91Ó°ÊÓ

Given data:

The ordinary speed is $u=\frac{2}{\sqrt{5}}c$.

The angle of the line in S is$\mathrm{θ}=45°$ .

(a)

Write the x-component of the ordinary velocity.

${\mathrm{Ï\dots }}_x=u\cos \mathrm{θ}$

Substitute $\frac{2}{\sqrt{5}}c$ for u and 45^o for $\mathrm{θ}$in the above equation.

$$\begin{gathered} {\mathrm{Ï\dots }}_x=\left(\frac{2}{\sqrt{5}}c\right)\cos \left(45°\right) \\ {\mathrm{Ï\dots }}_x=\left(\frac{2}{\sqrt{5}}c\right)\left(\frac{1}{\sqrt{2}}\right)×\left(\frac{\sqrt{2}}{\sqrt{2}}\right) \\ {\mathrm{Ï\dots }}_x=\frac{2\sqrt{2}}{2\sqrt{5}} \\ {\mathrm{Ï\dots }}_x=\sqrt{\frac{2}{5}}c \end{gathered}$$

Write the y-component of the ordinary velocity.

${\mathrm{Ï\dots }}_y=\mathrm{Ï\dots }\cos \mathrm{θ}$

Substitute $\frac{2}{\sqrt{5}}c$for $u$ and 45^o for $\mathrm{θ}$in the above equation.

$$\begin{gathered} {\mathrm{Ï\dots }}_y=\left(\frac{2}{\sqrt{5}}c\right)\sin \left(45°\right) \\ {\mathrm{Ï\dots }}_y=\left(\frac{2}{\sqrt{5}}c\right)\left(\frac{1}{\sqrt{2}}\right)×\left(\frac{\sqrt{2}}{\sqrt{2}}\right) \\ {\mathrm{Ï\dots }}_y=\frac{2\sqrt{2}}{2\sqrt{5}}c \\ {\mathrm{Ï\dots }}_y=\sqrt{\frac{2}{5}}c \end{gathered}$$

Therefore, the components ${\mathrm{Ï\dots }}_x$and ${\mathrm{Ï\dots }}_y$of the ordinary velocity are $\sqrt{\frac{2}{5}}c$.

(b)

Write the expression for the proper velocity.

$\mathrm{η}=\frac{u}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$

Write the x-component of the proper velocity.

${\mathrm{η}}_x=\frac{{\mathrm{Ï\dots }}_x}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$

Substitute $\sqrt{\frac{2}{5}}c$for $u_x$and $\frac{2}{\sqrt{5}}c$for $u$ in the above equation.

$$\begin{gathered} {\mathrm{η}}_x=\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{\sqrt{1-{\displaystyle \frac{{\left({\displaystyle \frac{2}{\sqrt{5}}}c\right)}^2}{c^2}}}} \\ =\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{\sqrt{1-{\displaystyle \frac{4}{5}}}} \\ =\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{\sqrt{{\displaystyle \frac{1}{5}}}} \\ =\sqrt{2}c \end{gathered}$$

Write the y-component of the proper velocity.

${\mathrm{η}}_y=\frac{{\mathrm{Ï\dots }}_y}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$

Substitute $\sqrt{\frac{2}{5}}c$for $u_y$and $\frac{2}{\sqrt{5}}c$for u in the above equation.

$$\begin{gathered} {\mathrm{η}}_y=\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{\sqrt{1-{\displaystyle \frac{{\left({\displaystyle \frac{2}{\sqrt{5}}}c\right)}^2}{c^2}}}} \\ {n}_y=\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{\sqrt{1-{\displaystyle \frac{4}{5}}}} \\ {n}_y=\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{\sqrt{{\displaystyle \frac{1}{5}}}}=\sqrt{2}c \end{gathered}$$

Therefore, the components of the proper velocity are$\sqrt{2}c$ .

(c)

Write the expression for the zeroth components of the 4-velocity.

${\mathrm{η}}^o=\frac{c}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}}$ …… (1)

It is known that:

$u=\sqrt{{{\mathrm{η}}_x}^2+{{\mathrm{η}}_y}^2}$ …… (2)

Substitute $\sqrt{2}c$for ${\mathrm{η}}_x$and $\mathrm{η}$in equation (2).

$$\begin{gathered} u=\sqrt{{\left(\sqrt{2c}\right)}^2+{\left(\sqrt{2c}\right)}^2} \\ =\sqrt{2c^2+2c^2} \\ =\sqrt{4}c \\ =2c \end{gathered}$$

Substitute $\sqrt{\frac{2}{5}}c$for $\mathrm{Ï\dots }$in equation (1).

$$\begin{gathered} \mathrm{η}°=\frac{c}{\sqrt{1-{\displaystyle \frac{{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}^2}{c^2}}}} \\ =\frac{c}{\sqrt{1-{\displaystyle \frac{4}{5}}}} \\ =\frac{c}{\sqrt{{\displaystyle \frac{1}{5}}}}=\sqrt{5}c \end{gathered}$$

Therefore, the zeroth component of the 4-velocity is $\mathrm{η}°=\sqrt{5}c$

(d)

Write the x-component of the ordinary velocity in .

