# Circumpolar Equation Derivation

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I am trying to understand where the circumpolar equation comes from.

$$delta geq frac{pi}{2} - l$$

I can understand that the sort of "rotation" of the reference frame that one experiences in horizon coordinates. I'm not sure where to go from the fact that a star's lowest altitude must be greater than 0.

It is pretty simple to show that $$a = l$$ and $$c = frac{pi}{2} - l$$. $$a$$ in this case is the altitude of the north celestial pole. What I need to do is show that the angle of the cone made by a circumpolar star with $$delta + l geq frac{pi}{2}$$.

You can deduce that $$a_{min} geq 0$$ for a star to be considered circumpolar. I just need to do this geometrically in terms of its declination from the celestial equator.

Edit: I think my latest attempt has found a solution.

For a star to be circumpolar, the star must, at the very least, come above the horizon.

$$0 leq a leq frac{pi}{2}$$

In the celestial sphere coordinate system, $$c$$ can be found to be

$$c = frac{pi}{2} - delta$$

In the horizon coordinate system, $$c$$ can be found to be

$$c = l - a_{min}$$

Therefore

$$l - a_{min} = frac{pi}{2} - delta$$ $$\$$ $$l + delta = frac{pi}{2} + a_{min}$$

$$a_{min}$$ can be set to 0 because it is the lower bound for circumpolarity.

$$l + delta = frac{pi}{2}$$

$$delta$$ must be in the cone made by the star's rotation about its declination angle. Therefore

$$frac{pi}{2} - l leq delta$$

There is nothing wrong with your explanation. You could add some explanatory notes like $$a_{min} = l-c$$ and $$a_{max}=l+c$$. Also, I suggest that you put the equation $$l+delta = frac{pi}2-a_{min}$$ on a separate line.

It is very good to figure out these equations on your own.



Here is an alternative explanation with less algebra:

• If you walk outside at night, you will notice that the north star has an elevation of $$l$$ degrees if your latitude is $$l$$.
• The north star does not move.
• The angular distance between stars does not change (in the short term!).
• In order for the a star to be circumpolar, it needs to be within $$l$$ degrees of the north star (otherwise it will dip below the horizon when its hour angle is 12 hours.)
• So, the angle $$heta$$ between the north star to a circumpolar star needs to be less than $$l$$. That angle is $$heta=90^circ-delta = pi/2 -delta$$ where $$delta$$ is the declination of the star.
• Finally, egin{align} heta &leq l pi/2 -delta &leq l pi/2 &leq delta + l pi/2 - l &leq delta. end{align}

## 2.10: Derivation of Wien's and Stefan's Laws

• Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

Wien's and Stefan's Laws are found, respectively, by differentiation and integration of Planck's equation. Neither of these is particularly easy, and they are not found in every textbook. Therefore, I derive them here.

## 1.16: Gaussian Quadrature - Derivation

• Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

In order to understand why Gaussian quadrature works so well, we first need to understand some properties of polynomials in general, and of Legendre polynomials in particular. We also need to remind ourselves of the use of Lagrange polynomials for approximating an arbitrary function.

First, a statement concerning polynomials in general: Let (P) be a polynomial of degree (n), and let (S) be a polynomial of degree less than (2n). Then, if we divide (S) by (P), we obtain a quotient (Q) and a remainder (R), each of which is a polynomial of degree less than (n).

That is to say: [frac

= Q + frac

. label <1.16.1> ag<1.16.1>]

What this means is best understood by looking at an example, with (n = 3). For example,

let [P = 5x^3 - 2x^2 + 3x + 7 label <1.16.2> ag<1.16.2>]

and [S = 9x^5 + 4x^4 - 5x^3 + 6x^2 + 2x - 3. label <1.16.3> ag<1.16.3>]

If we carry out the division (S ÷ P) by the ordinary process of long division, we obtain

For example, if (x = 3), this becomes

The theorem given by Equation ( ef<1.16.1>) is true for any polynomial (P) of degree (l). In particular, it is true if (P) is the Legendre polynomial of degree (l).

Next an important property of the Legendre polynomials, namely, if (P_n) and (P_m) are Legendre polynomials of degree (n) and (m) respectively, then

[int_<-1>^1 P_n P_m dx = 0 quad ext m = n. label <1.16.5> ag<1.16.5>]

This property is called the orthogonal property of the Legendre polynomials.

I give here a proof. Although it is straightforward, it may look formidable at first, so, on first reading, you might want to skip the proof and go on the next part (after the next short horizontal dividing line).

