1
$\begingroup$

Reading Misner-Thorne-Wheeler concerning the metric for spacelike and timelike displacements it seems to me that two different metrics must be distinguished, one metric for spacelike and one for timelike displacements - is this a wrong interpretation, or are there two metrics?

My question concerns the combination of three citations: On page 20, I read in the section IIA "Coordinate-free language":

…The proper distance $s_{AB}$ (spacelike separation) or proper time $\tau_{AB}$ (timelike separation) is given by …

On page 21, section IIB "Language of coordinates", shorter:

From any event A to any other nearby event B there is a proper distance $s_{AB}$ or proper time $\tau_{AB}$ given in suitable (local Lorentz) coordinates by $$ s_{AB}^2 = -\tau_{AB}^2 = -[x^0 (B) - x^0 (A)]^2 + [x^1 (B) - x^1 (A)]^2 + [x^2 (B) - x^2 (A)]^2 + [x^3 (B) - x^3 (A)]^2. $$

And finally, on page 305 (at equation 13.1):

In the language of coordinates, "metric" is a set of ten functions of position $g_{\mu\nu} (x^a) $, such that the expression $$ Δs^2 = -Δ\tau^2 = g_{\mu\nu} (x^\alpha)Δx^\mu Δx^\nu$$ gives the interval between any event $x^\alpha$ and any nearby event $x^\alpha + Δx^\alpha$.

The first citation explains (correctly) that proper distance corresponds to spacelike separation and proper time corresponds to timelike separation.

In the second citation it is said that for any event there is one of two possibilities, either a proper distance $s$ or a proper time $\tau$, and the square of $s$ equals the square of $\tau$ with opposite sign. This seems logic because for timelike displacements, $s^2$ would give a negative square, and $s$ itself would be imaginary, and vice versa.

But, with respect to the third citation this would mean that

$$ g_{\mu\nu} (x^\alpha) Δx^\mu Δx^\nu$$

is only referring to spacelike displacements, as the squared proper time is $Δ\tau^2$ and not $-Δ\tau^2$, and that it is not applying to timelike displacements where the metric has the opposite sign:

$$- g_{\mu\nu} (x^\alpha) Δx^\mu Δx^\nu$$

So my question is: Must we distinguish between one spacelike metric and one timelike metric?

$\endgroup$
2
$\begingroup$

Other answers already stated that in standard covariant formulation of general relativity there is only one metric. Nothing in MTW quotes suggests otherwise, only different notation for spacelike / timelike separation.

At the same time I would like to mention that in the nonrelativistic, Newtonian limit, spacelike and timelike displacements become completely different entities, corresponding to distinct physical concepts of Newtonian physics: time interval and (spatial) distance.

Consequently, in geometrization of Newtonian gravity, Newton–Cartan theory defined on a 4-manifold spacetime (similar to the spacetime of general relativity) there are in fact two separate metrics: termporal metric $t_{\alpha\beta}$, a nowhere vanishing symmetric, 2-covariant tensor field of rank 1 (in some formulations a 1-form differential of “universal time” is used instead), and (inverse) spatial metric $s^{\alpha\beta}$, a nowhere vanishing, symmetric, 2-contravariant tensor field (of rank 3).

Note: the discussion of Newton–Cartan theory in MTW book (chapter 12) does not use the same notation. For more detailed exposition on NC theory see e.g. book Topics in the Foundations of General Relativity and Newtonian Gravitation Theory by D. Malament OA pdf.

Also, there is a frame theory of J. Ehlers, a unified approach to both general relativity and Newtonian gravity best suitable for the discussion of Newtonian limits of general relativistic spacetimes. This theory also has two distinct metrics: temporal and spatial. For more details see a “Golden Oldies” republication (and translation from German) of 1981 paper:

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

This interpretation is wrong. There is only one metric. It is not positive-definite like a Euclidean metric is, so the squared spacetime interval can be positive, zero, or negative.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

$g_{\mu\nu}$ is a matrix used to calculate "intervals". The sign of a given interval depends on what it is about: either it is a distance in space or it is a duration of particle motion.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So, my question, does also the metric depend on what it is about: spacelike or timelike? $\endgroup$ – Moonraker Jan 23 at 17:59
  • $\begingroup$ No, in a flat space-time its matrix elements are just constant $\pm 1$/ $\endgroup$ – Vladimir Kalitvianski Jan 23 at 19:23
  • $\begingroup$ According to the quotation on page 305, the metric of $ Δs^2$ is $g_{\mu\nu}$, and the metric of $ Δ\tau^2$ seems to be $-g_{\mu\nu}$, meaning constant matrix elements but change of sign with $- g_{\mu\nu}$. $\endgroup$ – Moonraker Jan 23 at 20:02
1
$\begingroup$

Yes, you can talk about there being two metrics, which are coincident for a proper relativistic spacetime - but whose distinction becomes important when considering the classical limit. This is similar to the case for quantum mechanics, where the classical limit requires us to bifurcate the wave function to separate positional and momental wave functions, so that in $\hbar \rightarrow 0$ it reduces to the case of a classical probability scenario with classical Bayesian agent primed with incomplete information about the system (hence leading naturally to the interpretation of $\hbar$ as being an information resolution limit for the universe, just as $c$ is information speed limit).

In particular, we have indeed that

$$d\tau^2 = dt^2 - \frac{1}{c^2} (dx^2 + dy^2 + dz^2)$$

as the time-like metric, and

$$ds^2 = -c^2 d\tau^2 = (dx^2 + dy^2 + dz^2) - c^2 dt^2$$

as the space-like metric. When $c \rightarrow \infty$, we get that $d\tau^2 = dt^2$ and $ds^2 = dx^2 + dy^2 + dz^2$, i.e. two separate spatial and temporal metrics, corresponding to the distinct space and distinct time of Newtonian mechanics. (Note that for the spacelike case, we must ensure $dt \rightarrow 0$ so the interval remains spacelike, otherwise it blows up because you in effect now have a category error.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ One timelike and one spacelike metric - This is also my understanding of MTW, however it is written nowhere, and the other answers seem not to agree. Is there some literature supporting this interpretation? $\endgroup$ – Moonraker Jan 24 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.