$r$ and $R$ difference with Schwarzschild metric Why did Schwarzschild in his article about the solution of Einstein's equation outside a massive sphere use a variable $R$: $$R=(r^3+\alpha^3)^{1/3}$$ 

Whereas, nowadays the variable used is just the radius $r$? Doesn't that change the meaning of the equations?
 A: $\let\a=\alpha \let\b=\beta \let\th=\theta \def\ra{{(r^3+\a^3)}} 
\def\sa{{(r_1^3+\a^3)}}$
Let's begin to put some firm points.


*

*The coordinate we today call Schwarzschild's $r$ isn't his $r$, but
$R=\ra^{1/3}$.

*He calls $R$ an "auxiliary quantity". E.g. consider the form he gives to Kepler's third law (second-to-last equation of his paper):
$$n^2 = {\a \over 2\,\ra}$$
with $n=d\phi/dt$.

*Schwarzschild writes metric as
$$ds^2 = F\,dt^2 - (G + H\,r^2)\,dr^2 - 
G\,r^2 (d\th^2 + \sin^2\!\th\,d\phi^2) \tag1$$
with $F$, $G$, $H$ functions of $r$ (eq. 6).
From (1) we see that angular part of the metric is different from the
one we are accustomed to. In particular, surfaces $r=\rm const.$ don't
have area $4\pi r^2$, but $4\pi r^2 G$ and $G$ is not constant.
It's clear from these and other aspects of Schwarzschild's paper that
he considers $t$, $r$, $\theta$, $\phi$ as true physical space-time
coordinates. In particular, he constrains the "singularity" to $r=0$,
i,e, out of spacetime. He states that clearly when he writes condition
(13), having precisely that aim.
Of course both $r$ and $R$ are legitimate radial coordinates. They are
however not physically equivalent. In mathematical terms, using $r$
the spacetime manifold exhibits a singularity only at border $r=0$.
Using $R$ that singularity becomes the horizon ($R=\a$) and it makes sense
to ask oneself if points with $R<\a$ are to be included in spacetime. As
is well known, a thorough understanding of that issue would have taken about half a century.
OK, maybe this is history of physics. Or it's physics in its own
right?

Edit
From comments and answers I've read I deem necessary to expand
somewhat my answer, hoping it will help to solve several doubts and
misconceptions.
(A short historical note. Einstein's paper exhibiting the final form
of his equations is dated Nov 25th 1915. Schw. paper is dated January
13th 1916. Schw. died - by an autoimmune disease still incurable
today - on May 11th 1916. He was 43.)
First, there is a strong difference between our present way of
understanding spacetime in GR and the way of E.'s and S.'s times. I
already remarked that one century ago there still weren't clear ideas
about the meaning of spacetime coordinates. There still was a tendency
to consider them endowed of some physical significance by themselves.
In particular, although S. was well aware that spacetime in his
solution is curved and space sections are curved too (not euclidean),
he writes a formula for angular velocity in a circular orbit 
$$n^2 = {\a \over 2\,\ra}$$ and concludes

the angular velocity does not, as with Newton's law, grow without
  limit when the radius of the orbit gets smaller and smaller, but it
  approaches a determined limit 
  $$n_0 = {1 \over \a \sqrt2}.$$
  (For a point with the solar mass the limit frequency will be around
  $10^4$ per second).

(remember that S.'s $\a$ is what is known as "S. radius" and is
proportional to the mass of central body).

On the contrary, present-day approach, grounded on sounder
mathematical bases, is roughly the following.


*

*Spacetime is a semi-Riemannian manifold of dimension 4 and
signature $+---$.


1.1 A (real) manifold is a set wherein real coordinates may be
defined. This can be done in several ways. Coordinates are nothing but
labels for the set's points.
1.2 It's not required that a set of coordinates be able to cover the
whole manifold. More sets are allowed - it's only required that
together they cover the whole manifold and smoothly overlap between
them. Each set is named a card and the ensemble is named an atlas.
1.3 An easy instance. To define a sphere as a (2D) manifold a minimum
of two cards are required. Among geographers a lot of cartographic
projections are in use and a world's atlas is a good example of the
general idea.


*A Riemannian manifold is a manifold where a metric is defined.
Roughly, a formula giving the distance between (infinitesimally) near
points.


2.1 More exactly, the distance squared $ds^2$. So the metric must be
positive definite.
2.2 In a semi-Riemannian manifold an extension is allowed: $ds^2$ may
also be negative. (A contradiction? Not quite. It's enough to relax
the intuitive interpretation as a distance squared.) Minkowski's
spacetime of SR is already an instance of that: there are spacelike
intervals, timelike ones, and lightlike too.
2.3 The signature refers to how many independent displacements have
metric of each sign. In GR there are two conventions in use: $+----$
means timelike has positive $ds^2$, spacelike negative. $-+++$ is the
opposite. There is no real difference - it's only necessary to
consistently adhere to one convention. Mixing them in a calculation
leads to certain disaster.


*Leaving aside more sophisticated usages, the metric is the only way
we have to give coordinates a physical meaning. It allows to compute
the time a clock measures between events or the length of a space
interval and so on.


3.1 Assume a coordinate is called $t$. It's the initial of "time" in
English, of "temps" in French, of "tempo" in Italian, of "tiempo" in
Spanish ... but not of "Zeit" in German or "czas" in Polish. So why
should we assume that coordinate means time? It may (and usually will)
be, but not always. Only looking at the metric can we get a safe
answer.

