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Quadrupole tensor for spherical star (schutz ch9q28))

  • Thread starter Mmmm
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  • #1
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Homework Statement



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.

Homework Equations



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

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

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....
 

Answers and Replies

  • #2
dx
Homework Helper
Gold Member
2,011
18
When you change to spherical coordinates, x1x1 will become r2sin2(θ)cos2(φ), not r2.
 
  • #3
dx
Homework Helper
Gold Member
2,011
18
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)
 
  • #4
63
0
Oh dear..... yes you are right...
how silly of me :redface:
Thanks dx.
 

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