Thursday, January 20, 2011

Mastering Physics: A Breakdown in Simultaneity

Three enemy spacecraft have been causing trouble in the asteroid belt. They always travel in a line, evenly spaced apart, attempting to chase down local spacecraft to steal their goods. The local asteroid colonists have decided to set a trap to capture these three spacecraft. They'll get them to chase one of their fastest ships into an asteroid with a large hole in it and, once the three enemy ships are inside, close two giant trapdoors on each side of the asteroid to catch them. These spacecraft all travel close to the speed of light so the locals will have to take relativity into account. Intelligence about the enemy spacecraft reveals that, in their reference frame, they always travel 90 m behind their teammate, each spacecraft is 10 m in length, and their maximum velocity is 90% the speed of light (relative to the asteroids). The asteroid tunnel is only 215 m in length. In this problem we will analyze whether the locals will be able to capture the enemy spacecraft after taking into account relativity.

Part A
First, we will construct some of the coordinates in both the asteroid frame and the spacecraft frame. The coordinates for the asteroid frame are given by x, y, z, and t whereas those for the spacecraft frame are given by x', y', z', and t', respectively. Set the time that the front of the first spacecraft enters the asteroid to be t = t' = 0. Also, let the asteroid tunnel be oriented in the x direction so that throughout this problem z = z' and y = y' and we can ignore these coordinates.
If the spacecraft are traveling at 90% the speed of light, what is the total length L of the three-spacecraft team as observed from the asteroid?
Express your answer in meters to three significant figures.
ANSWER:

  L = 91.5
 \rm m
Part B
Now the locals can see that, taking into account relativity, the enemy spacecraft will be in a line that is only 91.5 m long when they're traveling at 90% the speed of light relative to the asteroid. For how long a time period will all three spacecraft be inside of the asteroid?
Express your answer in microseconds to two significant figures.
ANSWER:

  t = 0.46
 \mu {\rm s}
Part C
Suppose the trigger is set so that the moment the first enemy spacecraft enters the asteroid a signal is sent at the speed of light to the other end of the asteroid to trigger the rear trapdoor. How long a time t_trig will this signal take to get to the rear of the asteroid?
Express your answer in microseconds to two significant figures.
ANSWER:

  t_trig = 0.72
 \mu {\rm s}
The locals would like the two ends to close simultaneously in their rest frame, so the front trapdoor will be set to spring 0.72 \;\mu {\rm s} after the first enemy spacecraft passes whereas the rear trapdoor will spring the instant it gets the signal.
 
Part D
How close d_rear to the rear of the asteroid will the first enemy spacecraft be when the trapdoors close?
Express your answer in meters to two significant figures.
ANSWER:

  d_rear = 21
 \rm m
Part E
One of the local students had been studying relativity and began to question the plan. How would this all appear to the enemy spacecraft as they flew through the asteroid? Specifically, if the spacecraft were in an inertial frame (i.e., one with no acceleration) how long would the asteroid tunnel appear to be in their rest frame?
Express your answer in meters to three significant figures.
ANSWER:


93.7  \rm m
This is strange. It appears that, in the enemy spacecraft rest frame, there will never be a time that all three will be in the asteroid at the same time. As a result they probably aren't worried about becoming trapped in the asteroid.
 
Part F
After thinking about this the locals begin to argue. Half the group thinks that the plan will not work, because it won't be possible for all the enemy spacecraft to be in the asteroid at the same time. The other half still believes that the plan will work. It appears to be a paradox. An elderly physicist pipes up to try to resolve the paradox. He asks both groups of people to draw a space-time diagram to back up their argument that the plan will or will not work.
Using worldlines to indicate the path of a particle through space-time, each group presents its diagram. Red is used for the worldlines of the enemy spacecraft (with the first ship being the one on the right) and blue is used for the worldlines of the front and back of the asteroid (with the front of the asteroid being the blue line on the left).
Which statement about the two drawings (Diagram 1 and Diagram 2 ) is correct?
ANSWER:



Part G
The physicist then explains that the only way out of the apparent paradox the colonists find themselves in is to realize that events that are simultaneous in one reference frame are not simultaneous in another frame. In fact, events can only be simultaneous in one specific frame. To illustrate his point, the physicist draws a space-time diagram in the frame of the asteroid. The physicist then asks the student to draw in the worldlines of the three spacecraft as viewed from this frame (this is the ct' axis for each spacecraft). What is the angle theta that these lines make to the ct axis in the asteroid frame? (Note that the worldlines of the three ships are all parallel to each other.)
Express your answer in degrees to two significant figures.
ANSWER:

  theta = 42
 ^\circ
Part H
The physicist reminds the student that, because of Lorentz transformations, the x' axis is not parallel to the x axis. Use these transfomations and the fact that ct' = 0 along the x' axis to figure out the angle theta the x' axis makes with the x axis.



Express your answer in degrees to two significant figures.
ANSWER:

  theta = 42
 ^\circ
Part I
Now, the physicist draws a dashed line parallel to the x axis that represents the moment when both trapdoors slam shut on the asteroid in the asteroid's frame . In the spacecraft rest frame, what can you say about when the doors appear to shut? The two green dots mark the events where the doors shut.


ANSWER:


The result from the diagram that the physicist shows is that, in the speeding spacecraft's frame, the door at the rear end of the asteroid actually appears to close long before the door at the front of the asteroid. In fact, the first spacecraft must stop very quickly in order not to run into the door but won't have time to turn around and get out when the front door shuts. The colonists therefore have a viable plan.

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