# Schwarzschild's Metric 1916

#### PeterDonis

Mentor
within that metric a surface of constant $R$ would have surface area $4\pi R^2>4\pi r^2$ would it not?
Yes, but that's not why the metric is non-Euclidean. The relationship between $R$ and $r$ is irrelevant here. See below.

That would indicate $R^+ \times S^2$ has negative curvature and not be Euclidean
No. What indicates that a surface of constant $t$ has negative curvature is, as I said, that the coefficient of $dR^2$ in the metric is not 1. The radial coordinate in this metric is $R$, not $r$; you don't need to know anything about $r$ to determine that the spatial metric is non-Euclidean.

Put another way, what makes a surface of constant $t$ non-Euclidean in g(1916) is that a 2-sphere of constant $R$ has surface area $4 \pi R^2$, but the proper distance between a 2-sphere at $R$ and a 2-sphere at $R + dR$ is *greater* than $dR$, whereas if the space were Euclidean, that proper distance would be equal to $dR$. For small $dR$, the proper distance between the 2-spheres is $dR / \sqrt{1 - r_s / R}$. That is, the ratio of the proper distance between two nearby 2-spheres to the difference in their areas is larger than it would be if the space were Euclidean.

as in g(current).
No; g(current) is also non-Euclidean, and for the same basic reason. The only difference is that in g(current), $r$ plays the role that $R$ plays in g(1916): in g(current), the proper distance between two nearby 2-spheres at $r$ and $r + dr$ is larger than $dr$, but the surface area of a 2-sphere at $r$ is $4 \pi r^2$.

#### StateOfTheEqn

The metric representations are:

$$g=\left(\begin{array}{cccc} 1-r_s/r&0&0&0\\ 0&(1-r_s/r)^{-1}&0&0\\ 0&0&r^2&0\\ 0&0&0&r^2sin^2\theta \end{array}\right)$$

and
$$\overline g=\left(\begin{array}{cccc} 1-r_s/R&0&0&0\\ 0&(1-r_s/R)^{-1}&0&0\\ 0&0&R^2&0\\ 0&0&0&R^2sin^2\theta \end{array}\right)$$

Let $\overline{\textbf{x}}$ be a column vector w.r.t. the basis $(dt,dR,d\theta,d\phi)$ and $\textbf{x}$ be the same vector w.r.t the basis $(dt,dr,d\theta,d\phi)$. Then $\overline{\textbf{x}} = D\Phi \textbf{x}$. The vectors are assumed to be carrying a Minkowski signature. For example,

$$\textbf{x}=\left(\begin{array}{c} dt\\ idr\\ id\theta\\ id\phi \end{array}\right)$$

Let $<.,.>_g$ be the metric w.r.t. $g$ and $<.,.>_{\overline{g}}$ be the metric w.r.t. $\overline{g}$.

Then $$<\overline{\textbf{x}},\overline{\textbf{x}}>_{\overline{g}}=(\overline{\textbf{x}})^T(\overline{g}) (\overline{\textbf{x}})=(D\Phi\textbf{x})^T(\overline{g})(D\Phi\textbf{x})=\textbf{x}^T(D\Phi)^T(\overline{g})(D\Phi)\textbf{x} = <\textbf{x},\textbf{x}>_{(D\Phi)^T(\overline{g})(D\Phi)}$$

But a simple calculation shows $g \neq (D\Phi)^T(\overline{g})(D\Phi)$ unless $r=R$ but they are unequal by assumption. Therefore $g$ and $\overline g$ do not represent the same metric.
I would like to clarify the above somewhat:

The metric(s) above take an element in $T_p^*M$ to an element in $T_p^*M \otimes T_p^*M$. It is somewhat an abuse of terminology to call it a metric but it is common in GR. An actual metric would be $g:T_pM \times T_pM \rightarrow \mathbb{R}$. Re-doing the above calculation using the actual metric would be as follows:

Let $\textbf{x}=\left(\begin{array}{c} x^0\\ ix^1\\ ix^2\\ ix^3 \end{array}\right)$

and $\overline{\textbf{x}}=\left(\begin{array}{c} \overline{x}^0\\ i\overline{x}^1\\ i\overline{x}^2\\ i\overline{x}^3 \end{array}\right)$

Then $$<\overline{\textbf{x}},\overline{\textbf{x}}>_{\overline{g}}=(\overline{\textbf{x}})^T \overline{g} (\overline{\textbf{x}})=(D\Phi\textbf{x})^T\overline{g}(D\Phi\textbf{x} )=\textbf{x}^T(D\Phi)^T\overline{g}(D\Phi)\textbf{x}=<\textbf{x}, \textbf{x}>_{(D\Phi)^T\overline{g}(D\Phi)}$$

and the same result follows.

