What is the difference between standard and isotropic metrics?

[Emilie.Jung]

The metric
$$ds^2=-R_1(r)dt^2+R_2(r)dr^2+R_3(r)r^2(d\theta^2+sin^2d\phi^2)$$
when changed to
$$ds^2=-R_1(r)dt^2+R_2(r)(dr^2+r^2d\Omega^2)$$
upon setting $R_2(r)=R_3(r)$, the later metric holds the name of isotropic metric.

My question what is the difference between the first and the second metric, so that, the first is standard meanwhile the second is called isotropic? We know that isotropy means no preferred direction, but I cannot see how isotropy holds in the secondbut not the first.

Note: I already checked wikipedia, where it doesn't say much about what is an isotropic metric by itself.

 

[PeterDonis]

what is the difference between the first and the second metric,

First we have to get clear about what these two metrics are supposed to be describing, and what "changing" from one to the other means.

You haven't given a reference for where you got these metrics from (if you have one, it would be helpful to give it), so I'm not sure what physical situation you are thinking about. In general, it's not always possible to just set $R_2(r) = R_3(r)$; it's only possible in particular cases. Nor is it always possible in general to even write the metric in the first form.

One case in which it is possible to write the metric in the first form is a spherically symmetric spacetime. A general spherically symmetric spacetime only requires two functions in the metric, so in general we can write it in either of two forms. One is the first form you wrote, but with $R_3(r) = 1$; in other words,

$$
ds^2 = - R_1(r) dt^2 + R_2(r) dr^2 + r^2 d\Omega^2
$$

where I have used $d\Omega^2$ for the angular part of the metric for brevity. The other form, which I will write using $\bar{r}$ for the radial coordinate and $F_1$ and $F_2$ for the functions to make it clear that they are different from $R_1$ and $R_2$, is your second form:

$$
ds^2 = - F_1(\bar{r}) dt^2 + F_2(\bar{r}) \left( d \bar{r}^2 + \bar{r}^2 d\Omega^2 \right)
$$

The key thing to understand is that these two different formulas are supposed to describe the same spacetime, i.e., the same geometry. Each surface of constant time $t$ in this spacetime is an infinite set of 2-spheres, each labeled with a radial coordinate ($r$ or $\bar{r}$). The difference in the two forms of the metric is in how the radial coordinate labeling is chosen, i.e., what label is attached to a given 2-sphere.

To see the difference, we need to find some invariant that identifies each 2-sphere. Since the spacetime is spherically symmetric, the area $A$ of a given 2-sphere will be such an invariant. So a given 2-sphere with area $A$ will have two different radial coordinate labels, one assigned by the first metric ($r$) and one assigned by the second ($\bar{r}$).

From the first metric, we can see that $r$ is simply the "areal radius" of the sphere, i.e., $r = \sqrt{A / 4 \pi}$, so $A = 4 \pi r^2$. (The way we see this is by looking at the coefficient of the angular part of the metric; the area of the 2-sphere will be the integral of the angular part of the metric over the full range of $\theta$ and $\phi$, which gives a factor of $4 \pi$ times the coefficient.)

From the second metric, however, we can see that $A = 4 \pi \bar{r}^2 F_2(\bar{r})$, which means that $\bar{r} \sqrt{F_2(\bar{r})} = \sqrt{A / 4 \pi}$, so the same 2-sphere of area $A$ will have a label $\bar{r}$ in this metric that is different from its label $r$ in the other metric. We can't know specifically what $\bar{r}$ is in terms of $A$ unless we know $F_2(\bar{r})$. (To know $F_2$, we would have to know more about the specific physical situation.)

We know that isotropy means no preferred direction

As a matter of physics, yes, but that is not true in general of a spherically symmetric spacetime; the radial direction in such a spacetime is different from the tangential directions. Think, for example, of the spacetime surrounding a gravitating body like the Earth. Obviously things work differently in the radial direction than in the tangential directions; objects fall radially but not tangentially.

As a matter of coordinate choice, however, "isotropy" doesn't mean the spacetime itself has no preferred direction; it just means the coordinates don't. (More precisely, they don't have a preferred spatial direction locally; we'll see what that means in a moment.) The second form of the metric above clearly has this property: the function $F_2(\bar{r})$ multiplies the entire spatial part of the metric, so locally, the radial coordinate $\bar{r}$ doesn't work any differently from the other coordinates. We could underscore this by switching, locally, to Cartesian coordinates; we have $dx^2 + dy^2 + dz^2 = d\bar{r}^2 + \bar{r}^2 d\Omega^2$ (this is just the usual transformation between Cartesian and spherical polar coordinates), and clearly the $x$, $y$, and $z$ coordinates all work the same, with none of them picking out a preferred direction.

