Новое решение

Valter правка от 4 сент. 2026 г., 10:44

правка #20549 позже →

en/13.5.9.md +59 −0

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		+	### Statement
		+
		+	$13.5.9.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### 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.
		+
		+	### Solution
		+
		+	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
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,59 @@
			### Statement

			$13.5.9.$ [Insert the problem statement]

			### Solution

			### 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.

			### Solution

			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

			[Insert a concise answer or boxed result]