From the symmetry of the Legendre polynomials (see figure ( ext)), the following are obvious:

[int_<-1>^1 P_n P_m dx eq 0 quad ext m=n]

and [int_<-1>^1 P_n P_m = 0 quad ext m ext < or >n ext< is odd>.]

In fact we can go further, and, as we shall show,

[int_<-1>^1 P_n P_m dx = 0 quad ext m = n , ext < whether >m ext < and >n ext< are even or odd>.]

Thus (P_m) satisfies the differential Equation (see Equation 1.14.7)

which can also be written

[P_n frac left[ (1-x^2) frac ight] + m(m+1)P_m P_n = 0, label <1.16.8> ag<1.16.8>]

which can also be written

In a similar manner, we have

Subtract one from the other:

[frac left[ (1-x^2) left( P_n frac - P_m frac ight) ight] + [m(m+1) - n(n+1)]P_m P_n = 0. label <1.16.11> ag<1.16.11>]

[ left[ (1-x^2) left( P_n frac - P_m frac ight) ight]_<-1>^1 = [n(n+1) - m(m+1)] int_<-1>^1 P_m P_n dx. label <1.16.12> ag<1.16.12>]

The left hand side is zero because (1 &minus x^2) is zero at both limits.

[int_<-1>^1 P_m P_n dx = 0 . quad quad ext label <1.16.13> ag<1.16.13>]

I now assert that, if (P_l) is the Legendre polynomial of degree (l), and if (Q) is any polynomial of degree less than (l), then

I shall first prove this, and then give an example, to see what it means.

To start the proof, we recall the recursion relation (see Equation 1.14.4 &ndash though here I am substituting (l &minus 1) for (l)) for the Legendre polynomials:

The proof will be by induction.

Let (Q) be any polynomial of degree less than l. Multiply the above relation by (Qdx) and integrate from (&minus1) to (+1):

[l int_<-1>^1 P_l Q dx = (2l-1) int_<-1>^1 x P_ Q dx - (l-1) int_<-1>^1 P_ Q dx. label <1.16.16> ag<1.16.16>]

If the right hand side is zero, then the left hand side is also zero.

A correspondent has suggested to me a much simpler proof. He points out that you could in principle expand (Q) in Equation ( ef<1.16.14>) as a sum of Legendre polynomials for which the highest degree is (l-1). Then, by virtue of Equation ( ef<1.16.13>), every term is zero.

For example, let (l = 4), so that

and let [Q = 2(a_3 x^3 + a_2 x^2 + a_1 x + a_0 ) . label <1.16.19> ag<1.16.19>]

It is then straightforward (and only slightly tedious) to show that

[int_<-1>^1 P_ Q dx = left( frac<6> <5>- frac<2> <3> ight) a_2 label <1.16.20> ag<1.16.20>]

and that [int_<-1>^1 xP_ Q dx = left( frac<10> <7>- frac<6> <5> ight) a_2 . label <1.16.21> ag<1.16.21>]

But [7 left( frac<10> <7>- frac<6> <5> ight) a_2 - 3 left( frac<6> <5>- frac<2> <3> ight) a_2 = 0, label <1.16.22> ag<1.16.22>]

and therefore [int_<-1>^1 P_4 Q dx = 0 . label <1.16.23> ag<1.16.23>]

We have shown that [l int_<-1>^1 P_l Q dx = (2l - 1) int_<-1>^1 x P_ Q dx - (l - 1) int_<-1>^1 P_ Q dx = 0 label <1.16.24> ag<1.16.24>]

for (l = 4), and therefore it is true for all positive integral (l).

You can use this property for a parlour trick. For example, you can say: &ldquoThink of any polynomial. Don&rsquot tell me what it is &ndash just tell me its degree. Then multiply it by (here give a Legendre polynomial of degree more than this). Now integrate it from (&minus1) to (+1). The answer is zero, right?&rdquo (Applause.)

Thus: Think of any polynomial. (3x^2 - 5x + 7). Now multiply it by (5x^3 - 3x). OK, that&rsquos (15x^5 - 25x^4 - 2x^3 + 15x^2 - 21x). Now integrate it from (&minus1) to (+1). The answer is zero.

Now, let (S) be any polynomial of degree less than (2l). Let us divide it by the Legendre polynomial of degree (l), (P_l), to obtain the quotient (Q) and a remainder (R), both of degree less than (l). Then I assert that

This follows trivially from Equations ( ef<1.16.1>) and ( ef<1.16.14>). Thus

[int_<-1>^1 S dx = int_<-1>^1 (QP_l + R) dx = int_<-1>^1 Rdx . label <1.16.26> ag<1.16.26>]

Example: Let (S = 6x^5 - 12x^4 + 4x^3 + 7x^2 - 5x + 7). The integral of this from (&minus1) to (+1) is (13.86). If we divide (S) by (frac<1> <2>(5x^3 - 3x)), we obtain a quotient of (2.4x^2 - 4.8x + 3.04) and a remainder of (-0.2x^2 - 0.44x + 7). The integral of the latter from (&minus1) to (+1) is also (13.86).

I have just described some properties of Legendre polynomials. Before getting on to the rationale behind Gaussian quadrature, let us remind ourselves from Section 1.11 about Lagrange polynomials. We recall from that section that, if we have a set of n points, the following function:

(in which the (n) functions (L_i(x )), (i = 1,n), are Lagrange polynomials of degree (n-1)) is the polynomial of degree (n-1) that passes exactly through the (n) points. Also, if we have some function (f(x)) which we evaluate at (n) points, then the polynomial

is a jolly good approximation to (f(x)) and indeed may be used to interpolate between nontabulated points, even if the function is tabulated at irregular intervals. In particular, if (f(x)) is a polynomial of degree (n &minus 1), then the expression ( ef<1.16.28>) is an exact representation of (f(x)).

We are now ready to start talking about quadrature. We wish to approximate (int_<-1>^1 f(x) dx) by an (n)-term finite series

[int_<-1>^1 f(x) dx approx sum_^n c_i f(x_i), label <1.16.29> ag<1.16.29>]

where (-1 < x_i < 1). To this end, we can approximate (f(x)) by the right hand side of Equation ( ef<1.16.28>), so that

[int_<-1>^1 f(x) dx approx int_<-1>^1 sum_^n f(x_i) L_i (x) dx = f(x_i) int_<-1>^1 sum_^n L_i (x) dx . label <1.16.30> ag<1.16.30>]

Recall that the Lagrange polynomials in this expression are of degree (n &minus 1).

The required coefficients for Equation ( ef<1.16.29>) are therefore

Note that at this stage the values of the (x_i) have not yet been chosen they are merely restricted to the interval [&minus1 , 1].

Now let&rsquos consider (int_<-1>^1 S(x) dx), where (S) is a polynomial of degree less than (2n), such as, for example, the polynomial of Equation ( ef<1.16.3>). We can write

[int_<-1>^1 S(x) dx = int_<-1>^1 sum_^n S(x_i) L_i (x) dx = int_<-1>^1 sum_^n L_i (x) [Q(x_i) P(x_i) + R(x_i)] dx. label <1.16.32> ag<1.16.32>]

Here, as before, (P) is a polynomial of degree (n), and (Q) and (R) are of degree less than (n).

If we now choose the (x_i) to be the roots of the Legendre polynomials, then

[int_<-1>^1 S(x) dx = int_<-1>^1 sum_^n L_i (x) R(x_i) dx. label <1.16.33> ag<1.16.33>]

Note that the integrand on the right hand side of Equation ( ef<1.16.33>) is an exact representation of (R(x)). But we have already shown (Equation ( ef<1.16.26>)) that (int_<-1>^1 S(x) dx = int_<-1>^1 R(x) dx), and therefore

[int_<-1>^1 S(x) dx = int_<-1>^1 R(x) dx = sum_^n c_i R(x_i) = sum_^n c_i S(x_i) . label <1.16.34> ag<1.16.34>]

It follows that the Gaussian quadrature method, if we choose the roots of the Legendre polynomials for the (n) abscissas, will yield exact results for any polynomial of degree less than (2n), and will yield a good approximation to the integral if (S(x)) is a polynomial representation of a general function (f(x)) obtained by fitting a polynomial to several points on the function.

## Find torsion coefficient

When the balance bar is initially released and the moving balls approach the larger balls, the inertia of the smaller balls causes them to overshoot the equilibrium angle. The torsion coefficient must be calculated by measuring the resonant oscillation period of the wire.

### Oscillation period

This results in the torsion balance oscillating back-and-forth at its natural resonant oscillation period:

• T is the oscillation period in seconds
• &pi (small Greek letter pi) is 3.14.
• I is the moment of inertia of the torsion bar in kg-m 2
• &kappa is the torsion coefficient in newton-meters/radian.

Note: Since the balls are heavy lead, the mass of the bar is considered negligible and not a factor in the inertia.

### Solve for torsion coefficient

Square T = 2&pi&radic(I/&kappa) and solve for torsion coefficient:

### Moment of inertia

The moment of inertia of the smaller balls is:

Substitute inertia in the torsion coefficient equation:

## Reflection off moving object

One method to determine the velocity of an object is to reflect a wave off the object and measure the Doppler shift caused by the motion. In this case, both the velocity of the source and observer are zero: vS = 0 and vO = 0. The observer is usually nearby the source.

Waves moving toward moving object

Waves reflected off moving object

### Waves "observed" by moving object

Let vR be the velocity of the object, moving in the x-direction. The wavelength and frequency "observed" by the object are:

• &lambdaR is the observed wavelength of the moving object
• &lambdaS is the original source wavelength
• vR is the constant object velocity in the x-direction
• fR is the observed frequency at the moving object
• fS is the original source frequency

### Waves reflected to stationary observer

The object reflects the "observed" waves as if the object was a moving source.

Note: Although the motion is still in the positive direction, the wave is now moving in the negative direction. Thus, the sign of c must change.

#### Wavelength equation

The wavelength equation for a moving source and stationary observer is:

However, &lambdaR represents the reflected source wavelength and vR is the velocity of the reflecting object, acting as a source. Replace &lambdaS with &lambdaR and vS with vR in the equation. Also, change the sign of c since the wave is moving in the opposite direction.

Thus, the reflected wavelength equation is:

where &lambdaO is the wavelength measured by the stationary observer.

Subtract &lambdaOc and &lambdaSvR from both sides of the equaiton:

If the object is moving in the opposite direction, vR becomes negative, and the equation is:

#### Frequency equation

The frequency equation for a moving source and stationary observer is:

However fR represents the reflected source frequency and vR is the velocity of the reflecting object, acting as a source. Also, the sign of c changes.

The reflected frequency equation is:

Using the equation fR = fS(c &minus vR)/c, substitute for fR and then solve for vR:

## Teaching

This course presents an indepth treatment of the dynamics of astrophysical fluids. This includes both collisional and collisionless fluids, as well as neutral and charged fluids (plasmas). After a first-principles derivation of the various fluid equations (continuity, momentum and energy) that links continuum mechanics to kinetic theory, and a brief discussion of astrophysical equations of state, we will focus on specific types of flows, including inviscid barotropic flow, turbulent flow, viscous accretion flow, shocks, and spiral density waves. We then study various fluid instabilities (convective instability, thermal instability, interface instabilities, gravitational instability) with applications to astrophysics. Next we discuss numerical hydrodynamics, and end with a treatment of plasma physics, including plasma orbit theory, magneto-hydrodynamics (MHD), magnetic tension and Alfven waves, the Vlasov equation and the two-fluid model, magnetic reconnection and dynamos, and various astrophysical applications of plasma physics.

##### ASTR 610: Theory of Galaxy Formation [Yale F20,F18, S17, S15, F12]

This course is aimed at graduate students in physics or astronomy

This course prepares the student for state-of-the-art research in galaxy formation and evolution. Topics include Newtonian perturbation theory, the spherical collapse model, formation and structure of dark matter haloes (including Press-Schechter theory), the virial theorem, dynamical friction, cooling processes, theory of star formation, feedback processes, elements of stellar population synthesis, chemical evolution modeling, AGN, and supermassive black holes. The course also includes a detailed treatment of statistical tools used to describe the large scale distribution of galaxies and introduces the student to the concepts of galaxy bias and halo occupation modeling. During the final lectures we discuss a number of outstanding issues in galaxy formation.

##### ASTR 170: Introduction to Cosmology [Yale F13, S11]

This course is aimed at undergrad non-science majors

Cosmology is the study of the origin, structure and evolution of the Universe itself: the totality of phenomena of space and time. It is the oldest science, and addresses the biggest questions: How old is the Universe? Did time have a beginning? What is the Universe made of? What are Dark Matter and Dark Energy? Is the Universe finite, and if so, what do we find at the edge? Are we alone in the Universe? In this course we journey from the Ancient Greek world views of Ptolemy and Aristotle to the hot Big-Bang model of modern-day cosmology. Along the way we learn how stars shine, how black holes form, how galaxies take on their shapes, and how Copernicus, Galilei, Newton, Einstein, Hubble and others have transformed our geocentric views to one in which we inhabit just a small planet immersed in an infinite, expanding space-time that is 13.7 billion years old and consists of dark matter and dark energy, and in which galaxies are the product of quantum fluctuations.

