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My question is why, when expressing the Schwarzschild solution in isotropic coordinate system, the coordinate system is considered valid only outside the event horizon.

For simplicity, we assume that the source of gravity is a point mass.

The line element is expressed in isotropic coordinate system as follows. \begin{equation} ds^2=-\left(\frac{1-\frac{a}{r}}{1+\frac{a}{r}}\right)^2dt^2+\left(1+\frac{a}{r}\right)^4d{r}^2+\left(1+\frac{a}{r}\right)^4{r}^2(d\theta^2+\sin^2\theta d\phi^2).\tag{1} \end{equation} Then, we obtain the following through a coordinate transformation. \begin{equation} ds^2=-\left(\frac{1-\frac{a}{r}}{1+\frac{a}{r}}\right)^2dt^2+\left(1+\frac{a}{r}\right)^4(dx^2+dy^2+dz^2).\tag{2} \end{equation} Therefore, if we assume that the isotropic coordinate system is not valid inside the event horizon or especially in the vicinity of the center of the gravitational field, two possibilities can be considered:

  1. The Schwarzschild solution cannot be expressed in the following form in the vicinity of the center of the gravitational field. \begin{equation} ds^2=-F(r)dt^2+G(r)(dx^2+dy^2+dz^2).\tag{3} \end{equation}
  2. The Schwarzschild solution can be expressed in the avove form, but the metric is different from the isotropic coordinate system.

Regarding 1) For any coordinate system, the metric tensor matrix is a symmetric matrix, so an appropriate coordinate transformation exists to diagonalize the metric tensor matrix. Let $(t,x,y,z)$ be the coordinate system obtained in this way. Since the space is isotropic, the coefficients of $dx^2$, $dy^2$, and $dz^2$ can be taken to be equal. Now, let $A$ be a $3 \times 3$ rotation matrix. We define a coordinate transformation from $(t,x,y,z)$ to $(t,x',y',z')$ as follows. \begin{equation*} \begin{pmatrix} x' \\ y' \\ z' \\ \end{pmatrix} =A \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix}. \end{equation*} Then, the coefficients of each term in the line element are preserved because $A$ is an rotation matrix. \begin{align*} ds^2&=-F(x,y,z)dt^2+G(x,y,z)(dx^2+dy^2+dz^2), \\ &=-F(x,y,z)dt^2+G(x,y,z)(d{x'}^2+d{y'}^2+d{z'}^2). \end{align*} Taking into account that space-time is spherically symmetric, the line element is expressed in the same form in both coordinate systems. \begin{align*} ds^2&=-F(x,y,z)dt^2+G(x,y,z)(dx^2+dy^2+dz^2), \\ &=-F(x',y',z')dt^2+G(x',y',z')(d{x'}^2+d{y'}^2+d{z'}^2). \end{align*} Therefore, \begin{align*} F(x,y,z)&=F(x',y',z'), \\ G(x,y,z)&=G(x',y',z'). \end{align*} Since this holds for any rotation matrix $A$, $F$ and $G$ are radial functions. Consequently, possibility 1) is excluded.

Regarding 2) Assuming that the Schwarzschild solution is expressed in the form of (3), if we solve the Einstein's equations under a suitable coordinate transformation, we obtain (2), and hence (1). Therefore, the possibility of 2) is also excluded.

When considering this, I thought that the isotropic coordinate system is actually valid in the vicinity of the center of the gravitational field, and that the point where the radial coordinate $r=\sqrt{x^2+y^2+z^2}$ is $0$ corresponds to the center. So I wrote a paper and submitted it to a peer-reviewed journal, but it was rejected before it reached the referees.

By the way, there is a relationship between the radial coordinate $r'$ in the Schwarzschild coordinate system and the radial coordinate $r$ in the isotropic coordinate system, which is as follows. \begin{equation} r'=r\left(1+\frac{a}{r}\right)^2 \end{equation} Based on this relationship, it is clear that if the Schwarzschild coordinate is valid in the vicinity of the center of the gravitational field, then the isotropic coordinate system is not valid there. However, in deriving the expression for the Schwarzschild solution in Schwarzschild coordinate system, an assumption is made that the coefficient of $d\theta^2+\sin^2\theta d\phi^2$ is ${r'}^2$, which is an additional assumption on top of assuming spherical symmetry in spacetime. On the other hand, in my derivation of (1), no such assumption is made. Therefore, I think it is logically impossible to say that the Schwarzschild coordinate system is correct and the isotropic coordinate system is incorrect.

I would appreciate it if you could correct any logical or mathematical errors in my arguments. Please refrain from suggesting that the validity of isotropic coordinates is undermined by the correctness of the Schwarzschild coordinate system.

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    May 23, 2023 at 17:11
  • $\begingroup$ Hi Igarashi. Welcome to Phys.SE. Linking to private clouds, dropbox, etc, is for various reasons not acceptable on SE, cf. this meta post. $\endgroup$
    – Qmechanic
    Mar 17 at 13:22
  • $\begingroup$ @Qmechanic Thank you for letting me know. I've removed the link. $\endgroup$
    – Igarashi
    Mar 18 at 22:44

2 Answers 2

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  1. The metric in isotropic coordinates (1) and (2) has a (coordinate) singularity at $r=a$. Consequently, there is no a priori relationship between the metric for $r>a$ and $r<a$. They are both solutions of the vacuum Einstein equations, but may or may not be related.

