According to Einstein's theory, nothing can travel faster then light and also a black hole doesn't allow light to escape but recently scientists got an image of a black hole. Here I am confused: if light can't escape from black hole as it lost most of its mass then how did scientists get the image of a black hole? If we can only see things which can reflect back light and if light can't be reflected back from the black hole then how do we get the image? Or is it just an illusion? Please guide me.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ bbc.com/news/science-environment-47873592 , you mean this image? I think t is clear that all the light in the photos comes from outside the event horizon of the black hole, which is the point of no return. The name of the collaboration of telescopes is called "the event horizon" $\endgroup$– anna vCommented Jul 24, 2019 at 4:53
-
3$\begingroup$ Possible duplicate of Orange ring in a black hole image $\endgroup$– ProfRobCommented Jul 24, 2019 at 6:14
-
$\begingroup$ Related: physics.stackexchange.com/q/137837/25301 $\endgroup$– Kyle KanosCommented Jul 24, 2019 at 12:47
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
The image that has been captured by the Event Horizon Telescope is actually an image of the photon ring (the glowing region) that surround the dark shadow.
The black hole is located at the center of the dark disk and the disk radius is about 2.5 times that of the event horizon. The photon ring results from light rays that orbit around the black hole in the near-field region before escaping to infinity, where they arrive near a ring-shaped critical curve on the image plane. The shadow represents the interior of the critical curve.
The "photon ring" is a region of enhanced brightness near the critical curve that arises if optically thin matter emits from the region where unstable bound photon orbits exist. The light rays that comprise the photon ring can orbit many times through the emission region and thereby pick up extra brightness. Since the optical path lengths become arbitrarily long near the critical curve, the brightness can become arbitrarily large (neglecting absorption).
$\begingroup$
$\endgroup$
2
Look again at the image. There's a black spot in the center - that corresponds to the black hole. It's black because it's not reflecting light. The bright things that are around the black hole are not part of the black hole - they could e.g. be matter infalling into the black hole, which have been heated and therefore emit light.
-
$\begingroup$ This isn't quite right. For instance, why then can we not see matter in front of the black hole? $\endgroup$– ProfRobCommented Jul 24, 2019 at 5:55
-
$\begingroup$ @RobJeffries I imagine it's been processed out of the image? If you are able to write a better answer, feel free to do so and I'll delete this one. $\endgroup$– AllureCommented Jul 24, 2019 at 5:56
$\begingroup$
$\endgroup$
In the case of M87, the reason we can "see the black hole" is that it is surrounded by hot gas that is essentially transparent to the radiation it emits at mm wavelengths.
The radiation is emitted in all directions by the gas, but of course we only see light rays that come towards the Earth.
In the highly distorted spacetime around a black hole, light emitted near the black hole can be severely bent (and even loop right around) around the black hole. The net outcome is that we see a concentration of light rays that appear to come from a ring (defined by the "photon sphere") that surrounds the black hole.
No light emerges from the black hole itself (by definition) and very little light comes straight to us from light emitted from material directly in front of the black hole.
Note also: the bright ring is not an image of the accretion disk around the black hole. The accretion disk is a much larger structure than the scale of the image and is inclined to the line of sight. The effect seen is almost entirely determined by the geometry of light rays close to an event horizon.
A more detailed answer was already given in Orange ring in a black hole image
$\begingroup$
$\endgroup$
What we see in the image is light from the accretion disk and light from behind the black hole that is perturbed by it. This creates the shadow disk which has 2.5 times the diameter of the event horizon. There is no sign of the event horizon itself other than the fact that the image is consistent with its existence.
$\begingroup$
$\endgroup$
3
The bright glowing ring surrounding the black hole is known as the accretion disc.
The accretion disc is a superheated cloud of gas and cosmic dust spinning around the event horizon.
The gravity inside of a black hole is too powerful for any light to escape past the event horizon.
But the accretion disc itself is immensely bright and releases vast amounts of electromagnetic radiation, such as x-rays, radio waves and infrared beams into space.
The dark black hole like spot is NOT the black hole itself, It is a dark, round shadow, which is the black hole's event horizon.The event horizon is the point of no return, past which all matter, including light, falls towards the black hole’s infinitely compressed singularity.
In order to interpret the image of black hole better, you may want to watch this video
-
$\begingroup$ There is an incorrect statement here. The dark disk in the EHT image corresponds to the Einstein shadow, not to the event horizon. $\endgroup$– my2ctsCommented Jul 24, 2019 at 6:23
-
$\begingroup$ What is the Einstein shadow? Apparently, I have not yet encountered this word. $\endgroup$ Commented Jul 24, 2019 at 6:29
-
$\begingroup$ A large portion of this answer seems to be unattributed quotes from this news article. $\endgroup$– Chris ♦Commented Jul 24, 2019 at 11:13