<p>The gravitational focus you are talking about is actually a <em>minimum</em> value, defined by parallel rays of light from a very distant star just skimming past the Sun as they are bent according to General Relativity.</p>

<p>The general formula for such lensing is that light is bent through an angle (in radians) of
<span class="math-container">$$\alpha = \frac{4 GM}{c^2 r},$$</span>
where <span class="math-container">$M$</span> is the mass of the lens (assumed to be a point or spherically symmetric mass)  and <span class="math-container">$r$</span> is the closest approach of a light ray to the lensing mass.</p>

<p>To work out where a ring of rays will be focused is just a bit of trigonometry. 
<span class="math-container">$$ d_f \simeq \frac{r}{\alpha} = \frac{c^2 r^2}{4GM}$$</span></p>

<p>This focal distance is a minimum because it would be larger for a ring of rays that passed the lens with a larger value of <span class="math-container">$r$</span>.</p>

<p>For the Sun as a lens you use <span class="math-container">$M=2\times 10^{30}$</span> kg and <span class="math-container">$r=6.9\times 10^{8}$</span> m, and calculate <span class="math-container">$d_f = 540$</span> au.</p>

<p>White dwarf stars have a similar mass (actually most are about 60% the mass of the Sun, but Sirius B is almost exactly a solar mass), but have radii about the size of the Earth - i.e. a hundred times less than the Sun.</p>

<p>This means that the value of <span class="math-container">$d_f$</span> is going to be about 10,000 times less than 540 au. You can use the formula above to calculate it for any combination of mass and radius.</p>

<p>To use the telescope, you place detectors at your chosen focus and observe the bright "Einstein ring" of a distant source that is <em>exactly</em> behind the lens. The magnification factor (the increase in the amount of light collected from the source) then is <span class="math-container">$4\alpha/\theta$</span>, where <span class="math-container">$\theta$</span> is the angular size of the source without the lens.</p>

<p>For a white dwarf, the magnification at the minimum focus would be 100 times larger, because <span class="math-container">$\alpha$</span> is 100 times larger.</p>

<p>Note that the size of the image is modified by the ratio of the focal length to the source distance.
<span class="math-container">$$ x_i = x_o \frac{d_f}{d_o}$$</span>
Thus the image of a distant object will be 10,000 times smaller than if using the Sun, which is much more convenient!</p>

<p>e.g. Observe an Earth-like planet at 10 ly at a focus of 630 au (= 0.01 ly) from the Sun. The image diameter will be 12.5 km. That's a lot of CCD detectors! Using a white dwarf at a focal length that is 10,000 times smaller gives an image just 1.25 m across.</p>

<p>All this assumes that the telescope is perfectly pointed with the source right behind the lens. Any relative motion has to be corrected or the image will move through the focal plane very quickly (like a planet viewed with high magnification through a normal telescope).</p>

CC BY-SA 4.0
2019-09-27T11:07:25.617Z
Brilliant answer. It sounds like a Sirius B telescope might actually be worth building if and when we ever get to the Sirius system. An Earth-like planet in the Andromeda galaxy would still be a few pixels across for such a telescope, I think.
Steve Linton
CC BY-SA 4.0
2019-09-27T12:05:32.143Z
@SteveLinton $x_i = 1.25\times 10^6 \times 0.01 \times 10^{-4}/2\times 10^6 = 0.6$ microns. You'd be better off using the Sun and getting an image of size 6 mm.
ProfRob
CC BY-SA 4.0
2019-09-27T14:45:27.837Z
So the focussing star creates an lens whose power varies with radius. 
 That's more or less like a correcting element for spherical convex lens  (whose power is greatest at large radius).
Carl Witthoft
CC BY-SA 4.0
2019-09-30T23:57:17.357Z
If I'm reading this right, the white dwarf distance and image size would be adjustable with distance.    Further away would be better to avoid the ridiculous high orbital velocity at 0.054 AU and keeping the thing focused on one place.  Perhaps some kind of lower adjustment Lagrange orbit (kinda sorta), maybe 5 AU for L1, about 20 AU for L4/L5 could work to minimize adjustment, plus, it's good for solar power from Sirius A.  8 years is a long time to wait for info though, but not outrageously long.   The biggest obstacle, of-course, would be getting that much equipment 8 light years away.
userLTK
CC BY-SA 4.0
2019-09-27T08:21:22.010Z
2019-09-27T08:21:22.010Z
ProfRob
<p>I tried looking this up, but I couldn't find any formula on gravitational lensing distance.   I know that our Sun's is about 550 AU, though further distances work too, as it's not a single focus due to the gravitaional field diminishing over distance from the focusing body.</p>

<p>Is there a reasonably simple formula for calculating distance for a gravitational lens.   I'm specifically curious for white dwarf stars as there's one just 8 light years away and they see like a good object with a good lensing but not super tight focusing like a neutron star or black hole.   </p>

<p>For example, if a telescope was built using Sirius B as a focus, how far would the telescope have to be and how powerful might it be (Maybe how powerful should be a separate question but I'll leave it here for now?    </p>

<p>Would Sirius B's binary orbit be a hindrance or a benefit, allowing a greater area of focus? </p>

<p>Pure curiosity.   I don't expect we'll get there anytime soon.  </p>

<space-telescope><gravitational-lensing>
What is the gravitational lensing focal distance of a white dwarf star?
userLTK
# What is the gravitational lensing focal distance of a white dwarf star?

