Question

According to observer o, two events occur separated by a time interval-At=0.48 us and at locations separated by 45 m (a) According to observer O, who is in motion relative to O at a speed of u=0.970c in the positive x direction, what is the time interval between the two events? 146 (b)What is the spatial separation between the two events, according to 07 349.1496

Answer 1

Answer:

(a) $\Delta t'=1.38\mu s$

(b) $\Delta x'=-389.46m$

Explanation:

We use Lorentz transformations, since they relate the measures of a physical magnitude obtained by two different observers, these are:

$\Delta x'=\frac{\Delta x-u\Delta t}{\sqrt{1-\frac{u^2}{c^2}}}\\\Delta t'=\frac{\Delta t-\frac{u\Delta x}{c^2}}{\sqrt{1-\frac{u^2}{c^2}}}$

Here $\Delta x$ is the spatial separation according to O, $\Delta x'$ is the spatial separation according to O', $\Delta t$ is the time interval according to O, $\Delta t'$  is the time interval according to O', $u$ is the relative speed between the two observers and $c$ is the speed of light.  All we do now is write the quantities we were given, recall that $1\mu s=10^{-6}s$

(a)

$\Delta t'=\frac{0.48*10^{-6}s-\frac{0.97c(45m)}{c^2}}{\sqrt{1-\frac{(0.97c)^2}{c^2}}}\\\Delta t'=\frac{0.48*10^{-6}s-\frac{0.97(45m)}{(3*10^{8}\frac{m}{s})}}{\sqrt{1-0.97^2}}\\\Delta t'=1.38*10^-6 s=1.38\mu s$

(b)

$\Delta x'=\frac{45m-(0.97*3*10^8\frac{m}{s})0.48*10^{-6}s}{\sqrt{1-\frac{(0.97c)^2}{c^2}}}\\\Delta x'=-389.46m$

The minus sign means that the second event is closer for one observer and the first is closer for the other.