Problems on time dilation and length contraction 2

SummaryProblems on time dilation and length contraction 2

Problem :
If observer Bill, who is on a train moving with speed 0.6c, waves to Julie at four second intervals as measured in
Bill's frame, how long will Julie measure between waves?

Bill is in motion so we know that his seconds must be dilated (longer) with respect to Julie's seconds, by a factor γ. Thus Julie will measure more
seconds between waves. What is γ?

γ = = 5/4

Thus Julie measures 5/4×4 = 5 seconds between waves.

Problem :
Bill and Julie are both now on identical trains. Bill's train is moving to the right with velocity (/2)c with respect to Julie's train. Julie measures her train to be 100 meters long. How long does Julie
measure Bill's train to be? How long does Bill measure Julie's train to be?

Bill's train is in motion so we would expect it to appear contracted (shorter) by a factor γ to Julie.
What is γ? γ = = 2. Thus Julie will measure Bill's train to be 50
meters long. We know that Bill's train is identical, so because of the equivalence of frames and the
symmetry of the situation, we can say that Bill must measure his own train to be 100 meters long and Julie's
to be 50 meters long.

Problem :
What must be the average speed of a muon, a certain type of elementary particle, in order for it to travel 20
meters before it decays? The average rest lifetime of a muon is 2.60×10^{-8} seconds.

In the rest frame of the muon it has 2.60×10^{-8} seconds before it decays. In this time it must
travel 20.0 meters in the lab frame. In the lab frame, the muon is measured to be travelling with speed v
to the right (v is the speed we wish to find), so the muon sees the lab whizzing past to the left with speed
v. For the muon, it sees the lab contracted by a factor γ (which corresponds to v), so in its
frame it only has to travel a distance 20/γ in order to cover 20 meters as measured by an observer
in the lab. Thus the velocity required is v = = 202.60×10^{-8}. Solving this equation we find: v = = 1.72×10^{4} m/s.

Problem :
Consider the following scenario: two meter sticks, call the S_{A} and S_{B} are oriented parallel to the y
axis, some distance apart. The travel towards one another along the x-direction: that is, S_{A} one
moves in the positive x-direction and S_{B} moves in the negative x-direction (see ). S_{A} has paint brushes on its ends, pointing towards S_{B} such that if S_{B} is
longer than S_{A}, for example, it will leave paint marks on S_{B}. Show that there is no length
contraction in the y-direction (that is, the sticks both appear 1 meter long to each other)? (Hint: assume
this is not the case and derive a contradiction).

The crucial fact here is that if S_{A} sees S_{B} shorter than (or longer, or equal to) itself, then S_{B}
must also see S_{A} as shorter than itself. This arises from the equivalence of all inertial reference
frames. Moreover the factors by which each stick sees the other shorter or longer must be the same. First
assume, then, that S_{A} sees S_{B} to be longer than itself. Then S_{A} will paint marks on S_{B}. But
then, S_{B} must see S_{A} to be longer than itself, so its ends will miss S_{B} and no marks will be
painted. Hence we have a contradiction. If we assume that S_{A} sees S_{B} to be shorter than itself,
then S_{A} concludes no marks will be made, and S_{B} concludes it will be painted. Again a
contradiction. The only way out of this is if both sticks see each other to be the same length, in which case
they both agree the brushes will just touch edges of S_{B}.

Problem :
Imagine a train going through a tunnel. The train and the tunnel both have length l in their own frame.
The train moves through the tunnel with speed v. There is a bomb at the front of the train which is
designed to explode when the front of the train passes out the far end of the tunnel. However, there is a
disarming sensor located on the back of the train which will disarm the bomb just as the back of the train
enters the near end of the tunnel. Will the bomb explode?

The answer is yes, the bomb will explode. In the frame of the train, it sees the tunnel as having length
l /γ < l so the front of the train will pass out of the tunnel before the rear enters the tunnel (the train
has length l in its own frame). One might argue that in the frame of the tunnel, the train appears
contracted by the same factor and so in the tunnel frame the train is shorter than the tunnel by a factor
γ, so the rear of the train will enter the tunnel before the front passes out, and the bomb will be
disarmed. We appear to have a paradox. However, this second line of reasoning is false because it ignores
the finite time that any disarming signal must take to move from the rear of the train to the bomb at the
front. The fastest such a signal could move is at c. The bomb will be disarmed if and only if a signal
traveling at c emitted from the rear of the tunnel at the instant the rear of the train passes, reaches the far
end of the tunnel before the train does. Working still in the frame of the tunnel, the signal takes a time l /c,
and the train takes a time , since the front of the train is already a distance
l /γ (the length of the train) through the tunnel. For the bomb not to explode we need: l /c < , which simplifies to < , which is clearly false. The bomb
explodes.