Issue 
A&A
Volume 566, June 2014



Article Number  A87  
Number of page(s)  11  
Section  Galactic structure, stellar clusters and populations  
DOI  https://doi.org/10.1051/00046361/201322919  
Published online  19 June 2014 
Online material
Appendix A
A.1. Derivation of Eqs. ( 2 ) and ( 3 )
The kernel function J defined in the introduction is singular at an isolated point (ζ = 0,r = R) and is continuous elsewhere. This singularity is integrable in the principalvalue sense in Eqs. (2), and (3). Function J is scaleinvariant. In particular, J(r,R,ζ) = J(r/R,1,ζ/R), which means that J is effectively dependent only on two variables: r/R and ζ/R. This property allows us to represent J on a plane as in Fig. A.1.
Fig. A.1
Contour plots of tanh(J(r,R,ζ)) and . These functions are singular at r = R and ζ = 0 and continuous elsewhere. 

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Differentiation and taking limits can be interchanged with the integration only under particular conditions imposed on the function under the integration sign. If not said otherwise, we assume that these conditions are met. For this reason, the case of an infinitesimally thin disk (with f(z) = δ(z)) must be treated separately.
The expression for R∂_{R}Ψ(R,Z) in Sect. 2 involves an integral The integral can be written as a sum if the two summands exist. By substituting , the first integral can be rewritten as , and next, since ρ(r, − ζ) = ρ(r,ζ) and J(r,R, − ζ) = J(r,R,ζ), as . Finally, by renaming , we obtain
This proves Eq. (2).
The integral expression for the vertical gradient of rotation in the quasicircular orbit approximation can be proved by performing a partial differentiation of ℐ under the integration sign: Now, ∂_{Z}J(r,R,z ± Z) = ± ∂_{z}J(r,R,z ± Z), which implies that . Integration by parts under usual conditions gives , where ℬ = (limz → + ∞ρ(r,z)[J(r,R,z + Z) − J(r,R,z − Z)]) − (ρ(r,0^{+})[J(r,R,Z) − J(r,R, − Z)]) is a boundary term. The expression in the first round bracket in this term is zero for any finite Z,R,r, on account of the vanishing of the difference in the square bracket and the vanishing of ρ in the same limit (it suffices to assume a finite support of ρ). The expression in the second round bracket also vanishes for finite ρ(r,0) since J is an even function of the third argument. Hence, the boundary term ℬ also vanishes. Formally, one should also verify whether the integration with respect to r and taking the above limit z → + ∞ commute, as there is an integration over r present in the expression for R∂_{R}Ψ. When this requirement is met, we have which proves Eq. (3).
A.2. A special case: the exponential vertical profile
In the case of the vertical exponential falloff of the density profile, , the calculation of the integrals in Eqs. (2) and (3) can be simplified. Then, and . Now, both v_{φ}(R,Z) and ∂_{Z}v_{φ}(R,Z) can be expressed in terms of two integrals I_{+} and I_{−}: Namely, for , and
A.3. Qualitative properties of the vertical gradient of rotation and the presence of turnovers
Equation (3) involves an integral [ J(r,R,z − Z)− J(r,R,z + Z) ] that equals J(r,R,Z − z), owing to the symmetry of J and f. Now, consider  Z  >A> 0 large enough, beyond the main mass concentration, such that (by assumption ). For such Z, only a region  z  ≪  Z  contributes to , and we can use the approximation formula J(r,R,Z − z) − J(r,R,Z) ≈ −z∂_{Z}J(r,R,Z) to obtain . Since f(z) and zf(z) vanish at the infinity, integrating by parts gives . On the other hand, for f(z) = δ(z) and for Z ≠ 0, . Hence, we obtain an intuitively clear result: for all finitewidth thin disks with the same column mass density σ(r), the behavior of the vertical gradient at high enough altitudes is universal and the same as for an infinitesimally thin disk with surface mass density σ(r).
Another qualitative result is obtained in the limit Z → 0. For Z ≠ 0, , thus by continuity of as a function of Z, and the vertical gradient is zero at Z = 0, at least for the mass distributions for which the usual theorems on the continuity of functions defined by integrals Eq. (3) apply.
However, there is an exception from the above continuity behavior of the gradient lines at Z = 0. It is important to remember that the operation of taking various limits and the operation of integration are not interchangeable in general. In particular, an integral of a function sequence consisting of continuous functions with a parameter can result in a discontinuous function of that parameter. For f(z) = δ_{n}(z), where δ_{n} is a functional sequence representing the Dirac δ, the result of continuity of the gradient does not necessarily follow and we can have a nonzero value in the same limit, in which case the integral Eq. (3) is discontinuous at Z = 0. To give a simple example of what then may happen, consider a function sequence , then , with . Now, consider a limiting function g(x) = limn → + ∞g_{n}(x) and see if it is continuous at x = 0. For x = 0, g(0) ≡ limn → + ∞g_{n}(0) = 0, whereas for x ≠ 0, , thus, g(0) = 0 ≠ 1 = limx → 0g(x), therefore g(x) is discontinuous at x = 0. Now, think of the gradient lines in Fig. 7 – then the finite disk corresponds to the situation described by g_{n}(x) (h = n^{1} > 0), whereas the infinitesimally thin disk corresponds to the situation of discontinuous g(x) (h = 0).
Finally, we may try to understand the occurrence of the turnovers in the gradient lines for h > 0, such as those seen in Fig. 10 or Fig. 7. First, note that a gradient line must asymptotically converge to zero, which is the universal asymptotics property discussed earlier. Second, the gradient line starts from 0 at Z = 0, which we have also seen above. Now, we perform a mapping of the region 0 < Z < +∞ to an interval 0 < Z < 1 by means of a transformation Z → tanhZ. Then the transformed gradient lines are continuous for 0 ≤ Z ≤ 1 and vanish at the boundaries. Next, we apply the Rolle theorem on continuously differentiable functions that vanish on the boundaries of a compact and simply connected interval, and we infer that there must be at least one point inside the interval where the gradient line has a local minimum, which explains the presence of a turnover. The Rolle theorem does not apply to gradient lines of the infinitesimally thin disk, because of the discontinuity, and the analogous turnovers do not have to occur, which is the case in Figs. 7 or 10.
© ESO, 2014
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