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WendysRules

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## Homework Statement

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}. ##

## Homework 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##

## The Attempt at a Solution

Putting answers up here, as my work is below:[/B]

## 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!)

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