We left off with:
x(t')
= x o
+ ò t’ o v (t") dt"
Now
define:
d v(t’) =
v(t) - v o = (-e)E o/ m e ò t’’ o dt’ sin kx (t’)
è
d v(t’) = v o - (-e)E o/ m e ò t’’ o dt”’ sin kx (t’”)
Where: x = x o
+ v o (t"’)
v (t") =
v o + (e)E o/ m e k v o [ cos(kx o) + k v o t”) - cos(kx o)]
And the next to last integration yields:
x (t’) = x o + v o t’ - t’ eE o/ m e k
v o cos (kx o)
+ (e)E o/ m e k 2 v o 2 [sin (kx o) + k v o t’) – sin (kx o)]
Note the first integrand is of the form:
sin (kx o) + k v o t’)
+ D^ ]
Where:
D^ = t’ e E o/ m e v o cos (kx o)
+ eE o/ m e k
v o 2 [sin kx o) + k v o t’)
- sin kx o]
Given we seek the lowest order correction to E
o we can use Taylor expansion of the form:
sin (a +
D^) = sin
a +
D^ cos a
to lowest order in D^. Then:
d v(t) =
- e
E o/ m e ò t’ o dt [D^cos (kx o) + k v o t’)
+ sin (kx o + k v o t’)]
We then average d v over one wavelength
upon which the sine term disappears. The
other terms are evaluated using the identities:
<sin (u – a) cos (u – b)>u =
- ½ sin (a- b)
<sin (u – a) sin (u – b)>u =
<cos (u – a) cos (u – b)>u =
½ cos (a
- b)
Where < > means an average over one period
of the variable u. From this we find:
<d v(t) )>u =
(e E o/ m e ) 2 k/2 ò t’o dt’ {- 1/ k2 v o 2 sin (k v o t’) +
t’/ m e k
v o + cos (k v o t’)
Integration yields:
<d v(t) )>u = (e
E o/ m e ) 2 ½ k2 v o 2 | 2 | cos (k v o t) –
1] + k v o t sin
(k v o t) ]
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