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I have always been a little confused by the meaning of the "$t$" which appears in spacetime intervals or metrics in general relativity. I concluded that $t$ was just a mathematical thing which allow to label the "spacetime manifold" and only proper time $\tau$ had a physical meaning. On wikipedia I also found:

"But the coordinate time is not a time that could be measured by a clock located at the place that nominally defines the reference frame." (http://en.wikipedia.org/wiki/Coordinate_time)

I don't know if my point of view is correct or not, but the following two computations made me think that I'm missing something:

1) Time dilation: $$ \frac{dt}{d\tau} = \gamma $$ If $t$ is non-physical, what are the two clocks which one should "compare" to see this time dilation?

2) Gravitational redshift:

This is a computation I have seen in my lectures which in my opinion has no physical sense.

Consider an atomic transition at the surface of the earth, at $x^{\mu}_1$. The time interval measured by a stationary observer close to the atom is given by: $$d\tau_1 = \sqrt{g_{00}(x_1)} dx^0_1$$ Imagine now the same atomic transition but, say, 100 km above the surface of the earth at $x^{\mu}_2$. The time interval measured by an observer near the atom is: $$d\tau_2 = \sqrt{g_{00}(x_2)} dx^0_2$$ Since the physics of atomic transitions is the same in $x_1$ and $x_2$ then one should have: $$d\tau_1=d\tau_2$$ $$\frac{dx^0_1}{dx^0_2} = \frac{\sqrt{g_{00}(x_2)}}{\sqrt{g_{00}(x_1)}}$$ But what is the physical meaning of the quantity $\frac{dx^0_1}{dx^0_2}$? In my opinion the only way to compute gravitational redshift is to compare the proper interval measured by an observer in $x_1$ and one in $x_2$ for an atomic transition happening in $x_1$.

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  • $\begingroup$ The exact physical meaning of coordinate time depends on the details of the metric and of the coordinate system on that metric, but it should always be possible to define it in terms of some function of the proper time on a set of clocks filling space. See for example my answer here where I talked about the physical meaning of both Schwarzschild coordinates and Eddington-Finkelstein coordinates, which are both coordinate systems for the Schwarzschild metric describing a non-rotating, uncharged, eternal black hole. $\endgroup$ – Hypnosifl Jan 6 '15 at 22:21
  • $\begingroup$ Also, in the specific case of an inertial frame in flat spacetime, the coordinate time of an event is just the proper time on a member of a set of clocks at rest in that frame (whichever clock was next to the event when it happened)...though note that the coordinate time between two events requires looking at local readings on two different clocks if the events don't happen at the same position in that frame, so the definition of simultaneity is an issue (for inertial frames it's defined by Einstein synchronization). $\endgroup$ – Hypnosifl Jan 6 '15 at 22:24
  • $\begingroup$ @Worldsheep : won't you give us the address of the site in the Wikipedia where you saw that statement? $\endgroup$ – Sofia Jan 6 '15 at 22:26
  • $\begingroup$ @Sofia: Sorry I forgot it, I'll edit my post. $\endgroup$ – Worldsheep Jan 6 '15 at 22:27
  • $\begingroup$ @Hypnosifl: I see what you mean, and things seem a bit clearer now. But I still can't understand the meaning in my second example. The metric we considered was the Newtonian limit $g_{00} = 1+2\phi$. $\endgroup$ – Worldsheep Jan 6 '15 at 22:31
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Proper time represents the physical aging of a massive particle, and by this it is the only time which is to take into account for the physical description of a particle.

But coordinate time is not without physical meaning: There would be no detection of events without coordinate time. When two particles are traveling through the same place in space, their proper time will not provide the information if it happened simultaneously, i.e. that they encountered, i.e. that there is an event. For this information you need the Minkowski diagram of at least one of both particles, and by the way the Minkowski diagram of any observer includes the coordinates of both particles, providing the information if they did encounter or if they did not.

Minkowski diagrams are showing the coordinate time of all particles (with different simultaneities). In contrast, it is not possible to represent the proper time of two different frames in one diagram.

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  • $\begingroup$ I don't know Minkowski diagrams, I'll have a look. $\endgroup$ – Worldsheep Jan 7 '15 at 8:21
  • $\begingroup$ Minkowski space is one essential part of special relativity. Coordinate time is represented in Minkowski space/ Minkowski diagram. Without it you won't be able to answer your question. $\endgroup$ – Moonraker Jan 7 '15 at 8:30
  • $\begingroup$ I just never used Minkowski diagrams, but I know about Minkowski spacetime. $\endgroup$ – Worldsheep Jan 7 '15 at 8:34
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Coordinate time is merely a parameterization, only proper time is physical.

However, for every timelike curve (not necessarily a geodesic) you can choose a set of coordinates such that proper time is equal to coordinate time (proof). These are called comoving coordinates and are often used in cosmology.

The time coordinate in a comoving frame is physical by definition, because it evaluates to the proper time of the observer. When we quote the age of the universe, we're exactly using the comoving time of an observer travelling with the expansion since the Big Bang.

The confusing statement you quoted from Wikipedia is the product of a poor naming convention. If we define the barycentric frame of reference as the comoving frame with the sun, then coordinate time is precisely the time measured on a clock on the sun.

For reasons unknown to me, astronomers define the barycentric reference frame differently, possibly because it makes computations easier! But so long as there's a standard everyone agrees on, the choice of frame doesn't really matter.

Finally, I think that your lecture notes are misleading. Let's take your gravitational redshift example. The whole point of general relativity is that $d\tau_1\neq d\tau_2$ when you look at the same event from different frames! The laws of physics in each frame must be the same, but measurements of proper time can differ because they're an avatar of your perspective.

Here's a specific example (reference). Consider an atomic transition close to a black hole. Observer $A$ sits at rest relative to the atom, infinitely far away from the black hole and measures $d\tau_A$. Observer $B$ is comoving with the atom and measures $d\tau_B$.

To do a calculation we must choose some coordinate system. Let us pick Schwarzchild coordinates, which are defined to be the comoving coordinates of the observer at infinity. Therefore we have

$$d\tau_A=dt$$

Since observe $A$ is at rest relative to the atom, $dx^i$ must be zero for observer $B$. Therefore, using the Schwarchild metric in Schwarzchild coordinates

$$d\tau_B=dt\sqrt{1-\frac{r_s}{r}}$$

where $r_s$ is the Schwarzchild radius, and $r$ the distance of the atom from the black hole centre. We can see immediately that

$$d\tau_B<d\tau_A$$

which corresponds to a redshifting of the frequency as the emitted photon moving outwards against the gravitational field.

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  • $\begingroup$ Do let me know if you have any questions on this, btw! $\endgroup$ – Edward Hughes Jan 14 '15 at 15:37
  • $\begingroup$ Can you explain $d\tau_A = dt$? $\endgroup$ – Thomas Dec 29 '18 at 16:17
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[...] the meaning of the "$t$" which appears in spacetime intervals or metrics in general relativity. I concluded that $t$ was just a mathematical thing which allow to label the "spacetime manifold"

Foremost, coordinates merely provide distinct ("one-to-one") labeling of elements of a given set $\mathcal S$ by elements of $\mathbb R^n$ (i.e. by $n$-tuples of real numbers, for some suitable natural number $n$); and in particular of distinct events (i.e. of a particular "spacetime" set $\mathcal S$ under consideration). Formally, the assignment of coordinates is (merely) a map:

$c~:~ \mathcal S ~ \rightarrow ~ \mathbb R^n$.

Depending on further relations between elements of set $\mathcal S$ (geometric relations between events under consideration) there can be additional demands on coordinate assignments:

  • if elements (or subsets) of set $\mathcal S$ can be identified which are ordered (as a sequence) then a given coordinate assignment $c$ may or may not be monotonic, in one or in several coordinate tuple components, with respect to the "obvious order of real numbers";

  • if subsets of set $\mathcal S$ can be identified which constitute a topological space $T$ then a given coordinate assignment $c$ may or may not be compatible with $T$ in the sense of a homeomorphism with respect to the "obvious topology of real $n$-tuples". Thus the pair "$(~\mathcal S, T~)$" may or may not be a manifold; and if so, a given coordinate assignment $c$ may or may not be continuous.

