$13.5.9^*.$ 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:
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 $F$:
$$\frac{1}{F} = \frac{6GM}{c^2 R^2}\quad\implies\quad F = \frac{c^2 R^2}{6GM}$$
@@ -39,10 +37,10 @@Solution
Now, we estimate the numerical values in SI units:
$13.5.9^*.$ 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.
### Solution
### Solution
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:
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 $F$:
$$\frac{1}{F} = \frac{6GM}{c^2 R^2}\quad\implies\quad F = \frac{c^2 R^2}{6GM}$$
$$\frac{1}{F} = \frac{6GM}{c^2 R^2}\quad\implies\quad F = \frac{c^2 R^2}{6GM}$$
@@ -39,10 +37,10 @@Solution
Now, we estimate the numerical values in SI units:
Now, we estimate the numerical values in SI units:
