الانكسار الجوي

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الانكسار الجوي

{Note: If your browser does not distinguish between "a,b" and "α, β" (the Greek letters "alpha, beta") then I am afraid you will not be able to make much sense of the equations on this page.}

The apparent position of an object in the sky may be changed by several different physical effects. One of these is refraction.

The speed of light changes as it passes through a medium such as air. We define the refractive index of any transparent medium as 1/v, where v is the speed of light in that medium.

The speed of light in air depends on its temperature and its pressure, so the refractive index of the air varies in different parts of the atmosphere.

Make a simple model of the atmosphere as n layers of uniform air above a flat Earth, with a different velocity of light vi for each layer (i from 1 to n). Apply Snell's Law of Refraction at each boundary.

At the first boundary, sin(i1) / sin(r1) = v0 / v1 .
At the next boundary, sin(i2) / sin(r2) = v1 / v2 , and so on.

But, by simple geometry, r1 = i2, r2 = i3 and so on.

So we have
sin(i1) = (v0 / v1) sin(r1) = (v0 / v1) sin(i2) = (v0 / v1) (v1 / v2) sin(r2) = (v0 / v2) sin(r2) = .......... = (v0 / vn) sin(rn)

In other words, the refractive indices of the intervening layers all cancel out. The only thing that matters is the ratio between v0(which is c, the speed of light in vacuum) and vn (the speed in the air at ground level).

Now rn is the apparent zenith distance of the star, z',
and i1 is its true zenith distance, z.
So sin(z) = (v0 / vn) sin(z').

Refraction has no effect if a star is at the zenith (z=0). But at any other position, the star is apparently raised; the effect is greatest at the horizon.

Define the angle of refraction R by R = z - z'. Rearrange this as z = R + z'. Then sin(z) = sin(R) cos(z') + cos(R) sin(z').

We assume R will be small, so, approximately, sin(R) = R (in radians), and cos(R) = 0.

Thus, approximately,
     sin(z) = sin(z') + R cos(z').
Divide throughout by sin(z') to get
     sin(z)/sin(z') = 1 + R/tan(z')
which is to say,
     v0/vn = 1 + R/tan(z').
So we can write
     R = (v0/vn - 1) tan(z')
We write this as
     R = k tan(z')
where k = (v0/vn - 1)

Here v0 is c, the velocity of light in a vacuum, which is constant.
But vn depends on the temperature and pressure of the air at ground level.
At "standard" temperature (0°C = 273K) and pressure (1000 millibars),
     k = 59.6 arc-seconds.
The formula in the Astronomical Almanac is
     k = 16.27" P/(273+T)
where P is in millibars, and T is in °C.

تمرين:

A star is at Right Ascension 5h 0m and declination +26°20'.
The latitude φ is +56°20'.
Local Sidereal Time is 5h 0m.
Atmospheric pressure is 1050 millibars,
and the temperature is +5°C. What is the star’s true altitude?

How much will the star’s image be shifted by atmospheric refraction, and in which direction?

What will be the star’s Right Ascension and declination, corrected for refraction?

الحل:

LST = RA, so the star is on the local meridian, so altitude a = (90°-φ) + δ = 60°.

How much will the star’s image be shifted by atmospheric refraction, and in which direction?

Zenith distance z = 90°-altitude = 30°.
Angle of refraction R = k tan(z')
where k = 16.27" P/(273+T) = 16.27" x 1050 /278 = 61.45".

However, we don't know z', only z.
For a first, approximate answer, take z = z' = 30°.
This gives R = 35.5",
so z' = z+R = 30° 0' 35.5".

Now recalculate R = k tan(z') = 35.5" (unchanged, so no need to iterate further)

So the star is raised by 35.5".

What will be the star’s Right Ascension and declination, corrected for refraction?

Since the star is on the local meridian, the shift is only in declination. The apparent altitude is increased by 35.5", so the apparent declination is similarly increased. So the apparent coordinates are:
Right Ascension 5h 0m and declination +26°20'35.5".

At large zenith distances, this model is inadequate. The amount of refraction near the horizon is actually determined observationally. At standard temperature and pressure, refraction at the horizon (horizontal refraction) is found to be 34 arc-minutes.