##### ASTR 530: Galaxies [Yale F10]

This course is aimed at graduate students in physics/astronomy

This course provides the student with a survey of the content, structure, dynamics, formation and evolution of galaxies. After a detailed overview of the various components of galaxies (disk/spheroid, stars, gas, dark matter, supermassive black holes), their statistical properties (luminosity function, size distribution, color distribution, metallicity distribution), and the corresponding scaling relations, the course focusses on the physical processes underlying galaxy formation and evolution.

##### ASTR 5580: Extragalactic Astronomy [Utah S10]

This course is aimed at graduate students in physics or astronomy

##### Dynamics of Collisionless Systems [ETH S05]

This course is aimed at graduate students in physics/astronomy

## Circumpolar Equation Derivation - Astronomy

This reduces, by Ampere's law, to,

where is magnetic diffusivity. By simple use of a vector identity, using Gauss's law, and assuming constant 1 η this further reduces to,

which is known as the induction equation. This states that a local change in the magnetic field is due to both convection and diffusion. The magnetic Reynolds number is the ratio of the convective to the diffusive term,

and is an indication of the coupling between the plasma flow and the magnetic field. This can simply be approximated to,

for length scale, l 0 , and plasma speed, v 0

For large R m (ݏ), as found in most astrophysical cases, the convective term dominates in Eqn. 12, and the field lines move as if they are frozen into the plasma (Alfvén, 1943) with typical timescale, t c = l 0 /v 0 . For small R m (گ), typically found in laboratory plasmas, the diffusion term dominates and flux `leaks' with typical ohmic diffusion timescales then given by t d = l 0 2 /η. For typical solar photospheric values (v 0 10 m s -1 ,η

10 3 m 2 s -1 ) R m becomes less than unity (t d

100m, ohmic dissipation becomes important, and magnetic reconnection may occur.

## 5. Synthesis

[46] According to the estimates given in Table 4, the combined average annual sediment flux of the eight largest arctic rivers is 249 Mt/yr. By comparison, estimates provided in other papers for these eight rivers yield combined flux estimates of 165 Mt/yr [ Lisitzin, 1972 ], 175 Mt/yr [ AMAP, 1997 ], and 178 Mt/yr [ Walker, 1998 ]. In all cases, the majority of the difference between our estimate and the others comes from the Mackenzie River. In work by AMAP [1997] and Walker [1998] , the Mackenzie values are erroneously low due to propagation of a typographical error from Milliman and Syvitski [1992] . The Mackenzie value given by Lisitzin [1972] is also unrealistically low.

[47] The eight rivers that have been the focus of this paper contribute ∼65% of riverine freshwater inputs to the Arctic Ocean, but are they equally significant in terms of sediment flux? This is a difficult question to answer, largely because there is only limited data for smaller arctic rivers which may contribute disproportionally large amounts of sediment [ Milliman and Syvitski, 1992 ]. Gordeev et al. [1996] provide the most complete list of estimates, with values presented for 20 Eurasian arctic rivers. In addition, they give flux estimates for other, presumably ungauged, areas in the Eurasian arctic. Their estimate of total sediment flux in Eurasian arctic rivers is 115 Mt/yr, whereas our estimate from the sum of the 16 Eurasian arctic rivers presented in Tables 4 and 5 is 84 Mt/yr. Ungauged areas and extra rivers included in the Gordeev et al. [1996] compilation account for over half of the difference between our estimate and theirs. The remaining difference is due to higher estimates for some rivers given by Gordeev et al. [1996] as compared to our new estimates. Regardless of this difference, it is clear that many rivers make substantial contributions to the total sediment flux from Eurasia to the Arctic Ocean. In contrast, it is highly likely that the Yukon and Mackenzie rivers carry most of the river sediment from the North American Arctic because they drain the areas of tectonism and active alpine glaciation that are the great generators of fluvial sediment. Data from smaller rivers in North America are needed to confirm this.

[48] Although sediment yields vary greatly among arctic rivers (Tables 4 and 5), distinct geographical patterns are evident. The Yukon and Mackenzie rivers contribute only 21% of the combined annual water discharge of the eight largest arctic rivers (Table 1), but transport 73% of the suspended sediments. In contrast, the Yenisey, Lena, and Ob contribute 65% of the combined annual water discharge of the eight largest arctic rivers while transporting only 17% of the suspended sediments. Sediment yields in these three rivers have sometimes been considered anomalously low [ Milliman and Meade, 1983 ], but in fact their yields are generally in line with what has been observed in other lowland rivers [ Milliman and Syvitski, 1992 ].

