Zero (e.g., unexpectedly trivial) poynting vector

[bjnartowt]

Homework Statement

The Poynting vector is,
${\bf{S}} = \frac{{{\mu _0}}}{{16{\pi ^2}c}}\frac{{{{({\bf{\hat r}} \times {\bf{\ddot p}})}^2}}}{{{{\left| {\bf{r}} \right|}^2}}}{\bf{\hat r}}$ [I.1]

Consider two charged particles, one at the origin (charge q1 and mass m1) and the other (charge q2 and mass m2) passing by with a large speed v, large enough that the trajectory is a straight line. The distance of closest approach is a. Find the Poynting vector in the center of mass frame, with use of [I.1].

Homework Equations

see attached .pdf

The Attempt at a Solution

see attached .pdf

My question: both charges are moving at constant speed with respect to one another. I totally do not see how the p-double dot in [I.1], the "acceleration" of the dipole moment, will be anything but zero. In fact, here's how I compute it; for both r'[1] and r'[2] being position vectors of two charges moving with constant velocity, it is obvious that,

$${\bf{\ddot p}} = \frac{{{d^2}}}{{d{t^2}}}\left( {{q_1}{{{\bf{r'}}}_1} + {q_2}{{{\bf{r'}}}_2}} \right) = {q_1}{\bf{0}} + {q_2}{\bf{0}} = {\bf{0}}$$

(This result is in the attached .pdf; see Eq. [I.10]).


I just need a third opinion to help me see if I'm not going crazy! The problem shouldn't be this trivial.

 

[mark726]

I understand your confusion regarding the Poynting vector and the acceleration of the dipole moment in this scenario. However, it is important to note that the Poynting vector is not solely dependent on the acceleration of the dipole moment, but also on the distance between the two charges and their velocities. In this case, the distance of closest approach (a) and the large speed (v) of the charges will play a significant role in determining the Poynting vector.

Additionally, it is important to consider the frame of reference in which the Poynting vector is being calculated. In the given scenario, the center of mass frame is being used, where the two charges are moving with constant velocities with respect to each other. Therefore, the acceleration of the dipole moment in this frame will be zero, as you have correctly calculated.

In summary, while the acceleration of the dipole moment may be zero in this scenario, the Poynting vector can still be calculated using the given equation and taking into account the distance of closest approach and the velocities of the charges.

I hope this helps clarify your confusion. If you have any further questions, please feel free to ask.