DOC.
185
JANUARY
1916 187
The left-hand
side reads
fully
as
1
9
(
ut

(
dffa
,
d9ßr
dg"T\
\
2 dxa
\
V
dxT
dxv
dxß)j
.
The third
term
yields nothing,
because
gVT
dgvt/dxB
=
dlgg/dxB
=
0.
The first two
coincide
through
an
exchange
of
v
and
r,
such
that
they
can
be combined.
By
applying
(ß')
you
obtain
finally
d2gaT
dxadxT
From
(9)
thus
results
d2gaß
dxadxß
k(T T
i)
=
0...
(9a)
We execute
the
operation
d/dxv
on
(8)
and
allowing
for
(4)
obtain[14]
IdjT
+
t)
dxadxu \
{
a
j)
2
&xr
The left-hand
side
is
more
completely
(10)
__(
tv

(dJoß_
dgrß
_
dgaT\
2 dxa dxv
\ \
dxT
dxa
dxß
J
If
you exchange
in
the
first
term
a
and
v
as
well
as
ß
and
r,
then
disregarding
the
index,
it coincides with the
third.
Only
the
second
remains,
which
allowing
for
(ß1)
changes
into
1
d3goa/
2dxadxvdxa
That
is
why you
obtain instead
of
(10)[15]
d
(
d2gav
dxa
\dxadxv
(T +
i)
=
-2KAa...
(10a)
From
(9a)
and
(10a) Aa
=
0 results,
that
is,
according
to
(3)
of
the
conservation
law
of
matter,
as a
consequence
of
the
field
equations (2).-
You will
certainly
not encounter
any
more
problems
now.
Show
this
thing
to
Lorentz
as well,
who also does not
yet perceive
the
need for
the
structure
on
the
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