Question

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?

Short answer

(a) The velocity of the particle according to the relativistic is $0.84\mathrm{c}\hat{\mathrm{i}}$.

(b) The velocity of the particle according to the classical velocity transformation is $1.1\mathrm{c}\hat{\mathrm{i}}$.

(c) The velocity of the particle according to the relativistic is $0.21\mathrm{c}\hat{\mathrm{i}}$.

(d) The velocity of the particle according to the classical velocity transformation is $0.15\mathrm{c}\hat{\mathrm{i}}$.

Step-by-step solution

Describe the expression for the relativistic velocity of the particle

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.

Determine the particle’s velocity according to the relativistic

(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 $0.84\mathrm{c}\hat{\mathrm{i}}$.

Determine the particle’s velocity according to the classical velocity transformation

(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 $1.1\mathrm{c}\hat{\mathrm{i}}$.

Determine the particle’s velocity according to the relativistic

(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 $0.21\mathrm{c}\hat{\mathrm{i}}$.

Determine the particle’s velocity according to the classical velocity transformation

(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 $0.15\mathrm{c}\hat{\mathrm{i}}$.