Consider a Riemannian Manifold with a metric tensor $g_{\mu\nu}$ and coordinates $(t, x^i)$. Let us assume that the spacetime is stationary, so $\partial_t g_{\mu\nu} = 0$. At a fixed coordinate time slice $t = t_f$, imagine placing an observer at each spatial coordinate point, and that the observers remain there for all coordinate time. One can ask what the spatial metric looks like from the perpsective of these observers. There are two seemingly reasonable notions of this, and I am trying to understand their respective physical meanings and the potential relation between the two.

Pullback Metric: Because all of the observers sit on a hypersurface of fixed time, it seems reasonable to think of their common "spatial manifold", endowed with coordinates $y_i$, via the embedding $y_i\rightarrow (t = t_f, x_i = y_i)$. The natural metric on the hypersurface is then the pullback metric, given by

$$dl^2 = h_{ij}dy^i dy^j,\hspace{1 cm}h_{ij} \equiv g_{\mu\nu}\frac{\partial x^{\mu}}{\partial y_i}\frac{\partial x^{\nu}}{\partial y_j} = g_{ij}$$

Effective Spatial Metric: In this construction, observer $A$ at $(t_f, x^i)$ sends a light ray to nearby observer B whose spatial location is $x^i + d x^i$. Observer $B$ receives the signal at the event $(t_f + dt, x^i + dx^i)$, and sends a response. Observer $A$ measures the proper time $d\tau$ between sending the signal and receiving the response, and defines the corresponding spatial distance between themselves and the other observer as $dl = cd\tau$. One can workout that this spatial line element is given by

$$ dl^2 = \gamma_{ij}dx^i dx^j, \hspace{1 cm}\gamma_{ij} = g_{ij} - \frac{g_{0i}g_{0j}}{g_{00}}$$

Details of the derivation may be found in Landau & Liftshitz "The Classical Theory of Fields" (Section 84, page 233).

My confusion is that both of these notions of spatial metric seem extremely reasonable to me. This tells me that I don't actually understand what each one is physically capturing. Moreover, is there a physical connection between the two? I am vaguely aware of some mathematical relationship between the metrics related to second fundamental forms, but I am really more interested in understanding what is going on from a physical perspective as opposed to a purely mathematical one.

EDIT: Since the question as written is not getting much traction, I will try and make my ask more specific. I would be satisfied if the following confusion is resolved. Let me set up two "experiments" to measure the length between two points, which I think give different results.

First, suppose observer $A$ and $B$ are situated at spatial position $x^i$ and $x^i + dx^i$. Observer $A$ extends a tape measure to $B$. Each of them holds their respective ends of the tape measure in their hands, and at time $t = t_f$, $A$ marks how much of the tape measure is extended. Intuitively, it seems like this should measure distance as described by the pullback metric.

Second, suppose $A$ and $B$ trade light signals as in the construction of the effective spatial metric. $A$ takes their proper time between sending and receiving the signal, and multiples by $c$ to get the distance. This should measure distance as described by the effective spatial metric.

What is the way out of this contradiction? I lean towards the effective spatial metric being the correct notion of distance, but I cannot see why the tape measure "experiment" is obviously wrong or ill-posed.

  • $\begingroup$ Haven't you added some additional assumptions here? I.e. that the timeline killing vector is hypersurface orthogonal? (Also see physics.stackexchange.com/questions/597823/…) $\endgroup$
    – Eletie
    Jun 6, 2022 at 7:22
  • $\begingroup$ I'm not sure I understand. For the pullback case, I have computed the metric according to the embedding I specified. I don't think I used any information about the hypersurface normal or timelike killing vector there. For the effective spatial metric, I don't think I have made any assumptions about the normal vector to the hypersurface, I just cited the general formula (which I think coincides with the expression for $\gamma$ in the post you linked). Can you specify where the additional assumption is? $\endgroup$ Jun 6, 2022 at 19:19
  • $\begingroup$ No you're right, disregard that last comment. I believe the different results just imply that the choice of embedding is not unique and depends on your mapping. (For a static spacetime, the two different spatial metrics align. I think this is what I was thinking of when I made the comment before). Also it might be useful for you that when you do the 3+1 decomposition and obtain $h_{ab} = g_{ab} + n_{a} n_{b}$ and are in adapted coordinates with a timelike killing vector $\partial_t$, then it quickly follows that $n_{a} =0$ and you obtain $h_{ab} = g_{ab}$ $\endgroup$
    – Eletie
    Jun 6, 2022 at 20:47

1 Answer 1


If the spacetime in question is static - so $\partial_t g_{\mu\nu}=0$ and $g_{0i}=0$ in the appropriate coordinates - then the issue is resolved by the fact that $h_{ij}=\gamma_{ij}$.

