# Derivative of unit vector in spherical coords.

1. Feb 9, 2017

### WendysRules

1. The problem statement, all variables and given/known data
Given $d \vec r = dr \hat r + r d \theta \hat {\theta} + r \sin \theta d \phi \hat {\phi}.$ Find $d \hat r , d \hat {\theta} , d \hat {\phi}.$

2. Relevant equations
I know that $d \hat {e_j} = \omega^i_j \hat {e_i}$ and that $\omega_{ij}=- \omega_{ji}$ and $0 = d \sigma^i + \omega^i_j \wedge \sigma^j$ and $d^2(r) = 0$ and lastly, $\omega_{ij}= \hat e_i \cdot d \hat e_j$

3. The attempt at a solution
Putting answers up here, as my work is below:

$d \hat r = d \theta \hat \theta + sin\theta d\phi \hat \phi$
$d\hat \theta = -d\theta \hat r - cos \theta d \phi \hat \phi$
$d \hat \phi = -sin\theta d\phi \hat r + cos \theta d \phi \hat \theta$
Which looks... wrong?

Using the above, we can get:
$d \hat r = \omega^r_r \hat r + \omega^\theta_r \hat \theta + \omega^\phi_r \hat \phi$
$d \hat \theta = \omega^r_\theta \hat r + \omega^\theta_\theta \hat \theta + \omega^\phi_\theta \hat \phi$
$d \hat \phi = \omega^r_\phi \hat r + \omega^\theta_\phi \hat \theta + \omega^\phi_\phi \hat \phi$
We also must use our torsion condition thus... we get:
$0 = d(dr) + \omega^r_r \wedge dr + \omega^r_\theta \wedge r d \theta + \omega^r_\phi \wedge r \sin \theta d \phi$
$0 = d( r d \theta) + \omega^\theta_r \wedge dr + \omega^\theta_\theta \wedge r d \theta + \omega^\theta_\phi \wedge r \sin \theta d \phi$
$0 = d ( r sin\theta d \phi) + \omega^\phi_r \wedge dr + \omega^\phi_\theta \wedge r d \theta + \omega^\phi_\phi \wedge r \sin \theta d \phi$

Expanding out our derivatives, we get:
$0 = 0 + \omega^r_r \wedge dr + \omega^r_\theta \wedge r d \theta + \omega^r_\phi \wedge r \sin \theta d \phi$
$0 = dr \wedge d \theta + \omega^\theta_r \wedge dr + \omega^\theta_\theta \wedge r d \theta + \omega^\theta_\phi \wedge r \sin \theta d \phi$
$0 = sin\theta dr \wedge d \phi + rcos\theta d \theta \wedge d \phi + \omega^\phi_r \wedge dr + \omega^\phi_\theta \wedge r d \theta + \omega^\phi_\phi \wedge r \sin \theta d \phi$

Solving our first equation, I see that $\omega^r_r= 0, \omega^r_\theta = a d \theta, \omega^r_\phi = b d\phi$, and using our metric compatibility, we know that $\omega^\theta_r= - a d \theta , \omega^\phi_r = - bd\phi$ where a, b are unknowns so far.
From second torsion condition, I see that $\omega^\theta_\theta = 0 , \omega^\theta_\phi = cd\phi$ which forces $\omega^\phi_\theta = - cd\phi$ where c is an unknown. For my 3rd torsion condition, I see that $\omega^\phi_\phi = 0$ Thus, plugging these into my torsion condition I see...
The first one turns into:
$0 = 0 + a d \theta \wedge r d \theta + b d\phi \wedge r \sin \theta d \phi$ which does equal 0 for any value of a, b.
The second one turns into:
$0 = dr \wedge d \theta + - a d \theta \wedge dr + 0 + cd\phi \wedge r \sin \theta d \phi$ which only works if a=-1, but will work for any value of c. (For why, a=-1 works, notice that if I flip my basis, I get $0 = dr \wedge d \theta + (- a) - dr \wedge d\theta + 0 + cd\phi \wedge r \sin \theta d \phi$ which will only make the $dr \wedge d \theta$ vanish if a = -1. )
The third one turns into:
$0 = sin\theta dr \wedge d \phi + rcos\theta d \theta \wedge d \phi + - bd\phi \wedge dr + - cd\phi \wedge r d \theta + 0$ where only if $b = -sin\theta and c=-cos\theta$ will satisfy this condition. To see why, we notice that $0 = sin\theta dr \wedge d \phi + rcos\theta d \theta \wedge d \phi + - (-sin) d\phi \wedge dr + -(cos)d\phi \wedge r d \theta$ which becomes $0 = sin\theta dr \wedge d \phi + rcos\theta d \theta \wedge d \phi -sin\theta dr \wedge d \phi + r(cos\theta)d \theta \wedge d \phi$ which causes all of our terms to vanish, woo.

