Правка раздела «Solution»
en/13.5.9.md
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| ### Problem | |||
| $13.5.9^*.$ Large masses of matter in the Universe can focus light from distant objects, forming a "gravitational lens". Estimate the focal length of a spherical galaxy of radius $R \approx 20000 \text{ pc}$ and mass $M \approx 3 \cdot 10^{11} M_\odot$ (where $M_\odot$ is the mass of the Sun), assuming that the mass in the galaxy is distributed uniformly. | |||
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| ### Solution | |||
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| According to General Relativity, the deflection angle of a light ray passing a mass with an impact parameter $b$ is: | |||
| $$\alpha = \frac{4GM_{enc}}{c^2 b}$$ | |||
| Let's consider paraxial rays ($b \ll R$) to determine the focal length. For a ray passing close to the center, the path length inside the spherical galaxy is approximately equal to its diameter ($L \approx 2R$). | |||
| By symmetry, the gravitational effect on the photon is exerted only by the matter enclosed within a cylinder of radius $b$, whose axis coincides with the line of sight. The volume of this cylinder is approximately: | |||
| $$V_{cyl} \approx \pi b^2 \cdot 2R = 2\pi R b^2$$ | |||
| Since the density of matter in the galaxy is uniform ($\rho = \frac{M}{\frac{4}{3}\pi R^3}$), the enclosed mass within the cylinder is: | |||
| $$M_{enc} = \rho V_{cyl} = \frac{M}{\frac{4}{3}\pi R^3} \cdot 2\pi R b^2 = \frac{3M}{2R^2} b^2$$ | |||
| Substituting this mass into the deflection angle formula, we get: | |||
| $$\alpha(b) = \frac{4G}{c^2 b} \left( \frac{3M}{2R^2} b^2 \right) = \frac{6GM}{c^2 R^2} b$$ | |||
| From geometrical optics, a thin lens focuses paraxial rays if the deflection angle is proportional to the impact parameter: $\alpha = \frac{b}{F}$. Thus, the central part of the galaxy acts as an ideal lens with a focal length $F$: | |||
| $$\frac{1}{F} = \frac{6GM}{c^2 R^2} \quad \implies \quad F = \frac{c^2 R^2}{6GM}$$ | |||
| Now, we estimate the numerical values in SI units: | |||
| - $R \approx 20000 \text{ pc} \approx 6 \cdot 10^{20} \text{ m}$ | |||
| - $M \approx 3 \cdot 10^{11} M_\odot \approx 6 \cdot 10^{41} \text{ kg}$ | |||
| - $c \approx 3 \cdot 10^8 \text{ m/s}$ | |||
| - $G \approx 6.67 \cdot 10^{-11} \text{ m}^3/(\text{kg} \cdot \text{s}^2)$ | |||
| Calculating the focal length in meters: | |||
| $$F = \frac{(3 \cdot 10^8)^2 \cdot (6 \cdot 10^{20})^2}{6 \cdot 6.67 \cdot 10^{-11} \cdot 6 \cdot 10^{41}} \approx 1.35 \cdot 10^{26} \text{ m}$$ | |||
| Converting the distance back to parsecs ($1 \text{ pc} \approx 3 \cdot 10^{16} \text{ m}$): | |||
| $$F \approx \frac{1.35 \cdot 10^{26}}{3 \cdot 10^{16}} = 4.5 \cdot 10^9 \text{ pc}$$ | |||
| #### Answer | |||
| $$F \approx 10^9 \text{ pc}$$ | |||
| ещё строк без изменений 45 | |||
| ### Problem | ### Problem | ||
| $13.5.9^*.$ Large masses of matter in the Universe can focus light from distant objects, forming a "gravitational lens". Estimate the focal length of a spherical galaxy of radius $R \approx 20000 \text{ pc}$ and mass $M \approx 3 \cdot 10^{11} M_\odot$ (where $M_\odot$ is the mass of the Sun), assuming that the mass in the galaxy is distributed uniformly. | $13.5.9^*.$ Large masses of matter in the Universe can focus light from distant objects, forming a "gravitational lens". Estimate the focal length of a spherical galaxy of radius $R \approx 20000 \text{ pc}$ and mass $M \approx 3 \cdot 10^{11} M_\odot$ (where $M_\odot$ is the mass of the Sun), assuming that the mass in the galaxy is distributed uniformly. | ||
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| @@ -6,7 +6,7 @@Problem | |||
| ### Solution | ### Solution | ||
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| According to General Relativity, the deflection angle of a light ray passing a mass with an impact parameter $b$ is: | According to General Relativity, the deflection angle of a light ray passing a mass with an impact parameter $b$ is: | ||
