Undergrad Matter density in Weinberg's Cosmology book

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In Weinberg's Cosmology, the matter density function ρ(r) must be expressed as a power series of r² near the origin to maintain analyticity in a spherically symmetric context. The absence of odd powers of r is crucial because including them would disrupt the function's analyticity at r = 0, where the coordinate system becomes singular. The discussion highlights that while ρ(r) depends solely on r, the divergence in spherical coordinates affects the behavior of the function. Ultimately, the necessity for ρ(r) to be a function of r² arises from the divergence properties of the gradient and vector Laplacian at the origin. This ensures that the function remains well-defined and analytic in the vicinity of r = 0.
jouvelot
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Hi everyone,

On Page 72 of S. Weinberg's Cosmology book, it's mentioned, just after Equation 1.9.16, that, for the universe matter density ρ(r) to be an analytic function near the origin (spherical symmetry), it has to be given near r = 0 by a power series of r2. I'm not a math wizard, so can anyone explain this little detail to me, please (why no odd powers of r)?

Thanks in advance.

Pierre
 
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If it contained powers of ##r##, then it would not be analytic. Consider the function ##r## itself.
 
Hi,

Thanks a lot for taking the time to answer my question, but I still don't get it. Analytic means, to me, for a function to be locally identical to its (convergent) Taylor expansion. Why wouldn't this work for r near 0, using spherical coordinates?

Thanks.

Bye,

Pierre
 
You are looking at it as a one-dimensional function. It is a three-dimensional function. Note that the coordinate system is singular at r=0.
 
Hi,

Indeed, but since ρ(r) is supposed to only depend on r and the coordinate divergence on the gradient only comes from θ and φ, I assumed this wouldn't be a factor. But I guess a 0 doesn't remove the 1/r divergence ;)

Thanks.

Bye,

Pierre
 
jouvelot said:
Indeed, but since ρ(r) is supposed to only depend on r and the coordinate divergence on the gradient only comes from θ and φ, I assumed this wouldn't be a factor. But I guess a 0 doesn't remove the 1/r divergence ;)
Answering my own question, if the gradient has only a 1/r divergence for r=0, the vector Laplacian diverges in 1/r2, thus mandating ρ(r) to be a function of r2.
 
I always thought it was odd that we know dark energy expands our universe, and that we know it has been increasing over time, yet no one ever expressed a "true" size of the universe (not "observable" universe, the ENTIRE universe) by just reversing the process of expansion based on our understanding of its rate through history, to the point where everything would've been in an extremely small region. The more I've looked into it recently, I've come to find that it is due to that "inflation"...

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