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

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

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

Step-by-step 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-(u^2)/(c^2)}}\\\Delta t'=\frac{\Delta t-(u\Delta x)/(c^2)}{\sqrt{1-(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-(0.97c(45m))/(c^2)}{\sqrt{1-((0.97c)^2)/(c^2)}}\\\Delta t'=(0.48*10^(-6)s-(0.97(45m))/((3*10^(8)(m)/(s))))/(√(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(m)/(s))0.48*10^(-6)s}{\sqrt{1-((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.