Съдържание

IAO 2008, Теория, Група β — Задача 5. Алтернативна теория

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2008

XIII Международная астрономическая олимпиада / XIII International Astronomy Olympiad / XIII Olimpiade Internazionale di Astronomia. Италия, Триест, 13 – 21. X. 2008, Trieste, Italia. Round Th, Group B. · 13 октомври 2008 г. · 8 т.

Бележка

Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may have no real sense. In this case you have to write in your answer (in English or Russian): «situation is impossible – ситуация невозможна». Of course, this answer has to be explained numerically or logically.
Data from the table of planetary data may be used for solving every problem.

Note for jury and team leaders.

The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Элементы орбит. Физические характеристики некоторых планет, Луны и Солнца

Parameters of orbits.
Physical characteristics of some planets, Moon and Sun

Небесное тело, планета / Body, planet	Среднее расстояние от центрального тела / Average distance to central body, астр. ед. / astr. units	Среднее расстояние от центрального тела / Average distance to central body, млн. км / mln. km	Сидерический (или аналогичный) период обращения / Sidereal (or analogous) period, тропич. годах / troph. years	Сидерический (или аналогичный) период обращения / Sidereal (or analogous) period, средних сутках / days	Эксцентриситет / Eccentricity, e	Экваториальн. диаметр / Equat. diameter, km	Масса / Mass, $10^{24}$ kg	Средняя плотность / Av. density, g/cm³	Ускор. своб. пад. у пов. / Grav. acceler. at surf., m/s²	Макс. блеск, вид. с Земли / Max. magn. from Earth **)	Альбедо / Albedo
Солнце / Sun	1,6·10⁹	2,5·10¹¹	2,2·10⁸	8·10¹⁰		1392000	1989000	1,409		−26,8ᵐ
Меркурий / Mercury	0,387	57,9	0,241	87,97	0,206	4 879	0,3302	5,43	3,70		0,06
Венера / Venus	0,723	108,2	0,615	224,70	0,007	12 104	4,8690	5,24	8,87		0,78
Земля / Earth	1,000	149,6	1,000	365,26	0,017	12 756	5,9742	5,515	9,81		0,36
Луна / Moon	0,00257	0,38440	0,0748	27,3217	0,055	3 475	0,0735	3,34	1,62	−12,7ᵐ	0,07
Марс / Mars	1,524	227,9	1,880	686,98	0,093	6 794	0,6419	3,94	3,71		0,15
Юпитер / Jupiter	5,204	778,6	11,862	4 332,59	0,048	142 984	1899,8	1,33	24,86		0,66
Сатурн / Saturn	9,584	1433,7	29,458	10 759,20	0,054	120 536	568,50	0,70	10,41		0,68

**) Для Луны – в среднем противостоянии.
**) For Moon – in mean opposition.

Координаты на эпоху 2000
Epoch 2000 coordinates

	α	δ
Полярная звезда / Polar star	2ʰ 31ᵐ 48ˢ.7	89° 15′ 51″

Условие

Теорията за разширяващата се Вселена понастоящем е най-разпространената космологическа теория. Без каквито и да е съмнения към тази теория се придържат над 85% от астрофизиците. Съществуват обаче и алтернативни теории. В една такава теория се предполага, че Вселената е стационарна, а космологичното червено отместване се дължи не на ефекта на Доплер, а на „стареенето“ на фотоните, т.е. енергията на всеки фотон намалява с времето по следния закон: $E = E_0 \cdot 2^{-t/T_0}$ , където $E_0$ е първоначалната енергия на фотона (в момента на неговата поява), $t$ е времето, изминало от момента на неговата поява, $T_0$ е т. нар. период на полуразпад на фотона, аналогичен на времето за полуразпадане в ядрената физика. Оценете на колко е равен периодът на полуразпад на фотона $T_0$ (в години) в модела на „стареенето“ на фотоните, който би съответствал на наблюдателните астрономически данни.

Отговори

Решение

Официални скици на решенията, на оригиналния английски език.

Simplified direct solution — classical Doppler effect

The energy of every photon becomes smaller according to the law $E=E_0\cdot2^{-t/T_0}$ , which means that after a time $t=T_0$ the energy becomes smaller by a factor 2. $E_1=E_0/2$ , $h\nu_1=h\nu_0/2$ , $hc/\lambda_1=(hc/\lambda_0)/2$ , so $\lambda_1=2\lambda_0$ . The theory of an expanding Universe explains it by terms of the Doppler effect due to that the source of the photon is receding from us with the velocity $v$ .

There are two ways of solution presented later: in the classical theory and in the special theory of relativity. The solution according to the special theory of relativity is better but not all students know the formula of this theory (1 pt difference recommended in the evaluation for Group β, no difference for Group α).

According to the classical Doppler effect,

\lambda_1=\lambda_0(1+v/c),

2\lambda_0=\lambda_0(1+v/c),\qquad v/c=1,\qquad v=c.

In other words the red shift

z=1.

According to the Hubble law

v=H\cdot R,\qquad R=v/H.

Since $v=c$ ,

R=c/H.

The time passed since the emission of the photon is $t=R/c$ , so

t=R/c=c/H/c=1/H.

