I 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?

 

[Ibix]

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?

Solar irradiance is about $1.4\mathrm{kWm^{-2}}$, so about $60\mathrm{\mu W}$ falls on your annulus. Bear in mind that this is spread across half the sky (from 90 to 270 degrees deflection) if I'm following your argument, so the inverse square law at 400,000km takes this to something like $6\times 10^{-23}\mathrm{Wm^{-2}}$ at Earth's surface, or about magnitude 36.

 

[ohwilleke]

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

It depends upon how good your telescope is. For example, you can see huge numbers of stars with the James Webb Space Telescope in places where inferior older telescopes saw nothing but black emptiness. It would not be valuable from a geological stand point, because it would be distorted and because there are much less cumbersome ways to accurately make the same determinations.

 

[snorkack]

It depends upon how good your telescope is. For example, you can see huge numbers of stars with the James Webb Space Telescope in places where inferior older telescopes saw nothing but black emptiness. It would not be valuable from a geological stand point, because it would be distorted and

Distortion would not be inherent problem because gravity of black hole would act like a smooth curved mirror - mere geometric distortions could be interpreted.
But I have a suspicion that the image reflected from a black hole might have inherent limits of angular resolution due to diffraction of electromagnetic waves off the black hole itself. Can anyone confirm or deny?

 

[L Drago]

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?

No, light bends inside event horizon of black holes. So it won't affect our observation. Even light can't escape black holes immense gravity.

 

[Ibix]

No, light bends inside event horizon of black holes. So it won't affect our observation. Even light can't escape black holes immense gravity.

Gravitational lensing around stars has been observed, and it could also be seen near black holes. It's just not an effect that makes much difference to our astronomical observations unless we specifically go looking for it.

 

[L Drago]

Gravitational lensing around stars has been observed, and it could also be seen near black holes. It's just not an effect that makes much difference to our astronomical observations unless we specifically go looking for it.

It's so fascinating that if we go near a black hole time would go slower far us as black holes have very much gravity and the curvature in space time fabric will be very much. If we go to singularlity and somehow manage to come out we will be in future as time would nearly stop for us. But after crossing the event horizon we cannot come back as any physical object cannot travel from in speed of light. According to Einstein special relativity e = mc². We would require infinite energy and mass to travel in the speed of light or faster than that.

 

[Dale]

According to Einstein special relativity e = mc². We would require infinite energy and mass to travel in the speed of light or faster than that.

The famous $E=mc^2$ formula only applies for an object at rest. For a moving object you need to use the more general $m^2 c^2=E^2/c^2-p^2$ where $p$ is the momentum. That, combined with $v=c^2 p/E$ can show that an object with $0<m$ travels with $-c<v<c$ regardless of $E$.

 

[L Drago]

The famous $E=mc^2$ formula only applies for an object at rest. For a moving object you need to use the more general $m^2 c^2=E^2/c^2-p^2$ where $p$ is the momentum. That, combined with $v=c^2 p/E$ can show that an object with $0<m$ travels with $v<c$ regardless of $E$.

Okay thanks for the informations. But if we go inside a supermassive blackhole. For instance, take example of sagattorious A star. If we go to singularlity and somehow manage to come outside of it (though impossible according to Einstein special relativity). How to calculate home much years have passed on earth if we spend 1 hour there. I know time would pass a lot more slowly for us as more curvature in space time and slower the time. I just want to calculate. As I am just a seventh grader, kindly tell me a little easy method to calculate if possible.

 

[Dale]

If we go to singularlity and somehow manage to come outside of it (though impossible according to Einstein special relativity)

It is impossible according to general relativity, not just special relativity.

How to calculate home much years have passed on earth if we spend 1 hour there.

This question cannot be answered. You propose a scenario that is impossible in general relativity and ask what the calculation would show? Which equations should we use? The equations of general relativity cannot even be set up because the scenario is impossible. So what other theory could be used?

Newtonian gravity doesn't have time dilation or black holes or singularities at all, so the question doesn't apply to that theory either.

You simply cannot propose a scenario that is impossible in some theory and then use that theory to calculate anything about the scenario.

 

[L Drago]

It is impossible according to general relativity, not just special relativity.

This question cannot be answered. You propose a scenario that is impossible in general relativity and ask what the calculation would show? Which equations should we use? The equations of general relativity cannot even be set up because the scenario is impossible. So what other theory could be used?

