# Help me interpret this person's work please

1. Aug 18, 2014

### space-time

First, here is my source: http://www.thescienceforum.com/physics/30059-solving-einstein-field-equations.html [Broken]

As you can see, this person was showing how to solve at least one basic scenario for the Einstein field equations. Now I have a couple of questions about this person's work?

1. When he proposed that ansatz, why is it that only the first two terms of the space time interval had functions of r attached to them? Here is what I mean:

ds2 = B(r)c2t2 - A(r)dr2 - r2(dθ2+ sin2θdø2)

Notice that only the first two terms of this equation have unknown functions of r like B(r) or A(r) and the other terms don't. Why is this? Why do the unknown functions only go towards the first two terms?

2. I can tell that the space time interval for this scenario is very similar to that of spherical coordinates. Therefore, it is not too hard for me to imagine what the metric tensor for this particular scenario would look like. However, in general cases where the ansatz and the ultimate space time interval that you derive as your solution are not similar to commonly known coordinate systems, how exactly would you derive your metric tensor from your solution?

From what I have learned thus far, you are solving for the metric tensor when you solve the Einstein field equations. If you just derive the space time interval however, how do you actually get your metric tensor (and actually solve the equations) from this?

Last edited by a moderator: May 6, 2017
2. Aug 18, 2014

### Staff: Mentor

He is assuming a spherically symmetric solution, presumably because the underlying physical situation is spherically symmetric. Spherical symmetry implies that the physics at a given $r$ value is the same in all spatial directions.

You can read the components of the metric tensor off directly from the equation for the spacetime interval. The general form of the spacetime interval is $ds^2=g_{ij}dx^idx^j$ so you just have to look at the equation for the interval and pick out the coefficients for each term. If a particular term doesn't appear, then the coefficient is zero. For example, using Minkowski coordinates in flat spacetime we have $ds^2=-c^2dt^2+dx^2+dy^2+dz^2$ so we have $g_{00}=-c^2$ and $g_{11}=g_{22}=g_{33}=1$ and all the other components of the metric tensor in these coordinates are zero.

3. Aug 18, 2014

### space-time

You said that spherical symmetry implies that the physics at a given state r are the same in all spatial directions. However, the term dr2 involves spatial direction just like the final two terms in the parentheses do. I still don't understand then, why those final two spatial terms don't get the function A(r) like the previous spatial term dr2 does.

In other words, what is different about dr2 as opposed to r2(dθ2 + sin2θdø2)? Aren't all of these spatial terms and therefore warranting the A(r) function?

4. Aug 18, 2014

### Staff: Mentor

It's just a particular choice of coordinates, in particular the radial coordinate. The choice you describe in the OP is usually called the Schwarzschild radial coordinate; it is defined so that the area of a 2-sphere with radial coordinate $r$ is $4 \pi r^2$. That is what leads to the $A(r)$ coefficient only being attached to the $dr^2$ term in the line element.

There is another choice of radial coordinate, usually called the isotropic radial coordinate, which makes the line element look different; with isotropic coordinates, the $A(r)$ coefficient does indeed multiply all three spatial terms, i.e., the line element looks like this:

$$ds^2 = B(\bar{r}) c^2 dt^2 - A(\bar{r}) \left[ d \bar{r}^2 + \bar{r}^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right) \right]$$

where I have used $\bar{r}$ to make it clear that we are using a different definition of the radial coordinate. Isotropic coordinates are more intuitive in that they are isotropic--locally, all three spatial coordinates "work" the same, unlike Schwarzschild coordinates where the radial coordinate is different. However, they are less intuitive in that the area of a 2-sphere at $\bar{r}$ is *not* $4 \pi \bar{r}^2$, it's $4 \pi A(\bar{r}) \bar{r}^2$, so the $\bar{r}$ coordinate does not have a direct physical interpretation as the "radius" of the 2-sphere.

5. Aug 18, 2014

### Staff: Mentor

I'm sorry, I wasn't completely clear. If you are sitting at the center of a spherically symmetric space then everything looks the same whether you're looking up or down or left or right. That direction is specified by the $\theta$ and $\phi$ (left-right and up-down) coordinates. The $r$ coordinate is the distance away from you in whatever direction you're looking, and spherical symmetry means that no matter what direction you're looking (that is, no matter what the value of $\theta$ and $\phi$) what's going on at a given distance away from you (that is, at a given value of $r$) will be the same. Thus, the ansatz is written to say that the metric can change as as the distance from you changes ($dr$ is non-zero) but not as the direction ($d\theta$ and $d\phi$ are non-zero) changes. That's what makes $r$ different for this particular set of coordinates.

6. Aug 18, 2014

### Staff: Mentor

The distance from you changing (assuming you are at the center of the sphere) does not correspond to $dr \neq 0$. It corresponds to the value of $r$ that you use to evaluate the metric changing. What $dr \neq 0$ means is that the particular infinitesimal line element you are looking at has a component in the radial direction; but since the line element is infinitesimal, the value of $r$ at which the metric is evaluated does not change even if $dr \neq 0$.

Note that in the line element for isotropic coordinates that I wrote down, it is still true that "the metric can change as the distance from you changes, but not as the direction changes", because $A(\bar{r})$ is still a function of the radial coordinate--the fact that it multiplies the entire spatial metric instead of just $dr^2$ doesn't matter. (Changes in direction correspond to different values of $\theta$ and $\phi$, not $d\theta \neq 0$ or $d \phi \neq 0$--but note that the metric does change as $\theta$ changes, since $\sin^2 \theta$ appears in the line element, so it takes some more sophisticated math to show that changes in direction do not in fact change the metric--basically you need to exhibit the rotational Killing vector fields.)