Съдържание

IAO 2025, Теория, βγ — Задача 7. Distant galaxy

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2025

XXIX IAO — IV International Remote Astronomy Olympiad, Dhaka, Bangladesh, 6-15.XI.2025 — Theoretical round. Problems to solve · 6 ноември 2025 г. · 8 т.

Внимание: Бележка към темата

The separate table of constants and Hertzsprung–Russell diagram referenced by the solutions are not present in the inspected 2025 archive inventory. No replacement reference values have been invented.

Reminding

Reminding. Most of the problems are compiled in such a way that when writing solutions, minimal use of any language is required from the participant. So minimum texts are allowed in solutions of the problems (neither in the native language, nor in one of the official languages of the IAO). Pay more attention to formulas, drawings and graphs. It is permitted to write a few keywords in English (for example, “Yes”, “No”, “Assume”, “Answer”, “Average”, “Minimum”, “Maximum”, “Impossible situation”, etc.), as well as terms contained in the accompanying tables and names of objects.

Условие

The wavelengths $\lambda$ of the spectral lines of hydrogen obey the formula:

\frac{1}{\lambda} = R\left(\frac{1}{m^2} - \frac{1}{n^2}\right),

where $R = const$ , $m$ and $n$ are natural numbers, and $n > m$ . Lines with the same $m$ are combined in series. The most famous line is Hα, the first line (i.e., the line with the longest wavelength) in the Balmer series ( $m=2$ ).

Research of a distant galaxy showed us that the Hα line in its spectrum is observed at a wavelength of about the first line of the Brackett series ( $m=4$ ). Roughly (with an accuracy of 10%) estimate:

7.1. The receding velocity of the galaxy. [5 т.]

7.2. The distance to the galaxy. [3 т.]

Отговори

Решение

Sketches for solutions — for jury job ONLY

These solutions are written for jury members, so they contain a lot of explanations in the form of text. Students are instructed to use mostly formulas, calculations, drawings and graphs, writing texts is not allowed in the solutions. It is permitted to write a few keywords in English (for example, “Yes”, “No”, “Assume”, “Answer”, “Average”, “Minimum”, “Maximum”, “Impossible situation”, etc.), as well as terms contained in the accompanying tables and names of objects.

If you find typos, errors or missing points in the solution, please inform the organizing committee. Any suggestions for modifying the solutions are welcome, write justifications to gavrilovissp@list.ru and gavrilov@issp.ac.ru (to both addresses for reliability), if the suggestions are reasonable, the proposal can be approved.

Redshift and approximate answers

From the formula

\frac1\lambda=R\left(\frac1{m^2}-\frac1{n^2}\right)

it is easy to understand that the first line (that is, the line with the longest wavelength) in each series is the line for which $n=m+1$ . Therefore, for the ratio between the inverse wavelengths for Hα and the first line of the Brackett series, we can write:

\frac1{\lambda_\alpha}\bigg/\frac1{\lambda_B}=\left(\frac1{2^2}-\frac1{3^2}\right)\bigg/\left(\frac1{4^2}-\frac1{5^2}\right),

\lambda_B/\lambda_\alpha=\left(\frac{9-4}{36}\right)\bigg/\left(\frac{25-16}{400}\right)=\frac{500}{81}.

The wavelength $\lambda_B$ itself may not even be calculated, as it is possible not to know the wavelength $\lambda_\alpha$ . The redshift for this galaxy is

z=\left(\frac\lambda{\lambda_0}-1\right)=\frac{419}{81}\approx5.17.

If we use the Doppler formula from classical mechanics ( $v/c=(\lambda-\lambda_0)/\lambda_0$ ), we get the galaxy receding with a velocity of $z\cdot c=5.17c$ , greater than the speed of light. Therefore, since the redshift is large, we need to use the formulas of special relativity for an accurate answer. But to solve the problem as formulated (with an accuracy of 10%), it is not necessary to know these formulas, it is enough to realize that the relative velocity of the galaxy is close to the speed of light, i.e. it is moving away from us at a speed of about 300000 km/s (answer to 7.1.), and the distance to it can be found from the Hubble formula $v=H\cdot R$ . It is about

R=c/H=\frac{300000\,\mathrm{km/s}}{68\,\mathrm{km/s/Mpc}}\approx4400\,\mathrm{Mpc}\approx4.4\,\mathrm{Gpc}\quad\text{(answer to 7.2.)}.

Note: exact solution as supplied

The exact solution (not required by the condition of the problem):

Redshift

z=(\lambda-\lambda_0)/\lambda_0=(40525\,\text{Å}-6563\,\text{Å})/6563\,\text{Å}=5.17.

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

v/c=\frac{(z+1)^2-1}{(z+1)^2+1}.

Then for $z=5.17$ obtain

v=0.949c\approx285000\,\mathrm{km/s}\quad\text{(answer to 7.1.)}.

According to the Hubble formula ( $v=H\cdot R$ ), we obtain

R=v/H=\frac{285000\,\mathrm{km/s}}{68\,\mathrm{km/s/Mpc}}\approx4190\,\mathrm{Mpc}\approx4.2\,\mathrm{Gpc}\quad\text{(answer to 7.2.)}.

Transcription note: “exact solution” is the supplied sketch's label. Its special-relativistic velocity and linear Hubble-law distance are reproduced as the official approach, not independently replaced with a cosmological distance model.

Official grading criteria

Item	Criterion	pts
	Understanding that the first line in each series is the line for which n = m+1	1
	Calculation $\lambda_B/\lambda_\alpha$ and z	1.6
	Understanding receding velocity close to speed of light or calculating the receding velocity of the galaxy with exact formula	1.6
	Answer for the velocity	0.8
7.2.	Formula for the distance	1.8
7.2.	Calculation of the distance and final answer	1.2
	Total	8

General notes for the jury

Note. Jury members should evaluate the student’s solutions in essence, and not by looking on formal existence the mentioned sentences or formulae. The formal presence of the mentioned positions in the text is not necessary to give the respective points. Points should be done if the following steps de facto use these positions. (For example, except the final conclusion and artistic picture, all the points in solution of problem 2 may be in only one common formula.) Of course, fractions of the mentioned points should be awarded if only part of the item was done.

Note. Jury members may elaborate more detailed criteria, and also create criteria for other correct ways of the student’s solutions.

Note. Everywhere “formula” means “correct formula”, “calculation” means “correct calculation”, “picture” means “correct picture”, etc., and in all cases: based on correct facts and intermediate formulae or calculations.

Note. Everywhere final calculation means also using correct number of significant digits. Excessive number of significant digits, especially in estimational problems, will lead to subtraction of the points (column “digits”, minus 1-2 points).

Note. Solutions with absurd or almost absurd answers should be rated lower (maybe ignoring the scoring schemes below, and maybe down to zero, column “absurd”).

Any suggestions for modifying the criteria are welcome, write justifications to gavrilovissp@list.ru and gavrilov@issp.ac.ru (to both addresses for reliability), if the suggestions are reasonable, the proposal can be approved.

Оригинал в Архива: problems_t_67_en.pdf · официални решения: solutions_t_en.pdf · допълнителни данни: criteria_t_en.pdf