Answer to Question #205288 in Mechanics | Relativity for Safi Ullah

Question

A point moves with constant speed t in a direction which forms an angle with the x-axis of the frame of reference S'. Frame S' moves with a velocity V = Và relative to another frame of reference Ss. What is the angle 0 formed by the direction of motion of the point with the x-axis of S? What is the relationship between the two angles as v - c? f sin sin d Ans.: tan 0 = tan e

Answer 1

Solution:-

Wave from equation

$\psi=Ae^{{i2\pi\upsilon }{(t-\frac{r}{c})}}$ =$Ae^{i2\pi\upsilon(t-\frac{(xcos\theta+ysin\theta)}{c})}$

Similarly S' co-ordinate system

$\psi'=A'e^{{i2\pi\upsilon '}{(t'-\frac{r'}{c})}}$ =$A'e^{i2\pi\upsilon '(t'-\frac{(x'cos\theta'+y'sin\theta')}{c})}$

Number of frames does not depend motion of frame

$\upsilon (t-\frac{r}{c})=\upsilon'(t'-\frac{r'}{c})$

$\upsilon(t-\frac{xcos\theta+ysin\theta}{c})=\upsilon'(t'-\frac{x'cos\theta'+y'sin\theta'}{c})$

x,y,t value put

$\upsilon(\gamma(t'+\frac{\beta x'}{c})-\frac{\gamma (x'+\beta ct')cos\theta +y'sin\theta}{c})=\upsilon'(t'-\frac{x'cos\theta'+y'sin\theta'}{c})$

Comparison x' components

$\gamma v(cos\theta-\beta)=\upsilon'cos\theta'$

Comperision of y' cofficient

$vsin\theta=v'sin\theta'$

Comperision of t' components

$\gamma v(1-\beta cos\theta)=\upsilon'$

$\upsilon =\frac{\upsilon'}{\gamma(1-\beta cos\theta)}$

equation (1)and(2)

$tan\theta'=\frac{sin\theta}{\gamma(cos\theta-\beta)}$