# Time difference evaluation in a Michelson-Morley apparatus with arbitrary orientation

Historically, by assuming the validity of the Lorentz-Fitzgerald contraction, the null result obtained in the Michelson-Morley experiment can be explained. In particular light travels along the two arms of the apparatus (one positioned parallel to the direction of motion and one perpendicular) with a time difference given by: $$t_1-t_2= \frac{2(l_{10}-l_{20})/c}{(1-v^2/c^2)^{1 \over 2}}$$ Where $$l_{10}$$ and $$l_{20}$$ are the normal length of each arm i.e for the arms at rest. This relation is true also if we rotate the interferometer by 90° assuming the contraction. I want to prove that the relation holds for every angle of rotation. Assuming the contraction I reckon that the angle formed by an arm and the direction of $$\vec{v}$$ (the speed of the apparatus), measured with the system in motion and at rest would be different so we have the relations for each arm: $$(l_{10})^2=(l_1 \sin \theta)^2+\frac{(l_1 \cos \theta)^2}{1-v^2/c^2}$$ $$(l_{20})^2=(l_2 \cos \theta)^2 +\frac{(l_2 \sin \theta)^2}{1-v^2/c^2}$$ The first relation has been given by my textbook whereas the second has been derived accordingly by my self. I suppose that $$\theta$$ is the angle between the first arm and $$\vec{v}$$ with the system in motion. Using this relations I'm not able to obtain the time difference stated before. I think that light must travel with a certain angle different from $$\theta$$ in order for the ray to reach the mirror of the interferometer and bounce back compensating the motion of the system. For example, in the case of the first arm, taking the apparatus as our system of reference i think the speed will be, according to Galilean transformation: $$(v_c)^2=(c \cos \phi -v)^2+(c \sin \theta)^2$$ for the ray traveling towards the mirror and for the one that bounced back: $$(v_c)^2=(c \cos \phi +v)^2+(c \sin \theta)^2$$ Any hint or help in the solution process would be appreciated. Thank you in advance.