Chapter 25, Problem 41QAP

To determine

(a)

How much time would light, moving at speed c, need to travel from the partially silvered plate to one mirror and back again, suppose that the distance in the Michelson-Morley experiment from the partially silvered plate to either mirror is 20,000 m?

Explanation of Solution

Given info:

Formula used:

Calculation:

The Michelson−Morley experiment attempted to measure a difference between light travel times in two perpendicular legs of an interferometer: One leg was situated parallel to the ether wind and the other perpendicular to the ether wind. For the "parallel leg" of the interferometer, we need to calculate the speed of the light traveling with the wind and against the wind; the round-trip time on this leg is equal to the sum of the times of each one-way trip. For the "perpendicular leg," we need to calculate the component of the speed of light that lies along this axis; this speed will be the same for both portions of the one-way trip. Subtracting these two expressions for the travel times-the times for the "parallel leg" and the "perpendicular leg"-will give us an expression for the time difference between the legs in terms of the speed of the ether wind. Using this, we can calculate the time difference if the speed of the ether wind were equal to the orbital speed of Earth around the Sun, 0.01c, 0.1c, 0.5c, and 0.9c.
Parallel time:

$t_{parallel}=\frac{L}{c+v}+\frac{L}{c-v}=\frac{L(c-v+c+v)}{(c+v)(c-v)}=\frac{2Lc}{c^2-v^2}=\frac{2L}{c\left(1-{\left(\frac{v}{c}\right)}^2\right)}=\frac{2L{\gamma }^2}{c}$

Perpendicular time:

$t_{peprpendicular}=\frac{L}{\sqrt{c^2-v^2}}+\frac{L}{\sqrt{c^2-v^2}}=\frac{2L}{\sqrt{c^2-v^2}}=\frac{2L}{c\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{2L\gamma }{c}$

Time difference between legs:

$\begin{array}{l}{t}_{parallel}-t_{peprpendicular}=\frac{2L{\gamma }^2}{c}-\frac{2L\gamma }{c}=\frac{2L\gamma }{c}(\gamma -1)=\frac{2\times 20000}{3\times {10}^8}\gamma (\gamma -1)\\ t_{parallel}-t_{peprpendicular}=133\gamma (\gamma -1)\mu s\end{array}$

Speed:

$v=\frac{2\pi \times 1.5\times {10}^{11}}{365\times 24\times 60\times 60}=2.99\times {10}^{4}m/s$

Gamma:

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{2.99\times {10}^4}{3\times {10}^8}\right)}^2}}=\text{1}\text{.000000005}$

Time difference between legs:

$t_{parallel}-t_{peprpendicular}=133\times \text{1}\text{.000000005}\times (\text{1}\text{.000000005}-1)=7\times {10}^{-7}\mu s$

Conclusion:

Time difference between legs $=7\times {10}^{-7}\mu s$

To determine

(b)

How much additional time would be required if the ether existed, if the Galilean velocity transformation were valid, and if Earth moved relative to the ether along the direction from the partially silvered plate to this mirror at speed at 0.01c?

Explanation of Solution

Given info:

Formula used:

Calculation:

Gamma:

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{0.01c}{c}\right)}^2}}=\text{1}\text{.00005}$

Time difference between legs:

$t_{parallel}-t_{peprpendicular}=133\times \text{1}\text{.00005}\times (\text{1}\text{.00005}-1)=7\times {10}^{-3}\mu s$

Conclusion:

Time difference between legs $=7\times {10}^{-3}\mu s$

To determine

(c)

How much additional time would be required if the ether existed, if the Galilean velocity transformation were valid, and if Earth moved relative to the ether along the direction from the partially silvered plate to this mirror at speed at 0.1c?

Explanation of Solution

Given info:

Formula used:

Calculation:

Gamma:

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{0.1c}{c}\right)}^2}}=\text{1}\text{.005}$

Time difference between legs:

$t_{parallel}-t_{peprpendicular}=133\times \text{1}\text{.005}\times (\text{1}\text{.005}-1)=0.7\mu s$

Conclusion:

Time difference between legs $=0.7\mu s$

To determine

(d)

How much additional time would be required if the ether existed, if the Galilean velocity transformation were valid, and if Earth moved relative to the ether along the direction from the partially silvered plate to this mirror at speed at 0.5c?

Explanation of Solution

Given info:

Formula used:

Calculation:

Gamma:

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{0.5c}{c}\right)}^2}}=\text{1}\text{.155}$

Time difference between legs:

$t_{parallel}-t_{peprpendicular}=133\times \text{1}\text{.155}\times (\text{1}\text{.155}-1)=20\mu s$

Conclusion:

Time difference between legs $=20\mu s$

To determine

(e)

How much additional time would be required if the ether existed, if the Galilean velocity transformation were valid, and if Earth moved relative to the ether along the direction from the partially silvered plate to this mirror at speed at 0.9c?

Explanation of Solution

Given info:

Formula used:

Calculation:

Gamma:

$\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}=\frac{1}{\sqrt{1-{\left(\frac{0.9c}{c}\right)}^2}}=\text{2}\text{.294}$

Time difference between legs:

$t_{parallel}-t_{peprpendicular}=133\times \text{2}\text{.294}\times (\text{2}\text{.294}-1)=400\mu s$

Conclusion:

Time difference between legs $=400\mu s$