91Ó°ÊÓ

The ratio of the maximum value of the electric field to the maximum value of the magnetic field provides the velocity of the electromagnetic waves.

Mathematically, it can be expressed as follows,

$c=\frac{E_o}{B_o}$

Here$E_o$ and $B_o$is the maximum value of the electric and magnetic field,$c$ is velocity of the electromagnetic waves.

(a)

The plane wave (Electric field wave) is polarized and it’s electric field is expressed as,

$E(x,y,z,t)=E_o\cos (kx−\mathrm{Ó\neg }t)\stackrel{âˆ\S }{y}$

Here$k$ is equal to $k=E_o×B_o$,$k$ is called as propagation vector.

As,

$c=\frac{E_o}{B_o}$

$⇒B_o=\frac{E_o}{c}$

So, the magnetic field wave is,

$B(fx,y,z,t)=\frac{E_o}{c}\cos (kx−\mathrm{Ó\neg }t)\stackrel{âˆ\S }{z}$

(b)

The relativistic equations are given by,

${\stackrel{→}{E}}_x=E_x$

${\stackrel{→}{E}}_y=\mathrm{γ}(E_y−v{B}_z)$…… (1)

Here, $x$is the position of the event in frame $S$,$v$is the velocity of frame$S^{\text{'}}$relative to$S$, $c$is the speed of light in vacuum,$\mathrm{γ}$ is the Lorentz factor for a velocity $v$.

As, $B_o=\frac{E_o}{c}$, the magnetic field equation can be written as,

${\stackrel{→}{B}}_x=B_x$

${\stackrel{→}{B}}_y=\mathrm{γ}(B_y+\frac{v}{c^2}{E}_z)$

Substitute $E_o\cos (kz−\mathrm{Ó\neg }t)$for $E_y$and $\frac{E_o}{c}\cos (kz−\mathrm{Ó\neg }t)$for $B_z$in the equation (1).

role="math" localid="1658135705784" $\begin{array}{c}{\stackrel{→}{E}}_y=\mathrm{γ}(E_o\cos (kz−\mathrm{Ó\neg }t)−v\frac{E_o}{c}\cos (kz−\mathrm{Ó\neg }t))\\ =E_o\cos (kz−\mathrm{Ó\neg }t)\mathrm{γ}(1−\frac{v}{c})\end{array}$

$\mathrm{Î\pm }=\mathrm{γ}(1−\frac{v}{c})$ ……. (2)

Then,

$\stackrel{→}{E_y}=\mathrm{Î\pm }{E}_o\cos (kz−\mathrm{Ó\neg }t)$

Using the special theory of relativity, the expression for the Lorentz factor is given by,

$\mathrm{γ}=\frac{1}{\sqrt{1−{\left(\frac{v}{c}\right)}^2}}$ ……. (3)

Here$v$is the speed of the object and$c$is the speed of the light.

From equation (2)

$\mathrm{Î\pm }=\mathrm{γ}(1−\frac{v}{c})$

Substitute $\frac{1}{\sqrt{1−{\left(\frac{v}{c}\right)}^2}}$for$\mathrm{γ}$in the above equation,

$\mathrm{Î\pm }=\frac{1}{\sqrt{1−{\left(\frac{v}{c}\right)}^2}}(1−\frac{v}{c})$

Simplify the above equation,

$\begin{array}{c}\mathrm{Î\pm }=\frac{1}{\sqrt{(1−\frac{v}{c})(1+\frac{v}{c})}}(1−\frac{v}{c})\\ =\frac{\sqrt{1−\frac{v}{c}}}{\sqrt{1+\frac{v}{c}}}\end{array}$

The expression for the time in frame can be expressed as,

$t^{\text{'}}=\mathrm{γ}(t−\frac{v}{c^2}x)$

As, $x=\mathrm{γ}(\stackrel{¯}{x}+vt)$ and$t^{\text{'}}=\mathrm{γ}(\stackrel{¯}{t}+\frac{v}{c^2}\stackrel{¯}{x})$

Then,

$kx−\mathrm{Ó\neg }t=\mathrm{γ}(\stackrel{¯}{x}+vt)−\mathrm{Ó\neg }\left[\mathrm{γ}(t+\frac{v}{c^2}\stackrel{¯}{x})\right]$

