# I Coordinate transformations in GR

1. Aug 12, 2018 at 10:41 AM

### eoghan

Hi there
I'm studying GR and I am confused about coordinate transformations.

In my understanding, if I want to study a rotating reference system this is what I do.
In my inertial system the object trajectory is described by
$$x = r\cos(\theta - \omega t)\\ y = r\sin(\theta - \omega t)$$
This means that if I make the change of coordinates
$$t = \tau\\ x = r\cos(\theta - \omega \tau)\\ y = r\sin(\theta - \omega \tau)$$
then I switch to a frame where the object is at rest, i.e. I switch to the rotating reference frame. The problem is that this transformation is not a Lorentz transformation, so the metric in the rotating frame is not going to be Minkowskian, and indeed it is
$$ds^2 = (1-\omega^2r^2)d\tau^2-2\omega r^2 d\tau d\theta -dr^2 - r^2 d\theta^2$$

But why can't I do the same to describe a uniformly moving reference frame? I.e. if an object is in uniform motion along the x-axis, then it's trajectory is
$$x = vt$$
and so with the coordinate transformation
$$t = \tau\\ x = v\tau - x'$$
I switch to a reference frame where this body is at rest. The metric is obviously not Minkowskian
$$ds^2 = (1-v^2)d\tau'2 - dx'^2 + 2v dx' d\tau$$
Clearly this is not the metric of an inertial reference system, so what is this metric describing? How do I relate this metric to the metric of an inertial observer?

One answer I can come up with is that I'm not considering length contraction and time dilation, but I'm not considering them in switching to the Rindler coordinates either!
Another possibility is that in SR we are obliged to make only Lorentz transformations. But in GR we can do whatever coordinate transformations we want, so the transformation above should be allowed.

2. Aug 12, 2018 at 11:00 AM

### stevendaryl

Staff Emeritus
That's a perfectly good metric. But notice that a clock that is at "rest" in this coordinate system will show elapsed time:

$ds = \sqrt{1-v^2} d\tau$

So the time $\tau$ is not the time that clocks at rest keep. The nice thing about an inertial coordinate system is that for clocks at rest, $ds = dt$. So if you synchronize all your clocks, then the time on any clock will be equal to the coordinate time.

3. Aug 12, 2018 at 12:42 PM

### Orodruin

Staff Emeritus
Just to add that metrics are not something that ”belongs” to any observer. The metric is a (0,2) tensor in the spacetime. Its components will take different values in different bases.

There is nothing wrong with the coordinates you chose, just that you did not choose them in such a way that your holonomic basis is not orthogonal. You can perfectly well do this in Euclidean space too. Any observable you compute will still be the same (just more annoying to compute).

4. Aug 13, 2018 at 2:24 PM

### eoghan

I think this is what confuses me... what does the coordinate time represent? Or, you say that an object that doesn't move in space has a ds which is $\sqrt{1-v^2}d\tau$, so this is the time reported by a clock at rest. But what is not clear to me is: $ds$ is just an abstract quantity which is conserved in all reference systems...how do we go from this abstract object to say that it represents the real time separation between two events at rest? I mean, why do we say that $\sqrt{1-v^2}d\tau$ is the physical time and not $d\tau$? Because $\tau$ is the coordinate that I choose in my system to describe the time of the events...so isn't for me $\tau$ the physical time?

I think this is about a more general question about how to interpret a metric.... if I have a generic metric
$$ds^2 = A(x,t)dt^2 + B(x,t)dx^2$$
how do I interpret it? If I measure the distance between two points at the same time, the measuring device will give me $B(x,t)dx^2$ or $dx^2$?

5. Aug 13, 2018 at 3:11 PM

### Ibix

Usually, the time on some notional set of clocks (which may be creatively broken so that a clock doesn't always tick at the same rate, nor necessarily the same rate as its neighbour).
$ds^2$ is the distance-squared through spacetime associated with moving from (x,y,z,t) to (x+dx,y+dy,z+dz,t+dt). It's the generalisation of Pythagoras' Theorem to differential displacements in curved spaces, usually (in physics) with a non-Euclidean signature. It's not really an abstract quantity, any more than "the length of the hypotenuse" is an abstract quantity.
Well, the first thing to note is that usually in physics, $A$ will have the opposite sign to $B$.

Coordinates don't have to have a simple physical meaning. Take latitude and longitude, for example. If our longitude differs by one degree, we may be a hundred miles apart if we're on the equator, or jostling for space if we're near the pole. The metric is the tool to tell you how far apart we are given a coordinate difference. The metric on a sphere is $ds^2=R^2d\theta^2+R^2\sin^2\theta d\phi^2$. The coefficient of $d\theta$ is a constant, which tells you that a change in $\theta$ ($\phi$ held constant) always means the same thing in terms of how far we have to walk. But the coefficient of $d\phi$ depends on the latitude - so a change in $\phi$ means different things at different $\theta$ values.