| $$\alpha = \frac{4GM_{enc}}{c^2 b}$$ | $$\alpha = \frac{4GM_{enc}}{c^2 b}$$ | ||
| Let's consider paraxial rays ($b \ll R$) to determine the focal length. For a ray passing close to the center, the path length inside the spherical galaxy is approximately equal to its diameter ($L \approx 2R$). | Let's consider paraxial rays ($b \ll R$) to determine the focal length. For a ray passing close to the center, the path length inside the spherical galaxy is approximately equal to its diameter ($L \approx 2R$). | ||
| By symmetry, the gravitational effect on the photon is exerted only by the matter enclosed within a cylinder of radius $b$, whose axis coincides with the line of sight. The volume of this cylinder is approximately: | By symmetry, the gravitational effect on the photon is exerted only by the matter enclosed within a cylinder of radius $b$, whose axis coincides with the line of sight. The volume of this cylinder is approximately: | ||
| $$V_{cyl} \approx \pi b^2 \cdot 2R = 2\pi R b^2$$ | $$V_{cyl} \approx \pi b^2 \cdot 2R = 2\pi R b^2$$ | ||
| Since the density of matter in the galaxy is uniform ($\rho = \frac{M}{\frac{4}{3}\pi R^3}$), the enclosed mass within the cylinder is: | Since the density of matter in the galaxy is uniform ($\rho = \frac{M}{\frac{4}{3}\pi R^3}$), the enclosed mass within the cylinder is: | ||
| $$M_{enc} = \rho V_{cyl} = \frac{M}{\frac{4}{3}\pi R^3} \cdot 2\pi R b^2 = \frac{3M}{2R^2} b^2$$ | $$M_{enc} = \rho V_{cyl} = \frac{M}{\frac{4}{3}\pi R^3} \cdot 2\pi R b^2 = \frac{3M}{2R^2} b^2$$ | ||
| Substituting this mass into the deflection angle formula, we get: | Substituting this mass into the deflection angle formula, we get: | ||
| $$\alpha(b) = \frac{4G}{c^2 b} \left( \frac{3M}{2R^2} b^2 \right) = \frac{6GM}{c^2 R^2} b$$ | $$\alpha(b) = \frac{4G}{c^2 b} \left( \frac{3M}{2R^2} b^2 \right) = \frac{6GM}{c^2 R^2} b$$ | ||
| From geometrical optics, a thin lens focuses paraxial rays if the deflection angle is proportional to the impact parameter: $\alpha = \frac{b}{F}$. Thus, the central part of the galaxy acts as an ideal lens with a focal length $F$: | From geometrical optics, a thin lens focuses paraxial rays if the deflection angle is proportional to the impact parameter: $\alpha = \frac{b}{F}$. Thus, the central part of the galaxy acts as an ideal lens with a focal length $F$: | ||
| $$\frac{1}{F} = \frac{6GM}{c^2 R^2} \quad \implies \quad F = \frac{c^2 R^2}{6GM}$$ | $$\frac{1}{F} = \frac{6GM}{c^2 R^2} \quad \implies \quad F = \frac{c^2 R^2}{6GM}$$ | ||
| Now, we estimate the numerical values in SI units: | Now, we estimate the numerical values in SI units: | ||
| - $R \approx 20000 \text{ pc} \approx 6 \cdot 10^{20} \text{ m}$ | - $R \approx 20000 \text{ pc} \approx 6 \cdot 10^{20} \text{ m}$ | ||
| - $M \approx 3 \cdot 10^{11} M_\odot \approx 6 \cdot 10^{41} \text{ kg}$ | - $M \approx 3 \cdot 10^{11} M_\odot \approx 6 \cdot 10^{41} \text{ kg}$ | ||
| - $c \approx 3 \cdot 10^8 \text{ m/s}$ | - $c \approx 3 \cdot 10^8 \text{ m/s}$ | ||
| - $G \approx 6.67 \cdot 10^{-11} \text{ m}^3/(\text{kg} \cdot \text{s}^2)$ | - $G \approx 6.67 \cdot 10^{-11} \text{ m}^3/(\text{kg} \cdot \text{s}^2)$ | ||
| Calculating the focal length in meters: | Calculating the focal length in meters: | ||
| $$F = \frac{(3 \cdot 10^8)^2 \cdot (6 \cdot 10^{20})^2}{6 \cdot 6.67 \cdot 10^{-11} \cdot 6 \cdot 10^{41}} \approx 1.35 \cdot 10^{26} \text{ m}$$ | $$F = \frac{(3 \cdot 10^8)^2 \cdot (6 \cdot 10^{20})^2}{6 \cdot 6.67 \cdot 10^{-11} \cdot 6 \cdot 10^{41}} \approx 1.35 \cdot 10^{26} \text{ m}$$ | ||
| Converting the distance back to parsecs ($1 \text{ pc} \approx 3 \cdot 10^{16} \text{ m}$): | Converting the distance back to parsecs ($1 \text{ pc} \approx 3 \cdot 10^{16} \text{ m}$): | ||
| $$F \approx \frac{1.35 \cdot 10^{26}}{3 \cdot 10^{16}} = 4.5 \cdot 10^9 \text{ pc}$$ | $$F \approx \frac{1.35 \cdot 10^{26}}{3 \cdot 10^{16}} = 4.5 \cdot 10^9 \text{ pc}$$ | ||
| #### Answer | #### Answer | ||
| $$F \approx 10^9 \text{ pc}$$ | $$F \approx 10^9 \text{ pc}$$ | ||
| ещё строк без изменений 45 | |||