2

In general relativity, the coordinates $x^\mu$ we put on the manifold are arbitrary and need not have any physical interpretation. For this reason, it is said there is no objective notion of time in GR. However, at the same time there exists the coordinate-independent notion of "timelike", "null", and "spacelike" vectors. Doesn't this necessarily imply the existence of an objective time in the direction of the "most timelike" vector?

To make this more precise, consider an arbitrary manifold $(M,g_{\mu\nu})$ and on it choose a Riemannian metric $h_{\mu\nu}$. Consider all vectors $v^\mu$ satisfying $h_{\mu\nu}v^\mu v^\nu=1$ (or any constant). One of these vectors, $u^\mu$, will minimize $g_{\mu\nu}v^\mu v^\nu$ (using the convention $g_{\mu\nu}t^\mu t^\nu<0$ for timelike vector $t^\mu$) and we take this "most timelike" vector to be the direction of time. The vector field $u^\mu(p)$ constructed this way will generate a family of integral curves which we parameterize using their path length. Since the magnitude of a vector is independent of our choice of coordinates, this construction is coordinate-independent and so represents a unique, objective flow of time.

Where is the error in this construction? Is it possible that this merely corresponds to the proper time of an observer at rest in a given tangent space, equivalently to how the time axis is unique to a given reference frame in special relativity?

I feel like I'm missing something fundamental about how coordinates work in general relativity.

EDIT: My question is not a duplicate of the one asked here. I'm asking specifically about defining a unique time direction, not simply finding vectors that are timelike (which would be satisfied by any $v^\mu$ for which $g_{\mu\nu}v^\mu v^\nu<0$).

Kris Walker
  • 1,062
  • In mathematical gauge theory, the local coordinate transformation (diffeomorphism) can be regarded as the gauge transformation corresponding to Lie group GL(1,3; R). However, in SU(N) Yang Mill's theory these gauge transformation are not considered as physical observables, so should we interpret coordinate transformation in GR along same line of thought? I guess this should help : https://physics.stackexchange.com/q/4359/ – KP99 Jul 19 '21 at 15:36
  • 1
    I'm confused by your notation. The manifold $(M,g_{\mu\nu})$ has $g_{\mu\nu}$ as the metric. So what now is the $h_{\mu\nu}$ that you've introduced? – Brick Jul 19 '21 at 15:39
  • I don't think there is a preferred directionality of time in GR. The laws are invariant in both directions. The directionality is imposed by the entropy of the system – KP99 Jul 19 '21 at 15:41
  • Does this answer your question? How to determine "timelike"-ness without using a coordinate system? – Vincent Thacker Jul 19 '21 at 15:46
  • @Brick the physically relevant Lorentzian metric on $M$ is $g_{\mu\nu}$. The $h_{\mu\nu}$ is a Riemannian metric I've chosen on $M$ so that I can fix the length of the vector while minimizing $g_{\mu\nu}v^\mu v^\nu$. However, from the answer given, it seems the time direction will not be unique since $h_{\mu\nu}$ is totally arbitrary. – Kris Walker Jul 20 '21 at 04:12

1 Answers1

2

The choice of Riemannian metric $h_{\mu \nu}$ is itself arbitrary, since there are multiple inequivalent rank-2 non-degenerate tensors on an arbitrary space; and different choices of $h_{\mu \nu}$ will lead to different "preferred time directions."

For example, consider the following two rank-2 tensors on Minkowski space with coordinates $t$, $x$, $y$, $z$: $$ h^{(1)}_{tt} = h^{(1)}_{xx} = h^{(1)}_{yy}=h^{(1)}_{zz}=1 \quad \text{(all other components vanish)} $$ $$ h^{(2)}_{tt} = h^{(2)}_{xx} = \frac{5}{4} \qquad h^{(2)}_{xt} = h^{(2)}_{tx} = -1 \\ h^{(2)}_{yy}=h^{(1)}_{zz}=1 \qquad \text{(all other components vanish)} $$ In these coordinates, the vector which minimizes $\eta_{\mu \nu} v^{\mu} v^{\nu}$ subject to the constraint $h^{(1)}_{\mu \nu} v^{\mu} v^{\nu} = 1$ is $v^{\mu} = (1,0,0,0)$, while the vector which minimizes $\eta_{\mu \nu} v^{\mu} v^{\nu}$ subject to the constraint $h^{(2)}_{\mu \nu} v^{\mu} v^{\nu} = 1$ is $v^{\mu} = (\frac43, \frac23, 0, 0)$.

So different choices of the Riemannian metric will lead to different time directions. In effect, the Riemannian metric introduces a preferred geometric structure on the spacetime, which indirectly encodes the preferred time direction.

Michael Seifert
  • 48,772