  2. The coordinate transformation $r\to a^2/r$, maps the $r>a$ and $r<a$ into each other leaving the metric invariant. The $r<a$ region therefore describes just another copy of the $r>a$ region, not the interior of the black hole.

  3. In terms of the maximal analytic extension of the (exterior) Schwarzschild solution described by the Kruskal diagram, the two regions $r>a$ and $r<a$ can be thought of as describing regions I and III respectively, whereas the interior patch of the Schwarzschild metric in Schwarzschild coordinates describes region II or IV. So both the $r<a$ and the Schwarzschild interior patch describe valid analytic extensions of the exterior Schwarzschild metric beyond the horizon, they just describe different patches.

Since you explicitly asked, here is a list of logical and mathematical errors in your question (I will leave neutral how they affect your arguments):

  1. "Since the space is isotropic..." You are assuming without evidence that space inside the event horizon will be isotropic. It is at the very least a logical possibility that it isn't.
  2. "... the coefficients of each term in the line element are preserved because this coordinate transformation is linear..." Being linear is an insufficient condition for leaving the line element invariant. (Rotations leave it invariant because they are elements of SO(3), group of linear transformations that leaves the Euclidean line element invariant.)
  3. "... the center of the gravitational field ..." Have you considered the logical option that the field may not have a center?
  4. "I thought that the isotropic coordinate system is actually valid in the vicinity of the center of the gravitational field" As shown above, region near r=0 in isotropic coordinates, is isometric to the asymptotically flat region of spacetime at large r. How can it describe a center? At the very best you can argue that the metric describes some sort of wormhole (Note that Rosen and Einstein, would have beaten you to that interpretation by almost a century.)
  5. "However, in deriving the expression for the Schwarzschild solution in Schwarzschild coordinate system, an assumption is made that the coefficient of $d\theta^2+\sin2\theta d\phi^2$ is $r′^2$," This not an assumption, it is a coordinate choice that can be made for any spherically symmetric spacetime.
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  • $\begingroup$ I apologize for repeating, but please point out any logical or mathematical errors of the derivation of the Schwarzschild solution in isotropic coordinates in my question, without referring to the Schwarzschild coordinate system. $\endgroup$
    – Igarashi
    May 25, 2023 at 12:22
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    $\begingroup$ @Igarashi Your question does not contain a derivation of the Schwarschild solution in isotropic coordinates. It postulates it as Eq. 1 $\endgroup$
    – TimRias
    May 25, 2023 at 13:56
  • $\begingroup$ Regarding point 2, as you rightly pointed out, linearity of the coordinate transformation is not a sufficient condition for the invariance of the line element. It should be corrected to state that "$A$ is an orthogonal matrix"(I have corrected it). This can be derived from the transformation rules of second-order symmetric tensors and Rodriguez's rotation formula. I have further comments on other points, and I will provide them later. $\endgroup$
    – Igarashi
    May 27, 2023 at 2:00
  • $\begingroup$ orthogonal matrix -> rotation matrix $\endgroup$
    – Igarashi
    May 27, 2023 at 2:16
  • $\begingroup$ Regarding point 5, for example, consider a 2-dimensional manifold with the line element $ds^2=f(r)dr^2+g(r)d\theta^2$, where $f(r)>0$ and $g(r)=r/2+\sin r$. As $r\to 0$, $g(r)\to 0$, and as $r\to\infty$, $g(r)\to\infty$. However, $g(r)$ is not a monotonically increasing function. Therefore, if we assume a coordinate transformation from $r$ to $r'$ such that the line element is transformed to $ds^2=h(r')d{r'}^2+{r'}^2d\theta^2$, the $(r',\theta)$ coordinate system is only valid in the region where $g(r)$ is increasing because ${r'}^2$ is increasing. $\endgroup$
    – Igarashi
    May 27, 2023 at 12:19
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Equation for $r$ in terms of $r'$ (in the question as originally posed) has them the wrong way round. In notation $r$=isotropic, $r'$=Schwarzschild coordinate it should be $$r' = r (1 + a/r)^2 .$$ Notice that Schwarzschild $r'$ is then greater than $4a$ (which is the Schwarzschild radius) for all $r$. So no matter what value of isotropic coordinate you pick, you are always at an event in the outer part of the manifold.

Now concerning whether you can find isotropic-like coordinates in the inner part, i.e. near the curvature singularity, I have not investigated. But you should keep in mind that the curvature singularity is not at a spatial point (in this solution, i.e. vacuum and no rotation). It does not lie on a timelike line. It lies on a spacelike surface. Also, events separated only by $dr'$ now have timelike not spacelike separations, so such intervals do not represent distance in or out towards any spatial centre.

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  • $\begingroup$ Thank you. $r$ and $r'$ were reversed (I have corrected it). Since $r'>=4a$ for any $r$, if the Schwarzschild coordinates cover the region where $r'<4a$, then it is indeed correct to say that isotropic coordinates do not cover that region. But what if the Schwarzschild coordinates do not cover the region where $r'<4a$? And what if there exists a region in our spacetime, particularly inside a Schwarzschild black hole, where the coefficient of $d\theta^2+\sin^2\theta d\phi^2$ becomes a monotonically decreasing function of $r'$? That is the starting point of my question. $\endgroup$
    – Igarashi
    May 26, 2023 at 13:32

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