In Fig. 37-11, frame S^' moves relative to frame S with velocity $\mathbf{0}\mathbf{.}\mathbf{62}\mathbf{c}\hat{\mathbf{i}}$ while a particle moves parallel to the common x and x^' axes. An observer attached to frame S^' measures the particle’s velocity to be $\mathbf{0}\mathbf{.}\mathbf{47}\mathbf{c}\hat{\mathbf{i}}$. In terms of c, what is the particle’s velocity as measured by an observer attached to frame S according to the (a) relativistic and (b) classical velocity transformation? Suppose, instead, that the S^' measure of the particle’s velocity is $\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{47}\mathbf{c}\hat{\mathbf{i}}$. What velocity does the observer in Snow measure according to the (c) relativistic and (d) classical velocity transformation?

The relativistic speed observed from frame S is given by,

$\mathbf{v}\mathbf{=}\frac{\mathbf{u}\mathbf{+}{\mathbf{v}}^{\mathbf{\text{'}}}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{{\mathbf{uv}}^{\mathbf{\text{'}}}}{{\mathbf{c}}^{\mathbf{2}}}}}$ ……. (1)

Here, the speed of the light is c.

(a)

Substitute $0.47\mathrm{c}\hat{\mathrm{i}}$ for v^', and $0.62\mathrm{c}\hat{\mathrm{i}}$for u in equation (1).

$\begin{array}{rcl}\mathrm{v}& =& \frac{\left(0.47\mathrm{c}\hat{\mathrm{i}}\right)+\left(0.62\mathrm{c}\hat{\mathrm{i}}\right)}{1+{\displaystyle \frac{\left(0.47\mathrm{c}\hat{\mathrm{i}}\right)\left(0.62\mathrm{c}\hat{\mathrm{i}}\right)}{{\mathrm{c}}^2}}}\\ & =& \frac{0.47\mathrm{c}+0.62\mathrm{c}}{1+\left(0.47\mathrm{c}\right)\left(0.62\mathrm{c}\right)}\\ & =& 0.84\mathrm{c}\hat{\mathrm{i}}\end{array}$

Therefore, the velocity of the particle according to the relativistic is $0.84\mathrm{c}\hat{\mathrm{i}}$.

(b)

In the classical way, the velocity is calculated as follows.

$\begin{array}{rcl}\mathrm{v}& =& {\mathrm{v}}^{\text{'}}+\mathrm{u}\\ & =& 0.47\mathrm{c}\hat{\mathrm{i}}+0.62\mathrm{c}\hat{\mathrm{i}}\\ & =& 1.1\mathrm{c}\hat{\mathrm{i}}\end{array}$

Therefore, the velocity of the particle according to the classical velocity transformation is $1.1\mathrm{c}\hat{\mathrm{i}}$.

(c)

Substitute $-0.47\mathrm{c}\hat{\mathrm{i}}$ for v^', and $0.62\mathrm{c}\hat{\mathrm{i}}$for u in equation (1).

$\begin{array}{rcl}\mathrm{v}& =& \frac{\left(-0.47\mathrm{c}\hat{\mathrm{i}}\right)+\left(0.62\mathrm{c}\hat{\mathrm{i}}\right)}{1+{\displaystyle \frac{\left(-0.47\mathrm{c}\hat{\mathrm{i}}\right)\left(0.62\mathrm{c}\hat{\mathrm{i}}\right)}{{\mathrm{c}}^2}}}\\ & =& \frac{-0.47\mathrm{c}+0.62\mathrm{c}}{1+\left(-0.47\mathrm{c}\right)\left(0.62\mathrm{c}\right)}\\ & =& 0.21\mathrm{c}\hat{\mathrm{i}}\end{array}$

Therefore, the velocity of the particle according to the relativistic is $0.21\mathrm{c}\hat{\mathrm{i}}$.

(d)

In the classical way, the velocity is calculated as follows.

$\begin{array}{rcl}\mathrm{v}& =& {\mathrm{v}}^{\text{'}}+\mathrm{u}\\ & =& -0.47\mathrm{c}\hat{\mathrm{i}}+0.62\mathrm{c}\hat{\mathrm{i}}\\ & =& 0.15\mathrm{c}\hat{\mathrm{i}}\end{array}$

Therefore, the velocity of the particle according to the classical velocity transformation is $0.15\mathrm{c}\hat{\mathrm{i}}$.