2
$\begingroup$

My textbook has a table of coordinate transformations. Why are the coordinate variables and the vector components different? For example, for the conversion between spherical to cartesian for "z", it is equivalent to $R\cos(\theta)$, which I understand.

However, it is also said that the vector component of "z" is $A_R\cos(\theta)-A_\theta\sin(\theta)$ Why is this the case that they are different?

  • 0
    What is your definition of a vector? – 2017-02-24
  • 0
    @WilliamStagner The way they map points are basically the same as an absolute position, described by the coordinate variables. – 2017-02-24
  • 0
    A short answer is that the unit basis vectors in the spherical coordinate system change direction from point to point. See [this wikipedia article](https://en.wikipedia.org/wiki/Vector_fields_in_cylindrical_and_spherical_coordinates) for a gloss. – 2017-02-24

1 Answers 1

1

In the following development, note that the coordinate variables do not appear explicitly and have no influence on the result.

Note that a vector is independent of the choice of coordinate system. Let's represent a vector $\vec V$ using two different systems with unit basis vectors $(\hat e_1,\hat e_2,\hat e_3)$ and $(\hat e_1',\hat e_2',\hat e_3')$.

Then, we can write $\vec V=\vec V'$ as

$$\begin{align} \vec V&=\hat e_1V_1+\hat e_2V_2+\hat e_3V_3\\\\ &=\sum_{i=3}\hat e_iV_i \tag 1\\\\ =\vec V'&=\hat e_1'V_1'+\hat e_2'V_2'+\hat e_3'V_3'\\\\ &=\sum_{i=3}\hat e_i'V_i' \tag 2 \end{align}$$

We assume that each system is an orthogonal system, which means that $\hat e_i\cdot \hat e_j=\delta_{ij}$, where $\delta_{ij}$ is the Kronecker Delta and is equal to $1$ when $i=j$ and $0$ otherwise.

Forming the inner product of $\hat e_j$ with $(1)$ and $(2)$ and equating we find that

$$V_j=\sum_{i=1}^3 (\hat e_j\cdot \hat e_i')V_i' \tag 3$$

Now, let the unprimed system be the Cartesian system with unit basis vectors $((\hat e_1,\hat e_2,\hat e_3))=(\hat x,\hat y\hat z)$ and the primed system be the spherical system with unit basis vectors $(\hat e_1',\hat e_2',\hat e_3')=(\hat r,\hat \theta, \hat \phi)$.

Then, if $\hat e_j=\hat z$, then $(3)$ becomes

$$\begin{align} V_z&=(\hat z\cdot \hat r)V_r+(\hat z\cdot \hat \theta)V_\theta+(\hat z\cdot \hat \phi)V_\phi\\\\ &=\cos(\theta)V_r-\sin(\theta)V_\theta \end{align}$$

where we used $\hat z\cdot \hat e_i'=\cos(\text{angle between the unit vectors})$.

And we are done!