#### Mentz114

Gold Member
The two metric representations do not represent the same metric.

Let $(t,R,\theta,\phi)=\Phi (t,r,\theta,\phi) = (t,(r^3 + r_s^3)^{1/3},\theta,\phi)$ where $r_s$ is a constant > 0.

Then $$D\Phi=\left(\begin{array}{cccc} 1&0&0&0\\ 0&\partial_r R&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right)$$

The metric representations are:

$$g=\left(\begin{array}{cccc} 1-r_s/r&0&0&0\\ 0&(1-r_s/r)^{-1}&0&0\\ 0&0&r^2&0\\ 0&0&0&r^2sin^2\theta \end{array}\right)$$

and
$$\overline g=\left(\begin{array}{cccc} 1-r_s/R&0&0&0\\ 0&(1-r_s/R)^{-1}&0&0\\ 0&0&R^2&0\\ 0&0&0&R^2sin^2\theta \end{array}\right)$$

..
..
It looks like you have done a transformation of coordinates incorrectly and then proved it was incorrect.

Starting with
$ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$
and making the transformation
$r\rightarrow {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}},\ \ dr= \frac{{R}^{2}\,dR}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}}$
the result is

$ds^2={dt}^{2}\,\left( \frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}-1\right) +\frac{{dR}^{2}\,{R}^{4}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{4}{3}}\,\left( 1-\frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}\right) }+{d\phi}^{2}\,{\sin\left( \theta\right) }^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}+{d\theta}^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}$

This is the correct metric and has Ricci tensor $R_{ab}=0$.

#### StateOfTheEqn

It looks like you have done a transformation of coordinates incorrectly and then proved it was incorrect.

Starting with
$ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$
and making the transformation
$r\rightarrow {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}},\ \ dr= \frac{{R}^{2}\,dR}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}}$
the result is

$ds^2={dt}^{2}\,\left( \frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}-1\right) +\frac{{dR}^{2}\,{R}^{4}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{4}{3}}\,\left( 1-\frac{2\,m}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}}\right) }+{d\phi}^{2}\,{\sin\left( \theta\right) }^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}+{d\theta}^{2}\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}$

This is the correct metric and has Ricci tensor $R_{ab}=0$.
You start with $ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$ and show it can be written in terms of $R$. It seems that what you have shown is that the original Schwarzschild metric, what I have been calling g(1916), is not compatible with the above metric even when it is written in terms of $R$. The metric g(1916) does not have an event horizon at $r=2m$ but your metric does, even when written in terms of $R$. At $r=2m$, g(1916) has the metric
$ds^2=(1-1/\sqrt[3]{2})dt^2-(1-1/\sqrt[3]{2})^{-1}dR^2-4(2)^{2/3}m(d\theta^2+sin^2\theta d\phi^2)$.
In fact, for all $r>0$, $(1-2m/R)>0$ where $R=(r^3+8m^3)^{1/3}$ and therefore no event horizons.

#### Mentz114

Gold Member
You start with $ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$ and show it can be written in terms of $R$. It seems that what you have shown is that the original Schwarzschild metric, what I have been calling g(1916), is not compatible with the above metric even when it is written in terms of $R$. The metric g(1916) does not have an event horizon at $r=2m$ but your metric does, even when written in terms of $R$. At $r=2m$, g(1916) has the metric
$ds^2=(1-1/\sqrt[3]{2})dt^2-(1-1/\sqrt[3]{2})^{-1}dR^2-4(2)^{2/3}m(d\theta^2+sin^2\theta d\phi^2)$.
In fact, for all $r>0$, $(1-2m/R)>0$ where $R=(r^3+8m^3)^{1/3}$ and therefore no event horizons.
I don't know what you're trying to say but

1) if the 1916 metric is derived as a coordinate transformation from the Schwarzschild metric then it has the form I showed and makes no predictions different from Schwarzschilds metric.

2) the metric you show is not the Schwarzschild metric and is not the spherically symmetric vacuum solution.

