# Super Heavy Brown Dwarf

As I understand black holes, they're considered "black" because the escape velocity is, or exceeds, the speed of light. Alternatively, the escaping light is redshifted infinitely.

The Schwarzschild radius of an object is the radius of a sphere with equivalent mass for which the escape velocity is c.

Logically, this effect would occur on a continuum and wouldn't simply be state-dependent. That is, if you add a differential amount of mass to an object repeatedly it will (eventually) gradually redshift out of visibility and become a black hole. It won't simply vanish when the appropriate mass has been collected.

This suggests to me that, somewhere, there exist objects with enough mass to be highly redshifted but not enough to be "black."

Would it be possible to identify such an object from associated effects, such as gravitational lensing, or, without closer inspection, could it simply be mistaken for an extremely hot brown dwarf? (Or option C, is there a reason this couldn't happen?)

• "When the photon is emitted at a distance equal to the Schwarzschild radius, the redshift will be infinitely large, and it will not escape any finite distance from the Schwarzschild sphere." - Gravitational Redshift extracted direct from Wiki – user6760 Dec 26 '18 at 3:48
• @user6760 Yes, but if a photon is emitted just outside the Schwarzschild radius then the redshift would be finite and the photon would be hypothetically detectable. Yes? I'm asking about a massive object whose redshift is finite but extremely large. – CoilKid Dec 26 '18 at 3:56
• Give me some moment to imagine how a universe long photon being emitted would be like... – user6760 Dec 26 '18 at 4:12
• As a supplement to the great answer by @JohnRennie , you might be interested in Buchdal's theorem. For some recent related research, see arxiv.org/abs/gr-qc/0605097 and arxiv.org/abs/1606.03046. – Chiral Anomaly Dec 26 '18 at 5:43