Purcell's approach to Larmor's formula, assumption? In 'Electricity and Magnetism' By E.M.Purcell, a derivation is given of Larmor formula (a version of which can be found here). I will give a brief overview here:

  
*
  
*A particle is considered at initial, non-relativistic, constant velocity $v$. It produces a field away from the instantaneous location of the particle.
  
*It then undergoes a (de)acceleration, $a$, to rest and produces the typical electric field associated with a point particle. 
  
*By application of continuity of field lines and the assumption that the field lines produced in the region of (de)acceleration are straight the field in this region can be derived, and thus Larmor's formula.
  

Purcell, nor anywhere else I have looked provides a justification for the assumption in bold. What is a valid reason to assume this is true? 
 A: The assumption of straight field lines in the deceleration shell has to do with the geometry of the derivation's setup, as shown in the figure below (from Purcell simplified; please keep in mind that not all assumptions are mentioned in the caption):

1) The width of the shell is much less than the radius of the shell. This is because the field is observed at some time $T$ much later than the deceleration event, such that $T >> \tau$, where $\tau$ is the duration of the deceleration (from the initial velocity $v_0$ to rest). In other words, if the inner and outer radii of the shell are $R_1 = c(T-\tau)$, $R_2 = cT$, the relative width of the shell is 
$$
\frac{R_2-R_1}{R_2} = \frac{c\tau}{cT} <<1
$$
2) The distance between the directions of the field line before deceleration, outside the shell, and after deceleration, inside the shell, labeled $\;v_0 T \sin\theta\;$ in the figure, is also very small compared to the shell radii, this time because it is assumed that the initial velocity $v_0 $ is much smaller than the speed of light, $v_0/c << 1$: 
$$
\frac{v_0 T \sin\theta}{c(T-\tau)} < \frac{v_0}{c}\frac{1}{1- \frac{\tau}{T}} \approx \frac{v_0}{c} <<1
$$
This means that the corresponding arc subtended on the inner boundary of the shell may be safely approximated as a straight chord. 
3) What is not emphasized enough, perhaps, is that the shell width $c\tau$ can also be assumed small compared to a chord of length $\;v_0 T \sin\theta\;$ provided $T$ is large enough,
$$
\frac{c\tau}{v_0 T \sin\theta} < 1
$$
This is because the deceleration time $\tau$ is fixed, while the observation time $T$ may always be taken as large as needed. 
4) If we assume in addition that the deceleration field line segment is smooth and has relatively uniform curvature, then the last condition above implies that its (average?) curvature is necessarily less than the curvature of the inner arc. Since its length is also much smaller than the shell radii, it too may be safely approximated as a straight segment.
