Uniform distributions and solid angle dependence in random pulsar geometry generation
Hayden Tornabene, Yuping Huang, and Joel Weisberg April 6, 2016 We first derive an expression that yields a uniform distribution of magnetic axes on the surface of a sphere, specifically for α between 0 and π and π . We note that α that α is the angle between the rotation rotation axis and the magnetic magnetic axis as seen in Figure 1. The problem we consider consider only requires requires one dimension (quantified by α), α ), but we include the azimuthal (longitude) angle φ for the sake of completeness. completeness. Two alternate alternate techniques techniques are given given for generating generating the uniform magnetic magnetic axis distri distribut bution ion.. Next, Next, in order order to fully fully simula simulate te the geometry geometry,, the observe observerr is brough broughtt into into the picture by deriving the distribution of the angle ζ angle ζ between between the spin pole and the observers’ line of sight. Finally, we address a question from Simon Johnston considering the observer’s role.
Figure Figure 1: Geome Geometry try of a pulsar pulsar where where α is the angle between between the rotational rotational and magnetic magnetic axes. Figure Figure reference: http://inspirehep.net/record/1095127/plots .
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First First Approa Approac ch to Uniform Uniformly ly Distri Distribu bute ted d m - Inte Integra grati tion on and CDFs
There are two separate ways to consider the problem of a uniform distribution of magnetic axes m, as show show in Figure 2. The first solution considers considers the integratio integration n of the cumulativ cumulativee distribution distribution function. function. We begin with a brief explanation explanation of cumulative cumulative distribution distribution function, function, or cdf. A cdf describes the probability probability
that some real-valued variable X with X with some probability distribution will be found to have a value less than or equal to x . That is, F X (x) = P ( P (X ≤ x).
(1)
The probability that X that X lies lies within the interval (a,b] is given, P ( P (a < X ≤ b) = F X (b) − F X (a),
(2)
where α is the angle from the north rotational pole to the magnetic axis (or the polar angle) and φ is the azimuthal angle in the xy the xy-plane -plane from the x the x-axis -axis with 0 ≤ φ < 2π 2 π, and Ω is the solid angle. The probability that a point lies in an infinitesimal cone of solid angle Ω is given as, P (Ω) P (Ω)d d Ω,
where the normalization is,
α
(3)
P (Ω) P (Ω)d d Ω = 1. 1.
(4)
φ
The probability density function (pdf) for uniform density over a unit sphere is given as, 1 . (5) 4π As φ As φ and and α α are are chosen independently, we can write the pdf, f(Ω) as the product of the two probability density functions that describe φ and α. As solid solid angle is a functio function n of α and φ, for dΩ = sin α dφ dα, dα, the joint probability density function for φ and α is given, f (Ω) f (Ω) =
sin α . (6) 4π Thus the two marginal distributions, that is the probability density functions for each variable with the other variable ”integrated out,” are given as, f ( f (φ, α) =
sin α f (α) = f ( f (φ, α)dφ = dφ = ; 2 1 2π
(7)
α
0
π
f φ (φ) =
f ( f (φ, α)dα = dα =
0
(8)
2π
As the cdf and the pdf are related by the following,
F (x) = f ( f (x),
(9)
and from the definition definition in Equation Equation 2 and the fundamen fundamental tal theorem of calculus, calculus, for a uniform uniform generated generated set of α α and φ, φ , the cdf’s for F(α F( α) and F(φ F(φ) are defined as,
α
F α (α) =
0
F φ (φ) =
α
f ( f (α)dα = dα =
0
φ
f ( f (φ)dφ = dφ =
0
sin α 1 − cos α dα = ; 2 2
0
φ
1 φ dφ = . 2π 2π
(10)
(11)
Now that we have two cumulative distribution functions that describe α and φ, we consider two random variables, v and and u , that are uniform uniform over the interv interval al [0,1]. These variable variabless represen representt the random numbers numbers created by a uniform random number generator. In other words, we can define the random numbers in terms of the cdf where, v = v = F( α) and u = u = F( φ). Solve for α for α and φ and we get the following inverse cdf’s:
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v = F = F α (α) =
1 − cos α ; 2
α = F = F α 1 (v) = cos −
−1
(12)
(2v (2v − 1). 1).
(13)
And similarly for φ for φ,, φ = F = F φ 1 (u) = 2πu −
(14)
An incorrectly distributed array of points around a sphere concentrates values at the poles as shown in Figure 2 whereas a correct distribution is truly uniform with the consideration of the inverse cdf’s, shown in Figure 3.
