Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 1 down vote favorite

So, i'm going over the Thorne's derivation of the quadrupolar radiation term, and they write the core term as:

$$ \frac{3 r_i r_j - 2 r^2 \delta_{ij}}{4 r^5} $$

But if i try to obtain this term by Covariant deriving the dipole term;

$$ \nabla_{ij}{ \frac{1}{r}} = - \nabla_{j}{ \frac{r_i}{2 r^3}} $$

i am left with:

$$ \frac{3 r_i r_j - 2 r^2 \delta_{ij}}{4 r^5} - \frac{\Gamma^i_{kj}r_k}{2r^3} $$

Where the last term seems to be dismissed in the text with no good reason. Is there a reason why this term is being ignored?

share|improve this question
The quadropolar form is for flat space. – Ron Maimon Jul 10 '12 at 20:11
@RonMaimon, that i agree. But in the linear approximation, terms linear in $\Gamma$ are not ignored in the neighbourhood of a stellar object, only quadratic terms should be dismissed – lurscher Jul 10 '12 at 20:15
I see. You're asking how does the Schwartschild term deform the outgoing quadropole radiation--- the question of ougoing radiation is usually to expand the power in each outgoing multipole at infinity. The outgoing power also has an $\epsilon$ in it, it's weak presumably in the approximation you are using, so the geometrical correction from $\Gamma$ to the outgoing radiation profile looks second order to me, maybe you are thinking of a strong gravitational radiation case? What's the system? – Ron Maimon Jul 10 '12 at 20:32
i think your point is that this $\Gamma$ factor will multiply with the retarded source that already has linear terms, so it is second order as you point out – lurscher Jul 10 '12 at 20:44
yes, that's exactly it, but this might be an interesting and universal second order suppression of outgoing radiation, just by the ratio of the sphere area in Schwartzschild at the scale of the wavelength of the outgoing radiation to the same area in flat space (using coordinates where r is radial length, not sphere area). – Ron Maimon Jul 10 '12 at 21:56

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.