Does light bend around black holes?

[benorin]

Does light bend around black holes like a satellite "slingshotting" around Jupiter? If so, would there be a need to recalculate the true positions of stars from all the bending of light around black holes on the way to the telescope?

 

[russ_watters]

Yes: https://en.wikipedia.org/wiki/Gravitational_lens

The impact is only significant very, very near a black hole. Otherwise it's no different than trying to look around/past a star or planet.

 

[AlexB23]

Does light bend around black holes like a satellite "slingshotting" around Jupiter? If so, would there be a need to recalculate the true positions of stars from all the bending of light around black holes on the way to the telescope?

Yep, it does. Light travels the shortest distance, and the shortest distance is a bend around a black hole.

In 1919, during an eclipse of the sun, gravitational lensing occurred, moving the apparent position of the stars. The same concept happens around black holes.

Source:
https://en.wikipedia.org/wiki/Eddington_experiment

 

[benorin]

So is there a configuration of black holes that we could use to see the other side of the earth like shaving the back of your head with mirrors? lol

 

[AlexB23]

So is there a configuration of black holes that we could use to see the other side of the earth like shaving the back of your head with mirrors? lol

I am not sure of that, but I do know black holes have photon rings, which is light that orbits the black hole.

Photon ring info:
https://arxiv.org/abs/1907.04329

 

[phinds]

Does light bend around black holes like a satellite "slingshotting" around Jupiter?

Well, that depends. If you apply Euclidean Geometry to spacetime (where is does NOT apply), then yes, it curves/bends. If you use the actual geometry of spacetime (pseudo-Reimannian) then it travels in a straight line (a "geodesic").

Since we humans are used to using Euclidean, then yeah, we say it "bends" but be sure you understand what that really means.

 

[Orodruin]

In 1919, during an eclipse of the sun, gravitational lensing occurred, moving the apparent position of the stars. The same concept happens around black holes.

The gravitational lensing always occcurs. The reason the eclipse was needed was that without it the background stars would be drowned out by sunlight.

 

[Ibix]

So is there a configuration of black holes that we could use to see the other side of the earth like shaving the back of your head with mirrors? lol

Light can u-turn around one black hole, so one in front and one behind is enough in principle, same as conventional mirrors. But the aberration is horrible and the image will be badly distorted.

In practice, given the distances involved to known black holes, the answer is no.

 

[AlexB23]

The gravitational lensing always occcurs. The reason the eclipse was needed was that without it the background stars would be drowned out by sunlight.

Yes, that is true, I could have elaborated more, as the eclipse was needed so the stars could be visible.

 

[PeroK]

Yep, it does. Light travels the shortest distance, and the shortest distance is a bend around a black hole.

Classically, Fermat's principle states that light travels between two points along the path that requires the least time. That's not necessarily the path with the shortest distance. In a vacuum it would, however, be a straight line.

In curved spacetime, light follows null geodescics, these being paths of zero spacetime distance. There is, however, a fundamental difference between null paths and paths of least spatial distance.

 

[Vanadium 50]

You do not need a black hole to do gravitational lensing, Any meass will do.

To turn light completely around so you can see tbe back of your head takes a LOT of mass. This may be theoretically possible but never actually happen,.

 

[jbriggs444]

To turn light completely around so you can see tbe back of your head takes a LOT of mass. This may be theoretically possible but never actually happen,.

One could use existing known black holes. However, all of those are far enough away that any light you detected (good luck with that after two black hole rim shots on a 3000+ light year round trip) now would have left Earth long before your head existed. Any light emerging from the back of your head now would arrive back at your eyes at a time well after your eyes had ceased to exist.

 

[benorin]

what about a 3000 year old snapshot of the Earth? What would the detail describe? Continents? Would it be valuable from a geological standpoint?

 

[Ibix]

what about a 3000 year old snapshot of the Earth? What would the detail describe? Continents? Would it be valuable from a geological standpoint?

Continents move at centimetres per year. There's no meaningful change in 3000 years, even if you could resolve a distorted image at that distance.

 

[Vanadium 50]

what about a 3000 year old snapshot of the Earth?

How do you plan to install a mirror 1500 years ago?

 

[benorin]

@Vanadium 50 read post #12

 

[benorin]

What about spectroscopy at 3000 years old?

 

[PeroK]

@Vanadium 50 read post #12

Everything is moving relative to everything else, so there's no realistic possibility of light from the Earth naturally returning to Earth.

 

[benorin]

No "trick shot," @PeroK ?

 

[Ibix]

What about spectroscopy at 3000 years old?

