- #1
dchartier
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Moved from technical forum, so no template
Hello,
I'm self-studying Ta-Pei Cheng's Relativity, Gravitation and Cosmology. Problem 2.8 is the following:
Rather than just apply straightforward length contraction (which I've gotten burned doing in the past in less simple questions), I decided to solve the problem by setting out the events and using the Lorentz transform. Here's my solution. Is my reasoning sound? Thanks!
__________________________________
We'll place the clock in ship ##A## and let ship ##B## be the other ship passing by. Let frame ##O## be ship ##A##'s rest frame and frame ##O'## be ship ##B##'s rest frame, with frame ##O'## moving at ##v = 1.25 \times 10^8## m/s relative to frame ##O##. Let the two frames' origins coincide at ##t=t'=0## and their axes be parallel. We have two events in frame ##O##, with ##E_0## corresponding to when the front of the ship coincides with the clock (which we will designate as occurring at ##t=0##) and ##E_1## when the rear of ship ##B## coincides with the clock:
$$
\begin{array}{ccccc}
E_0 = (t_0,x_0) = (0,0) & & \mbox{and} & & E_1 = (t_1,x_1) = (\Delta t, 0),
\end{array}
$$
where ##\Delta t = 9.1 \times 10^{-8}## sec.
In frame ##O'##, the front of ship ##B## is at ##x'=0## and the rear is at ##x' = -L'_B##. Because this is ship ##B##'s rest frame, the front and rear positions are constant. In frame ##O'##, the two events correspond to:
$$
\begin{array}{ccccc}
E'_0 = (t'_0,x'_0) = (0,0) & & \mbox{and} & & E'_1 = (t'_1,x'_1) = (\Delta t', -L'_B).
\end{array}
$$
Using the Lorentz transform for event ##E'_1##, we have:
$$
x'_1 = -L'_B = \gamma(x - vt) = \gamma(0 - v \Delta t) = -\gamma v \Delta t = -12.5 \mbox{m}.
$$
Ship ##B##'s length is therefore ##12.5## m in its rest frame. Note that this is the same as simply undoing length contraction in frame ##O##, where ship ##B##'s measured length is ##L_B = v \Delta t = L'_B/\gamma##, so ##L'_B = \gamma v \Delta t##.
I'm self-studying Ta-Pei Cheng's Relativity, Gravitation and Cosmology. Problem 2.8 is the following:
Two spaceships traveling in opposite directions pass one another at a relative speed of ##1.25 \times 10^8## m/s. The clock on one spaceship records a time duration of ##9.1 \times 10^{-8}## s for it to pass from the front end to the tail end of the other ship. What is the length of the second ship as measured in its own rest frame?
Rather than just apply straightforward length contraction (which I've gotten burned doing in the past in less simple questions), I decided to solve the problem by setting out the events and using the Lorentz transform. Here's my solution. Is my reasoning sound? Thanks!
__________________________________
We'll place the clock in ship ##A## and let ship ##B## be the other ship passing by. Let frame ##O## be ship ##A##'s rest frame and frame ##O'## be ship ##B##'s rest frame, with frame ##O'## moving at ##v = 1.25 \times 10^8## m/s relative to frame ##O##. Let the two frames' origins coincide at ##t=t'=0## and their axes be parallel. We have two events in frame ##O##, with ##E_0## corresponding to when the front of the ship coincides with the clock (which we will designate as occurring at ##t=0##) and ##E_1## when the rear of ship ##B## coincides with the clock:
$$
\begin{array}{ccccc}
E_0 = (t_0,x_0) = (0,0) & & \mbox{and} & & E_1 = (t_1,x_1) = (\Delta t, 0),
\end{array}
$$
where ##\Delta t = 9.1 \times 10^{-8}## sec.
In frame ##O'##, the front of ship ##B## is at ##x'=0## and the rear is at ##x' = -L'_B##. Because this is ship ##B##'s rest frame, the front and rear positions are constant. In frame ##O'##, the two events correspond to:
$$
\begin{array}{ccccc}
E'_0 = (t'_0,x'_0) = (0,0) & & \mbox{and} & & E'_1 = (t'_1,x'_1) = (\Delta t', -L'_B).
\end{array}
$$
Using the Lorentz transform for event ##E'_1##, we have:
$$
x'_1 = -L'_B = \gamma(x - vt) = \gamma(0 - v \Delta t) = -\gamma v \Delta t = -12.5 \mbox{m}.
$$
Ship ##B##'s length is therefore ##12.5## m in its rest frame. Note that this is the same as simply undoing length contraction in frame ##O##, where ship ##B##'s measured length is ##L_B = v \Delta t = L'_B/\gamma##, so ##L'_B = \gamma v \Delta t##.
