Question

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship.
(a) Use the Lorentz coordinate transformations to calculate x and tas measured by Stanley for the event of turning the light.
(b) Use the time dilation formula to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Express your answer in seconds.

Answer 1

Answer:

a) x = 2 × 10⁹m and t = 8.34s

b)t = 8.34s

Explanation:

Part a

The Lorentz factor is given by

$\gamma =\frac{1}{\sqrt{1-\frac{u^2}{c^2} } }$

$\gamma = \frac{1}{\sqrt{1-(\frac{0.8c}{c})^2 } } \\\\ = 1.667$

From Lorentz transformation we have

$x = \gamma (x'+ut')\\\\ x = \gamma (0 + ut') \\\\ x= \gamma ut'$

substitute the values

$x = (1.667)(0.8c)(5.0s)\\\\x=(1.667)(0.8\times 3.0 \times 10^8m/s)(5.0s)\\\\x=2.0\times10^9m$

From Lorentz transformation we have

$t = \gamma (t'+\frac{ux'}{c^2} ) \\\\t = \gamma(t'+\frac{u(0)}{c^2} )\\\\t= \gamma t'$

$t = (1.667)\times 95.0s)\\\\t=8.34s$

Part b

From the concept of timr dilation we have,

$\Delta T = \gamma \Delta T_0$

$= 1.667\times 5.0\\\\=8.34s$

The value of time found is same as part a

Answer 2

Answer:

a) $x=2*10^{9} m$ and $t=8.35 s$

b) t = γt', so it is 8.35 s.

Explanation:

a) The equation of Lorentz transformations is given by:

$x=\gamma(x'+ut')$  

x' and t' are the position and time in the moving system of reference, and u is the speed of the space ship. x is related to the observer reference.

• x' = 0
• t' = 5.00 s
• u =0.800 c, c is the speed of light 3*10⁸ m/s

$\gamma=\frac{1}{\sqrt{1-(u/c)^{2}}}$

$\gamma=\frac{1}{\sqrt{1-(0.800c/c)^{2}}}$

$\gamma=\frac{1}{\sqrt{1-(0.800)^{2}}}$

$\gamma=1.67$

$x=1.67(0+0.800c*5.00)$

$x=2*10^{9} m$

Now, to find t we apply the same analysis:

$t=\gamma(t'+\frac{ux'}{c^{2}})$                        

but as x'=0 we just have:

$t=\gamma(t')$

$t=1.67*5.00=8.35 s$

b) Here, Mavis reads 5 s on her watch and Stanley measured the events at a time affected by the Lorentz factor, in other words t = γt', if we see it is the same a) part. So the time interval will be equal to 8.35 s.

I hope it helps you!