Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 320 / 3 / b

Table of contents
- Solution b

Solution

Put

Ψ = - Φ = GM (r^{p} + a^{p})^{- 1/ p}

. The spherical Poisson equation gives the hypervirial model density

ρ (r) = \frac{( p + 1 ) M a ^{p}}{4 π} \frac{r ^{p - 2}}{( r ^{p} + a ^{p} ) ^{2 + 1/ p}} = \frac{( p + 1 ) M a ^{p}}{4 π ( GM ) ^{2 p + 1}} r^{p - 2} Ψ^{2 p + 1} .

For the proposed hypervirial distribution function, write

E = - E = Ψ - v^{2} /2

and

n = (3 p + 1) /2

. Direct velocity integration gives

ρ = A r^{p - 2} \int_{0}^{2Ψ} v^{p} (Ψ - v^{2} /2)^{n} d v \int_{0}^{2 π} d φ \int_{0}^{π} sin^{p - 1} α d α = A 2^{(p + 1) /2} π^{3/2} \frac{Γ ( p /2 ) Γ ( 3 ( p + 1 ) /2 )}{Γ ( 2 p + 2 )} r^{p - 2} Ψ^{2 p + 1} .

Matching this expression to the density fixes

A = \frac{( p + 1 ) a ^{p} Γ ( 2 p + 2 )}{2 ^{(p + 5) /2} π ^{5/2} G ^{2 p + 1} M ^{2 p} Γ ( p /2 ) Γ ( 3 ( p + 1 ) /2 )} .

This coefficient is positive for

0 < p \leq 2

, so the distribution function self-consistently generates the stated density and potential.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version