Newtonian gravity doesn't have time dilation or black holes or singularities at all, so the question doesn't apply to that theory either.

You simply cannot propose a scenario that is impossible in some theory and then use that theory to calculate anything about the scenario.

But if there were a planet or star with gravitational force as high or close to a black hole then it will be possible to calculate right.

 

[Orodruin]

But if there were a planet or star with gravitational force as high or close to a black hole then it will be possible to calculate right.

There is no such thing. A black hole does not correspond to a particular gravitational strength so this statement is not very meaningful. The gravitational strength, or more accurately, the spacetime curvature caused by the black hole, depends on the mass of the black hole. The same holds for any gravitating body, but a planet or star will be much bigger meaning that the gravitation inside it is weaker than it would be inside a black hole.

 

[Vanadium 50]

@L Draco you are a) highjacking someone elses thread anf b) posting things that are not so.

 

[L Drago]

There is no such thing. A black hole does not correspond to a particular gravitational strength so this statement is not very meaningful. The gravitational strength, or more accurately, the spacetime curvature caused by the black hole, depends on the mass of the black hole. The same holds for any gravitating body, but a planet or star will be much bigger meaning that the gravitation inside it is weaker than it would be inside a black hole.

Even if it is like that. Can you please describe a formula to calculate if we spend 1 hour in Sun how much time would be spent on earth. Calculate even if there is a minute difference by equation.

 

[Dale]

But if there were a planet or star with gravitational force as high or close to a black hole then it will be possible to calculate right.

You can calculate the gravitational time dilation for a clock at rest at any altitude above the event horizon. Below the event horizon there cannot be any clocks at rest, and there is no beyond the singularity at all.

For a clock hovering at an altitude above the event horizon the gravitational time dilation factor is: $$\frac{1}{\gamma}=\sqrt{1-\frac{2GM}{rc^2}}$$ where $\gamma$ is the time dilation factor, and $r$ is the areal radius above the spherical mass $M$, and $G$ and $c$ are the usual gravitational constant and speed of light respectively.

 

[L Drago]

You can calculate the gravitational time dilation for a clock at rest at any altitude above the event horizon. Below the event horizon there cannot be any clocks at rest, and there is no beyond the singularity at all.

For a clock hovering at an altitude above the event horizon the gravitational time dilation factor is: $$\frac{1}{\gamma}=\sqrt{1-\frac{2GM}{rc^2}}$$ where $\gamma$ is the time dilation factor, and $r$ is the areal radius above the spherical mass $M$, and $G$ and $c$ are the usual gravitational constant and speed of light respectively.

Thanks a lot for this information. By this we can calculate relative times hopefully.

 

[Dale]

Thanks a lot for this information. By this we can calculate relative times hopefully.

Yes. For any clock at rest at any altitude above the event horizon.

 

[jbriggs444]

Thanks a lot for this information. By this we can calculate relative times hopefully.

The formula that @Dale gives works for the exterior of a spherically symmetric mass or black hole. In the interior of a spherically symmetric mass, things are a bit trickier because the layers above you do not contribute fully to your time dilation.

 

[jedishrfu]

With respect to light orbits of a blackhole, there is a great book by Kip Thorne calledThe Science of Interstellar where he explains the physics of black holes and how lght is bent by them so as to orbit the black hole many times before exiting in a new direction.

https://www.amazon.com/Science-Interstellar-Kip-Thorne/dp/0393351378/?tag=pfamazon01-20

If you are interested, you might enjoy this book.

 

[L Drago]

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:
View attachment 352634
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).

That's true even light can't escape the immense gravitational pull of a black holes and bends in the event horizon of a black hole.

 

[Orodruin]

That's true even light can't escape the immense gravitational pull of a black holes and bends in the event horizon of a black hole.

First of all, I think this description of things is a bit sensationalist. There is nothing inherently immense of the gravitational pull of a black hole. The tidal forces, which is what you would notice in free fall, of a supermassive black hole at the event horizon are practically negligible whereas they would be extreme for small black holes. What you really should be discussing is spacetime curvature.

I also don’t understand what you think this statement adds to this conversation. Most people in this discussion understand very well what a black hole is and how general relativity works to the extent of actually being able to compute physical observables.