper time alone.

Finally, Lorentz transformations from one frame to another amount simply to Minkowski "rotations" between different frame bases, while Poincaré transformations also include space-time translations.

So you can say that the "absoluteness" of time is equivalent to the "absolute" existence of a time-like coordinate, or of a defining time-like 4-vector, in any reference frame. And all this is perfectly consistent with the absence of an "absolute time" in the Galilean sense, since "time" is always only "relative to a frame".

## protected by Qmechanic♦Mar 3 '17 at 11:30

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