Adding two polar vectors

Is there a way of adding two vectors in polar form without first having to convert them to cartesian or complex form?

Answer

Here is another way forward that relies on straightforward vector algebra. Let r1 and r2 denote vectors with magnitudes r1 and r2, respectively, and with angles ϕ1 and ϕ2, respectively.

Let r be the vector with magnitude r and angle ϕ that denotes the sum of r1 and r2. Thus,

r=r1+r2

From the definition of the inner product we have

r1r2=r1r2cos(ϕ2ϕ1)

and

r1r=r1rcos(ϕϕ1)

Using (1) and (2), we find r2 can be written

r2=rr=(r1+r2)(r1+r2)=r1r1+r2r2+2r1r2r21+r22+2r1r2cos(ϕ2ϕ1)

and thus r is given by

\bbox[5px,border:2px solid #C0A000]{r=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}}\tag 4

Using (1), (3), and (4), yields

\begin{align}
\vec r_1\cdot \vec r&=r_1r\cos(\phi-\phi_1)\\\\
&=\vec r_1\cdot (\vec r_1+\vec r_2)\\\\
&=r_1^2+r_1r_2\cos(\phi_1-\phi_2)\\\\
\end{align}

whereupon solving for \cos (\phi-\phi_1) reveals

\cos(\phi-\phi_1)=\frac{r_1+r_2\cos(\phi_2-\phi_1)}{\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}}\tag 5

We can easily obtain the expression for \sin(\phi-\phi_1) by applying the cross product

\hat z\cdot(\vec r_1 \times \vec r)=\hat z\cdot(\vec r_1 \times \vec r_2)

which after straightforward arithmetic yields

\sin(\phi-\phi_1)=\frac{r_2\sin(\phi_2-\phi_1)}{\sqrt{r_1^2+r_2^2+2r_1r_2\cos(\phi_2-\phi_1)}} \tag 6

Dividing (5) by (6) and inverting shows that

\bbox[5px,border:2px solid #C0A000]{\phi =\phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),r_1+r_2\cos(\phi_2-\phi_1)\right)} \tag 7

where the function \operatorname{arctan2}(y,x) is described in this article.

Equations (4) and (7) provide the polar coordinates of \vec r strictly in terms of the polar coordinates of \vec r_1 and \vec r_2. And the development of (4), (5), and (6) did not appeal to Cartesian coordinates.


NOTE:

In a parallel development, we can express the sum of two complex numbers z_1=r_1e^{i\phi_1} and z_2=r_1e^{i\phi_2} in terms of their magnitudes and arguments.

First, recall that the inner product of two complex numbers is given by

\begin{align}
\langle z_1,z_2 \rangle &=z_1 \bar z_2\\\\
&=r_1r_2e^{i(\phi_1-\phi_2)}
\end{align}

where \bar z denotes the complex conjugate of z.

Next, we let z=re^{i\phi}=z_1+z_2 be the sum of z_1 and z_2. The magnitude of z is given by

\begin{align}
r&=\sqrt{\langle z,z \rangle}\\\\
&=\sqrt{\langle z_1+z_2,z_1+z_2 \rangle}\\\\
&=\sqrt{r_1^2+r_2^2+r_1r_2\left(e^{i(\phi_1-\phi_2)}+e^{-i(\phi_1-\phi_2)}\right)}\\\\
&=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}
\end{align}

Therefore, we have

\bbox[5px,border:2px solid #C0A000]{r=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}} \tag 8

Finally, we find the argument of z by taking the inner product of z and z_1. To that end, we write

\begin{align}
\langle z,z_1 \rangle &=rr_1e^{i(\phi-\phi_1)}\\\\
&=\langle z_1+z_2,z_1 \rangle \\\\
&=r_1^2+r_1 r_2 e^{i(\phi_2-\phi_1)}
\end{align}

which reveals that

e^{i(\phi-\phi_1)}=\frac{r_1+r_2e^{i(\phi_2-\phi_1)}}{\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}} \tag 9

whereupon inverting yields

\bbox[5px,border:2px solid #C0A000]{\phi =\phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),r_1+r_2\cos(\phi_2-\phi_1)\right)}

Equations (8) and (9) provide the polar coordinates of z strictly in terms of the polar coordinates of z_1 and z_2. Again, this development did not appeal to Cartesian coordinates.

Attribution
Source : Link , Question Author : lash , Answer Author : Mark Viola

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