Complex current density of sphere confusion

[Shinobii]

Hello,

I am a bit confused on how to get this into the proper form,

$$
\begin{eqnarray}
\vec{J} &=& \vec{v}\rho \\
&=& (\vec{\omega} \times \vec{r}')\rho_o \Theta(R-r') \\
&=& \omega r' \sin(\theta')\rho_o \Theta(R-r')[\cos(\phi')\hat{y} - \sin(\phi')\hat{x}] \\
&=& \vdots \\
J_x + i J_y &=& -i \sqrt{\frac{3}{2 \pi}}\frac{q \omega r'}{R^3}\Theta(R-r')Y_{1,1}(\theta', \phi')
\end{eqnarray}
$$

I have tried expanding and such but the algebra does not work out. What am I missing conceptually? I know that the term in the brackets should equate to,

$$
[\cdots] = i e^{i \phi'}
$$

which I can then do some algebra to get the result into spherical harmonics. How does the term in brackets equate to [itex]i e^{i \phi'}?[/itex] Do I simply multiply the [itex]\hat{y}[/itex] term by i?

Hope you can shed some light on this matter!

EDIT

Yep, turns out I simply multiple [itex] \hat{y} [/itex] term by i. I just missed a minus sign! Sorry for the silly post!

 

[eljose79]

Hello,

It looks like you are working on a problem involving angular momentum and spherical harmonics. The term in brackets is a vector quantity, and it can be written in terms of the complex exponential as you have noted. In order to get the correct form, you just need to make sure that the vector components are multiplied by the appropriate imaginary unit, in this case i. So yes, you are correct in your solution - multiplying the \hat{y} term by i will give you the desired result.

In general, when dealing with complex quantities, it is important to keep track of the imaginary units and make sure they are properly accounted for in calculations. I'm glad you were able to figure it out and sorry for the confusion! Let me know if you have any further questions. Good luck with your research!