Derivative of Unit Vector in a Rotating Frame

[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.

 

[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!?

 

[AlephZero]

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

 

[Jhenrique]

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

OH GOD!

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

 

[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.

 

[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.

 

[PeroK]

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:

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

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}

 

[Jhenrique]

look this

The last equation in book is wrong?

 

[PeroK]

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

 

[AlephZero]

The last equation in book is wrong?

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

 

[D H]

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.

 

[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...

 

[Sunfire]

Sorry if this is out of place:

time derivative of a unit vector in a rotating frame is
(http://en.wikipedia.org/wiki/Rotating_reference_frame)

\frac{d\vec{e}}{dt}=\vec{\omega} \times \vec{e}