Physics 105B Problem Set 8
May 27, 2008
Je
ff
Schonert:
schonert (at) physics.ucsb.edu
Taylor 15.17
a)
Let
Δ
t
denote the time of event 2 minus the time of event 1 (and similarly for
Δ
t
). The
time interval in
S
is related to the time interval
S
by
Δ
t
=
γ
Δ
t

β
c
Δ
x
We are given that
Δ
t
= 0 and
Δ
x
=
a
, so plugging in gives
Δ
t
=

γβ
a/c
.
b)
Now consider the same problem except in a frame
S
that is moving with a speed of the
same magnitude of
S
but opposite direction. This means we only need to replace
β
with

β
in the above formula (
γ
stays the same). Then we have
Δ
t
=
γ
Δ
t
+
β
c
Δ
x
=
γβ
a
c
So in
S
the events are simultaneous, in
S
event 1 happens after event 2, and in
S
event 1
happens before event 2.
Taylor 15.22
Since
S
is moving along the
x
axis of
S
with velocity
.
9
c
, the two relevant velocity addition
formulas are (
V
=
.
9
c
)
v
x
=
v
x
+
V
1 +
v
x
V/c
2
,
v
y
=
v
y
γ
(1 +
v
x
V/c
2
)
Now plug in
V
=
.
9
c
,
v
y
=
.
9
c
, and
v
x
= 0. Doing this gives
v
x
=
.
9
c
and
v
y
=
.
392
c
. The
magnitude of the velocity vector is
v
=
(
.
392
c
)
2
+ (
.
9
c
)
2
=
.
98
c
The angle that this vector makes with the
x
axis in
S
is
θ
= arctan
v
y
v
x
= arctan
√
.
19
= 23
.
6
◦
1
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Taylor 15.48
a)
According to (15.64),
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 Spring '08
 MARTINIS
 Physics, mechanics, Momentum, Special Relativity

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