Chapter 2, Problem 5P

To determine

The time difference after rotating $90°$ to Michelson interferometer in Michelson Morley experiment.

Answer to Problem 5P

The time difference after rotating $90°$ to Michelson interferometer will be:

$\Delta t\prime =\frac{2}{c}\left(\frac{{\ell }_2}{1-v^2/c^2}-\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)$

Explanation of Solution

As from figure 2.2, total time ($t_1$) taken by the light to travel from source to target (From A to C) and from target to source (From C to A), velocity of the light from A to C is $c+v$ because ether is flowing along with light and from C to A is $c-v$ because ether is flowing opposite to light. Therefore,

$\begin{array}{l}{t}_1=\frac{{\ell }_1}{c+v}+\frac{{\ell }_1}{c-v}\\ =\frac{2c{\ell }_1}{c^2-v^2}\\ =\frac{2{\ell }_1}{c}\left(\frac{1}{1-v^2/c^2}\right)\end{array}$        (I)

Where, ${\ell }_1$ is the optical length from A to C, v is the velocity of the ether, and c is the velocity of the light.

As from figure 2.2, total time ($t_2$) taken by the light to travel from source to target (From A to D) and from target to source (From D to A).

When light travel from A to D, ether will deflect the direction of the light, thus light will not reach to the mirror $m_2$ at D. Therefore, by using the vector triangle, the speed of the light will be $\sqrt{c^2-v^2}$ to reach the mirror $m_2$ as shown in figure 1.

Therefore,

$\begin{array}{l}{t}_2=\frac{2{\ell }_2}{\sqrt{c^2-v^2}}\\ =\frac{2{\ell }_2}{c}\frac{1}{\sqrt{1-v^2/c^2}}\end{array}$        (II)

Where, ${\ell }_2$ is the optical length from A to D.

The time difference between two journey before rotating the Michelson interferometer is as follows:

$\Delta t=t_2-t_1$

Substituting the values of $t_1$ and $t_2$ from equation (I) and (II) in above equation. Therefore,

$\begin{array}{l}\Delta t=t_2-t_1\\ =\frac{2{\ell }_2}{c}\frac{1}{\sqrt{1-v^2/c^2}}-\frac{2{\ell }_1}{c}\left(\frac{1}{1-v^2/c^2}\right)\\ =\frac{2}{c}\left(\frac{{\ell }_2}{\sqrt{1-v^2/c^2}}-\frac{{\ell }_1}{1-v^2/c^2}\right)\end{array}$        (III)

After rotating $90°$ to Michelson interferometer the optical length l₂ and l₁ will be interchanged and rest of the quantities will be same and direction of the flow of the ether will be in the direction of the l₂.

Therefore, from Galilean transformation velocity of the light along the l₂ (from A to D)  will be $c+v$ and for return trip(from D to A) will be $c-v$. Therefore, total time taken by light to complete the round-trip journey from A to D is $t{\prime }_2$

$\begin{array}{l}t{\text{'}}_2=\frac{{\ell }_2}{c+v}+\frac{{\ell }_2}{c-v}\\ =\frac{2c{\ell }_2}{c^2-v^2}\\ =\frac{2{\ell }_2}{c}\left(\frac{1}{1-v^2/c^2}\right)\end{array}$        (IV)

When light travel from A to C, ether will deflect the direction of the light, thus light will not reach to the mirror $m_1$ at C.

Therefore, by using the vector triangle, the speed of the light will be $\sqrt{c^2-v^2}$ to reach the mirror $m_1$ at C as shown in figure 2:

Thus, total time taken by light to complete the round-trip journey from A to C is $t{\prime }_1$.

$\begin{array}{l}t{\text{'}}_1=\frac{2{\ell }_1}{\sqrt{c^2-v^2}}\\ =\frac{2{\ell }_1}{c}\frac{1}{\sqrt{1-v^2/c^2}}\end{array}$        (V)

From equation (IV) and (V) time difference between the two journey is $\Delta t\text{'}$

$\begin{array}{l}\Delta t\prime =t_2^{\prime }-t_1^{\prime }\\ =\frac{2{\ell }_2}{c}\left(\frac{1}{1-v^2/c^2}\right)-\frac{2{\ell }_1}{c}\left(\frac{1}{\sqrt{1-v^2/c^2}}\right)\\ =\frac{2}{c}\left(\frac{{\ell }_2}{1-v^2/c^2}-\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)\end{array}$

Conclusion:

Therefore, the time difference after rotating $90°$ to Michelson interferometer will be:

$\Delta t\prime =\frac{2}{c}\left(\frac{{\ell }_2}{1-v^2/c^2}-\frac{{\ell }_1}{\sqrt{1-v^2/c^2}}\right)$.