# Derivative of unit vector

1. Jan 30, 2014

### Jhenrique

Before anyone thinks I didn't numerous attempts before opening this topic, take a look at my rough draft of mathematics in the annex.

So, a simple question. How derivate an unit vector wrt any variable? I can derivate any unit vector wrt θ or φ, obivious, but how derivate the vector φ wrt to x, for example? What is the rule? What is formula? I already searched and I not found.

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Last edited by a moderator: Jan 30, 2014
2. Jan 31, 2014

### Jhenrique

see as the calculation of a derivative of an unit vector is doubtful:

from http://mathworld.wolfram.com/SphericalCoordinates.html follows that:
$$\frac{d\hat{r}}{dr}=\vec{0}$$
$$\frac{d\hat{r}}{d\phi}=\vec{\phi}$$
by chain rule
$$\frac{d\hat{r}}{d\phi}=\frac{d\hat{r}}{dr}\frac{dr}{d\phi}$$
$$\vec{\phi}=\vec{0} \frac{dr}{d\phi}$$
what is an absurd!
The vec $\vec{\phi}$ isn't zero. If I can't believe in the chain rule thus I will believe in what!?

3. Jan 31, 2014

### AlephZero

None of that is correct. The Wolfram page says $$\frac{\partial\hat r}{\partial r} = \vec 0$$ not $$\frac{d\hat r}{dr} = \vec 0.$$

4. Jan 31, 2014

### Jhenrique

OH GOD!

Wich the difference between take the partial derivative versus the total derivative of an unit vector?

5. Jan 31, 2014

### PeroK

You have to think about the quantity you are differentiating and what it is a function of:

$$\hat r(\phi, θ) = (sin \phi cosθ, sin \phi sinθ, cos \phi)$$

So, it's not a function of r at all. Hence: $$\frac{\partial\hat r}{\partial r} = 0$$

And, it is a function of two variables. So, the derivatives wrt θ and ø will be partial.

Until you get used to multivariables, perhaps it's best to put them in each time you are differentiating. So, always write:

$$\hat r(\phi, θ)$$
So that you know it's a function of two variables.

6. Feb 1, 2014

### Jhenrique

By wolfram page (http://mathworld.wolfram.com/CylindricalCoordinates.html)

I can derivate the unit vector r by the christoffel's symbols and the derivative will be:

$\frac{\partial \hat{r}}{\partial \theta}=\frac{1}{r}\hat{\theta}$

or by the identity that exist in the page:

$\frac{\partial \hat{r}}{\partial \theta}=\hat{\theta}$

And this is more thing that makes me angry, and without understand why these equations do not coincide.

7. Feb 1, 2014

### PeroK

The second equation is correct. I don't know how you got the first equation. Instead, we have:
$$\vec r = r \hat r$$
$$\frac{\partial{\vec r}}{\partial θ} = r \frac{\partial{\hat r}}{\partial θ} = r \hat θ$$

So: $$\hat θ = \frac{1}{r} \frac{\partial{\vec r}}{\partial θ}$$

8. Feb 1, 2014

### Jhenrique

look this

The last equation in book is wrong?

9. Feb 1, 2014

### PeroK

Are you sure that relates to the cylindrical co-ordinate system?

Last edited: Feb 1, 2014
10. Feb 1, 2014

### AlephZero

Nothing in your image says that page of the book is about spherical coordinates. The fact that it only talks about two unit vectors and two Christoffel symbols, not three, suggests to me that it is about something else.

This is a simple way to get the right answers, without tying yourself in knots with fancy notation.
http://www.csupomona.edu/~ajm/materials/delsph.pdf [Broken]

Last edited by a moderator: May 6, 2017
11. Feb 1, 2014

### D H

Staff Emeritus
I would say that unless it's addressing some rather weird coordinate system (ellipsoidal, maybe?), it's wrong. If it's addressing either spherical or cylindrical coordinates, it's wrong.

12. Feb 1, 2014

### Jhenrique

But the coordinate system chosen by the author no matter because the factor 1/r no appears in any derivative of unit vector in cylindrical or spherical system. Conclusion, the book is wrong...!?...

PS: however, the factor 1/r appears a lot of times in wolfram page...

I don't know what is correct or wrong wrt to this christoffel's symbols...

Last edited: Feb 2, 2014
13. Feb 3, 2014