An observer in reference frame S measures two events to be...

An observer in reference frame S measures two events to be simultaneous. Event A occurs at the point (50.0 m, 0, 0) at the instant 9:00:00 Universal time on January 15, 2008. Event B occurs at the point (150 m, 0, 0) at the same moment. A second observer, moving past with a velocity of $0.800 \mathrm{ci},$ also observes the two events. In her reference frame $\mathrm{S}^{\prime},$ which event occurred first and what time interval elapsed between the events?

Key Concepts

Space-Time Event Ordering

In special relativity, the ordering of events (whether one event occurs before another) can depend on the observer's frame of reference, particularly for events that are separated by distances such that their spatial separation exceeds the product of the speed of light and the time interval between them. For space-like separated events, different observers may disagree on the sequence in which events occur.

Relativity of Simultaneity

The relativity of simultaneity is a principle in special relativity stating that events which occur simultaneously in one reference frame may not be simultaneous in another frame moving relative to the first. This concept underscores how the temporal order of events can depend on the motion of the observer, especially when the events are spatially separated.

Lorentz Transformation

The Lorentz transformation is a set of equations that relate the space and time coordinates of events as seen in one inertial frame to those in another moving at a constant velocity relative to the first. It quantitatively describes how measurements of time and space change between observers and forms the mathematical foundation of many relativistic phenomena.

Reference Frames

A reference frame is a coordinate system used to describe the position and time of events. In physics, particularly in special relativity, different observers may be in different inertial frames that move relative to one another, and events that occur are described differently in each frame according to their relative motion.

Transcript

00:01 In this exercise, we have two event a and b and two reference frames, s and s prime.

00:09 S -prime moves with a velocity of 0 .8 times the speed of light in the x direction relative to the s -reference frame.

00:23 In s -reference frame, event a occurs at the spatial coordinates 50 meters, 0 -0, and as a, event b occurs at coordinate 150 meters 0 .0.

00:37 The first coordinate is x, the second is y, the 30s, and events a and b happen at the same time in s coordinate.

00:52 And are going to find which event happened first in the s -prime coordinate, and also what is the separation of time between the two events.

01:03 In order to find this, we need to use lawrence transformations.

01:09 And actually, the only one that we're going to need to use is the transformation for time.

01:14 So consider that we have these two reference frames.

01:19 And we want to find what is the time coordinate t prime of the s prime reference frame as a function of the coordinate t and x of the s reference frame.

01:33 So basically we just need to use this equation.

01:35 To prime is equal to gamma t minus v over c squared x where gamma is equal to one over the square root of one minus v squared over c squared okay so basically we can write the coordinate in the s prime system for event a and this is equal to gamma t a minus v over c squared times xa xa is the x coordinate of the event a in s coordinate system okay and notice that we can simply disregard the other coordinates since they're both zero they're in both event a and b for the s s coordinate system and the the s prime coordinate system moves with a that is in the x direction and t b prime is gamma times t b minus v over c squared xb okay so t a prime minus t b is equal to gamma times t a minus t b minus b minus v squared x a minus xb.

03:25 Now notice that tae minus tb is zero since both events happen at the same time in the s coordinate system.

03:33 So we have that t a prime minus db prime is equal to gamma v over c square times xb minus xa.

03:46 Okay we have the t prime minus t b prime is equal to one over the square root of one minus v squared over c squared, that's just 0 .8 squared, since the speed is 0 .8, times v, which is 0 .8, times c divided by c squared...

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