Quadrupole tensor for spherical star (schutz ch9q28))

[Mmmm]

Homework Statement

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

Homework Equations

(9.78)
I^{jk}=\int T^{00}x^ix^jd^3x

(9.73)
\overline{I}_{jk}=I_{jk}-\frac{1}{3}\delta_{jk}I^l_l

The Attempt at a Solution

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

so for I^{11} I should have
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
=\int^r_0\int^{2\pi}_0 2 \rho(r,t)r^4 d\phi dr
=\int^r_0 4\pi \rho(r,t)r^4dr
=4\pi \int^r_0 \rho(r,t)r^4 dr


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

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...

 

[dx]

When you change to spherical coordinates, x¹x¹ will become r²sin²(θ)cos²(φ), not r².

 

[dx]

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)

 

[Mmmm]

Oh dear... yes you are right...
how silly of me
Thanks dx.