Съдържание

IAO 2007, Теория, Група β, руски — Задача 1. Галактика.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2007

Задачи теоретического тура · 29 септември 2007 г.

Общее замечание

Не исключено, что не во всех задачах вопросы поставлены корректно. Некоторые вопросы (возможно, главный вопрос задачи, возможно – подвопрос) могут не иметь смысла. В этом случае следует написать в ответе (по-русски или по-английски): «ситуация невозможна – situation is impossible». Естественно, ответ должен быть подкреплён вычислениями или логическими рассуждениями.

Материалы источника

Числовая таблица тел Солнечной системы не обнаружена среди 32 архивных файлов 2007 года. Лист bgtheo6.jpg содержит только переводы заголовков колонок. Официальные эскизы решений сохранены на предоставленном английском языке с обозначением утрат по краям снимков. Полная отдельная фотография самолёта и Луны приложена к задаче 4.

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 for8pts) may be shorter.

Source photograph damage

The second and fourth official solution photographs lose some initial letters or words at the left edge. Ellipses inside the affected solution text mark those source losses. Legible formulas, results and grading remarks are retained; missing letters have not been silently supplied. The solution for question4 contains no explanatory drawing, although the question requests one.

Printed source annotations

The galaxy key prints the same final logarithm1+HL₂/c in its classical and relativistic final lines, despite different numerical totals22.46 and22.82. The sail key repeats Δp and uses the printed multiplication/division notation. The source results and formulas are reproduced without scientific correction; punctuation and multiplication spacing are normalized only for mathematical rendering.

Сопоставление листов

Снимок theo.jpg содержит рядом группы α и β и повёрнут относительно ориентации файла. Отдельный betatheo.jpg повторяет задачи 1–5 группы β; это второй снимок того же издания, а не дополнительный комплект задач.

Условие

Болометрическая (полная) звёздная величина галактики, видимая с расстояния $L_1=3\ \text{МПк}$ , равна $m_1=6{,}88^{\mathrm m}$ . Найдите болометрическую звёздную величину $m_2$ этой галактики, видимую с расстояния $L_2=3\ \text{ГПк}$ .

Решение

A simple short solution that may be written by somebody is:

m_2=m_1+5^{\mathrm m}\lg(L_2/L_1)=6.88^{\mathrm m}+15^{\mathrm m}=21.88^{\mathrm m}.

This short solution would be correct, for example, for stellar magnitudes of a star for $L_1=3$ pc and $L_2=3$ kpc.

Note for jury: the above part of solution should render not more than 2 pts.

But the second distance done in the problem is of cosmological scale. So we have to take into account cosmological effects, red shift. According to the Hubble law the galaxy is moving away with a speed of

v=H\cdot L,

where L is distance to the galaxy, and H is the Hubble constant.

Note for jury: The value of the Hubble constant is not defined. Student may use any value in the interval 50–100 (km/s)/Mpc.

We will use the value H = (70 km/s)/Mpc. For $L_1=3$ Mpc this effect is not important (and the Galaxy can move with velocities larger than 210 km/s due to other reasons). But for $L_2=3$ Gpc the speed of the galaxy is about

v_2=70\ \mathrm{(km/s)/Mpc}\times3000\ \mathrm{Mpc}=210\,000\ \mathrm{km/s},

i.e. 0.7 of the light speed. So we have to take into account that every photon emitted by stars in the galaxy reach us with lower energy. So the visible stellar magnitude is larger due to this diminishing of the energy of all photons.

For every photon

E_0=h\nu_0=hc/\lambda_0

and

E_2=h\nu_2=hc/\lambda_2,

E_2/E_0=\lambda_0/\lambda_2.

Since the ratios $\lambda_0/\lambda_2$ are equal for every photon, the ratio $E_2/E_0$ for the whole galaxy is equal to this ratio for any photon. Therefore we may conclude that $\Delta m$ for the whole galaxy may be calculated as

\Delta m=-2^{\mathrm m}.5\lg(E_2/E_0)=-2^{\mathrm m}.5\lg(\lambda_0/\lambda_2).

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).

According to the classical Doppler effect,

\lambda_2=\lambda_0(1+v_2/c),

and

\Delta m=-2.5^{\mathrm m}\lg(\lambda_0/\lambda_2)=2.5^{\mathrm m}\lg(\lambda_2/\lambda_0)=2.5^{\mathrm m}\lg(1+v_2/c).

For $v_2=0.7c$

\Delta m=2.5^{\mathrm m}\lg(1+0.7)=2.5^{\mathrm m}\lg(1.7)=0.58^{\mathrm m}.

m_2=m_1+5^{\mathrm m}\lg(L_2/L_1)+2.5^{\mathrm m}\lg(1+HL_2/c)=22.46^{\mathrm m}.

According to the Doppler effect in special theory of relativity,

\lambda_2=\lambda_0(1+v_2/c)/(1-(v_2/c)^2)^{1/2},

and

\Delta m=-2.5^{\mathrm m}\lg(\lambda_0/\lambda_2)=2.5^{\mathrm m}\lg(\lambda_2/\lambda_0)=2.5^{\mathrm m}\lg[(1+v_2/c)/(1-(v_2/c)^2)^{1/2}].

For $v_2=0.7c$

\Delta m=2.5^{\mathrm m}\lg[(1+0.7)/(1-0.7^2)^{1/2}]=2.5^{\mathrm m}\lg(2.38)=0.94^{\mathrm m}.

m_2=m_1+5^{\mathrm m}\lg(L_2/L_1)+2.5^{\mathrm m}\lg(1+HL_2/c)=22.82^{\mathrm m}.

Данные

Данные из таблицы от телах Солнечной Системы могут быть использованы в любой задаче.

Оригинал в Архива: betatheo.jpg · официални решения: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: photo.JPG · допълнителни данни: theo.jpg