(Asked by: username_0 on 2019-09-26)

I tried looking this up, but I couldn't find any formula on gravitational lensing distance. I know that our Sun's is about 550 AU, though further distances work too, as it's not a single focus due to the gravitaional field diminishing over distance from the focusing body.

Is there a reasonably simple formula for calculating distance for a gravitational lens. I'm specifically curious for white dwarf stars as there's one just 8 light years away and they see like a good object with a good lensing but not super tight focusing like a neutron star or black hole.

For example, if a telescope was built using Sirius B as a focus, how far would the telescope have to be and how powerful might it be (Maybe how powerful should be a separate question but I'll leave it here for now?

Would Sirius B's binary orbit be a hindrance or a benefit, allowing a greater area of focus?

Pure curiosity. I don't expect we'll get there anytime soon.

**Tags:** space-telescope, gravitational-lensing


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## 1 answers

### Answer by username_1 on 2019-09-27. Score: 11 | Accepted answer

The gravitational focus you are talking about is actually a *minimum* value, defined by parallel rays of light from a very distant star just skimming past the Sun as they are bent according to General Relativity.

The general formula for such lensing is that light is bent through an angle (in radians) of
$$\alpha = \frac{4 GM}{c^2 r},$$
where $M$ is the mass of the lens (assumed to be a point or spherically symmetric mass) and $r$ is the closest approach of a light ray to the lensing mass.

To work out where a ring of rays will be focused is just a bit of trigonometry.
$$ d\_f \simeq \frac{r}{\alpha} = \frac{c^2 r^2}{4GM}$$

This focal distance is a minimum because it would be larger for a ring of rays that passed the lens with a larger value of $r$.

For the Sun as a lens you use $M=2\times 10^{30}$ kg and $r=6.9\times 10^{8}$ m, and calculate $d\_f = 540$ au.

White dwarf stars have a similar mass (actually most are about 60% the mass of the Sun, but Sirius B is almost exactly a solar mass), but have radii about the size of the Earth - i.e. a hundred times less than the Sun.

This means that the value of $d\_f$ is going to be about 10,000 times less than 540 au. You can use the formula above to calculate it for any combination of mass and radius.

To use the telescope, you place detectors at your chosen focus and observe the bright "Einstein ring" of a distant source that is *exactly* behind the lens. The magnification factor (the increase in the amount of light collected from the source) then is $4\alpha/\theta$, where $\theta$ is the angular size of the source without the lens.

For a white dwarf, the magnification at the minimum focus would be 100 times larger, because $\alpha$ is 100 times larger.

Note that the size of the image is modified by the ratio of the focal length to the source distance.
$$ x\_i = x\_o \frac{d\_f}{d\_o}$$
Thus the image of a distant object will be 10,000 times smaller than if using the Sun, which is much more convenient!

e.g. Observe an Earth-like planet at 10 ly at a focus of 630 au (= 0.01 ly) from the Sun. The image diameter will be 12.5 km. That's a lot of CCD detectors! Using a white dwarf at a focal length that is 10,000 times smaller gives an image just 1.25 m across.

All this assumes that the telescope is perfectly pointed with the source right behind the lens. Any relative motion has to be corrected or the image will move through the focal plane very quickly (like a planet viewed with high magnification through a normal telescope).

> **Comments:**
> **username_2**: Brilliant answer. It sounds like a Sirius B telescope might actually be worth building if and when we ever get to the Sirius system. An Earth-like planet in the Andromeda galaxy would still be a few pixels across for such a telescope, I think.  
> **username_1**: @username_2 $x_i = 1.25\times 10^6 \times 0.01 \times 10^{-4}/2\times 10^6 = 0.6$ microns. You'd be better off using the Sun and getting an image of size 6 mm.  
> **username_3**: So the focussing star creates an lens whose power varies with radius.   That's more or less like a correcting element for spherical convex lens  (whose power is greatest at large radius).  
> **username_0**: If I'm reading this right, the white dwarf distance and image size would be adjustable with distance.    Further away would be better to avoid the ridiculous high orbital velocity at 0.054 AU and keeping the thing focused on one place.  Perhaps some kind of lower adjustment Lagrange orbit (kinda sorta), maybe 5 AU for L1, about 20 AU for L4/L5 could work to minimize adjustment, plus, it's good for solar power from Sirius A.  8 years is a long time to wait for info though, but not outrageously long.   The biggest obstacle, of-course, would be getting that much equipment 8 light years away.  

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