1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quadrupole tensor for spherical star (schutz ch9q28))

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    28. Calculate the quadrupole tensor [itex]I_{jk}[/itex] and its traceless counterpart [itex]\overline{I}_{jk}[/itex] (Eq. (9.78)) for the following mass distributions.
    (a) A spherical star whose density is [itex]\rho(r, t)[/itex]. Take the origin of the coordinates in Eq. (9.73) to be the center of the star.

    2. Relevant equations

    [tex]I^{jk}=\int T^{00}x^ix^jd^3x[/tex]


    3. The attempt at a solution

    Well, first of all [itex]T^{00}=\rho[/itex]
    so 9.78 becomes
    [tex]I^{jk}=\int \rho(r,t)x^ix^jd^3x[/tex]
    and then it should just be a simple case of integrating using spherical polars....right?

    so for [itex]I^{11}[/itex] I should have
    [tex]I^{11}=\int^r_0\int^{2\pi}_0\int^{\pi}_0 \rho(r,t)r^2 r^2sin\theta d\theta d\phi dr[/tex]
    [tex]=\int^r_0\int^{2\pi}_0 2 \rho(r,t)r^4 d\phi dr[/tex]
    [tex]=\int^r_0 4\pi \rho(r,t)r^4dr[/tex]
    [tex]=4\pi \int^r_0 \rho(r,t)r^4 dr[/tex]

    and to me all seems to have gone well, but the answer in the back of the book is
    [tex]=\frac{4\pi}{3} \delta^{ij} \int^r_0 \rho(r,t)r^4 dr[/tex]

    which is close, but not the same. If I look at other components of I they are nothing like this at all! even the components with different j & k don't vanish!

    I'm obviously completely misunderstanding something here....
  2. jcsd
  3. May 20, 2009 #2


    User Avatar
    Homework Helper
    Gold Member

    When you change to spherical coordinates, x1x1 will become r2sin2(θ)cos2(φ), not r2.
  4. May 20, 2009 #3


    User Avatar
    Homework Helper
    Gold Member

    Just to clarify a little bit, x1 = x = r⋅sin(θ)cos(φ), so x1x1 = [r⋅sin(θ)cos(φ)]2.

    The transformation equations from Cartesian to spherical are

    x = r⋅sin(θ)cos(φ)
    y = r⋅sin(θ)sin(φ)
    z = r⋅cos(θ)
  5. May 21, 2009 #4
    Oh dear..... yes you are right...
    how silly of me :redface:
    Thanks dx.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook