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whose elasticity is the same, and which differ only in their densities, the squares of the velocities of propagation are inversely proportional to these densities; that is,

${\frac {D'}{D}}={\frac {v^{2}}{v'^{2'}}}$

D and D' being the densities of the æther in a vacuum and in the body, and $v,v'$ the corresponding velocities. From the above we easily deduce the relations

$D'=D{\frac {v^{2}}{v'^{2}}},\ D'-D=D{\frac {v^{2}-v'^{2}}{v'^{2}}},$

the latter of which gives the excess of density of the interior æther.

It is assumed that when the body is put in motion, only a part of the interior æther is carried along with it, and that this part is that which causes the excess in the density of the interior over that of the surrounding æther; so that the density of this moveable part is D'-D. The other part which remains at rest during; the body's motion has the density D.

The question now arises, With what velocity will the waves be propagated in a medium thus constituted of an immoveable and a moveable part, when for the sake of simplicity we suppose the body to be moving in the direction of the propagation of the waves?

Fresnel considers that the velocity with which the waves are propagated then becomes increased by the velocity of the centre of gravity of the stationary and moving portions of æther. Now $u$ being the velocity of the body,

${\tfrac {D'-D}{D'}}u$

will be the velocity of the centre of gravity of the system in question, and according to the last formula this expression is equal to

${\frac {v^{2}-v'^{2}}{v{}^{2}}}u$

Such, then, is the quantity by which the velocity of light will be augmented; and since $v'$ is the velocity when the body is at rest,

$v'+{\frac {v^{2}-v'^{2}}{v{}^{2}}}u$ and $v'-{\frac {v^{2}-v'^{2}}{v{}^{2}}}u$

will be the respective velocities when the body moves with and against the light.

By means of these expressions the corresponding displacement of the bands in our experiment may be calculated in exactly the