Съдържание

IAO 2007, Теория, Група β, английски — Задача 1. Galaxy.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2007

Theoretical round. Problems to solve · 29 септември 2007 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.

Source availability

The numerical table of planetary data is not present in the32-file2007 archive inventory. The file bgtheo6.jpg contains only a translation of column headings, without numerical planetary rows. No table from another year has been substituted. The separately archived airplane/Moon photograph is included in full with question4.

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.

Условие

The bolometric (total) stellar magnitude of a galaxy visible from a distance of $L_1=3$ Mpc is $m_1=6^{\mathrm m}.88$ . Find the bolometric stellar magnitude $m_2$ of this galaxy visible from a distance of $L_2=3$ Gpc.

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}.

Data

Data from the table of planetary data may be used for solving every problem.

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