Comparing that to what I said at the start of the paragraph: it's the coefficient times the coordinate differential that's the physically significant thing, not the coordinate differential.

Taking that to your example, if you have a clock that remains at some constant x coordinate, then the time it shows will be $\int_0^Tds=\int_0^T\sqrt{A(x,t)}dt$ (assuming it was zeroed at t=0 and you ask at t=T). On the other hand, if you want to know how long is a ruler you need to get $\int\sqrt{B(x,t)}dx$, integrating from the x coordinate of one end of the ruler to the other. You do need to be sure that a surface of constant $t$ represents your notion of simultaneity, of course.

Hope that makes sense.

6. Aug 13, 2018 at 3:16 PM

### Staff: Mentor

In general, it represents nothing, or at least nothing of any physical significance. We usually choose our coordinates in such a way that the coordinate time difference between two points on the worldline of an observer at rest (spatial coordinates unchanging) in that coordinate system is equal to the proper time along that worldline... but we don't have to, and if we don't then the coordinate time has no particular physical significance.

7. Aug 13, 2018 at 4:01 PM

### pervect

Staff Emeritus
This might be obvious and something you already know, but I should mention that if you wanted a Minkowskii metric, you should use the Lorentz transform, not the one you used.

Mathematically, I'd say that at any point, you have vectors $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial t}$. Which I'll write as $\partial_x = \frac{\partial}{\partial x}$ and $\partial_t = \frac{\partial}{\partial t}$.

Are you familiar with vectors as partial derivative operators? If not, Carol has a discussion in his online lecture notes on GR. I'll assume you're not familiar with the notion, and use less-precise words to try to describe what I mean - but if you are familiar with the notion, or can find an explanation of it, it might help some of your confusion. Carol's lecture notes on General relativity (available online for free) have a discussion of this for instance. On to the informal discussion.

Informally, one can think of a unit vector $\partial_x$ as a little arrow that points in the direction where x changes and t does not, with the length of the arrow representing how fast the coordinate x changes with a unit "physical length", in the given direction. What do I mean by a "physical length"? It's just the Lorentz interval, the quantity that's the same for all observers.

We use the metric to calculate the "physical length", i.e. the Lorentz interval, from the infinitesimal coordinate changes dx and dt. I'll give the formulas, which are hopefully familar, below.

Similarly $\partial_t$ is a little arrow that points in the direction where t changes and x does not, and it's length is proportional to how fast t changes with proper time.

You used the symbol $\tau$ for your coordinate, but your coordinate is not proper time, the notation is a bit confusing to me but I'll attempt to go with it.

I'll give the formula used to calculate length of vectors that I mentioned earlier. We can find the squared length of a vector $\vec{u}$ with components $u^a$ as the absolute value of $g_{ab} u^a u^b$. Timelike vectors have one sign for the squared length, , spacelike vectors have another. Exactly what sign depends on the sign convention you use. So we can see from your metric that $\partial_x$ has a unit length, but $\partial_t$ does not.

We can find the dot product of two vectors u and v with components $u^a$ and $v^b$ repsectively as $g_{ab} u^a v^b$. Doing this, we find that in the Minkowskii metric, $\partial_x$ and $\partial_t$ are orthogonal, while in your metric, the vectors $\partial_{x'}$ and $\partial_\tau$ are not.

The physical significance of this is that your coordinates do not respect the Einstein clock sychnronization convention, basically because you used the Galilean transform rather than the Lorentz transform. This is what makes the vectors not orthogonal in your metric as opposed to the Minkowskii case where they are orthogonal. We can note that the non-orthogonality is manifest as the term 2 v dx' d$\tau$ in your line element.
$ds^2 = (1-v^2)d\tau'2 - dx'^2 + 2v dx' d\tau$. The non-unit length of $\partial_\tau$ which people have rightly mentiond as clocks (which keep proper time) not ticking at the same rate as your coordinate called $\tau$ is given by the $(1-v^2) d\tau^2$ term in your line element. If your coordinate time represented the time kept by a physical clock (proper time), this term would be one. But it's not, so it doesn't.

8. Aug 13, 2018 at 4:29 PM

### Orodruin

Staff Emeritus
Just to expand a bit: There is nothing that requires you to even have a coordinate that is time-like. Using four linearly independent space-like directions works perfectly well. The bottom line is, coordinates are not physical. You can only make sense of predictions when you compute invariant quantities, which includes using the metric tensor for several purposes.