[49] Variations in sediment concentration as a function of water discharge among the eight largest arctic rivers (Figure 4) also reflect geographical patterns. The drainage basins of the Mackenzie, Yukon, and Kolyma rivers share features of geology and climate that set them apart from the drainage basins of the Yenisey, Lena, Ob', Pechora, and S. Dvina rivers [ Gordeev et al., 1996 Semiletov et al., 2000 ]. This division is broadly reflected in Figure 4, although the Pechora is an obvious exception. In any case, the distribution of rivers in Figure 4 reminds us that simply grouping rivers according to their continental affiliations can disguise functional differences.

[50] Although we have stated that we are addressing sediment flux to the Arctic Ocean, in fact we are using this phrase rather loosely. Instead, we are evaluating sediment flux in the downstream reaches of major arctic rivers, much of which may be retained in the marginal filter [ Lisitzin, 1995 ]. The distribution of this sediment in deltas, estuaries, and the broad shelf of the Arctic Ocean is often unclear [ Bauch et al., 2001 ]. As pointed out earlier, there is considerable disagreement about the proportion of Lena River sediment that reaches the Laptev Sea, with estimates ranging from 10 to nearly 100% [ Alabyan et al., 1995 Are and Reimnitz, 2000 Rachold et al., 2000 ]. For the Mackenzie, it has been estimated that about half of the river's suspended sediment is transported through the extensive Mackenzie Delta [ Macdonald et al., 1998 ], but it seems unlikely that a significant portion of the suspended sediment from the Yenisey and Ob' rivers is transported through their lengthy estuaries on annual timescales [ Meade et al., 2000 ]. Still less likely is a significant contribution of sediment from the Yukon River to the Arctic Shelf. Thus, whereas the flux estimates provided in this paper allow assessment of sediment flux from a large percentage of the pan-arctic watershed, further research will be needed to determine how much of this sediment actually reaches the sea.

[51] Variation among published sediment flux estimates for individual rivers (Table 2) can largely be attributed to differences in the years of record included or use of data from different sampling stations. Because sediment flux is highly variable from year to year, establishing reliable average annual values requires integration over at least decadal time frames. Trends in sediment flux over time are not evident, and thus in most cases long-term averages of sediment flux provide best contemporary estimates. Notable exceptions are the Yenisey and Kolyma Rivers, where stepwise shifts accompanying dam construction and a change in sampling location, respectively, make it necessary to use only more recent flux data to represent present conditions.

[52] Long-term increases in water discharge have already been detected at the pan-arctic scale [ Semiletov et al., 2000 ]. Given the dependence of sediment flux on water discharge, we would suspect that sediment flux might be increasing as well. The absence of identifiable long-term trends in sediment flux is likely linked to the variability in the data. Records of sediment flux are much shorter than those of water discharge, and frequently lack values for winter months when changes in water discharge are most evident. Longer-term data sets, and reduction in variation induced by sampling and data handling, will be needed to determine if long-term changes are indeed occurring.

[53] At present, it is unclear to what extent inconsistencies in sampling and data handling contribute to variations in the sediment data. A unique feature of arctic rivers that greatly complicates accurate determination of sediment flux is ice breakup. During the breakup period, suspended sediment sampling is very dangerous if not impossible, yet sediment fluxes may be substantial during these periods. We must somehow figure out a way to reasonably account for sediment fluxes during the breakup period. In the meantime, we must acknowledge this deficiency in current sediment flux estimates for large arctic rivers. A further confounding factor is that sample collection and flux calculation methods often vary among rivers and perhaps over time. Ideally, standard methods would be used throughout the pan-arctic catchment. Perhaps the closer cooperation emerging among arctic nations will facilitate standardization of sediment methods as well as protocols for other hydrologic and water quality parameters. At any rate, for sediment flux to be a useful metric of global change in the future, monitoring must continue and artifacts introduced by sampling and data handling must be minimized.

[54] Fluxes of water and waterborne constituents from arctic rivers to the ocean provide an integrative signal of processes occurring in their watersheds. Shifts in these fluxes over time give clues about natural and anthropogenic changes in the Arctic. Increases in water discharge may be linked to anthropogenic increases in greenhouse gases and associated climate change [ Miller and Russell, 2000 ]. Waterborne constituents, such as nutrients and suspended sediments, provide information about alterations in biogeochemical processes accompanying climate and land-use changes. Compared to water discharge, however, analytical challenges and shorter time series of constituent data have made interpretation of long-term trends more difficult [ Holmes et al., 2000 , 2001 Zhulidov et al., 2000 ]. Thus, for many of these constituents our current challenge is not so much to identify historical trends but instead to establish a reliable contemporary baseline against which to evaluate future changes. In this paper, we have established contemporary sediment flux estimates for the eight largest arctic rivers. Together these values provide a baseline for sediment flux at the pan-arctic scale. This large-scale perspective is essential for understanding the effects of global change on the Arctic System as a whole.

