DOC.

185

JANUARY

1916 187

The left-hand

side reads

fully

as

1

9

(

ut

aß

(

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

aß

(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