Now let's come back to S. He makes use of two radial coordinates: $r$
and $R$. But there's no doubt about which he takes as the "physical"
radius. His paper's title says

On the Gravitational Field of a Mass Point according to Einstein's
  Theory

A "mass point". It's obvious that he locates that mass at $r=0$, that
$r$ can take all real positive values, that he'll require no
singularity appears for positive $r$. At paper's end, as I noted
above, he computes the revolution period of a planet as a function of
radius and comments on a peculiar result; that period doesn't go to
zero with $r$. On the contrary, it goes to a non-zero limit of about
$0.1$ ms. He doesn't ask himself what's the meaning of $r$ (physical
distance from the central mass?) nor what time would be that $0.1$ ms
- which clock would measure it.
No wonder: GR had just been born and E. himself wasn't much clearer
about such matter. But after a century and a lot of valuable work of
eminent theoreticians we can and must have sounder ideas.
As to $R$, I repeat that S. calls it an "auxiliary" quantity. In
modern terms we would consider it a radial coordinate as good as $r$ -
metric can be written both in terms of $R$ (S.'s eq. (14), exactly the
same universally denoted today as "S.'s metric") and in terms of his
$r$. S. doesn't write the latter but you can see it here:
$$ds^2 = \left[1 -\a\,\ra^{-1/3}\right] dt^2 -
{r^4 \over \ra\,\left[\ra^{1/3} - \a\right]}\,dr^2 -
\ra^{2/3} (d\th^2 + \sin^2\!\th\,d\phi^2).\tag2$$
You can use eq. (2) to answer e.g. the following question: "Once fixed
$t$ and $r$ you're left with a 2D surface (a sphere). What's its area?"
The answer isn't $4\pi r^2$, but the more complicated
$4\pi\,\ra^{2/3}$. You could use $R$ instead and then (from S.'s
metric) you'd find $4\pi R^2$, which is the same. Another question
could be: "What's the radius of that spehere?" S. wouldn't have
hesitated. His answer would have been
$$\int_0^r\!
{r_1^2 \over \sqrt{\sa \left[\sa^{1/3} - \a\right]}}\,dr_1.$$
(The integral looks intimidating, but it's easily solved through a
substitution - can you see it?) Surely S. would have preferred to use
his "auxiliary quantity" writing the required radius as
$$\int_\a^R\!\sqrt{R_1 \over R_1 - \a}\,dR_1.$$
He wouldn't have worried about the lower $\a$ limit, which isn't the
origin of $R$ coordinate. To him the real radial coordinate was $r$.
It's up to us to be worried: if $r$ and $R$ are on equal footings as
radial coordinates, where is the space origin? At $r=0$ or at $R=0$?
A mathematician's answer would be straight: if you use $r$ then that
coordinate works for all $r>0$ - only at $r=0$ metric (2) is
singular as the coefficient of $dr^2$ vanishes. Instead if you want to
use $R$ with S.'s metric you must keep $R>\a$ since metric becomes
singular at $R=\a$. 
In both cases - the mathematician would continue - this doesn't mean
that your manifold ends there. It could, or it could go on - it's
your choice and it doesn't depend on the radial coordinate you
initially assumed. It's true that $r$ suggests that $r=0$ is the end
of your manifold whereas $R$ naturally leads to think that there is
"something out there", but here mathematicians take a different view.
They would tell us "what you're looking for is whether the manifold defined by the chart you have built allows for an extension or not. If it doesn't, this is the end of our argument. If it does, it's to you to decide if you want to give a physical meaning to the extension, or not."
Well, the answer is that the extension exists. How can we say that?
Simply by finding other coordinates which are able to cover a region
wider than the original one. This was actually done, in several ways,
and opened physicists a novel world. A physical meaning to that world
was given in 1939, when a paper  by Oppenheimer and Snyder ["On Continued Gravitational Contraction"] 
(https://journals.aps.org/pr/abstract/10.1103/PhysRev.56.455)
introduced the idea we today know as gravitational collapse.
A: I Think that this is just coordinate transformation:
This  is the Schwarzschild line element that  usual use:
$$ds^2={{\it dt}}^{2} \left( 1-{\frac {\alpha}{r}} \right) -{{\it dr}}^{2}
 \left( 1-{\frac {\alpha}{r}} \right) ^{-1}-{d\vartheta }^{2}{r}^{2}-{
d\varphi }^{2}{r}^{2} \left( \sin \left( \vartheta  \right)  \right) ^
{2}\tag 1
$$
The singularity of the line element is for $r=\alpha$
and  $det(G)=-{r}^{4} \left( \sin \left( \vartheta  \right)  \right) ^{2}$
with 
$R=\sqrt [3]{{r}^{3}+{\alpha}^{3}}$
$\quad\Rightarrow\quad$ $r=\sqrt [3]{-{\alpha}^{3}+{R}^{3}}$ and
$dr={\frac {{R}^{2}{\it dR}}{ \left( -{\alpha}^{3}+{R}^{3} \right) ^{2/3}}
}
$
you get for the  transformed line element equation (1)
$ds^2={{\it dt}}^{2} \left( 1-{\frac {\alpha}{\sqrt [3]{{\alpha}^{3}+{R}^{3
}}}} \right) -{R}^{4}{{\it dR}}^{2} \left( -{\alpha}^{3}+{R}^{3}
 \right) ^{-4/3} \left( 1-{\frac {\alpha}{\sqrt [3]{-{\alpha}^{3}+{R}^
{3}}}} \right) ^{-1}-{d\vartheta }^{2} \left( -{\alpha}^{3}+{R}^{3}
 \right) ^{2/3}-{d\varphi }^{2} \left( -{\alpha}^{3}+{R}^{3} \right) ^
{2/3} \left( \sin \left( \vartheta  \right)  \right) ^{2}
$
The singularity is now for $R=2^{1/3}\alpha$ and the determinant is unchanged.
so you get the same physical information from this line element. 
