Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q65E

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

Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

You fire a light signal at 60°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 ux'uy'=-0.8c,06c.

(b) The velocities for the light beam for the moving westward observer are (ux',uy')=(0,c).

See the step by step solution

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.

u'x=ux-v1-uxvc2 ...(i)

Here,ux is the velocity component in the x direction, v is the velocity of frame S'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.

uy'=uyγv(1-uxvc2) ...(ii)

The expression to calculate the Lorentz factor is given by,

γv=11-(vc)2

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,

ux=-c cos θuy=c sin θ

Calculate the component of velocity in x direction relative to frame S'.

Substitute-c cosθ for ux into equation (i).

ux'=-c cos θ-v1+c cosθvc2ux'=-c cos θ-v1+v cos θc ...(iii)

Calculate the component of velocity in y direction relative to frame S'.

Substitute c sinθ for uy into equation (ii).

uy'=c sinθ1--c cosθvc2uy'=c sin θγv1+v cos θc

Substitute 11-v/c2 for γv into above equation.

uy'=c sinθ11-vc21+v cosθcuy'=c1-vc2sinθ1+v cosθc

Hence the component of velocity of light beam in x and y directions are -c cosθ-v1+v cosθc and c1-vc2 sinθ1+vcosθcrespectively.

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

Calculate the x component of the velocity,

Substitute 60°for θ and +0.5 c for v into equation (iii).

role="math" localid="1659329161430" ux'=-c cos60°-0.5 c1+0.5 c cos60°cux'=-0.5c-0.5 c1+0.25ux'=-1c1.25ux'=-0.8c

Calculate the component of the velocity,

Substitute 60°for θ and 0.5c for v into equation (iv).

uy'=c1-0.5cc2sin60°1+0.5 ccos60°1+0.5×0.5uy'=c1-0.250.8661+0.5×0.5uy'=c0.8660.8661+0.25uy'=0.6c

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

(ux',uy')=(-0.8c,0.6c).

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 θ and -0.5c for v into equation (iii).

ux'=-c cos 60°-(-0.5c)1+(-0.5c)cos60°cux'=-0.5c+0.5c1-0.25ux'=00.75ux'==0

Calculate the y component of the velocity,

Substitute 60° for θ and -0.5c for v into equation (iv).

uy'=c1--0.5cc2sin60°1+-0.5ccos60°cuy'=c1-0.250.8661-0.5×0.5uy'=c0-.8660.8660.75uy'=c

So, the velocities for the light beam for the moving westward observer are (ux',uy')=(0,c).

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.