Figu Figure re 2: Inco ncorre rrect two-di o-dime mens nsio iona nall (α and φ) uniform distribution. poin points ts are are cen central trally ly conc concen entr trat ated ed when when the the inco incorr rrec ectt dist distri ribu buti tion on is used used.. http://www.bogotobogo.com/Algorithms/uniform distribution sphere.php .
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Note that the Figu Figure re refe refere renc nce: e:
Secon Second d Appro Approac ach h to Uniform Uniformly ly Distri Distribu buted ted m - Differen Differenti tial alss
One may alternativ alternatively ely consider the problem problem in terms terms of differentials. differentials. This formulation formulation comes directly from Yuping Huang, and we include it here for the sake of completeness. We consider some vector m pointing in space with a tilt α tilt α with with respect to the z-axis z -axis of an orthogonal (x,y,z ( x,y,z)) coordinate system. We can consider m as the magnetic vector and z as the spin axis. axis. Thus Thus the angle angle α is the angle formed between m and z and φ and φ is the longitudinal angle. For Ω, the solid angle at m where, as in the first approach, dΩ = sin α dφ dα.
(15)
We again denote f denote f (Ω) (Ω) has the probability density function (pdf) of Ω. As Ω is to be uniformly distributed around a sphere, then f (Ω) f (Ω) d dΩ Ω = C.
(16)
This equation says that if we take some infinitesimal patch of solid angle dΩ, no matter where one picks this patch, the probability the vector points in that direction is the same, namely the constant C , where C = = 1/(4π (4π ).
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Figure 3: Correct two-dimensional (α (α and φ) φ ) uniform distribution. Figure reference: Same as Fig. 2. As α As α and and φ φ are are chosen independently, again, the pdf f pdf f (Ω) (Ω) may be written as the product of pdf’s describing f ( f (α) and f and f ((φ). From Eq. 16, we then have, f ( f (φ)f ( f (α)sin α dα dφ = dφ = C. C.
(17)
In order for the product of these two PDFs to be equal to some constant C , we notice that both f ( f (φ)dφ and dφ and f ( f (α)sin α dα will dα will also have to be constant. For A For A and B constant, f ( f (φ)dφ = dφ = A, A,
(18)
and f ( f (α)sin α dα = dα = B B..
(19)
We want to find a function f ( f (α) (particularly a pdf) that satisfies the above constant equation. (Note that the same can be done for φ.) One may may propose propose that that f ( f (α) = B sin α satisfies the equation, and indeed it does. This means that that the probability probability of α being at a certain angle is proportional the the value of sine at that that angle. angle. (This (This is not saying saying sin α is uniformly distributed, as this would require an equation with pdf, f (sin f (sin α)d(sin α) = D). D ). Instead, we manipulate Equation 19 and find, f ( f (α) d( d (− cos α) = B.
(20)
So we have found that ( − cos α) is uniformly distributed, given the following consideration: If we consider a distribution f ( f (− cos α) instead of f (α), this pdf, along with φ would generate a uniform distribution where again, α again, α ∈ [0,π [0,π], Ω can be completely determined by φ by φ and and -(cos α). Equation Equation 15 becomes, becomes, dΩ = d( d (− cos α)dφ,
(21)
and thus Equation 17 becomes, f ( f (φ)f ( f (− cos α) d( d (− cos α)dφ = dφ = C. C.
(22)
The separation of variables again gives, f ( f (− cos α) d( d (− cos α) = B = B..
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(23)
Hence, (-cos α) is uniformly distributed on [-1,1], and since it is an odd function, cos α is similarly uniformly distributed. Notice that the above two approaches approaches yield identical identical results. results. Howeve However, r, the first approach approach provides provides a explicit function for α (and φ) that may be used in some randomized statistical routine aimed at probing possible pulsar geometry (for v and u some random value of α and φ respective respectively). ly). The second approach approach gives a more geometric argument, useful for conceptualizing the uniform distribution in terms of probability density functions.
