DOC.
382
SEPTEMBER
1917 379
For
reasons
of
symmetry,
we
have
U2
=
U3
=
0.
(21a)
At
an
arbitrary point
(21)
and
(21a)
give
Ua
=
axa/r2r
(a
=
1, 2, 3).
From
this
it
follows
divSl
=
0,
(22)
and
because outside
of
the
body
E
=
0 as well,
(1)
yields
W*
=
0
and
finally, (10)
yields
t44
=
0.
(23)
The
gravitational
field
therefore would have
no
energy
outside
of
the
body,
and
this
does not
agree
with
your equation
(11)
in
“Approximative
Integration of
the
Field
Equations.”
Continuation
[Fortsetzung],
28
Sept.
I
had
just
written
all this down when
this
idea shot
through my
head: The
entire
inconsistency
could
lie
in
the
fact
that
we are
using
different coordinate
systems.
I
also
think that this
is
so, even
though I
don’t
obtain
exactly
the
same
result
as
you
for
your
coordinate
system.
As
you
know,
your
coordinate
system
is
characterized
by
the
velocity
of
light being equal
in all
directions,[24]
and
this
requirement agrees
with
p
=
u
in
(16)
and
(16a).
The
gravitational field
in such
a
coordinate
system
is
also examined
by
Droste
(dissertation,
p.
20).
Droste’s
formula
(31) gives
p
=
u=(
1
+
fV
w2
-
1------~2.
(24)
v
irJ
r(1
+
*)
So
this
takes
the
place of
(17).
For
U,
at
a
point
where
x2
=
x3
=
0,
because
p
=
u,
(20)
initially produces
U1
=
-4wp'
-
2w'p.
(25)
In order
to
achieve
the
same
degree
of
accuracy
as
in
your paper,
V
,
0L
1
+
2?
1
“
w
=
1--
2
r
must here be inserted for
p
and
w.
For
p'
and
w',
the
expressions
obtained
from
this
through
differentiation
may
not
be
used,
however,
but rather the
following
more
precise
ones:
p'=a/2r2
(1+a/4r)
w'=
a/2r2
(1-a/2r)