However, when plugging this back to get my derivatives of my unit vectors, I get:
$d \hat r = d\theta \hat \theta + sin\theta d\theta \hat \phi$
$d \hat \theta = -d\theta \hat r + cos\theta d\theta \hat \phi$
$d \hat \phi = -sin\theta \hat r + -cos\theta \hat \theta$
Which looks... wrong? I don't think this is right, maybe there isn't a sum over the i's? But then if I use the fact that $\omega_{ij}=-\omega_{ji}$ I see that $\omega_{12} = -d \theta$ and $\omega_{21}= d\theta$, $\omega_{13} = -sin\theta d \phi$ and $\omega_{31} = sin\theta d\phi$ $\omega_{23}=cos\theta d\phi$ and $\omega_{32} = -cos\theta d\phi$ which confirms metric compatibility, since for all i,j did $\omega_{ij}=- \omega_{ji}$ (can provide work if needed, this post is getting long!)

Last edited: Feb 9, 2017
2. Feb 9, 2017

### Orodruin

Staff Emeritus
Why do you think it is wrong? Apart from the fact that you have missed the differentials in the last expression and only used $d\theta$ everywhere for the others.

3. Feb 9, 2017

### WendysRules

Oops haha, I got lazy and copy and pasted wrong terms since the post got long. On paper I have:
$d \hat r = d \theta \hat \theta + sin\theta d\phi \hat \phi$
$d\hat \theta = -d\theta \hat r - cos \theta d \phi \hat \phi$
$d \hat \phi = -sin\theta d\phi \hat r + cos \theta d \phi \hat \theta$ I'll edit this into my first post.

EDIT: Just realized that I'm not talking about curvature.

I guess it should make sense given that these are only my *first* derivatives, so my $d(\hat r)$ should'nt vanish

4. Feb 9, 2017

### Orodruin

Staff Emeritus
You must have terms involving $\hat r$. Compare with the case of polar coordinates in the plane.

5. Feb 9, 2017

### WendysRules

Okay, I think i found my confusion after I woke up this morning. The problem I did before this had us basically compute the same things, however, our line element was $ds^2=r^2d \theta^2 + r^2sin^2 \theta d \phi^2$ which would be a 2d sphere with a basis ${rd\theta, rsin\theta d\phi}$ in this scenario, I took it as saying r=constant, and when i expanded my derivatives in my torsion condition, they vanished i.e $d(r d\theta) = dr \wedge d\theta + r d^2 \theta = 0+0$ because dr=0 since r=const. Since it was late, I expected that for my 3d sphere which wouldnt be right.

However, now im not sure if taking r to be const. on my 2d sphere is the way to go... (sorry for any typos in latex, am on phone)

6. Feb 9, 2017

### WendysRules

I think ive convinced myself that r=const for sure, and that my calculations are correct. When we look at our $\Omega_{ij} = d\omega_{ij} + \omega_{ik} \wedge \omega_{kj}$ from our 3sphere and 2sphere, i noticed that on the 3sphere that $\Omega_{\theta \phi} = -2sin\theta d\theta \wedge d \phi$ and that $\Omega_{\phi \theta} = 2 sin\theta d \theta \wedge d \phi$ and on the 2 sphere, theyre the same just without the factor of 2! Thus, it makes sense that the curvature would basically be the same since the underlying geometry would be the same.