T_0=t=(1/70)\cdot\mathrm{s\cdot Mpc/km}=(1/70)\cdot\mathrm s\cdot3.26\cdot10^6\,\mathrm{years}\cdot3\cdot10^5\,\mathrm{km/s/km}\approx14\cdot10^9\,\mathrm{years}.

In the theory of an expanding Universe this period is equal to cosmological age of the Universe.

Simplified direct solution — special relativity

According to the Doppler effect in special theory of relativity the red shift is also $z=1$ but

\lambda_2=\lambda_0(1+v/c)/(1-(v/c)^2)^{1/2}.

2\lambda_0=\lambda_0(1+v/c)/(1-(v/c)^2)^{1/2},

2(1-(v/c)^2)^{1/2}=(1+v/c),

4(1-(v/c)^2)=(1+v/c)^2,

4(1-v/c)(1+v/c)=(1+v/c)^2,

4(1-v/c)=(1+v/c),

v/c=0.6.

According to the Hubble law

v=H\cdot R,\qquad R=v/H.

Since $v=0.6c$ ,

R=0.6c/H.

The time passed since the emission of the photon is $t=R/c$ , so

t=R/c=0.6c/H/c=0.6/H.

T_0=t=0.6(1/70)\cdot\mathrm{s\cdot Mpc/km}=0.6(1/70)\cdot\mathrm s\cdot3.26\cdot10^6\,\mathrm{years}\cdot3\cdot10^5\,\mathrm{km/s/km}\approx8.4\cdot10^9\,\mathrm{years}.

In the theory of an expanding Universe this time is roughly the cosmological age of the Universe.

Answer: in both cases (the classical theory or the special theory of relativity) the parameter of the so-called half-decay period of a photon is about $10^{10}$ (10 billion) years that is of the order of the cosmological age of the Universe according to the theory of an expanding Universe.

Solution with derivatives and jury notes

This method is more precise than the “direct solution” but the math apparatus is not possible for most students.

For small values, if $\Delta\lambda\ll\lambda_0$ using derivatives of parts of equation ( $E=E_0\cdot2^{-t/T_0}$ ) one can find:

dE=d(E_0\cdot2^{-t/T_0})=dt\cdot(-\ln2/T_0)\cdot(E_0\cdot2^{-t/T_0})=dt\cdot(-\ln2/T_0)\cdot E,

dE/E=-\ln2\cdot dt/T_0.

Using a small time $t$ in comparison with $T_0$ we may write

\Delta E/E=-\ln2\cdot t/T_0.

Since

\Delta E/E=-\Delta\lambda/\lambda=-v/c,

-v/c=-\ln2\cdot t/T_0.

According to the Hubble law

v=H\cdot R,

H\cdot R/c=\ln2\cdot t/T_0.

Since the time passed since the emission of the photon is $t=R/c$ ,

finally we have

T_0=\ln2/H.

T_0=\ln2\cdot(1/70)\cdot\mathrm{s\cdot Mpc/km}=0.69\cdot(1/70)\cdot\mathrm s\cdot3.26\cdot10^6\,\mathrm{years}\cdot3\cdot10^5\,\mathrm{km/s/km}=9.7\cdot10^9\,\mathrm{years}\approx10\cdot10^9\,\mathrm{years}.

In the theory of an expanding Universe this period is about cosmological age of the Universe.

Answer: the parameter of the so-called half-decay period of a photon is about $10^{10}$ (10 billion) years that is of order of cosmological age of the Universe according to the theory of expanding Universe.

Note: students of group β should take the Hubble constant $H_0$ themselves. It may be in the range from 50 till 100 km/s/Mpc. The answers for these values will appear $\sqrt2$ larger or $\sqrt2$ smaller.

Note: the results are depend on the objects (or times) that we use as references (like for the nearest universe (solution with derivatives) or objects with the red shift 1). Nevertheless it does mean that “the situation is impossible”. It means only that at least one of the formulae (formula of Hubble $v=HR$ or formula of aging of photons with a “half-decay” $E=E_0\cdot2^{-t/T_0}$ ) is not coincide reality.

Бележка върху българския ръкопис

В долния край на листа се вижда дописан отговор с оценка $1{,}3\cdot10^{10}$ години. Знаменателят под единицата е отрязан от края на сканирания оригинал; не е възстановен по догадка.

Partly clipped handwritten answer at the bottom of the Bulgarian manuscript

Бележка от източника

No separate detailed rubric or Bulgarian official solution is present in the full year inventory. The English source explicitly supplies sketches, with a general reference to 8 points; all applicable supplied reasoning and jury notes are retained.

Бележка от източника

The Bulgarian sheet has a bottom-edge annotation with an unreadable clipped denominator. The visible estimate 1.3 × 10¹⁰ years is preserved in its original crop, not silently replaced by the official estimates.

Бележка от източника

The official key gives differing direct and differential approximations, recommends a one-point difference for beta relativistic versus classical answers, and prints a square-root-of-two range statement for H=50–100. Its final discussion literally says “it does mean” the situation is impossible; that wording is retained without editorial reversal.

Оригинал в Архива: betatheobg5.pdf · официални решения: theosol.pdf · допълнителни данни: betatheo.pdf · допълнителни данни: IAO2008-data.pdf