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Addi Additio tiona nall issue issue - The The role of the observ observer er via the the spin-pol spin-pole e to line-of-sight angle ζ
The above two sections dealt with alternate, equivalent techniques for uniformly distributing a set of magnetic axes on the face of a sphere with a fixed spin axis; characterizing the location of each magnetic axis with the two angles (α, ( α, φ). Howeve However, r, in order to fully simulate simulate an observer observer’s ’s view, the observer observer must be brought into into the analysis! Specifically Specifically,, we wish to randomly align the spin axes of a set of spheres with respect respect to us. We do so via ζ , the angle between between the north spin-pole and the observer observer’s ’s line of sight. To visualize this, take a globe (with a flashlight representing the beam and magnetic axis) and rotate the spin axis orientation randomly in space to simulate a set of spin axes oriented randomly with respect to the observer observer.. Note how the line of sight sight changes with changing changing spin axis. We note that a uniform distribution distribution of ζ ζ simulate simulatess a random random set of pulsar spin axis orientatio orientations ns with respect to the observer. observer. It is crucial to note that this procedure is similar, but not the same as randomly placing magnetic axes on the surface the pulsar, as discussed above in Sections 1 and 2). It may seem intuitive to find the distribution of the parameter β rather β rather than of ζ , ζ , where β is β is the angle between the magnetic the magnetic axis axis and the line of sight. (The parameter β is β is shown in Figure 1.) However, while ζ is distributed uniformly, β uniformly, β has has a much more complicated distribution, since β = ζ = ζ − α,
(24)
and ζ and α each each have have their own distribu distributio tions. ns. (Of course course a value alue of β β can easily be derived from Equation 24, once a particular pair of α and η has been selecte selected.) d.) It can be shown shown that the cos 1 factor that we discovered in deriving the α distribution α distribution in Sections 1 and 2 may be applied to ζ to ζ in in the same way as we are randomly distributing it across a concentric shell just outside the pulsar. The geometry argument is identical where the ζ the ζ vector vector represents the radius of a sphere we are trying to randomly distribute. Instead of a rigorous derivation of the ζ distribution, ζ distribution, let us consider another reiteration of the following plausibility argument argument:: Imag Imagine ine a sphere sphere (the pulsar) with another shell just outside the pulsar surface. surface. Now imagine imagine that the line of sight is fixed on the inner sphere. We may uniformly distribute the spin axis around the shell just outside the pulsar in the same way that we randomly distribute the magnetic axis. This is identical to a fixed line of sight on the outer shell while the pulsar sphere below it randomly orients itself. −
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Aligne Aligned d vs. vs. Orthogon Orthogonal al Rotato Rotators rs
In an email on September 29, 2015 to Joel Weisberg and Hayden Tornabene, Simon Johnston articulated the following question when considering randomized pulsar geometries: There is I suppose the further question of the illumination of the sky which also depends on alpha and which goes to the question of detectability but not sure what you want to do about that. The extra issue to my thinking comes from the fact that an aligned rotator only illuminates a tiny fraction of the celestial sphere whereas an orthogonal rotator illuminates a large fraction. Therefore (all other things being equal which they may not be) you should detect more orthogonal rotators than aligned rotators. This also comes down to solid angle so I presume there’s some sort of cosine dependence there too.
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This problem speaks to the the angle between the spin axis and the magnetic axis, α. α . When α When α = = 0, we have an aligned rotator, rotator, or a pulsar where the magnetic and spin axes are the same vector. vector. As the pulsar spins, the area swept out by the beam is at its smallest for α = 0. When When α = π/2, π/2, we have what is called an orthogonal rotator. As α As α increases from 0 to π/ π /2, the area swept out by the beam becomes larger, implying that, if all else is the same, we should see more orthogonal orthogonal rotators rotators than aligned aligned rotators. Simply consider consider the ratio of the areas swept out by the beam. It seems to us that if both ζ and α and α are randomized uniformly and correctly, then this problem is addressed in the statistica statisticall routine. The above question question is simply simply saying that the range of possible ζ for ζ for a detected aligned rotator is much smaller than the range of possible ζ for ζ for an orthogonal orthogonal rotator. rotator. In other words, words, the aligned rotator must be pointed right at the earth (one possible ζ ) ζ ) to ensure detection of the pulsar, whereas the orthogonal rotator has a much larger range of possible ζ ζ values values that ensure detection. detection. Think back back to the shell around the sphere of the pulsar in Section 3, where the line of sight is being randomized. It would seem that no bias should be made on ζ , ζ , that is to say no additional scaling function (like a cos 1 ) is needed given that we are implicitly considering whether or not we see the pulsar when we uniformly distribute ζ . Any scaling function beyond the cos 1 , which ensures a uniform distribution of ζ of ζ ,, would incorrectly bias the routine designed to produce n produce n random pulsar geometries. −
−
Bibliography [1] Thanks Thanks to K. Hong and his blog post on bogotobogo.co bogotobogo.com: m: http://w http://www.bogot ww.bogotobogo.c obogo.com/Al om/Algorit gorithms/ hms/ uniform distribution sphere.php . Figures 2 and 3 are taken from this reference. [2] Fiigure 1 image credit: http://inspirehep.net/record/1095127/plots
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