

Special Relativity T
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Question:
My question is the T/U paradox as explained in Taylor's spacetime
physics (it is also one of the scenarios in the program rellab
by physics academic software.
the paradox is as follows
a U made of steel has a detonator switch connected to the bottom of the
uU and connected to an explosive. TA T made of the same steel just fits in t
U when they are at trest wrt eraacth other. The T is moved far away from
the U and accelerated to v=.9c. The T From the reference fromaeame of the U
the T is lorentz contracted and the T is too short to reach the detonator
From the reference frame od f the he T the U is arms are contracted and the
T can strike the switch. Which occurs?
I have been able to work my way through the other typical problems
like the barn/pole paradox , and the runner train paradox. This one has
me stumped.
Replies:
What a beeyootiful problem. The answer to the paradox is that the bomb
goes off, since all observers agree after the collision that U and T are at
rest and fit exactly into one another. Relativity does not alter the nature
of events, just the order and interval between them. Two points are key to
understanding what T and U would deduce from their observations: the first
is that the speeding, shortened other object *lengthens* when it slows down
during the collision. The second is that neither T nor U see as simultan
eous (1) the crossbar of the T hitting the ends of the U and (2) the point
of the T hitting the bottom of the U. From T's point of view, 2 precedes 1,
while U concludes 1 happens before 2. There is another frame of reference
(moving with the center of mass) in which both events occur at the same
time and in which T and U are the same length always.
One always talks in relativity class about what observers *deduce* in
their respective frames of reference. But it is fun to imagine what T and U
would actually *see* according to our own habits of perception as the
collision unfolds. To be concrete, let us imagine T and U at rest are each
100 million miles long (about the distance from Earth to the Sun), and they
approach each other at a speed such that each would deduce with all the
relativity math that the other is half its rest length. This speed turns
out to be 87% of the speed of light c, or about 161,000 miles/sec. There
are four interesting points of view: (A) sitting on the ends of the U, (B)
sitting at the bottom of the U, (C) sitting on the point of the T, and (D)
sitting on the crossbar of the T. Suppose everybody somehow synchronizes
their watches so they all agree the point of the T passes the ends of the U
at 5:00 pm exactly. Now as 5:00 pm comes on A sees the T approaching fast.
Suppose A measures the length and speed of T naively, the way you usually
would do it. To measure the length we suppose mile markers along the route
T takes; A just notes which mile markers he sees next to the front and back
of the T at a certain moment and subtracts. To measure the speed of T he
starts his stopwatch when he sees the point pass a mile marker, then stops
it when he sees the point pass the next, and divides. What will he see?
The T will look 373 million miles long and will look like it is approaching
at 6.5 times c, or about 1.2 million miles/sec. Whoa! How can this be?
Because this is a *naive* measurement, A has forgotten to take into account
the time it takes the image of the front and back of the T to get to him at
c. This is the way we normally do things  if you are standing at the
finish line timing a footrace you start the watch when you see the runners
start and stop it when they go past you. You do not add to the elapsed time
the time it took for the image of the start to get to you. But you should,
strictly speaking. If a runner ran *at* the speed of light, he would get to
you *at the same time* as the image of him starting, and you would (wrongly)
conclude by the naive measurement method that he covered the course in no
time at all. Still, that is the way our habits of perception are set up.
Of course if A takes the speed of light into account and does his
relativity math correctly, he will conclude that, contrary to his naive
impression, the T is 50 million miles long and approaching at 87% of the
speed of light.
Now at 5:00 pm A sees the point of the T pass him. The crossbar is 373
million miles distant and still approaching at 6.5 times c, but the point
is now receding from A, apparently at 0.5 times c or 86,000 miles/sec.
Thus the T appears to shrink at the rate of 1.1 million miles/sec. At 5:05
pm the crossbar reaches A, hits the ends of the U, and stops. The point has
now reached a point 27 million miles away from A, or about 1/3 the distance
to the bottom of the U, so the T appears to be only 27 million miles long.
The point continues to recede, so the T appears to A to be growing at
86,000 miles/sec. At 5:19 pm the T has grown to its rest length of 100
million miles long, and A sees it hit the bottom of the U and stop. Now
everything is motionless and the U and T fit perfectly together.
Now let us consider it from the point of view of B (at the bottom of the
U). B sees again a huge (373 million mile long) T approaching at 6.4 times
c. B sees the point pass A at 5:09 pm (9 minutes after A sees it happen,
because it takes 9 minutes for the image to go the 100 million miles
between A and B). Barely a minute later, at 5:10 pm, the point of the T
reaches B, hits the bottom of the U and stops. (Note that the point gets
to B only a minute after B sees it leave A, even though light itself takes
9 minutes to get from A to B. The *naive* (wrong) conclusion is that the
point is traveling a lot faster than light.) Now the crossbar is 373
million miles distant, and still approaching at 6.5 times c, so B sees the
T shrinking at a rate of 1.2 million miles/sec. This goes on for 227
seconds until the T has shrunk to its rest length of 100 million miles. At
this point, 5:14 pm, B sees the crossbar of the T hit the ends of the U and
stop. Everything is motionless and the T and U fit perfectly. Everybody
goes out for a beer and compares notes.
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Update: June 2012

