Intuitive methods for representation of Cartesian Coordinates in terms of Spherical Coordinates as basis [closed]

I was going through Griffith's Electrodynamics and came upon an example, where he used that, $$\cos\theta \ \hat{r} - \sin\theta \ \hat{\theta} = \hat{z}$$

Now I admit I was confused for a while with this, but I got the relationship on writing out the spherical unit vectors in the terms of cartesian coordinates.

Which is: $$\hat{e_r} = \sin\theta \cos\phi \ \mathbf{\hat{i}} +\sin\theta \sin\phi \ \mathbf{\hat{j}} +\cos\theta \ \mathbf{\hat{k}} \\ \hat{e_\theta}=\cos\theta \cos\phi \ \mathbf{\hat{i}} +\cos\theta \sin\phi \ \mathbf{\hat{j}} - \sin\theta \ \mathbf{\hat{k}} \\ \\ \ \\$$

All that done we one can verify the results easily. Now, my actual question is, how to determine this intuitively? Like, when we use the area element we can use our physical reasoning to find it (By taking infinitesimal displacements in the $$\theta \ \text{and} \ \phi$$ direction). The volume element in a similar way.

I know this may be suited for Math StackExchange, but since physicists are expert at shorthands like these, hence my question here.

• Since your intuition is likely quite different from mine, asking for an intuitive answer seems a difficult thing to do. – Jon Custer Sep 17 '19 at 18:26

It is all about drawing pictures with circles, arrows, angles, rectangles and right triangles.
Then apply the geometrical definitions of sine and cosine and see what you get.

• In the picture above consider the point marked as $$\color{lime}{\bullet}$$ (at radius $$r$$, angle $$\theta$$).
• Attached to this point draw the unit vectors $$\hat{r}$$, $$\hat{z}$$ and $$-\hat{\theta}$$.
By definition all three have length $$1$$.
• Now consider the tilted red rectangle.
• Its diagonal is $$\hat{z}$$.
• By applying the geometrical definition of sine in one right triangle, and of cosine in the other right triangle you find:
Its two sides (the red arrows) are $$\hat{r}\cos\theta$$ and $$-\hat{\theta}\sin\theta$$.
• Finally, the diagonal is the vector sum of these two sides:
$$\hat{z} = \hat{r}\cos\theta -\hat{\theta}\sin\theta$$

This geometric approach with pictures is usually much less error-prone than the algebraic approach with cartesian coordinates.

But be assured: To develop the ability for finding the needed pictures you will need practice, practice, practice...

The way to "determine this intuitively" is to draw pictures. Lots of pictures.

After drawing and redrawing the same basic pictures hundreds of times in different situations, you won't need to draw them any more.

IMHO one of the huge downsides of vector notation, from an educational point of view, is that it leads students in the direction of the sort of algebraic manipulations in your OP, rather than thinking visually "the distance you move up is this much of the radial distance, minus this much of the circumferential..." which is pretty obvious from a scribbled drawing that looks roughly like a circle and two lines.

• That method is pretty neat. But, how do I figure out the coefficients? By how much do I move along the circumference and the radius? Is there any way to relate circumferential distance to the problem? – Anjan Sep 16 '19 at 20:44
• @Anjan by drawing little arrows, angles, and using trigonometry. – Javier Sep 16 '19 at 20:53

It's obvious that $$\hat z$$ doesn't have a component along $$\hat\phi$$, so it can be written as some linear combination of $$\hat r$$ and $$\hat\theta$$,

$$\hat z=A\hat r+B\hat\theta.$$

The two coefficients are easily determined by taking dot products:

$$A=\hat z\cdot\hat r;$$ $$B=\hat z\cdot\hat\theta.$$

But the dot product between two unit vectors is just the cosine of the angle between them. By drawing a little diagram it is obvious that the angle between $$\hat z$$ and $$\hat r$$ is $$\theta$$, so

$$A=\cos\theta,$$

and the angle between $$\hat z$$ and $$\hat\theta$$ is $$\theta+\pi/2$$, so

$$B=\cos{(\theta+\pi/2)}=-\sin\theta.$$

Drawing pictures is a good start, but it only works in 2 and 3 dimensions. Also, it may lead you to errors, so it is good to have a brute-force algebraic method.

