Astronomy 210 Spring 2011
Homework Set #8: Solutions
1.
The random walk of light inside the Sun.
(a)
[5 points]
At each step, each particle has an equal chance of going to the left or to the
right.
This symmetry means that the entire collection (“ensemble”) of particles, and the
average motion of a single particle, has not net displacement left or right.
Thus We must
have
(
D
N
)
= 0 for all
N
.
To infer this in a more mathematical way, we can use mathematical induction. The particle
starts (at “step
N
= 0”) at the origin:
D
0
= 0. A particle at step
N
has distance
D
N
, with
some (asyet unkown) average
(
D
N
)
. The next step has a 50% chance of being at distance
D
N
+1
=
D
N
+
ℓ
, and a 50% chance of being at distance
D
N
+1
=
D
N
−
ℓ
. The average of
these is
(
D
N
+1
)
=
(
D
N
±
ℓ
)
=
1
2
(
(
D
N
)
+
ℓ
) +
1
2
(
(
D
N
) −
ℓ
) =
(
D
N
)
(1)
And thus we infer that
(
D
N
+1
)
=
(
D
N
)
=
(
D
N
−
1
)
=
. . .
=
(
D
0
)
=
D
0
= 0
(2)
(b)
[5 points]
The first step has equal probabilities of
D
1
=
±
ℓ
, and thus both possibilties have
D
2
1
=
ℓ
2
. Taking the average of these two equal values just gives the same value:
(
D
2
1
)
=
ℓ
2
.
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 Spring '08
 Fields
 Astronomy, Dn, Barnard

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