By contrast, the first form of the metric does not have this local isotropy property in the coordinates; we can't do the same transform to Cartesian coordinates locally with this form of the metric, because the function $R_2(r)$ only multiplies the $dr^2$ term. So even locally, the radial coordinate $r$ in this form of the metric works differently from the others.

But globally, this spacetime clearly does have a preferred direction, the radial direction; we can see that because the metric coefficients are functions of the radial coordinate but not the other coordinates. So as we move radially, from one 2-sphere to another, the geometry changes. That is true for both forms of the metric.

 

[Emilie.Jung]

Thank you @PeterDonis for your self-contained answer. I have one question with the concept of "local" coordinates,

The second form of the metric above clearly has this property: the function $F_2(\bar{r})$multiplies the entire spatial part of the metric, so locally, the radial coordinate $\bar{r}$ doesn't work any differently from the other coordinates.

I can't seem to understand this or imagine it. How could one spherically symmetric spactime have two different radial coordinates? I am imagining here the radial coordinate as the "radius" of the sphere. Am I visualizing this the wrong way because it seems to me that I am confused about the very concept of local coordinates?

 

[DrGreg]

How could one spherically symmetric spactime have two different radial coordinates?

"Radial" here means "in the radial direction". Different radial coordinates could be scaled differently, i.e. they needn't be measuring proper distance -- indeed they won't unless the coefficient of $dr^2$ in the metric is 1.

 

[vanhees71]

Isn't a coordinate independent definition for "isotropic" that there exists a symmetry of the spacetime which is isomorphic to the rotation group $\mathrm{SO}(3)$? Then it's not just a property of specific coordinates but a covariant property of the pseudo-Riemannian manifold itself.

 

[PeterDonis]

How could one spherically symmetric spactime have two different radial coordinates?

Because coordinates are just labels; they don't have any inherent physical meaning. Their physical meaning comes from the metric, i.e., from how the coordinate labels are related to actual physical distances and times. The two forms of the metric are just different ways of labeling 2-spheres, i.e., different relationships between the coordinate, $r$ or $\bar{r}$, and the physical properties of the 2-sphere that it labels, like its area (which is what I used previously as an invariant labeling each 2-sphere) or its physical radius.

I am imagining here the radial coordinate as the "radius" of the sphere.

It isn't. This is one of those intuitions from everyday experience that has to be discarded in GR. Note that neither of the radial coordinates we've discussed, $r$ or $\bar{r}$, gives you the actual physical radius of the 2-sphere that it labels. The actual physical radius of a 2-sphere would be given by

$$
\rho = \int_0^r \sqrt{g_{rr}} dr'
$$

where $g_{rr}$ is the coefficient of the $dr^2$ term in the metric, and the $r$ in the upper limit of integration is the radial coordinate ($r'$ is a dummy integration variable that ranges from 0 to the radial coordinate). So for the two forms of the metric, we would have

$$
\rho = \int_0^r R_2(r') dr' \neq r
$$

$$
\rho = \int_0^{\bar{r}} F_2(r') dr' \neq \bar{r}
$$

In other words, the metric coefficients $R_2$ and $F_2$ not being $1$ are telling you that the physical radius of a 2-sphere is not the same as the coordinate $r$ or $\bar{r}$ that labels it.

 

[PeterDonis]

Isn't a coordinate independent definition for "isotropic" that there exists a symmetry of the spacetime which is isomorphic to the rotation group $\mathrm{SO}(3)$? Then it's not just a property of specific coordinates but a covariant property of the pseudo-Riemannian manifold itself.

I have usually seen this property of a spacetime referred to as "spherical symmetry", not "isotropy", because such a symmetry only needs to apply about one point, not about every point. (Actually, it's more complicated than that when you start thinking about cases like Schwarzschild spacetime, where there is an SO(3) symmetry group--a set of 3 spacelike Killing vector fields everywhere with the appropriate commutation relations--but no "center" point.) Whichever word we use, you're correct that this property is a property of the spacetime and is independent of coordinates.

Isotropic coordinates, however, as I said in post #2, are something different, and you can construct them, locally, even in spacetimes that are not isotropic or spherically symmetric. Basically, isotropic coordinates just means that, locally, you can transform the spatial part of the metric to Cartesian coordinates, as I described in post #2.

 

[vanhees71]

I see. If you assume that the manifold is isotropic around any point, you should end up with something more like the Robertson-Walker space-times which are of maximal symmetry of a spatial hypersurface that is either like a three-sphere, a flat space, or a three-hypersphere.

 

[Emilie.Jung]

@PeterDonis I understand, but if so, what are they? How can I explain them?

 

[Emilie.Jung]

I just read your post to Vanhees71 and understood what it means more clearly so no need to answer my previous question. Thank you @PeterDonis .