DOC.
188
FEBRUARY
1916
191
Meanwhile,
in order
to
acquaint myself
with
your energy tensor,
I
have been
working
on
the
incompressible liquid sphere problem
(energy
tensor
-p, -p, -p,
+Po).
I would
not
have done
so,
had
I
known
that
it would
cause me so
much
trouble.[6]
The
line element in
the
interior reads:
ds2
=
(-C-a°2-
C-a)2[da2
+
sin2a{dd2
+
sin2Mp2)\
d, p
are
the
normal
polar
coordinates,
o
would be related
to
the
radius vector
according
to:
(
^PO
2
o
9
/
1
.
.
I-J\
r
=
-
cos a0{(T-
~sin2a)-
-sinV,1.0
whereby a0
is
determined
by
the
sphere’s
radius
r0 according
to
3
/KPo\2
Q
9
.
1
.
.
1.0
(
-
I
ro =
-
cos
o-0(ö-o
-
2
sin
2c7o)
- ^
sin
ao-
For
pressure
p,
. .
/3
cos ao
-
cos
a\
(Po
+
p)
I-j-)
=
const.
applies.
The line element outside
is
the
old
one,
where
only
in
R3
=
r3 +
p
is
p
=
a3 not
valid,
but
is
instead:
cr
Kpo
a
-
m P
= r
(1-Z)
m
=
3Z-
1
2
sin0
ao
3
cos
uo
2 sin3
(j0
(cj0
- -
sin2j0).
The
strange thing is that,
for
a
finite
globe
of
coso0
=
1/3,
the
pressure
at
the
center
(o
= 0)
becomes
infinite;
smaller
spheres
of
a
given
mass are
not
possible.
The transition
to
an
infinitesimal
radius,
where
p
=
a3
must
then
result,
also
results in
physically meaningless
solutions-I
always
thought I
had
miscalculated
before
realizing
this.
Do
you
see
concretely
where the limit
cosa0
=
1/3
comes
from?
The
square
bracket
is
the
line element in
spherical geometry! Thus, spherical
geometry governs
within
the
sphere.
The interior of
the
sphere
is not
the entire
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