9. Aug 13, 2018 at 5:00 PM

### Staff: Mentor

The quick answer (but do check out the pitfalls in the next section) is that it will give you $B(x,t)dx^2$. An easy way to see this is to imagine measuring the distance between two points on the curved surface of the earth: do it once with both points on the equator, then again with both points on the Tropic of Cancer. The coordinate difference is the change in longitude between the two points and is the same for both pairs, but the measured distance is different; the metric coefficient makes the difference.

There are some pitfalls in doing this measurement in four dimensional spacetime though. First, the interval between the two events must be spacelike instead of timelike, so that "at the same time" makes sense. Furthermore, the answer will clearly depend on your choice of time coordinate, because that's what determines what "at the same time" means.

Then you need to figure out how to actually do the measurement. This is easy in a flat spacetime with both points stationary in your chosen coordinate system - just lay meter sticks end to end between the two points. It's a bit trickier if the points are the positions of moving objects at a particular moment, but still reasonably straightforward in flat spacetime - just mark the positions of the two objects at the desired time, and then measure the distance between the stationary marks at our leisure.

It is much trickier in a curved spacetime. We want the length of a string that lies on a (spacelike) curve of constant $t$ and passes through both points at the desired time (more precisely, intersects the world lines of the two points at the events with the same time coordinate). But clearly we can't just stretch a string between the two events, because we can't pull the end of a string along a timelike path, and because there's no reason why the geodesic path followed by a stretched string is necessarily the curve of constant $t$ that we want. Instead, we have to position the string ahead of time so that it as it follows its own path through the curved spacetime it ends up lying on the desired spacelike curve. Then at the right moment we mark the two points on the string, roll the string back up, and measure the length between the two points. This entire procedure will be easier in some spacetimes than others (for example, it's equivalent to using a sounding line outside the event horizon in a Schwarzschild spacetime) but may be excruciatingly difficult in an arbitrary curved spacetime with a perverse choice of $t$ coordinate.

Last edited: Aug 13, 2018 at 11:06 PM
10. Aug 13, 2018 at 5:36 PM

### Orodruin

Staff Emeritus
This is actually present in your example of the two-dimensional sphere as well. The metric is $ds^2 = R^2(d\theta^2 + \sin^2(\theta) d\varphi^2)$. The "distance" $R\sin(\theta) \Delta \varphi$ between two points on the same latitude is the distance along that latitude. It is not the distance on the 2-sphere itself (which is shorter and part of a great circle, not a non-equatorial latitude). The same goes in the case of spacetime. The "distance" one computes will be the distance along the space-like simultaneity hypersurface, which is not necessarily a geodesic in the full 4-dimensional spacetime.

I like the sphere. Many of the pitfalls you enter into in GR can actually be traced back to similar issues on the sphere. Including coordinate singularities, Killing vectors, circles that do not have a circumference of $2\pi$, rotation due to parallel transport, etc, etc.

11. Aug 13, 2018 at 7:20 PM

### stevendaryl

Staff Emeritus
Coordinate time just means a method for assigning times (real numbers) to events in space. In an "inertial coordinate system", the way that you assign times to events is this:
• Have clocks stationed at rest throughout space.
• Synchronize them (using slow clock transport, or light-signal synchronization, etc.)
• Let the time for an event be equal to the clock time for the nearest clock.
The point of the metric is that if you have a coordinate system $x,y,z,t$, then the clock time $ds$ (the elapsed time shown on a clock) for a clock to move from the point $x,y,z,t$ to the point $(x+dx, y+dy, z+dz, t+dt)$ is given by: $(ds)^2 = \sum_{ij} g_{ij} (dx^i) (dx^j)$. If the clock is at "rest" in the coordinate system, that means that the only coordinate that is changing is $dt$, so $(ds)^2 = g_{tt} (dt)^2$.

No, $ds$ is not abstract. It's the elapsed time on a clock that travels from the point $x,y,z,t$ to the point $x+dx, y+dy, z+dz, t+dt$.

That's what the significance of $ds$ is. It's the invariant time.

Because $ds$ is what you can actually measure with clocks.

Think about clocks in the real world. As long as their batteries don't run out, the elapsed time will continue to increase smoothly. But not so coordinate time. When you cross over into a different time zone, the coordinate time shifts by an hour. But the time on your clock changes smoothly.

