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$47M, 052,) WEE EL é. AWPM, Length Contraction & Time Dilation Consider “observers” (Ahthur and (B)erbara riding on two railroad trains which are
on parallel tracks. Suppose that the trip is veryr smooth (not elutraklj and that there
are no landmarks. Arthur thinks his train is at rest and Barbara's train is moving
past him in the positive x—direetion. However, Barbara. believes she’s at rest and
Arthur is Ineving post her {in the negative redirection). Unless the trains accelerate
or deeelerate there’s no way to resoive this disagreement; Arthur and Barbara are
equally good (“inertial”) observers. Suppose they each observe two “events”, say two
ﬂashes of light. Arthur may determine that the events occurred at different locations
separated by £191 and at different times separated by a time interval Arr. For the
some two events Barbara ﬁnds {MB and Am}. : According to Einstein / Lorentz, the reiations between the distance and time intervals
that Arthur and Barbara measure is: An,“ = “(£3.33 + 2:55.153) ; Ah 2 “(($33 + rigors/(:2).
where: 7 = (1 —— 182T”? and {3 m We. Suppose that Barbara ﬁnds the light source is at rest in her system (on her train], so
that A2}; = 0. Perhaps it is a clock which emits a ﬂesh of light every second. The
time interval She measures for the two flashes is At}; 5 in. Simple substitution in the
above er‘iuetions shows that Arthur ﬁnds a longer time interval between ﬂashes: ﬁts = 7513 : Ne J‘Lrther siso sees the flashes occur at different locations (Barbara‘s clock is moving),
separated by: Arr/1 m oﬁtd m veto. Arthur finds a lenger time interval between ﬂashes and concludes that Barbara‘s
clock is running slow. Note that if Arthur were carrying the clock and Barbara
were timing the ﬂashes, she would also measure a time interval of ﬁfty) and she would
coneiude that Arthur’s clock is running slow. Let us now consider measurements of the length of Barbara’s train. Barbara can
measure it by timing or light beam she bounces off a mirror at. the back of the train.
{et en 2 m .41) when she is at the front (at .123 m 0}. Then, for Barbara, A33 m —LO.
Since Barbara sees Arthur moving at Speed t} and she knows that her train is L0 in
Eength, she cemeteries it will take a. time interval 33133 = Lu/U for her train to pass
Arthur. Then, for Arthur: Arm a L = vAtA v2: 70053:}; + raging/(,3) = 71,00 _ £2) : Lg/“f. 30, Arthur will measure L < L0 and conclude that Barbara’s train is shorter than
Barbara says it is. Once again, if Barbara measured Arthur’s train, she would ﬁnd it
shorter than Arthur said it was. To summarize: moving clocks run slow; moving lengths are shorter. ...
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 Spring '11
 L
 Lorentz Transformation, Barbara ﬁnds, Arthur ﬁnds

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