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Expert-verified Found in: Page 67 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # You fire a light signal at ${\mathbf{60}}{\mathbf{°}}$north of west (a) Find the velocity component of this, signal according to an observer moving eastward relative to you at half the speed of light. From them. determine the magnitude and direction of the light signal's velocity according to this other observer. (b) Find the component according to a different observer, moving westward relative to you at half the speed of light.

(a) The velocities for the light beam for the moving eastward observer are $\left({u}_{x}^{\text{'}}{u}_{y}^{\text{'}}\right)=\left(-0.8c,06c\right)$.

(b) The velocities for the light beam for the moving westward observer are $\left({u}_{x}^{\text{'}},{u}_{y}^{\text{'}}\right)=\left(0,c\right)$.

See the step by step solution

## Step 1: Given data

The observer is moving eastward related to the half of the speed of the light.

## Step 2: Determine the formulas to calculate the velocity of the component, magnitude and direction according to the observer and component according to different observer.

The expression to calculate the velocity transformation for velocity of object in direction is given as follows.

${{{\mathbit{u}}}^{{\mathbf{\text{'}}}}}_{{\mathbf{x}}}{\mathbf{=}}\frac{{\mathbf{u}}_{\mathbf{x}}\mathbf{-}\mathbf{v}}{\mathbf{1}\mathbf{-}\frac{{\mathbf{u}}_{\mathbf{x}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}$ ...(i)

Here,${{\mathbit{u}}}_{{\mathbf{x}}}$ is the velocity component in the x direction, v is the velocity of frame ${{\mathbit{S}}}^{{\mathbf{\text{'}}}}$relative to S, and c is the velocity of the light.

The expression to calculate the velocity transformation for velocity of object in y direction is given as follows.

...(ii)

The expression to calculate the Lorentz factor is given by,

## Step 3: Calculate the velocity of the component, magnitude and direction according to the observer.

Since the light beam is moving with the x axis, so the component of velocity in x and direction y is given by,

${u}_{x}=-ccos\theta \phantom{\rule{0ex}{0ex}}{u}_{y}=csin\theta$

Calculate the component of velocity in x direction relative to frame ${S}^{\text{'}}$.

Substitute$-c\mathrm{cos}\theta$ for ${u}_{x}$ into equation (i).

...(iii)

Calculate the component of velocity in y direction relative to frame ${S}^{\text{'}}$.

Substitute $c\mathrm{sin\theta }$ for ${u}_{y}$ into equation (ii).

Substitute $1}{\sqrt{1-{\left(v/c\right)}^{2}}}$ for ${\gamma }_{v}$ into above equation.

Hence the component of velocity of light beam in x and y directions are and $\frac{c\sqrt{1-{\left(\frac{v}{c}\right)}^{2}}\mathrm{sin}\theta }{\left(1+\frac{v\mathrm{cos}\theta }{c}\right)}$respectively.

Since the observer is moving to right, the velocity,$v=+0.5c$

Calculate the x component of the velocity,

Substitute ${60}^{°}$for $\theta$ and +0.5 c for v into equation (iii).

role="math" localid="1659329161430"

Calculate the component of the velocity,

Substitute $60°$for $\theta$ and 0.5c for v into equation (iv).

So, the velocities for the light beam for the moving eastward observer are.

$\left({u}_{x}^{\text{'}},{u}_{y}^{\text{'}}\right)=\left(-0.8c,0.6c\right)$.

## Step 4: Calculate the component according to a different observer, moving westward relative to you at half the speed of light.

Since the observer is moving to left, the velocity,$V=-0.5c$

Calculate the x component of the velocity,

Substitute $60°$ for $\theta$ and $-0.5c$ for v into equation (iii).

Calculate the y component of the velocity,

Substitute $60°$ for $\theta$ and -0.5c for v into equation (iv).

So, the velocities for the light beam for the moving westward observer are $\left({u}_{x}^{\text{'}},{u}_{y}^{\text{'}}\right)=\left(0,c\right)$. ### Want to see more solutions like these? 