#### Bill_K

You start with $ds^2={r}^{2}\,{\sin\left( \theta\right) }^{2}\,{d\phi}^{2}+{r}^{2}\,{d\theta}^{2}+\frac{1}{1-\frac{2\,m}{r}}{dr}^{2}-\left( 1-\frac{2\,m}{r}\right) \,{dt}^{2}$ and show it can be written in terms of $R$.
It's called a coordinate transformation. "Writing it in terms of R" is the way you do a coordinate transformation. The 1916 paper never wrote the metric explicitly as Mentz114 does, rather left it implicitly in terms of the function R(r).

In fact, for all $r>0$, $(1-2m/R)>0$ where $R=(r^3+8m^3)^{1/3}$ and therefore no event horizons.
The event horizon is located at rcurrent = 2m, while in the 1916 coordinates it is located at r1916 = 0. The geometry is exactly the same.

#### StateOfTheEqn

There is only one unique spherically symmetrical solution to the vacuum field equations (see Birkhoff's theorem), and it's the one Schwarzschild described. His description differs only in appearance from how the Schwarzschild solution is commonly presented, due to the use of an unusual radial coordinate. Changing coordinates doesn't change any of the physical properties, metrical relations, or predictions of the solution.
The proof of the Birkhoff Uniqueness Theorem depends on defining $r$ as the 'area radius'. That is, $r=area(S^2)/4\pi$. However, a negatively curved space $\mathbb{R} \times S^2$ can have $area(S^2)/4\pi$ greater than the measured scalar radius $\rho$. In Schwarzschild's 1916 paper, $area(S^2)/4\pi=R=({\rho}^3+r_s^3)^{1/3}>\rho$. At present I am not sure what this says about uniqueness. In his paper Schwarzschild states:
Die Eindeutigkeit der Lösung hat sich durch die vorstehende Rechnung von selbst ergeben.(The uniqueness of the solution resulted spontaneously through the present calculation.) English translation at http://arxiv.org/pdf/physics/9905030v1

#### StateOfTheEqn

It's called a coordinate transformation. "Writing it in terms of R" is the way you do a coordinate transformation. The 1916 paper never wrote the metric explicitly as Mentz114 does, rather left it implicitly in terms of the function R(r).
His coordinate transformation is to an equivalent representation of his starting metric. But that misses the point. A coordinate transformation between equivalent representations of the same metric must be a local isometry. There is no local isometry transforming the representation which I have been calling g(current) to what I have been calling g(1916). His coordinate transformation is a local isometry (it seems) but it does not produce the g(1916) representation.

The event horizon is located at rcurrent = 2m, while in the 1916 coordinates it is located at r1916 = 0. The geometry is exactly the same.
There is a difference. There is zero volume inside a sphere with radius = 0 but there is non-zero volume inside a sphere with radius = 2m if m>0.

#### StateOfTheEqn

I don't know what you're trying to say but

1) if the 1916 metric is derived as a coordinate transformation from the Schwarzschild metric then it has the form I showed and makes no predictions different from Schwarzschilds metric.

2) the metric you show is not the Schwarzschild metric and is not the spherically symmetric vacuum solution.

The original is at http://de.wikisource.org/wiki/Über_das_Gravitationsfeld_eines_Massenpunktes_nach_der_Einsteinschen_Theorie and an English translation at http://arxiv.org/pdf/physics/9905030v1

The forward to the English translation states:
This fundamental memoir contains the ORIGINAL form of the solution of Schwarzschild’s problem.
It is regular in the whole space-time, with the only exception of the origin of the spatial co-ordinates; consequently,
it leaves no room for the science fiction of the black holes.

#### PeterDonis

Mentor
There is zero volume inside a sphere with radius = 0 but there is non-zero volume inside a sphere with radius = 2m if m>0.
But "radius" is not necessarily the same as $r$, the coordinate that appears in equations. That depends on how $r$ is defined. In what you have been calling g(1916), if you compute the surface area of a sphere at $r = 0$ (which is the same as $R = r_s$), with $r$ defined the way Schwarzschild does it in his paper, then that surface area is *not* zero, and therefore the volume enclosed by that sphere is not zero either.

In other words, you can't just arbitrarily say that $r$ is the "radius"; you have to actually look at the metric to see what it says about the geometric meaning of $r$. If you look at the metric you are calling g(1916), it is obvious that $R$ is the areal radius, *not* $r$, because the area of a 2-sphere at $R$ is $4 \pi R^2$, but the area of a 2-sphere at $r$ is *not* $4 \pi r^2$.

