Question 6.TC.3: Light Deflection by a Moving Mirror Reflection of light by a......
Light Deflection by a Moving Mirror
Reflection of light by a relativistically moving mirror is not theoretically new. Einstein discussed the possibility or worked out the process using the Lorentz transformation to get the reflection formula due to a mirror moving with a velocity ν. This formula, however, could also be derived by using a relatively simpler method. Consider the reflection process as shown in Fig. 6 – 8, where a plane mirror M moves with a velocity ν = ve_{x} (where e_{x}is a unit vector in the x-direction) observed from the lab frame F. The mirror forms an angle Φ with respect to the velocity (note that Φ ≤90°, see Fig. 6 – 8). The plane of the mirror has n as its normal. The light beam has an incident angle α and reflection angle β which are the angles between n and the incident beam 1 and reflection beam 1′ ,respectively in the laboratory frame F. It can be shown that
\sin\alpha-\sin\beta={\frac{v}{c}}\sin\phi\sin(\alpha+\beta). (1)
3A. Einstein’s Mirror
About a century ago Einstein derived the law of reflection of an electromagnetic wave by a mirror moving with a constant velocity ν = – ve_{x} (see Fig. 6 – 9). By applying the Lorentz transformation to the result obtained in the rest frame of the mirror, Einstein found that:
\cos\beta=\frac{\left[1+\left(\frac{v}{c}\right)^{2}\right]\cos\alpha-2\,\frac{v}{c}}{1-2\,\frac{v}{c}\cos\alpha+\left(\frac{v}{c}\right)^{2}}. (2)
Derive this formula using Eq. (1) without Lorentz transformation!
3B. Frequency Shift
In the same situation as in 3A, if the incident light is a monochromatic beam hitting M with a frequency f, find the new frequency f’ after it is reflected from the surface of the moving mirror. If α = 30° and v = 0.6 c in Fig. 6 – 9, find frequency shift Δf in percentage of f.
3C. Moving Mirror Equation
Fig. 6 – 10 shows the positions of the mirror at time t_{0} and t. Since the observer is moving to the left, the mirror moves relatively to the right. Light beam 1 falls on point a at t_{0} and is reflected as beam 1′. Light beam 2 falls on point d at t and is reflected as beam 2′. Therefore, \overline{ab} is the wave front of the incoming light at time t_{0}. The atoms at point are disturbed by the incident wave front \overline{ab} and begin to radiate a wavelet. The disturbance due to the wave front \overline{ab} stops at time t when the wavefront strikes point d.
By referring to Fig. 6 – 10 for light wave propagation or using other methods, derive Eq. (1) .
3A. Einstein’s Mirror
By taking \phi={\frac{\pi}{2}} and replacing v with – v in Eq. (1), we obtain
\sin\alpha-\sin\beta=-{\frac{v}{c}}\sin(\alpha+\beta). (3)
This equation can also be written in the form of
\left(1+{\frac{v}{c}}\cos\beta\right)\sin\alpha=\left(1-{\frac{v}{c}}\cos\alpha\right)\sin\beta. (4)
The square of this equation can be written in terms of a squared equation of cos β, as follows,
\left(1-2\,{\frac{v}{c}}\cos\alpha+{\frac{v^{2}}{c^{2}}}\right)\cos^{2}\!\beta+2\,{\frac{v}{c}}\,(1-\cos^{2}\alpha)\cos\beta
+2\,\frac{v}{c}\cos{\alpha}-\left(1+\frac{v^{2}}{c^{2}}\right)\cos^{2}\!\alpha=0\,, (5)
which has two solutions,
(\cos\beta)_{1}={\frac{2\,{\frac{v}{c}}\cos^{2}\!\alpha-\left(1+{\frac{v^{2}}{c^{2}}}\right)\cos\alpha}{1-2\,{\frac{v}{c}}\cos\alpha+{\frac{v^{2}}{c^{2}}}}} (6)
and
(\cos\beta)_{2}={\frac{-2\,{\frac{v}{c}}+\left(1+{\frac{v^{2}}{c^{2}}}\right)\cos\alpha}{1-2\,{\frac{v}{c}}\cos\alpha+{\frac{v^{2}}{c^{2}}}}}. (7)
However, if the mirror is at rest (v = 0) then cos α = cos β; therefore the proper solution is
