Does light bend around black holes?

[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.