If the spacetime is merely stationary, then the metric generally has spatiotemporal cross-terms. If this is the case, then a generic light signal does not propagate symmetrically. For example, if we have a metric of the form $$\mathrm ds^2 = -c^2\mathrm dt^2 + 2 a \mathrm dt\mathrm dx + \mathrm dx^2\tag{$\star$}$$ Then one can see that the equation of motion for light rays is simply $$ \dot x(t) = \frac{-a \pm \sqrt{a^2+c^2}}{2}$$ The round-trip time for a light ray bounced back and forth between observers whose spatial coordinates differ by an infinitesimal distance $\delta$ can be straightforwardly shown to be $\Delta t = 2\delta \sqrt{c^2+a^2}/c^2$ and therefore your effective distance between the observers is $\delta\sqrt{1+a^2/c^2}$.

On the other hand, computing the pullback metric simply gives $\mathrm dl^2 = \mathrm dx^2$, and so this yields a distance $\delta$.

From a geometrical standpoint, the idea is the following. Armed with your time coordinate $t$, let's say you pick some spacetime point (i.e. an event) $p$ and want to construct the spacelike hypersurface on which it lies. Consider the following options:

  1. We could choose the surface $t=\mathrm{const}$. If we do this, then we effectively set $\mathrm dt=0$ in the line element and the resulting spatial metric is the pullback metric: $$\mathrm ds^2 = -c^2\mathrm dt^2 + 2a\mathrm dt \mathrm dx + \mathrm dx^2 \overset{\mathrm dt=0}{\longrightarrow} \mathrm dl^2 = \mathrm dx^2$$
  2. We could choose the surface whose tangent vectors are all orthogonal to $\partial_t$. In our 2D example, this yields $$g(\partial_t, \underbrace{\alpha \partial_t + \beta \partial x}_{=\mathbf v}) = -\alpha c^2 + \beta a = 0 \implies \beta = \frac{\alpha c^2}{a}$$ $$\implies \mathbf v \propto a \partial_t + c^2\partial_x$$ The set of such points in $\mathbb R^2$ corresponds to $t= \frac{a}{c^2} x+\mathrm{const}$. From here, $\mathrm dt = \frac{a}{c^2} \mathrm dx$ and so $$\mathrm ds^2 = -c^2\mathrm dt^2 + 2a\mathrm dt \mathrm dx + \mathrm dx^2 \overset{\mathrm dt=\frac{a}{c^2}\mathrm dx}{\longrightarrow} \mathrm dl^2 = \left(1+\frac{a^2}{c^2}\right)\mathrm dx^2$$

I lean towards the effective spatial metric being the correct notion of distance, but I cannot see why the tape measure "experiment" is obviously wrong or ill-posed.

It isn't. The major takeaway is that in a static spacetime, one can choose coordinates such that $\partial_t g_{\mu\nu}=0$ and that there are no spatiotemporal cross terms. In these coordinates, the pullback metric and "effective spatial metric" coincide, and this is a reasonable definition of the spatial distance between points. On the other hand, if there are cross terms, then these choices of spacelike hypersurface do not coincide.

  • $\begingroup$ Thanks for the response! To be clear, should I take your final comment to mean that there is not a unique notion of "physical spatial distance" for a non-static spacetime? If I report the spatial distance between two points, is that just ill-defined unless I also report how I measured it? $\endgroup$ Jun 8, 2022 at 1:16
  • $\begingroup$ @bittermania Yes, that’s right. And one could argue that there’s not a unique choice of distance even in a static spacetime (although one may seem to be more natural than the others). In general, distances in GR don’t really mean anything unless you specify what notion of distance you have in mind. $\endgroup$
    – J. Murray
    Jun 8, 2022 at 2:46

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