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This question already has an answer here:

Basically a neutron/quark star that approaches it's Schwarzschild radius but doesn't surpass it. Its gravitational force is great enough to not let light escape from it, but not great enough to become a black hole.

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marked as duplicate by John Rennie black-holes Jan 15 at 17:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ If light can't escape, then it is a black hole, but the deeper a light source falls into a gravity well, the more the photons that escape will be red-shifted when they climb out. For any given light source, there will be some amount of red shift that makes the photons effectively undetectable even though, technically speaking, they do "escape." $\endgroup$ – Solomon Slow Jan 15 at 16:43
  • $\begingroup$ Maybe you can look this, en.wikipedia.org/wiki/… $\endgroup$ – Reign Jan 15 at 16:46
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If it is not a black hole, then light can escape from it. Why? Because a black hole is defined as something light cannot escape from.

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The answer is "no" because only for a black hole can light never escape. However, this question deserves a fuller answer because of a thing called the "photon sphere".

At the Schwartzchild radius of a compact body - whose mass is all inside the radius - light can no longer escape. However, at a distance which is 1.5 times the Schwartzchild radius there is a thing called the photon sphere. Any light incident on a massive body that comes from the outside of the photon sphere and enters it will invariably be "sucked into" the body. Light that originates within the photon sphere, but outside the Schwartzchild radius, and is directed outwards, may still escape however. This opens up the possibility that there exist objects so massive that their mass is within a photon sphere but not within the Schwartzchild radius. Such objects would likely be neutron stars.

Now the fun part - what do we mean by "invisible"? If by "invisible" we mean "I shine light on it and zero of it is ever reflected" then the answer to your question is "yes"!

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