## Historical Astronomy: Concepts: Kepler's 3rd Law Derivation

Kepler's 3rd Law is often called the Harmonic Law, and states that, for each planet orbitting the sun, its sidereal period squared divided by the cube of the semi-major axis of the orbit is a constant.

This is easy to show for the simple case of a circular orbit. A planet, mass m, orbits the sun, mass M, in a circle of radius r and a period t. The net force on the planet is a centripetal force, and is caused the force of gravity between the sun and the planet. Therefore we can write:

The mass of the planet cancels out. The speed of the planet going around the sun is just distance/time which is (2 pi r) / t. Substituting this makes the above equation:

Note that everything on the right is a constant, so that t 2 /r 3 is a constant for every planet in the solar system. This generalizes to any orbiting system. For example, if we look at t 2 /r 3 for a satellite orbiting the earth and the moon, we would get the same number, and we would use the mass of the earth in the equation. (Kepler himself showed that the moons of Jupiter also obeyed the Harmonic Law.)

It turns out that if we do the more formal derivation, with the two bodies orbiting about their center of mass in ellipses, you end up with

where r is the average distance between the objects, which would be the semi-major axis. Note that in the case of the planets around the sun, the mass of the sun is so much larger than the masses of any planet, that the result is basically a constant for all the planets going around the sun.

The radar range equation represents the physical dependences of the transmit power, which is the wave propagation up to the receiving of the echo signals. The power Pe returning to the receiving antenna is given by the radar equation, depending on the transmitted power PS , the slant range R , and the reflecting characteristics of the aim (described as the radar cross-section σ ). At the known sensibility of the radar receiver, the radar equation determines the achieved by a given radar theoretically maximum range. Furthermore one can assess the performance of the radar set with the radar range equation (or shorter: the radar equation ).

#### Argumentation/Derivation

First, we assume, that electromagnetic waves propagate under ideal conditions, i.e. without dispersion.

Figure 1: Nondirectional power density diminishes as geometric spreading

Figure 1: Nondirectional power density diminishes as geometric spreading

If high-frequency energy is emitted by an isotropic radiator then the energy propagates uniformly in all directions. Areas with the same power density, therefore, form spheres ( A= 4 π R² ) around the radiator. The same amount of energy spreads out on an incremented spherical surface at an incremented spherical radius. That means: the power density on the surface of a sphere is inversely proportional to the square of the radius of the sphere.

So we get the equation to calculate the Non-directional Power Density Su

• PS = transmitted power [W]
• Su = nondirectional power density
• R1 = range from transmitter antenna to the aim [m]

Figure 2: The antenna gain multiplied by the undirected power density gives the directed power density

Figure 2: The antenna gain multiplied by the undirected power density gives the directed power density

Since a spherical segment emits equal radiation in all direction (at constant transmit power) if the power radiated is redistributed to provide more radiation in one direction, then this results in an increase of the power density in direction of the radiation. This effect is called antenna gain. This gain is obtained by directional radiation of the power. So, from the definition, the directional power density is:

• Sg = directional power density
• Su = nondirectional power density
• G = antenna gain

Of course, in reality, radar antennas aren't ”partially radiating” isotropic radiators. Radar antennas must have a small beamwidth and an antenna gain up to 30 or 40 dB. (e.g. parabolic dish antenna or phased array antenna).

The target detection isn't only dependent on the power density at the target position, but also on how much power is reflected in the direction of the radar. In order to determine the useful reflected power, it is necessary to know the radar cross-section σ . This quantity depends on several factors. But it is true to say that a bigger area reflects more power than a smaller area. That means:

An Airbus offers more radar cross-section than a sporting aircraft at the same flight situation. Beyond this the reflecting area depends on the design, surface composition and materials used.

If the previously mentioned is summarized, the reflected power Pr (at the destination, i.e. the radar receiver) results from the power density Su , the antenna gain G , and the very variable reflection area σ :

• Pr = reflected power [W]
• σ = radar cross-section [m²]
• R1 = range, distance antenna - aim [m]

Simplified a target can be regarded as a radiator in turn due to the reflected power. In this case, the reflected power Pr is the emitted power.