You can find the unit vectors using gradient. i.e. for any function f

$$\boldsymbol{\nabla}f=\mathbf{\hat{r}}\partial_r f + \frac{1}{r}\boldsymbol{\hat{\theta}}\partial_\theta f+\frac{1}{r\sin\theta}\boldsymbol{\hat{\phi}}\partial_\phi f$$

So by letting $$f=r$$

$$\mathbf{\hat{r}}\propto\boldsymbol{\nabla}{r}=\boldsymbol{\nabla}{\sqrt{x^2+y^2+z^2}}$$

But now you can evaluate the left expression in Cartesian basis

$$\mathbf{\hat{r}}\propto\frac{2x\mathbf{\hat{x}}+2y\mathbf{\hat{y}}+2z\mathbf{\hat{z}}}{2\sqrt{x^2+y^2+z^2}}=\frac{x}{r}\mathbf{\hat{x}}+\frac{y}{r}\mathbf{\hat{y}}+\frac{z}{r}\mathbf{\hat{z}}$$

From your definition of the spherical coordinates you will know that $$x=r\cos\phi\sin\theta$$ etc so:

$$\mathbf{\hat{r}}=\cos\phi\sin\theta\mathbf{\hat{x}}+\sin\phi\sin\theta\mathbf{\hat{y}}+\cos\theta\mathbf{\hat{z}}$$

Where I dropped the proportionality sign as it is easy to check normalization. Next

$$\mathbf{\hat{\theta}}\propto r\mathbf{\nabla}{\theta}$$

So

$$\boldsymbol{\hat{\theta}}\propto\frac{r}{-\sin\theta}\boldsymbol{\nabla}{\cos\theta}=\frac{-r}{\sin\theta}\boldsymbol{\nabla}{\frac{z}{r}}=\frac{-r}{\sin\theta}\boldsymbol{\nabla}{\frac{z}{\sqrt{x^2+y^2+z^2}}}=\frac{-r}{\sin\theta}\left(\frac{\mathbf{\hat{z}}}{r}+z\frac{2x\mathbf{\hat{x}}+2y\mathbf{\hat{y}}+2z\mathbf{\hat{z}}}{-2r^3}\right)$$

finally

$$\boldsymbol{\hat{\theta}}\propto\frac{-r}{\sin\theta}\left(\frac{\mathbf{\hat{z}}}{r}-\frac{\cos\phi\sin\theta\cos\theta\mathbf{\hat{x}}+\sin\phi\sin\theta\cos\theta\mathbf{\hat{y}}+\cos^2\theta\mathbf{\hat{z}}}{r}\right)=\cos\phi\cos\theta\mathbf{\hat{x}}+\sin\phi\cos\theta\mathbf{\hat{y}}-\sin\theta\mathbf{\hat{z}}$$

I will leave you to try the one for $$\phi$$. The key in this gymnastics was to know the expression for the gradient in spherical coordinates. This you can work out from the line element, i.e. the square distance due to infinitessimal displacement in spherical polars is $$ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2$$

Even easier thing to do is this. Let's say you want to know $$\mathbf{\hat{x}}.\boldsymbol{\hat{\theta}}$$:

$$\mathbf{\hat{x}}.\boldsymbol{\hat{\theta}}=\boldsymbol{\hat{\theta}}.\boldsymbol{\nabla}x=\boldsymbol{\hat{\theta}}.\left(\mathbf{\hat{r}}\partial_r x + \frac{\boldsymbol{\hat{\theta}}}{r}\partial_\theta x + \frac{\boldsymbol{\hat{\phi}}}{r\sin\theta}\partial_\phi x\right)=\frac{1}{r}\partial_\theta x=\frac{1}{r}\partial_\theta \left(r\cos\phi\sin\theta\right)=\cos\phi\cos\theta$$

Here is the vector diagram with $$\sin^2\theta + \cos^2 \theta = 1$$.

The components of the $$\hat r\,/\,\hat \theta$$ can be found by projecting $$\hat r\,/\,\hat \theta$$ onto the $$z$$ axis, and onto the $$xy$$ plane followed by projection onto the $$x$$ axis and the $$y$$axis.