Съдържание

IAO 2008, Теория, Група β — Задача 5. Alternative theory.

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Всички материали от IAO 2008

XIII International Astronomy Olympiad · Theoretical round. Problems to solve · 13 октомври 2008 г.

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

The theory of an expanding Universe is the most popular and most believable modern cosmological model. They say more than 85% of astrophysicists follow this theory without any hesitance. Nevertheless some alternative theories exist as well. One of these theories proposes that the Universe is stable and the cosmological red shift appears not due to the Doppler effect, but due to “aging of photons”, i.e. the energy of every photon becomes smaller according to the law $E=E_0\cdot2^{-t/T_0}$ , where $E_0$ = initial energy at the appearance of a photon, $t$ = time of life since appearance and $T_0$ = the so-called half-decay period of a photon, analogous of the half-decay parameter in nuclear physics. Estimate the value of the half-decay period $T_0$ (in years) in the model of “aging of photons” that corresponds to astronomical observations.

Отговори

Решение

5αβ. Alternative theory. The simplified “direct solution”.

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.

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. 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_0=-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.

Теми и трудност

Оценена трудност: 4/5.

Галактики и космология · Светлина и астрономически измервания · Разширение на Вселената · Доплерово отместване и червено отместване

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