I don't think you are grasping quite how bad a mirror a black hole is. I would expect that there is a null path connecting the Earth back to itself via orbits around two arbitrarily chosen black holes. But the acceptance angle for light that does a half orbit (give or take) is small, and that light is spread across half the sky after the interaction, not spread over the same angle as before the reflection like you would expect with a mirror. So you're looking at orders of magnitude of dimming on each interaction, even before you worry about accretion discs and any radiation they put out.

I think the task (even under the best conceivable assumptions) is more comparable to trying spectroscopy on a direct view of a planet at several million light years.

 

[snorkack]

I don't think you are grasping quite how bad a mirror a black hole is. I would expect that there is a null path connecting the Earth back to itself via orbits around two arbitrarily chosen black holes. But the acceptance angle for light that does a half orbit (give or take) is small, and that light is spread across half the sky after the interaction, not spread over the same angle as before the reflection like you would expect with a mirror. So you're looking at orders of magnitude of dimming on each interaction, even before you worry about accretion discs and any radiation they put out.

Effectively a black hole looks like a circular convex mirror - or rather a series of concentric such mirrors.
For an extreme example to derive what the more realistic examples might look like... Suppose that there were a 1 million solar mass black hole 500 AU from Sun. (We know there isn´t, because it would leave obvious tidal gravity influences in Solar system). What would be its visual magnitude in backscattered sunlight?

 

[Ibix]

Effectively a black hole looks like a circular convex mirror - or rather a series of concentric such mirrors.

I'm not quite sure what you are imagining here. I would expect that if you were perfectly aligned between the sun and a non-rotating black hole, you could (in principle) see a ring of sunlight around the hole, because I think there's an infall approach that causes light to u-turn. Verifying that is relatively straightforward, although numerical integration is needed. You just have to see what an open orbit grazing the photon sphere does: if it's at least a 180 degree turn then you can get back-scattered light from some infall.

Working out how much power is returned is rather harder, because you need to think about ray bundles and how they spread. In principle that's easy, but I suspect it's very sensitive to how good your numerical integration is, because the angles are so small.

 

[snorkack]

I'm not quite sure what you are imagining here. I would expect that if you were perfectly aligned between the sun and a non-rotating black hole, you could (in principle) see a ring of sunlight around the hole, because I think there's an infall approach that causes light to u-turn. Verifying that is relatively straightforward, although numerical integration is needed. You just have to see what an open orbit grazing the photon sphere does: if it's at least a 180 degree turn then you can get back-scattered light from some infall.

Working out how much power is returned is rather harder, because you need to think about ray bundles and how they spread. In principle that's easy, but I suspect it's very sensitive to how good your numerical integration is, because the angles are so small.

Right. Maybe a logical approach would be to start from a bundle of parallel rays travelling towards a Schwarzschild black hole, and figure out their fate. You could mark the rays, for example with wavelength (gravitational deflection of electromagnetic waves should be independent of wavelength so long as it is much smaller than Schwarzschild radius) and the setup has axial symmetry around the path directly to singularity.
If you describe the original aim of the rays as the geometric aim past the black hole, without any deflection, calling the geometric target r, and Schwarzschild radius as R, what happens to the rays?
Clearly rays with r=0 fall straight in the event horizon.
Rays with r<R also are intercepted by event horizon even without deflection.
Some rays with r>R must also be intercepted by event horizon because although their geometric target is outside R, they are deflected inwards on approach, and do encounter event horizon.
On the other hand, rays with r>>R will be only weakly deflected.
There will be some r value at which light is deflected by exactly 180 degrees. There will be another, smaller r value at which light is deflected by exactly 540 degrees. And series of such r values, which stay above a certain minimum.
Are the distances r(180), r(540) etc. to r(∞), as functions of R, complicated functions that cannot be integrated other than numerically? Or are they by any luck simple algebraic expressions, which can be and have been integrated analytically?

 

[snorkack]

Found a source for analysis
https://www.nature.com/articles/s41598-021-93595-w
The distance I called r is there called b₀. And the impact parametre below which light is absorbed is quoted as (√27)/2*R.

 

[Orodruin]

On this topic, here is an animation I created some years back. It shows how an idealised Schwarzschild black hole would bend the (far) background - here represented by a star map based on Earth's location. The animation shows how the view would look for a stationary observer and the elapse of the video represents different positions of that observer.



Consider this frame for example:

The black circle in the middle represents the optical size of the black hole, i.e., if you send a light signal in that direction, it will end up inside the black hole. By time reversal, no light can come from that direction (unless emitted by something in between the black hole and the observer). The yellow circle is an Einstein ring - it is light from an object that is directly behind the black hole (in this case Sadalmelik, aka $\alpha$-Aquarii).

