f*******d 发帖数: 339 | 1 一般说来,如果总电荷不为零,通过选择坐标有可能使有限系统的电偶极矩为零, 因为
\vec p = \int \vec x \rho(x)
如果不为零, 可以取
\vec x' = \vec x - \vec x0
\vec p' = \int \vec x \rho (x) - \vec x0 \int rho (x)
所以只要选
\vec x0 = (\int rho(x) \vec x)/\int \rho(x) = (\int rho(x) \vec x) / Q_tot
就足以令电偶极矩为零。这里的条件是总电荷不为零, 如果总电荷为零就没有办法了。
电四极矩有五个独立分量,独立的坐标变换有六个(三个平移三个旋转), 因此也可以
选座标
使之为零, 具体公式我就不写了,总之是写出其变换方程,然后求解。
但是不能同时使电偶极距为零。 如果总电荷和电偶极矩都恰为零, 那么上述平移变换不
起作用,
也无法使电四极矩为零。 总之, 最低一级不为零的多极矩是与坐标无关的,不可能消去
。 | g******i 发帖数: 251 | 2 This was a homework problem when I studied the E&M of Jackson.
The conclusion is if q is not 0, we can find a suitable coordinate so that
p=0. But if q is not 0, Q(i,j) can not be 0.
Further more, the first nonvanishing multipole are independent of the origin
of the coordinate axes, but the values of all higher multipole moments do in
general dependent on the choice pf origin.
You can prove above by descibing the relations of multipole moments before and
after changing the origin of the coordina |
|