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Derivative of unit vector in spherical coords.

  1. Feb 9, 2017 #1
    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. jcsd
  3. Feb 9, 2017 #2

    Orodruin

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    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.
     
  4. Feb 9, 2017 #3
    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
     
  5. Feb 9, 2017 #4

    Orodruin

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    You must have terms involving ##\hat r##. Compare with the case of polar coordinates in the plane.
     
  6. Feb 9, 2017 #5
    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)
     
  7. Feb 9, 2017 #6
    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.
     
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