 

[snorkack]

On this topic, here is an animation I created some years back. It shows how an idealised Schwarzschild black hole would bend the (far) background - here represented by a star map based on Earth's location. The animation shows how the view would look for a stationary observer and the elapse of the video represents different positions of that observer.

The black circle in the middle represents the optical size of the black hole, i.e., if you send a light signal in that direction, it will end up inside the black hole. By time reversal, no light can come from that direction (unless emitted by something in between the black hole and the observer). The yellow circle is an Einstein ring - it is light from an object that is directly behind the black hole (in this case Sadalmelik, aka $\alpha$-Aquarii).

Are there circles that reflect the Sun behind the viewer? As explained, they have to be outside the optical edge ((√27)/2*R), but there must be infinitely many of them, converging to the optical edge of the hole.

 

[Orodruin]

Are there circles that reflect the Sun behind the viewer? As explained, they have to be outside the optical edge ((√27)/2*R), but there must be infinitely many of them, converging to the optical edge of the hole.

Those would not be visible at this resolution.

 

[snorkack]

Those would not be visible at this resolution.

I am not sure about the resolution mattering here. Sun is very bright compared to stars.
Analysing the Nature link...

In Fig. 2 the results can be seen for positive (where the angle ϕ is the unwrapped deflection angle) and negative perturbations (with the angle ϕ being defined as the angle orbited around the black hole at the time when photons cross the event horizon). For large perturbations (|δ₀|>10−2) the relationship between angle and distance is not simply exponential. However in the small perturbation regime (|δ₀|<10−2) a tight exponential relationship is visible.

And the exponent:

Inverting this expression for δ₀ implies that to achieve another orbit requires being a factor of f=e^2π=535.60±0.45 closer to the optical edge of the black hole.

At 1π (which is backscattering), a small deviation from exponential line is visible.
https://www.nature.com/articles/s41598-021-93595-w/figures/2
If anything, it is upwards. Meaning that the exponential line goes towards bigger δ₀ when approaching 0π.
Unfortunately, there does not seem to be any tabulation for the region around 1π. I expect that the δ₀ is somewhere around e^-π, which is around 0.04...0.05, but the deviation from exponentiality has some play here.
In any case: the second ring should be about 536 times dimmer than the first ring. But considering Sun is 10¹⁰ times brighter than brightest fixed stars, the first ring still should be brighter than the fixed stars!

 

[Orodruin]

Sun is very bright compared to stars.

The Sun is not in the image … it is a projection of background stars only.

 

[snorkack]

Everything is moving relative to everything else, so there's no realistic possibility of light from the Earth naturally returning to Earth.

The light from Earth obviously naturally returns to Earth, and that despite everything moving. Look at earthlight on Moon at crescent Moon. It is there despite Moon orbit.
And then reflect on how dim it is.
It is partly because Moon is black. But only partly.
If you replaced Moon with a Moon sized perfect mirror, like a polished aluminum ball, then the Moon would be brighter than it now is. But there would be a single dazzling spot too small to resolve by naked eye - the glint/image of Sun. The rest of Moon would be reflecting the black sky, stars and Earth. All of which would leave shrunken images, and therefore dimmed.
Now replace Moon with Moon mass black hole.
The black hole of Moon mass would orbit the Earth just as Moon does, and raise equal tides on Earth.
The Schwarzschild radius of Moon would be... about 100 μm. That is, 0,1 mm.
The visual size of a Moon-sized black hole? Well, sticking to exactly 100 μm Schwarzschild radius, the photon sphere radius would be exactly 3/2 Schwarzschild radii being 150 μm, and the visual radius exactly √3 times photon sphere radius, which rounds to 260 μm.
Visually, a Moon sized black hole would have diametre of 0,52 mm. Compare these 520 μm to 3500 km diametre of Moon.
Yes - a black hole would reflect light.
A spherical mirror the size of Moon would deflect light by over 90 degrees over a disc 1200 km radius - half the disc area. (The other half of disc would receive incident light at under 45 degrees from horizon and therefore deflect it by less than 90 degrees).
Going by the approximation δ₀=Re^-θ, the black hole 100 μm Schwarzschild radius would deflect light by 90 degrees, that is π/2, about Re^-π/2 from outer edge... which is approximately 21 μm. That is where this approximation is breaking down, so a bit more but not much more. And deflect light by 270 degrees about Re^-3π/2 from outer edge, which comes out as about 0,9 μm.
So, 0,02 mm wide reflective ring with inner diametre 0,52 mm, and outer diametre 0,56 mm. How bright do you figure it is? How much detail does reflection in it show?