Since the echoes encounter the same conditions as the transmitted power, the power density yielded at the receiver Se is given by:

Figure 3: Connection between
equations (3) and (4)

Figure 3: Connection between
equations (3) and (4)

• Se = power density at receiving place
• Pr = reflected power [W]
• R2 = range aim - receiving antenna [m]

At the radar antenna, the received power Pe is dependent on the power density at the receiving site Se and the effective antenna aperture AW .

The effective antenna aperture arises from the fact that an antenna suffers from losses, therefore, the received power at the antenna is not equal to the input power. As a rule, the efficiency of the antenna is around 0.6 to 0.7 (Efficiency Ka ).

Applied to the geometric antenna area, the effective antenna aperture is:

The power received, Pe is then calculated:

The transmitted and reflected waves have been seen separately. The next step is to consider both transmitted and reflected power: Since R2 (aim - antenna) is equal to the distance R1 (antenna - aim) then

Another equation, which will not be derived here, describes the antenna gain G in terms of the wavelength λ .

Solving for A , antenna area, and replacing A into equation 9 after simplification it yields:

Solving for range R , we obtain the classic radar equation:

All quantities that influence the wave propagation of radar signals were taken into account at this equation. Before we attempt to use the radar equation in the practice for example to determine the efficiency of radar sets, some further considerations are necessary.

For given radar equipment most sizes ( Ps, G, λ ) can be regarded as constant since they are only variable parameters in very small ranges. The radar cross-section, on the other hand, varies heavily but for practical purposes, we will assume 1 m².

The smallest received power that can be detected by the radar is called PEmin . Smaller powers than PEmin aren't usable since they are lost in the noise of the receiver. The minimum power is detected at the maximum range Rmax as seen from the equation.

An application of this radar equation is to easily visualize how the performance of the radar sets influences the achieved range.

#### Influences on the Maximum Range of a Radar Set

All considerations, when calculating the radar equation, were made assuming that the electromagnetic waves propagate under ideal conditions without disturbing influences. In practice, a number of losses should be considered since they reduce the effectiveness of the radar considerably.

First, the radar equation is extended by including the loss factor Lges.

This factor includes the following losses:

• L D = internal attenuation factors of the radar set on the transmitting and receiving paths
• L f = fluctuation losses during the reflection
• L Atm = atmospheric losses during propagation of the electromagnetic waves to and from the target

High-frequency components, such as waveguides, filters and also a radome, generate internal losses. For a given radar set this loss is relatively constant and also easily measured.

Atmospheric attenuation and reflections at the Earth's surface are permanent influences.

#### Influence of the Earth's Surface

An extended but less frequently used form of the radar equation considers additional terms, like the Earth's surface but does not classify receiver sensitivity and atmospheric absorption.

• Kα = Loss factor in place of Lges.
• Az = Effective reflection surface in place of σ
• ti = Pulse length
• nR = Noise figure of the receiver
• d = Clarity factor of the display
• Re = Distance of the absorbing medium
• K = Boltzmann's constant
• T0 = absolute temperature in K
• γ = Reflected beam angle
• δR = Break-even factor

#### Radar Reflections from Flat Ground

The trigonometric representation shows the influence of the Earth's surface. The Earth plane surrounding radar antenna has a significant impact on the vertical polar diagram.
The combination of the direct and re-reflected ground echo changes the transmitting and receiving patterns of the antenna. This is substantial in the VHF range and decreases with increasing frequency. For the detection of targets at low heights, a reflection at the Earth's surface is necessary. This is possible only if the ripples of the area within the first Fresnel zone do not exceed the value 0.001 R (i.e.: Within a radius of 1000 m no obstacle may be larger than 1 m!).

Figure 4: Detour of ground reflections

Figure 4: Detour of ground reflections

Specialized Radars at lower (VHF-) frequency band make use of the reflections at the Earth's surface and lobing to maximize cover at low levels. At higher frequencies, these reflections are more disturbing. The following picture shows the lobe structure caused by ground reflections. Normally this is highly undesirable as it introduces intermittent cover as aircraft fly through the lobes. The technique has been used in ATC ground mounted radars to extend the range but is only successful at low frequencies where the broad lobe structure permits adequate cover at higher elevations.

Figure 5: A vertical pattern diagram with influences of ground reflections

Figure 5: A vertical pattern diagram with influences of ground reflections

Figure 5: A vertical pattern diagram with influences of ground reflections

Increasing the height of the antenna has the effect of making finer the lobing pattern. A fine-grained lobing structure is often filled in by irregularities in the ground plane. Specifically, if the ground plane deviates from a flat surface then the reinforcement and destruction pattern resulting from the ground reflections breaks down. Avoidance of lobe effects is one of the prime considerations when selecting radar location and the height of the antenna.