534 DOCS.
512,
513
APRIL
1918
waited
with the
submission
until
you
had
reported to
me
your position
on
the
objection;
but
I
did
not
consider
myself
entitled
to
hold back
your
manuscript
for
so
long.
With
cordial
regards,
yours
very truly,
A.
Einstein.
513. To Hermann
Weyl
[Berlin,
19 April
1918]
Dear
Colleague,
My
third
message
since
yesterday![1]
There
is
an error
in
my
letter
of
yester-
day.
The indicated
solution
was
physically meaningless,
because
the
(positive)
constant
M
implies
a
negative
mass.[2]
Another
solution
that
is not
symmetric
around the
equator,[3]
without
a
point
mass
and
without
negative densities,
is
obtained
as
follows:
The
solution
is
in
three
parts
according
to
the
following
model:[4]
void
I
II
III
equator
The two
boundary
conditions are
I:
I-
~2
h2
16r
II:
_____
r
6
III:
-
1
-
2I~2+
h2
-
Mi
-~
=
~L1.
Ti
U
M
3
i
Choose
r1
and
u1.
Then
M
is
determined
by
the
first
equation.
If
one now
chooses
r2
r1,
u2 can
always
be chosen
(with a
positive
sign)
in such
a
way
that the
second
boundary
condition
is
also satisfied.
In
this
way we
have
a
non-“elliptic”
type of
solution
that
uses no negative masses.
Cordial
regards,
yours,
A.
Einstein.
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Extracted Text (may have errors)


534 DOCS.
512,
513
APRIL
1918
waited
with the
submission
until
you
had
reported to
me
your position
on
the
objection;
but
I
did
not
consider
myself
entitled
to
hold back
your
manuscript
for
so
long.
With
cordial
regards,
yours
very truly,
A.
Einstein.
513. To Hermann
Weyl
[Berlin,
19 April
1918]
Dear
Colleague,
My
third
message
since
yesterday![1]
There
is
an error
in
my
letter
of
yester-
day.
The indicated
solution
was
physically meaningless,
because
the
(positive)
constant
M
implies
a
negative
mass.[2]
Another
solution
that
is not
symmetric
around the
equator,[3]
without
a
point
mass
and
without
negative densities,
is
obtained
as
follows:
The
solution
is
in
three
parts
according
to
the
following
model:[4]
void
I
II
III
equator
The two
boundary
conditions are
I:
I-
~2
h2
16r
II:
_____
r
6
III:
-
1
-
2I~2+
h2
-
Mi
-~
=
~L1.
Ti
U
M
3
i
Choose
r1
and
u1.
Then
M
is
determined
by
the
first
equation.
If
one now
chooses
r2
r1,
u2 can
always
be chosen
(with a
positive
sign)
in such
a
way
that the
second
boundary
condition
is
also satisfied.
In
this
way we
have
a
non-“elliptic”
type of
solution
that
uses no negative masses.
Cordial
regards,
yours,
A.
Einstein.

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