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Lets look at bit closer at the averages one can
take by considering a (localized) swarm of summer flies "circling"
around like crazy, so that the ensemble looks like a small cloud of smoke. |
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"Localized" means that the swarm
maintains a defined form (on average), so all flies are always inside this
defined volume in space. |
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In the case of the charge carriers in a piece of
metal, it simple implies that the carriers stay inside the piece of metal. |
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First we notice
again that while the individual fly moves around quite fast, its average
vector velocity <vi>t, averaged over time t,
must be zero as long as the swarm as an ensemble doesn't move. |
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In other words, the flies, on average, move just
as often to the left as to the right etc. The net current flowing through some
surface produced by all flies at any given
instance, or by one individual fly after
sufficient time is obviously zero for any reference surface you care to chose. This is
illustrated schematically below. |
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Shown are 13 velocity vectors
of an individual fly; the chain of vectors closes so <vi>t = 0. |
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The average of the magnitude of the velocity of an
individual fly, <|vi|>t = <vi>t, however, is obviously
not zero - the fly, after all, is buzzing
around at high (average) speed. Note the details in
the equation above: Only the underlining of v is
different! |
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If we define <vi>t as follows, we have a simple way of
obtaining the average of the magnitude (we take only the positive root, of
course) . |
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v2 is a scalar, and the
(positive) square root of v2 gives always the
(positive) magnitude of v; i.e. |v| |
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This is an elegant and workable definition, but beware:
<(v2)1/2> is not the
same as
(<v2>)1/2!
Lets try it with a few arbitrary numbers Þ
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| |v| = |
3 |
4 |
6 |
| <(v2)1/2> = |
(3 + 4 + 6)/3 = 13/3 = 4,333... |
| (<v2>)1/2 = |
[(9 + 16 + 36)/3]1/2 =
20,331/2 = 4,51 |
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If we have <v>t =
<(v2)1/2>t , we may also calculate the average
(over time) of the velocity components in
x, y, and z -direction,
<vx>t , <vy>t , <vz>t , of an individual fly
for a truly random movement. (We drop the index "i" now to make life
easier). |
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Again, the vector averages <vx> and so on of the vector components must be = 0 because in a
truly random movement the components in + x and - x
direction and so on must cancel on average. |
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Since the magnitude |A|of a
vector A
is
given by the square root of the scalar product of the vector with itself .
We have |
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| A · A |
= |
Ax ·
Ax + Ay · Ay +
Az · Az = A2 |
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| A |
= |
|A| |
= (A2)½ |
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Since |
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| <v2>t |
= |
<vx2>t + <vy2>t + <vz2>t , |
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and since in a truly
random movement we have |
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we end up with |
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| <v2>t |
= |
3 <vx2> |
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| <vx2>t |
= |
= 1/3 <v2> . |
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From this we finally get
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| <vx>t |
= |
<(vx2)½>t |
= |
(1/3)½ · <(v2)½>t |
= |
<v>t
3½ |
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In real life, however, the fly swarm
"cloud" often moves slowly around - it has a finite
drift velocity
vD. |
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In consequence, <vi>t is not zero, and <vi, +x>t (= average velocity component in
+x direction) in general is different from <vi, –x>t. |
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Note that the drift velocity by definition is an
average over vectors; we do not use the <
> brackets to signify that anymore. Note
also, that the drift velocity of the fly
swarm and the drift velocity of an individual fly must be identical if the swarm is to
stay together. |
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Without prove, it is evident that vD, i,
x = <vi ,+x>t - <vi,
–x>t and so on. In
words: The magnitude of the component of the average drift velocity of fly
number i in x-direction is given by the difference of the
average velocity components in +x and –x
direction. |
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This induces us to look now at the
ensemble, the swarm of flies. What can we
learn about the averages taken for the ensemble from the known averages of individual flies? |
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As long as every fly does - on average - the same
thing, the vector average over time of the
ensemble is identical to that of an individual fly - if we sum up a few
thousand vectors for one fly, or a few
million for lots of flies does not make any
difference. However, we also may obtain this average in a different way: |
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We do not average one fly in
time obtaining <vi>t , but at any given time all flies in space. |
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This means, we just add up the
velocity vectors of all flies at some moment in time and obtain <ve>r , the ensemble average. It is evident (but not
easy to prove for general cases) that |
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i.e. time average =
ensemble average. The new subscripts "e" and
"r" denote ensemble and space, respectively. This is a simple
version of a very far reaching concept in stochastic physics known under the
catch word "ergodic
hypothesis". |
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This means that in
"normal" cases, it doesn't matter how averages are taken. This is the
reason why text books are often a bit unspecific at this point: It is
intuitively clear what a drift velocity is and we don't have to worry about how
it is obtained. It also allows us to drop all indices from now on whenever they
are not really needed. |
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In our fly swarm example, the drift velocity
<vD>
= <vi> is usually much smaller than the average <vi> of
the velocity magnitudes of an individual fly. |
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The magnitude of <vD> is
the difference of two large numbers - the average velocity of the
individual flies in the drift direction
minus the average velocity of the individual flies in the direction opposite to the
drift direction. |
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This induces an asymmetry: From a knowledge of the drift velocity
only, no inference
whatsoever can be made with regard to <vi, +x> ,
<vi, –x> or <vi>
whereas knowlegde of <vi, +x> and <vi,
–x> tells us all there is to know
in x-direction |
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This teaches us a few things: |
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1. Don't confuse <v> with
<v>. The
first quantity - for our flies - is zero or small, whereas the second quantity
is large; they are totally different "animals". |
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2. This means in other words: Don't confuse
the property of the ensemble - the drift
velocity vD of the ensemble or swarm - with the properties of
the individuals making up the ensemble.
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© H. Föll (Electronic Materials - Script)