\cos\beta_{2}=\frac{-2\,\frac{v}{c}+\left(1+\frac{v^{2}}{c^{2}}\right)\cos\alpha}{1-2\,\frac{v}{c}\cos\alpha+\frac{v^{2}}{c^{2}}}. (8)
3B. Frequency Shift
The reflection phenomenon can be considered as a collision of the mirror with a beam of photons each carrying an incident and reflected momentum of magnitude
{p}_{f}=\frac{h f}{c}\ \mathrm{and}\ {p}_{f}^{\ ^{\prime}}=\frac{h f^{\prime}}{c}. (9)
The conservation of linear momentum during its reflection from the mirror for the component parallel to the mirror appears as
p_{f}\sin\alpha={p_{f}}^{\prime}\sin\beta\ \mathrm{or}\ f^{\prime}\sin\beta =f^{\prime}\,\frac{\left(1-\frac{v^{2}}{c^{2}}\right)\mathrm{sin}\,\alpha}{\left(1+\frac{v^{2}}{c^{2}}\right)-2\,\frac{v}{c}\mathrm{cos}\,\alpha}= f sinα. (10)
Thus
f^{\prime}=\frac{\left(1+\frac{v^{2}}{c^{2}}\right)-2\,\frac{v}{c}\cos\alpha}{\left(1-\frac{v^{2}}{c^{2}}\right)}f. (11)
For α = 30° and v = 0.6 c,
\cos\alpha=\frac{1}{2}\sqrt{3}\,,\,\,1-\frac{v^{2}}{c^{2}}=0.64,\,\,1+\frac{v^{2}}{c^{2}}=1.36, (12)
so that
{\frac{f^{\prime}}{f}}={\frac{1.36-0.6{\sqrt{3}}}{0.64}}=0.5. (13)
Thus, there is a decrease of frequency by 50% due to reflection by the moving mirror.
3C. Relativistically Moving Mirror Equation
Fig. 6 – 10 shows the positions of the mirror at time t_{0} and t. Since the observer is moving to the left, system is moving relatively to the right. Light beam 1 falls on point a at t_{0} and is reflected as beam 1′. Light beam 2 falls on point d at t and is reflected as beam 2′. Therefore, \overline{ab} is the wave front of the incoming light at time t_{0}. The atoms at point are disturbed by the incident wave front \overline{ab} and begin to radiate a wavelet. The disturbance due to the wave front \overline{ab} stops at time t when the wavefront strikes point d. As a consequence
\overline{{{a c}}}=\overline{{{b d}}}=c(t-t_{0}). (14)
From this figure we also have {\overline{{e d}}}={\overline{{a g}}}\,, and
\sin\alpha=\frac{\overline{b d}+\overline{d g}}{\overline{a g}},\,\sin\beta=\frac{\overline{a c}-\overline{a f}}{\overline{a g}-\overline{e f}}. (15)
Fig. 6 – 11 displays the beam path 1 in more detail. From this figure it is easy to show that
{\overline{{d g}}}={\overline{{a e}}}={\frac{{\overline{{a o}}}}{\cos\alpha}}={\frac{v(t-t_{0})\sin\phi}{\cos\alpha}} (16)
and
\overline{{{a f}}}=\frac{\overline{{{a o}}}}{\cos{\beta}}=\frac{v(t-t_{0})\sin{\phi}}{\cos{\beta}}. (17)
From the triangles aeo and afo we have \overline{e o}=\overline{{{ao}}} tan α and \overline{o f}=\overline{{{a o}}}\tan\beta. Since \overline{{{e f}}}=\overline{{{eo}}}+o f, then
\overline{{{e f}}}=v(t-t_{0})\sin\phi(\tan\alpha+\tan\beta). (18)
By substituting Eq. (14), (16), (17), and (18) into Eq. (15) we obtain
\sin\alpha=\frac{c+v\displaystyle\frac{\sin\phi}{\cos\alpha}}{\displaystyle\frac{\overline{ag}}{t-t_{0}}}, (19)
and
\sin\beta={\frac{c-v{\frac{\sin\phi}{\cos\beta}}}{{\frac{{\overline{{{a g}}}}}{t-t_0}-v\mathrm{sin}\;\phi\;(\tan\alpha+\tan\beta)}}}. (20)
Eliminating \frac{\overline{ag}}{t-t_{0}} from the two equations above leads to
v\sin\phi(\tan\alpha+\tan\beta)= c\left(\frac{1}{\sin\alpha}-\frac{1}{\sin\beta}\right)+v\sin\phi{\Bigl(}\frac{1}{\sin \alpha\cos \alpha}+\frac{1}{\sin\beta\cos\beta}\Bigr). (21)
By collecting the terms containing v sin Φ, we obtain
{\frac{v}{c}}\sin\phi{\Big(}{\frac{\cos\alpha}{\sin\alpha}}+{\frac{\cos\beta}{\sin\beta}}{\Big)}={\frac{\sin\alpha-\sin\beta}{\sin\alpha\sin\beta}} (22)
or
\sin\alpha-\sin\beta={\frac{v}{c}}\sin\phi\;\sin(\alpha+\beta). (23)