Modern texts usually define what they call $r$ as the areal radius, but that just means that the modern $r$ plays the same role as Schwarzschild's $R$ did, as I said before. It does *not* mean that the $r$ in g(current) is the same as the $r$ in g(1916), even though they happen to be designated by the same lower-case letter.

#### StateOfTheEqn

Modern texts usually define what they call $r$ as the areal radius, but that just means that the modern $r$ plays the same role as Schwarzschild's $R$ did, as I said before. It does *not* mean that the $r$ in g(current) is the same as the $r$ in g(1916), even though they happen to be designated by the same lower-case letter.
I mention that in post #57 of this thread. I grant that if $r$ in g(current) is really equal to $R$ in g(1916) then they are equivalent representations of one metric. Schwarzschild states quite clearly that $r=\sqrt{x^2+y^2+z^2}$ on a line just before eqn(6) in http://arxiv.org/pdf/physics/9905030v1. Furthermore his $R=(r^3+r_s^3)^{1/3}$ where $r_s=2m$. So, if we transform $r$ in g(current) to the $R$ of g(1916) we get equivalent metric representations. Then there is no event horizon for any $r=\sqrt{x^2+y^2+z^2} >0$.

#### Bill_K

Schwarzschild states quite clearly that $r=\sqrt{x^2+y^2+z^2}$
OMG, I'm sure he does! But that does not mean that r is in any sense a "radius". Did you think that x, y and z were Cartesian coordinates?? They have no such meaning. It does not exclude the possibility that r = 0, and even r < 0 is possible.

Last edited:

#### Mentz114

Gold Member
OK, I will if I have the time. But there is only one spherically symmetric static vacuum solution, and it has an event horizon (singularity) in holonomic coordinates which is not present in some local frame bases. So what would I learn ?

#### StateOfTheEqn

OK, I will if I have the time. But there is only one spherically symmetric static vacuum solution, and it has an event horizon (singularity) in holonomic coordinates which is not present in some local frame bases. So what would I learn ?
You might learn that his original metric representation is not the same as the one currently used. See if you can find a local isometry that transforms the one into the other. A coordinate transformation between metric representations must be a local isometry if the metrics being represented are the same metric. If you can find such an isometry I would really like to know about it.

#### PeterDonis

Mentor
Schwarzschild states quite clearly that $r=\sqrt{x^2+y^2+z^2}$
Yes, he does. So what? That just raises the question of what $x$, $y$, and $z$ actually mean, geometrically. You can't assume that they are normal Euclidean Cartesian coordinates, any more than you can assume that $r$ is the actual radius. You have to look at the metric. Since the metric makes it clear that $r$ is not the radius, then that implies that $x$, $y$, and $z$ aren't standard Euclidean Cartesian coordinates either.

his $R=(r^3+r_s^3)^{1/3}$ where $r_s=2m$. So, if we transform $r$ in g(current) to the $R$ of g(1916) we get equivalent metric representations.
Yes.

Then there is no event horizon for any $r=\sqrt{x^2+y^2+z^2} >0$.
Only for Schwarzschild's definition of $r$. But the region $r > 0$, with Schwarzschild's definition of $r$, is not the entire spacetime. Schwarzschild appears to have assumed that it was, but he never proved it; and in fact that assumption is false. One way to see that it's false is to note, as I said before, that the area of the 2-sphere at $r = 0$ ($R = r_s$) is *not* zero.

With the modern definition of $r$, there is indeed an event horizon at $r = 2M$, which corresponds to $R = r_s$ in Schwarzschild's notation.

#### StateOfTheEqn

Only for Schwarzschild's definition of $r$. But the region $r > 0$, with Schwarzschild's definition of $r$, is not the entire spacetime. Schwarzschild appears to have assumed that it was, but he never proved it; and in fact that assumption is false. One way to see that it's false is to note, as I said before, that the area of the 2-sphere at $r = 0$ ($R = r_s$) is *not* zero.
Clearly $4\pi R^2/4\pi r^2 \rightarrow \infty$ as $r \rightarrow 0$ but that could be due to the negative spatial curvature growing without bound near the central singularity (at $r=0$).

Last edited:

#### Bill_K

A coordinate transformation between metric representations must be a local isometry if the metrics being represented are the same metric.
I don't know where on Earth you got this idea.

#### PeterDonis

Mentor
Clearly $4\pi R^2/4\pi r^2 \rightarrow \infty$ as $r \rightarrow 0$
So what? What does this mean, physically?

but that could be due to the negative spatial curvature growing without bound near the central singularity (at $r=0$).
Negative spatial curvature of what? What is $r$ supposed to represent? You can't just wave your hands and say it's a "radius" of something. You have to actually look at the metric and compute things from it. Where in any such computation does $4 \pi r^2$ arise?

