This is a question about coordinate time versus clock time / observed time, which I want to understand because I am teaching a GR course. Consider the Schwarzschild metric for specificity. I understand that the coordinate time $t$ is the "time measured by an observer at infinity". Now there is a trivial sense in which this is true: if one takes $r\to\infty$ and considers a stationary (relative to the coordinates) clock there then $d\tau = dt$. But this would be true even if one were to introduce a new global time coordinate $t'=t(1+e^{-\lambda|r-r_0|})$, which would induce an arbitrary "bulge" in $g_{00}(r)$ in the vicinity of some $r_0$, because it would still be the case that $\lim_{r\to\infty}g_{00}(r) = 1$. So, I presume, there is some less trivial meaning to the statement that $t$, the usual time coordinate in the Schwarzschild metric is the "time measured by an observer at infinity".

Is it that the observer at infinity assigns a time $t$ to an event $(t,r,\theta,\phi)$ if he sends to that event, and receives from that event, a light signal at times (clock time for the observer at infinity) $t-T$ and $t+T$ respectively, for some $T$? If it is this, is that equivalent to saying that the observer at infinity observes a gravitational redshift of photons emitted from "lower down" in the gravitational well and uses that observation to define the rate of time change at those lower points? If so, under which conditions are these the same? Or, more generally, under what conditions does $\sqrt{g_{00}(A)/g_{00}(B)}$ give the gravitational redshift between points $A$ and $B$. My example above, with $t'$, seems to show that it is not always given by this expression, even for a stationary metric, because of the arbitrariness of coordinate time. This is not addressed in the book I have been using, which just says the gravitational redshift is $\sqrt{g_{00}(A)/g_{00}(B)}$ for stationary metrics, with no qualifications.

(Apologies if this is addressed elsewhere, but I couldn't find it by keyword search.)

  • 1
    $\begingroup$ Related: physics.stackexchange.com/a/146611/123208 Although the question there asks about Eddington–Finkelstein coordinates, that answer first gives an operational way of determining Schwarzschild coordinates. $\endgroup$ – PM 2Ring Sep 1 '19 at 5:08
  • $\begingroup$ What is "clock time / observed time"? Observed by whom and where using whose clock and to measure what time intervals? $\endgroup$ – safesphere Sep 1 '19 at 5:56
  • $\begingroup$ The Schwarzschild fame is based on the assumption that the metric at infinity is Minkowski. $\endgroup$ – safesphere Sep 1 '19 at 6:50

The meaning of the coordinate t is much broader than that. In fact we might want to talk about events that never send or receive signals from infinity . After all signals to and from infinity will literally take infinite time to “get there”. The coordinate t in a stationary spacetimes is often chosen to be adapted to the Killing field i.e. adapted to precisely represent the time translation invariance present in the spacetime. We might use different time coordinates but quite generically it will be only one with the very special feature that all components of the metric are independent of t. When that is the case one can simplify calculations considerably. For instance if we know that a certain geodesic connecting the events A and B is given by $[t (\lambda) , r(\lambda), \theta (\lambda), \phi (\lambda) ]$ where $\lambda$ is the parameter along the geodesic. Then the points obtained by a change in the coordinate t by an amount $\Delta$ will be connected by a geodesic given by $[t (\lambda) + \Delta , r(\lambda), \theta (\lambda), \phi (\lambda) ]$. This will not work if your time coordinate is not adapted to the Killing field.

The gravitational red shift being talked about in this context, i.e. that of a stationary spacetime refers to the ( proper) time between two signals, both connecting two stationary observers as measured by each one of them. The ratio between those two ( proper) time intervals ( the one measured by the stationary observer at A and the stationary observer at B) is given by the expression you wrote. The change of time coordinate you want consider is one where the coordinates will no longer be adapted to the Killing field ( i.e. the feature used to characterize the time translation invariance of the spacetime). In other words if you make the change of coordinates you propose the metric will not be invariant under the translations in the new coordinate t and the expression for the relationship between the two proper time intervals will become much more complicated. One important point is that the change of coordinates you propose will not only affect the “tt” component of the metric, but also the “tr “and “rr “ components.

  • 1
    $\begingroup$ Welcome to the site! You need to wrap your equations in dollar signs to enable MathJax LaTeX. $\endgroup$ – PM 2Ring Sep 1 '19 at 7:11

You raise good points, which are often not clear in pedagogy. Indeed, there is a big difference between the statements:

  • "at $r_0 \gg 2M$, $\quad t$ is the proper time", and
  • "at any event $(t,r,\theta,\phi)$, $\quad t$ is the proper time at infinity `right now' ''.

The latter is a claim about global simultaneity, hence depends on the definition or convention chosen. The former is local, and unambiguously correct. In addition to your example, Eddington-Finkelstein "time" and Gullstrand-Painleve time concur with Schwarzschild $t$ as $r \rightarrow \infty$, but globally are very different.

There is a purely geometric motivation for $t$. This is based on the unique Killing vector field which is timelike and normalised at infinity, and future-pointing in our region of spacetime. This is simply $\partial_t$. Then the hypersurfaces $t = \textrm{const}$ are everywhere orthogonal to this Killing vector field (let's stick to $r > 2M$ for this paragraph). Hence $t$ is the "time at infinity", with additionally this static foliation or simultaneity choice. Also unique observers are defined by having 4-velocity parallel to the Killing vector. These are often called stationary observers but I prefer static observers which is more specific.

Your radar-like procedure of obtaining an event's $t$-coordinate is correct, I think. It relies on the time symmetry of spacetime, and also that $t$ itself respects this symmetry. $|dt/dr|$ is the same for ingoing and outgoing photons. Misner, Thorne & Wheeler $\S23.3$ have a good discussion. The redshift factor is defined as $V := \sqrt{g(\partial_t,\partial_t)}$. This describes time-dilation, redshift of photons etc, as compared between measurements of various static observers.


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.