Substitute for and in the above expression.

Write the y-component of the ordinary velocity in .

……. (3)

Here, is the Lorentz contraction which is given by,

Substitute in the above expression.

Substitute for, for and for in equation (3).

Therefore, the components and of the ordinary velocity in are and respectively.

(e)

Write the x-component of the proper velocity in .

$\stackrel{¯}{{\mathrm{η}}_x}=\mathrm{γ}\left({\mathrm{η}}_x-\mathrm{β}\mathrm{η}°\right)$ …… (4)

Here, $\mathrm{β}$is the Lorentz factor which is given by,

$\mathrm{β}=\frac{v}{c}$

Substitute $\mathrm{γ}=\sqrt{\frac{5}{3}},\sqrt{2}c,\mathrm{β}=\frac{v}{c},\mathrm{η}°=\sqrt{5}c$and $v=\sqrt{\frac{2}{5}}c$in equation (4).

$$\begin{gathered} \stackrel{¯}{{\mathrm{η}}_x}=\sqrt{\frac{5}{3}}\left(\sqrt{2}c-\frac{v}{c}\sqrt{5}c\right) \\ \stackrel{¯}{{\mathrm{η}}_x}=\sqrt{\frac{5}{3}}\left(\sqrt{2}c-\frac{\left(\sqrt{{\displaystyle \frac{2}{5}}}c\right)}{c}\sqrt{5}c\right) \\ \stackrel{¯}{{\mathrm{η}}_x}=\sqrt{\frac{5}{3}}\left(\sqrt{2}c-\sqrt{2}c\right)=0 \end{gathered}$$

Write the y-component of the ordinary velocity in $\stackrel{¯}{S}$.

$\stackrel{¯}{{\mathrm{η}}_x}=\stackrel{¯}{{\mathrm{η}}_y}$

Substitute $\sqrt{2}c$for ${\mathrm{η}}_y$in the above expression.

$\stackrel{¯}{{\mathrm{η}}_y}=\sqrt{2}c$

Therefore, the components $\stackrel{¯}{{\mathrm{η}}_x}$and $\stackrel{¯}{{\mathrm{η}}_y}$of the proper velocity in $\stackrel{¯}{S}$are 0 and $\sqrt{2}c$respectively.

Write the value of .

$\stackrel{¯}{u}=\sqrt{{\stackrel{¯}{u_x}}^2+{\stackrel{¯}{u_y}}^2}$

Substitute 0 for $\stackrel{¯}{u_x}$and $\sqrt{\frac{2}{3}}c$for $\stackrel{¯}{u_y}$in the above equation.

$$\begin{gathered} \stackrel{¯}{u}=\sqrt{{\left(0\right)}^2+{\left(\sqrt{\frac{2}{3}}c\right)}^2} \\ =\sqrt{\frac{2}{3}c} \end{gathered}$$

Calculate the R.H.S value.

$\frac{\left(\sqrt{{\displaystyle \frac{2}{3}}}c\right)}{\sqrt{1-{\displaystyle \frac{{\left(\sqrt{{\displaystyle \frac{2}{3}}}c\right)}^2}{c^2}}}}=\frac{\left(\sqrt{{\displaystyle \frac{2}{3}}}c\right)}{\sqrt{1-\left({\displaystyle \frac{2}{3}}\right)}}=\left(\sqrt{\frac{2}{3}}c\right)×\left(\sqrt{3}\right)=\sqrt{2}c$

Write the value of $\stackrel{¯}{\mathrm{η}}$.

Substitute 0 for$\stackrel{¯}{{\mathrm{η}}_x}$ and $\sqrt{2}c$for$\stackrel{¯}{{\mathrm{η}}_y}$ in the above equation.

$$\begin{gathered} \stackrel{¯}{\mathrm{η}}=\sqrt{{\left(0\right)}^2+{\left(\sqrt{2}c\right)}^2} \\ \stackrel{¯}{\mathrm{η}}=\sqrt{2}c \end{gathered}$$

Hence, L.H.S = R.H.S

Hence verified.

Therefore, it is proved that $\stackrel{¯}{\mathrm{η}}=\frac{\stackrel{¯}{u}}{\sqrt{1-{\displaystyle \frac{{\stackrel{¯}{u}}^2}{c^2}}}}$.