$\begin{array}{c}kx−\mathrm{Ó\neg }t=\mathrm{γ}(k−\frac{\mathrm{Ó\neg }v}{c^2})\stackrel{¯}{x}−\left[(\mathrm{Ó\neg }−kv)\stackrel{¯}{t}\right]\\ =\stackrel{¯}{k}\stackrel{¯}{x}−\stackrel{¯}{\mathrm{Ó\neg }}t\end{array}$

Here,

$\begin{array}{c}\stackrel{¯}{x}=\mathrm{γ}(k−\frac{\mathrm{Ó\neg }v}{c^2})\\ =\mathrm{γ}k(1−\frac{v}{c})\\ =\mathrm{Î\pm }k\end{array}$

And

$\begin{array}{c}\stackrel{¯}{\mathrm{Ó\neg }}=\mathrm{γ}\mathrm{Ó\neg }(1−\frac{v}{c})\\ =\mathrm{Î\pm }\mathrm{Ó\neg }\end{array}$

Therefore the equation of electric filed is $\stackrel{¯}{E}(\stackrel{¯}{x},\stackrel{¯}{y},\stackrel{¯}{z},\stackrel{¯}{t})={\stackrel{¯}{E}}_o\cos (\stackrel{¯}{k}\stackrel{¯}{x}−\stackrel{¯}{\mathrm{Ó\neg }}\stackrel{¯}{t})\stackrel{âˆ\S }{y}$.

Therefore the equation of magnetic field is $\stackrel{¯}{B}(\stackrel{¯}{x},\stackrel{¯}{y},\stackrel{¯}{z},\stackrel{¯}{t})=\frac{\stackrel{¯}{E_o}}{c}\cos (\stackrel{¯}{k}\stackrel{¯}{x}−\stackrel{¯}{\mathrm{Ó\neg }}\stackrel{¯}{t})\stackrel{âˆ\S }{z}$.

The expression for the relativistic electric field is ${\stackrel{¯}{E}}_o=\mathrm{Î\pm }{E}_o$.

The propagation constant is $\stackrel{¯}{k}=\mathrm{Î\pm }k$.

The angular velocity is $\stackrel{¯}{\mathrm{Ó\neg }}=\mathrm{Î\pm }\mathrm{Ó\neg }$.

The constant $\mathrm{Î\pm }=\sqrt{\frac{1−\frac{v}{c}}{1+\frac{v}{c}}}$.

(c)

The expression for the frequency of the wave in$\stackrel{→}{S}$ frame is,

$\stackrel{¯}{\mathrm{Ó\neg }}=\mathrm{Ó\neg }\sqrt{\frac{1−\frac{v}{c}}{1+\frac{v}{c}}}$

The expression for the wavelength of the wave in $\stackrel{→}{S}$frame is,

$\begin{array}{c}\stackrel{¯}{\mathrm{λ}}=\frac{2\mathrm{Ï€}}{\stackrel{¯}{k}}\\ =\frac{2\mathrm{Ï€}}{\mathrm{Î\pm }k}\\ =\frac{\mathrm{λ}}{\mathrm{Î\pm }}\end{array}$

The expression for velocity of wave in$\stackrel{→}{S}$frame is,

$\begin{array}{c}\stackrel{¯}{\mathrm{λ}}=\frac{\stackrel{¯}{\mathrm{Ó\neg }}}{2\mathrm{Ï€}}\stackrel{¯}{\mathrm{λ}}\\ =\frac{\stackrel{¯}{\mathrm{Ó\neg }}}{2\mathrm{Ï€}}\left(\frac{\mathrm{λ}}{\mathrm{Î\pm }}\right)\\ =\left(\frac{\mathrm{Ó\neg }}{2\mathrm{Ï€}}\right)\mathrm{λ}\\ =c\end{array}$

Thus the velocity of light is same in all inertial frames of references.

(d)

As the intensity is directly proportional to the square of the amplitude,

Then the ratio of the intensity is,

$\begin{array}{c}\frac{\stackrel{¯}{I}}{I}=\frac{{\stackrel{¯}{E}}_o^2}{E_o^2}\\ ={\mathrm{Î\pm }}^2\\ =\frac{1−\frac{v}{c}}{1+\frac{v}{c}}\end{array}$

Therefore the ratio of the intensity is $\frac{1−\frac{v}{c}}{1+\frac{v}{c}}$.

Thus amplitude, frequency and intensity of the light wave all reduce to zero as we move faster and faster.

It will get so faint thus we won’t be able to see it, and it becomes red-shifted. Even your light vision goggles won’t help. But the light is moving relative to you.