5
$\begingroup$

The magnitude of a vector in Cartesian coordinates is just the length of the vector. But what about a vector in spherical coordinates and cylindrical coordinates? Do we just take the root of its dot product with itself?

$\endgroup$

2 Answers 2

4
$\begingroup$

The magnitude of a vector in spherical coordinates is quite tricky, as you need to distinguish between points in $\mathbb R^3$ and vectors in $\mathbb R^3$. For example: The point $(r=0, \theta =0, \phi = 1) $ technically does not exit, but if it did it would be at a distance of 0 units from the origin. But the vector $\pmb{ \hat \phi }$ does exist, and has magnitude 1, like all unit vectors.
Now for the magnitude of a vector in spherical coordinates (in cylindrical coordinates it will be similar):
Starting with $\mathbf r = r\pmb{ \hat r }+ \phi\pmb{ \hat \phi }+\theta\pmb{ \hat \theta}$, and plugging in the following: $$\pmb{\hat r} = \sin\theta\cos\phi \pmb{\hat x} + \sin\theta\sin\phi\pmb{\hat y} + \cos\theta \pmb{\hat z}$$ $$\pmb{\hat \theta} = \cos\theta\cos\phi \pmb{\hat x} + \cos\theta\sin\phi\pmb{\hat y} - \sin\theta \pmb{\hat z}$$ $$\pmb{\hat \phi} = -\sin\phi \pmb{\hat x} + \cos\phi \pmb{\hat y}$$ (Taken from the back of Introduction to Electrodynamics 4th edition by David J. Griffiths. )
we get $$\mathbf r = r(\sin\theta\cos\phi \pmb{\hat x} + \sin\theta\sin\phi\pmb{\hat y} + \cos\theta \pmb{\hat z}) + \phi(-\sin\phi \pmb{\hat x} + \cos\phi \pmb{\hat y}) + \theta(\cos\theta\cos\phi \pmb{\hat x} + \cos\theta\sin\phi\pmb{\hat y} - \sin\theta \pmb{\hat z})$$ after rearranging as multiples of the rectangular unit vectors, we can find the magnitude of r by taking the root of its dot product with itself, or equivalently by taking the root of the squares of the xyz-components. Then after a lot of simplification, we come to the following: $$|\mathbf r| = \sqrt{r^2+\phi^2 +\theta^2}$$ What a surprise! (To me at least) If you do not believe it, I suggest you work it out yourself.

$\endgroup$
1
  • 1
    $\begingroup$ This is the correct answer and agrees with the answer to a problem I found elsewhere that I struggled with for a while before finding my way here. Thank you. $\endgroup$ Commented Feb 17, 2022 at 6:52
3
$\begingroup$

The "magnitude" of a vector, whether in spherical/ cartesian or cylindrical coordinates, is the same.

Think of coordinates as different ways of expressing the position of the vector. For example, there are different languages in which the word "five" is said differently, but it is five regardless of whether it is said in English or Spanish, say.

Think of the coordinates then as different languages in which you are expressing the position of the vector. In each language, you must know how to find the magnitude.

For example, in cartesian coordinates $(x,y,z)$ you would take the dot product with itself and then take the square root i.e. $\sqrt{x^2+y^2+z^2}$.

In spherical coordinates, one of the coordinates is the magnitude! Recall $(r,\theta,\phi)$ are the Spherical coordinates, where $r$ is the distance from the origin, or the magnitude. You can see here.

In cylindrical coordinates $(r,\theta,z)$, the magnitude is $\sqrt{r^2+z^2}$. You can see the animation here.

$\endgroup$
3
  • $\begingroup$ Yes but the spherical coordinates vector has three components, the theta unit vector, phi unit vector and the magnitude unit vector $\endgroup$
    – user29463
    Commented Sep 10, 2019 at 5:21
  • $\begingroup$ So I thought just take the sum of square of the components and find its root $\endgroup$
    – user29463
    Commented Sep 10, 2019 at 5:22
  • $\begingroup$ The sum of squares of the Cartesian components gives the square of the length. Also, the spherical coordinates doesn't have the magnitude unit vector, it has the magnitude as a number. For example, $(7 , \frac \pi 2, \frac \pi 2)$ makes sense in spherical coordinates. The first coordinate contains the length of the vector. $\endgroup$ Commented Sep 10, 2019 at 5:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .