# Proof 1: Angular Separation in the Sky

Quote from Fahim Rajit Hossain on December 13, 2019, 3:10 pmAngular Separation

Spherical trigonometry is important to compute distances to astrophysical objects. However, due to the very large radial distances, the angular separation is small and thus, small-angle approximations (Taylor Series expansions) can be used to evaluate the functions of these small angles.Here,

We're considering mainly Equatorial Coordinate System

Where each symbol denotes,[latex] Latitude\,of\,an\,observer, = \phi [/latex]

[latex] Declination\,of\,a\,star = \delta [/latex]

[latex] Right\,Ascension\,of\,a\,star = \alpha [/latex]

Angular Separation between two object [latex] = \Delta\theta [/latex]Consider the diagram above:

[latex] \frac {sin(\Delta\theta)} {sin(\Delta\alpha)} = \frac {sin[90^\circ - (\delta +(\Delta \delta))]}{sin(\phi)} [/latex]

[latex] sin(\Delta \alpha) \, cos(\delta + (\Delta \delta)) = sin(\Delta \theta) \, sin(\Delta) [/latex]

Using small angle approximation

[latex]\Delta \alpha = \Delta \theta \, \frac{sin\phi}{cos\delta} [/latex]then

[latex] cos[90^\circ - (\delta +(\Delta \delta))] = cos(90^\circ - \delta) cos(\Delta \theta) + sin(90^\circ - \delta) sin(\Delta \theta) cos\phi [/latex]or

[latex] sin (\delta +(\Delta \delta)) = sin \delta cos(\Delta \theta) +cos\delta sin(\Delta \theta) cos\phi [/latex]finally,

[latex] \boxed {(\Delta \theta)^2 = (\Delta \alpha \, cos\delta)^2 + (\Delta \delta)^2 } [/latex]

Angular Separation

Spherical trigonometry is important to compute distances to astrophysical objects. However, due to the very large radial distances, the angular separation is small and thus, small-angle approximations (Taylor Series expansions) can be used to evaluate the functions of these small angles.

Here,

We're considering mainly Equatorial Coordinate System

Where each symbol denotes,

Angular Separation between two object

Consider the diagram above:

Using small angle approximation

then

or

finally,

Quote from Md Mahmudunnobe on December 15, 2019, 9:16 pm#1

So one very important thing to keep in mind that, we can only use the

small-angle approximationfor two points who are quiteclose to each other.For example, (Dhaka and Rajshahi) or at maximum (Dhaka and Delhi).

But we cannot use the approximation for larger distances such as (Dhaka and Seoul) on the globe or celestial sphere.

#2

One easier (at least to me) interpretation of using the small angle approximation is the following.

When two points are close enough, we can say that the line joining them ([latex]\Delta\theta[/latex] or AB in the picture) is approximately a line instead of an arc. Because of their closeness, the surface of the sphere can be approximated as flat for that small region.

So, we can consider [latex]\Delta \text{ANB}[/latex] as a normal triangle, not a spherical triangle and apply Pythagoras' Theorem.

The only difference is, the perimeter of the small circle for point N and B is smaller than the perimeter of the celestial equator. We need to express the angular distance NB with a similar scale of the angular distance at the celestial equator. For that reason, the angular distance NB is not equal to only [latex]\Delta \alpha[/latex], but NB = [latex]\Delta \alpha \cos (\delta+\Delta \delta)[/latex], [because the radius at declination [latex]\delta+\Delta \delta, r = R \cos (\delta+\Delta \delta) [/latex] ].

As [latex]\Delta \delta [/latex] is small enough, [latex] \cos(\delta+\Delta \delta) \approx cos(\delta) [/latex].

[latex ]\therefore (\Delta \theta)^2 = \text{AN}^2 + \text{NB}^2\\ \quad = (\Delta \delta)^2 + (\Delta \alpha \cos (\delta))^2[/latex]

#3

When the two points are large enough, we can no longer use the small angle approximation and need to consider [latex]\Delta \text{ANB}[/latex] as a speherical triangle.

In that case, in order to find the angular distance AB, we need to make a spherical triangle with A, B, and the north pole, P: [latex]\Delta \text{APB}[/latex].

Where,

[latex] AP = 90^{\circ} - (\delta+\Delta \delta)[/latex]

[latex] BP = 90^{\circ} - \delta[/latex]

[latex] \angle A = \Delta \alpha[/latex]

Using the cosine rule of a spherical triangle in [latex]\Delta \text{APB}[/latex], we can say that,

[latex]\cos AB = \cos BP \cos AP + \sin BP \sin AP \cos \angle A[/latex]

[latex]\cos \theta = \cos (90^{\circ} - \delta) \cos (90^{\circ} - (\delta+\Delta \delta)) + \sin (90^{\circ} - \delta) \sin (90^{\circ} - (\delta+\Delta \delta)) \cos (\Delta \alpha) [/latex]

[latex]\cos \theta = \sin\delta \sin (\delta+\Delta \delta) + \cos\delta \cos (\delta+\Delta \delta)\cos (\Delta \alpha) [/latex]

[latex]\therefore \theta = \cos^{-1} \left( \sin\delta \sin (\delta+\Delta \delta) + \cos\delta \cos (\delta+\Delta \delta)\cos (\Delta \alpha) \right) [/latex]

Let me know if I did any mistakes or if you have any questions.

Thanks

#1

So one very important thing to keep in mind that, we can only use the **small-angle approximation** for two points who are quite **close to each other. **

For example, (Dhaka and Rajshahi) or at maximum (Dhaka and Delhi).

But we cannot use the approximation for larger distances such as (Dhaka and Seoul) on the globe or celestial sphere.

#2

One easier (at least to me) interpretation of using the small angle approximation is the following.

When two points are close enough, we can say that the line joining them ( or AB in the picture) is approximately a line instead of an arc. Because of their closeness, the surface of the sphere can be approximated as flat for that small region.

So, we can consider as a normal triangle, not a spherical triangle and apply Pythagoras' Theorem.

The only difference is, the perimeter of the small circle for point N and B is smaller than the perimeter of the celestial equator. We need to express the angular distance NB with a similar scale of the angular distance at the celestial equator. For that reason, the angular distance NB is not equal to only , but NB = , [because the radius at declination ].

As is small enough, .

#3

When the two points are large enough, we can no longer use the small angle approximation and need to consider as a speherical triangle.

In that case, in order to find the angular distance AB, we need to make a spherical triangle with A, B, and the north pole, P: .

Where,

Using the cosine rule of a spherical triangle in , we can say that,

Let me know if I did any mistakes or if you have any questions.

Thanks