  • if there is a (suitably generalized) metric $s$ available for set $\mathcal S$ then a given coordinate assignment $c$ may or may not be compatible in the sense of $s$ being differentiable or even affine, separately for any one coordinate tuple component (e.g. for "$t$", for "$r$", or for "$\phi$" etc.) with respect to the "obvious metric of real numbers".

In general relativity coordinates are assigned to events (generally) differentiable, or even smooth, wrt. the (given) spacetime intervals $s^2$; within any sufficiently "small" coordinate patch.

Moreover, the name "$t$" is not usually given to just any coordinate tuple component, but (only, if applicable) to one which is monotonous with respect to the sequence of elements of time-like curves, and monotonous with respect to the sequence of space-like hypersurfaces, and even affine with respect to the durations $\tau A_{\circ P}^{\circ Q} \equiv \sqrt{-s^2[~\varepsilon_{AP}, \varepsilon_{AQ}~]}$ of suitable participants $A$ (but, importantly, therefor not affine collectively to the durations of each and any participant).

1) $\frac{dt}{d\tau} = \gamma.$ If $t$ is non-physical [...]

Well, in the context in which this equation is derived, $t$ is not just any (arbitrary, one-to-one but otherwise "non-physical") coordinate assignment. With a more explicit and appropriate notation the equation appears as

$$\frac{\tau P_{\circ A}^{\circledS Q \circ A}}{\tau A_{\circ P}^{\circ Q}} = \frac{1}{\sqrt{1 - (\beta_{PQ}[~A~])^2}},$$

where

  • $P$ and $Q$ denote two suitable participants at rest to each other
  • "$P_{\circ A}$" denotes participant $P$'s indication of having been met and passed by participant $A$, and
  • "$P^{\circledS Q \circ A}$" denotes participant $P$'s indication simultaneous to participant $Q$'s indication of having been met and passed by participant $A$.

2) Consider an atomic transition at the surface of the earth, at [...]. The time interval

... say: the duration of any one oscillation period ...

measured by a stationary observer close to the atom is given by: $d\tau_1 = [...]$

... where it is of course completely irrelevant for the duration of any one oscillation period of the atom under consideration (at the surface of the earth) whether and how it might be labelled with coordinates.

Imagine now the same atomic transition but, say, 100 km above the surface of the earth at [...]. The time interval [oscillation period duration] measured by an observer near the atom is: $d\tau_2 = [...]$.

Since the physics of atomic transitions is the same [for these two separate atoms] then one should have: $d\tau_1 = d\tau_2$.

Rigth: that's what we mean by the oscillation period durations of these two atoms being equal; or for short: these two atoms being equal
(in terms of the measure which is most relevant here, and or course regardless of any particular sprinkling of these atoms with coordinate labels).

But what is the physical meaning of the quantity $\frac{dx^0_1}{dx^0_1} = \frac{\sqrt{g_{00}(x_2)}}{\sqrt{g_{00}(x_1)}}$ ?

As far as
- the oscillation period durations of the two atoms are separately constant, and - the coordinates are assigned such that both $g_{00}(x_1)$ and $g_{00}(x_2)$ are constants

then the "physical meaning" of the coordinates is that they are affine with respect to the durations of either atom, respectively.

But the additional, given or measurable fact that $d\tau_1 = d\tau_2$ does not further constrain the value of $\frac{\sqrt{g_{00}(x_2)}}{\sqrt{g_{00}(x_1)}}$.

In my opinion the only way to compute gravitational redshift is to compare the proper interval measured by an observer in $x_1$ and one in $x_2$ for an atomic transition happening in $x_1$.

In my opinion the most important and relevant chronometric comparison is between ping durations (cmp. my answer there: "An accelerating train ...", PSE/q/38377;
especially for observer pairs whose mutual ping duarions are (separately) constant, i.e. who are "chronometrically rigid to each other".

Only in reference to the unequal ping durations

  • of an observer "at the surface of the earth" (from having stated a signal indication, until having seen that the companion "100 km above the surface of the earth" had ssen this signal indication), and

  • of an observer "100 km above the surface of the earth" (from having stated a signal indication, until having seen that the companion "on the surface of the earth" had ssen this signal indication),

could they even conclusively determine that their separate atoms had equal oscillation period durations, in the trial(s) under consideration.

In particular: the number of oscillation periods which were counted "at the surface of the earth" in the course of one "ping period (100 km up, and back)"
is not equal to the number of oscillation periods which were counted "100 km above the surface of the earth" in the course of one "ping period (all the way down, and back)".

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In general, the coordinates used to write an arbitrary metric should be seen as labels of space-time points. Some coordinates may be related to familiar things, but some will not be so (at least not in a simple way) so beware of trying to find too much familiarity in them.

Now, let's consider Minkowski space-time and the meaning of the coordinate $t$ of an inertial observer. What does $t$ mean? well, it is just the time given by clocks that happen to be stationary respect to you.

Now, if you expect this time given by clocks stationary respect to you to be a faithful way to order events everywhere in spacetime you will be disappointed, and it is easy to see why. You can easily see that two inertial observers might disagree on the $\Delta{}t$ of two events. Just consider two any events and using a general Lorentz transformation and going to another frame you will see that $\Delta{}t$ can change. If these events are space-like separated different observers may even disagree on the sign of $\Delta{}t$.

So, for some observers it's as if one event happened first, and then the other, and for other observers it's the opposite. So, which event did happen first? the answer is that this question is really meaningless. Time IS local. The time given by a clock only makes sense for he who carries it in the point he is carrying it. This is the sense in which relativity kills the concept of absolute time. Asking any question with the word "while" in it is really meaningless.

The Lorentz coordinate $t$, (and also the rest $x$,$y$,$z$) just tell you how you are causally connected with the rest of space-time points, how the universe appears to you, or maybe better said, how events in the rest of space-time points can affect you. For example, if for some observer two events are simultaneous ($\Delta{}t=0$) and spatially equally separated from the observer, information of both this events (transmitted via photons for example) will reach the observer at the same time. Nonetheless, another Lorentz observer will know of one before the other (and as argued above, different Lorentz observers may even disagree on which one arrived first, even though all these observers are located at the the same point!(of course, since these Lorentz obervers have different speed they will not tay in the same point waiting for the signal to come so the effect of the movement must also be considered)). Different Lorentz observers are just caussally connected in different ways to the rest of events.

Now, if you wanna compare the elapsed times for different observers (or objects or whatever) that happened to be together once and are reunited after a while you just have to compare the proper times, ie the time given by clocks that have been with them the whole time.

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  • $\begingroup$ Ok thank you. In some sense you agree with me, the only "physical" sensitive quantity is proper time... Clock which are with me... $\endgroup$ – Worldsheep Jan 6 '15 at 23:10
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Proper time is the time measured by a clock which is traveling along an inertial path. Coordinate time can be thought of as the time axis on a space time diagram; so as an example the coordinate time can be thought of as the time measured by an outside observer of the clock. These times are meant to be compared since in relativity since time passes at different rates for different inertial paths. If two objects are co-moving then the two times are equal. Hope this helps Good Luck!

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  • $\begingroup$ What do you mean by: "by an outside observer of the clock"? $\endgroup$ – Worldsheep Jan 7 '15 at 18:56
  • $\begingroup$ @Worldsheep "Coordinate time" is time as measured by someone who is in the reference frame implicitly defined by the diagram. If you assume that the diagram is an image of what a camera is seeing, then coordinate time is the proper time for the camera. $\endgroup$ – Robert Mastragostino Jan 27 '15 at 0:47

protected by Qmechanic Jan 30 '16 at 18:31

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