All infalling matter-energy appears to an external observer as frozen in time at the event horizon. Why then is this horizon not extremely bright due to radiation that is able to escape radially? So should all black holes not appear as a bright sphere?
2 Answers
As Ben says, black holes that have accretion disks shine very brightly indeed due to heating of the matter in the disk as it falls inwards. However not all black holes have accretion disks. For example the black hole at the centre of our galaxy does not, and I suspect you're asking why it is no longer shining brightly.
If you take a glowing object. e.g. a metal sphere heated to white heat, and drop it into a black hole you'll see its light fade as it approaches the event horizon and in the final stages of the approach to the horizon it would fade to black. This happens because the light emitted by the sphere loses energy as it climbs out of the gravitational well. This effect is called gravitational redshift. This question, Gravitational Redshift around a Schwarzschild Black Hole, has already asked how to calculate the redshift for an object falling into a black hole, but the answer is a bit technical. To summarise, the redshift is given by:
$$ z = \frac{f-f_o}{f_o} = \frac{1}{\sqrt{1 - 2GM/c^2r}} - 1 $$
where $r$ is the distance from the centre of the black hole, $f$ is the original frequency and $f_o$ is the observed frequency. The event horizon is at $r = c^2/2GM$, and at this distance the red shift goes to infinity i.e. the frequency of the light you see falls to zero.
So matter falling into a black hole may well be heated by friction as it falls through the accretion disk, however this is only temporary and as it reaches the event horizon you will see it red shift to infra-red, then radio and eventually to nothing.
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Not all black holes have large amounts of infalling matter. A quasar is an example that does, and it does appear extremely bright to an external observer -- often 100 times brighter than an entire galaxy.