If you have two nearby events (locations in spacetime), then there are three possible relationships between them:
1. They could have a timelike separation. That means that it is possible for a clock moving slower than the speed of light to travel at constant velocity from one event to the next. In that case, $|ds|$ gives the elapsed time on the clock.
2. They could have a spacelike separation. That means that the events are so far apart, and occur so close together in time, that it is impossible even for light to travel from one event to the other. In that case, $|ds|$ gives the spatial distance between the events (in a frame in which the events are simultaneous)
3. They could have a lightlike separation. That means that it is possible for a light signal to travel from one to the other. In that case, $|ds| = 0$.
So for timelike separations, $|ds|$ is the time between the events, as measured in an inertial coordinate system in which the events have the same spatial coordinates. For spacelike separations, $|ds|$ is the distance between the events, as measured in an inertial coordinate system in which the events take place at the same time.

It might help to think a little about the role of metrics in ordinary geometry if you use curvilinear coordinates. For example, if you use polar coordinates $r, \theta$, then the distance between the point with coordinates $r, \theta$ and the point $r + dr, \theta + d\theta$ is given by $ds^2 = dr^2 + r^2 d\theta^2$

12. Aug 14, 2018 at 12:31 PM

### Staff: Mentor

Actually, this is too strong--in almost any curved spacetime it will be impossible to do this except for one (or a very small number) of observers at rest in the coordinates. FRW spacetimes, used to describe the universe as a whole, are an exception; in these, in standard FRW coordinates, coordinate time is the same as proper time for all observers at rest in the coordinates.

13. Aug 14, 2018 at 3:19 PM

### PAllen

I wonder about this. Consider a timelike congruence and a foliation of spacetime. Pick one hypersurface of the foliation (reference hypersurface). Assign spatial coordinates on it. Now define the coordinates of any intersection of a congruence world line with any other hypersurface of the foliation as: (spatial coordinates for that congruence world line in the reference hypersurface, proper time along the congruence world line measured from the reference hypersurface). This achieves what Nugatory proposed, and should be possible for 'very large regions' of any spactime.

[edti: actually, the foliation is overkill. All you need is a reference spacelike hypersurface and a timelike congruence. ]

Last edited: Aug 14, 2018 at 3:34 PM
14. Aug 14, 2018 at 11:20 PM

### Staff: Mentor

Hm. I think you're right, the statement as I made it was too strong. It is possible to construct coordinates that meet @Nugatory's requirements. However, in almost all cases, such coordinates will be nothing like "usual"--i.e, we do not "usually" choose our coordinates this way, so that part of @Nugatory's statement was still not accurate.

I'll illustrate what I mean by doing your suggested construction in the exterior region of Schwarzschild spacetime (i.e., the region outside the horizon). We pick any hypersurface orthogonal to the timelike Killing vector field as our reference hypersurface, and the integral curves of the timelike KVF as our timelike congruence. Then our new coordinates are $T, r, \theta, \phi$, where $r, \theta, \phi$ are the same as in standard Schwarzschild coordinates, and $T$ is defined such that $dT = \sqrt{1 - 2M/ r} dt$ (where $t$ is the standard Schwarzschild time coordinate). With this definition, the time coordinate $T$ will indeed be the same as proper time for an observer whose worldline is at rest in our new chart (this will be one of the integral curves of the timelike KVF).

However, this chart is certainly not the "usual" coordinate chart we use in this spacetime. Why not? Because, as should be evident with a little bit of reflection, surfaces of constant coordinate time in this chart are very, very different from those in the standard Schwarzschild chart. In fact, I think that for sufficiently positive or negative values of $T$, the surfaces of constant $T$ are not even spacelike everywhere.

15. Aug 15, 2018 at 5:11 AM

### Staff: Mentor

Mmmmm...... I did say "the worldline of an observer at rest" not "all observers".

16. Aug 15, 2018 at 6:16 AM

### Orodruin

Staff Emeritus
Assuming I did the algebra correctly (a big if ...), the radial coordinate becomes time-like when
$$T^2 > \frac{4r^2(r-R_S)}{R_S^2}$$

17. Aug 15, 2018 at 9:36 AM

### Staff: Mentor

Yes. And, for example, in the "usual" coordinate chart on Schwarzschild spacetime, for observers at rest at any finite value of $r$, coordinate time is not the same as their proper time. So this statement of yours, for that case, is not correct:

More generally, in almost any curved spacetime, the "usual" coordinates do not have the property you are describing. There might be some coordinates that do, but they aren't the usual ones. The only example I can think of where the usual coordinates, the standard ones that appear in all the textbooks, have the property you describe is FRW spacetime.

18. Aug 15, 2018 at 9:38 AM

### Orodruin

Staff Emeritus
Minkowski space ...

19. Aug 15, 2018 at 9:45 AM

### Staff: Mentor

Ok, so I left out one. Sue me.

20. Aug 15, 2018 at 9:49 AM

### martinbn

Minkowski is included in FLRW, $k=0$ and $a(t)=1$.