#### StateOfTheEqn

I don't know where on Earth you got this idea.
If your coordinate transformations between metric representations are not local isometries (i.e. metric preserving) then what do they mean? Bear in mind that what is losely called a metric in GR is only a metric representation (one for each coordinate system). There is a family of metric representations for each metric and they are pairwise transformable into each other by local isometries. Does that help?

#### Mentz114

Gold Member
If your coordinate transformations between metric representations are not local isometries (i.e. metric preserving) then what do they mean? Bear in mind that what is losely called a metric in GR is only a metric representation (one for each coordinate system). There is a family of metric representations for each metric and they are pairwise transformable into each other by local isometries. Does that help?
Coordinate transformation can change the components of the metric but all scalars formed by tensor contractions are invariant. If any of these invariants is different between two spacetimes, they are not the same spacetime.

For the two metrics I gave earlier, the second one gives the K-invariant
$\frac{16\,{m}^{2}\,{\left( 2\,m\,{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{2}{3}}-{R}^{3}+8\,{m}^{3}\right) }^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{10}{3}}\,{\left( {\left( {R}^{3}-8\,{m}^{3}\right) }^{\frac{1}{3}}-2\,m\right) }^{2}}+\frac{32\,{m}^{2}}{{\left( {R}^{3}-8\,{m}^{3}\right) }^{2}}$
If we substitute $(r^3-(2m)^3)^{1/3}$ for $R$ we get $48m^2/r^6$. So for any calculation, whether we start with $R$ or $r$ will give the same answer.

Last edited:

#### Bill_K

If your coordinate transformations between metric representations are not local isometries (i.e. metric preserving) then what do they mean?
They mean that the equations of physics are generally covariant. Yes, the components of the metric will change. gμν is a tensor and is naturally expected to change under coordinate transformations. By your definition it would be unacceptable to transform, say, from rectangular coordinates to polar.

Generally, solutions of Einstein's equations have no isometries.

#### StateOfTheEqn

It seems to me the key to understanding the comparison between g(current) and g(1916) is in comparing the spatial curvature. In both g(current) and g(1916), $R^2:=area(S^2)/4\pi$ so $R$ is known as the area radius. Defining $R$ this way, we can show the spherically symmetric solution is unique (Birkhoff Uniqueness Theorem). However, that conceals a difficulty. Solutions of the required form can have different spatial curvatures but be treated as equivalent in the proof of Birkhoffs theorem.

This can happen in the spherically symmetric space as follows: If $r$ is the Euclidean distance we could have $R<r$ (positive spatial curvature), $R=r$ (zero spatial curvature), or $R>r$ (negative spatial curvature). In g(1916), Schwarzschild derived a solution with negative spatial curvature where $R=(r^3+r_s^3)^{1/3}$ and $r_s=2GM$.

What about a Black Hole event horizon? In g(1916), $R=r_s$ only when the Euclidean distance $r=0$, that is, at the central singularity itself. In g(current), when $R=r_s$ the value of $r$ is left undefined because the spatial curvature (and therefore the relation of $R$ to $r$) is left undefined. So, in g(current), we have no way of knowing the value of $r$ when $R=r_s$. It could be zero, as in g(1916), which would imply no event horizon, except at the central singularity itself.

#### StateOfTheEqn

By your definition it would be unacceptable to transform, say, from rectangular coordinates to polar.
Incorrect. That transformation is a local isometry and therefore preserves the metric (but not the metric representation which is obviously different).

#### StateOfTheEqn

Generally, solutions of Einstein's equations have no isometries.
Both g(current) and g(1916) have rotational isometries in (at least) one of the angles.

#### stevendaryl

Staff Emeritus
Incorrect. That transformation is a local isometry and therefore preserves the metric (but not the metric representation which is obviously different).
Whether a transformation preserves the metric depends on what the new metric is. If you have one patch describing using coordinates $x^i$ and with metric tensor with components $g_{ij}$, and you transform to another patch described using coordinates $x^\mu$ and metric tensor components $g_{\mu \nu}$, then it is metric-preserving if the metric components are related by:

$g_{\mu \nu} = \frac{\partial x^i}{\partial x^\mu} \frac{\partial x^j}{\partial x^\nu} g_{ij}$

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving