Conference on the Michelson-Morley experiment held at the Mount Wilson observatory Pasadena, California February 4 and 5, 1927

CONFERENCE ON MICHELSON-MORLEY EXPERIMENT 377

Fig. 14

the direction 9) while the portion of the wave front at E advances to M\ then BM^r is the position of the equivalent fixed mirror. Denote the angle MBM^! by p; then

GM'

where GM' is perpendicular to BM.

GM' — MM' sin 2a = r sin 0; BG—BM-\-MM' cos 2a;

EM r

cos 2a ^{; P} EM+MM' *

378 E. R. HEDRICK

Therefore

_ r sin 0 _ rß sin 0

^P~BM+MM' cos ₂a~ EMß , '

¹ MM’ß cos 2a

cos 2 a

rß sin 0 cos 2a

r—rß sin 0 sin 2a * But we have tan a = i — £; hence, to terms of the second order,

. £²

sin a=i--, cos 2a = H—.

2 2

Substituting these values in the expression for tan p, we have tan p = ß² sin 0 (cos 0 —sin 0) , q.p.

Now if we denote the angle CA^fT by <£ and the angle CA^rT^r by \f/, we have (remembering that 0 and yf/ are negative angles)

<£+p=2a —p or $=2(p— a)

^ = 27 — 180° .

Thus the positive angle

T'A'T = (j) — \f/=2p— 2a— 27+180° .

To determine the tangent of this angle, we find

tan ( — 2a) = ^

tan (27 — 180°)

i-(i+£)²’

₌ 2(i+0

i — (i+£)² ’

and therefore

*£2

tan ( — 2a — 27+180°) =-— q.p.

4 — f⁴

From this we obtain

* /, /\ £²+²ß² sin 0 (cos 0 —sin 0)

i — 2ß²^² sin 0 (cos 0 — sin 0) ^?

since tan 2p = 2ß² sin 0(cos 0 —sin 0) to terms of the second order. Substituting for £ and reducing, we have finally

tan (cj)—\p)=ß² cos 20 .