4
$\begingroup$

In A Very Short Introduction: Black Holes by Katherine Blundell, the author discusses the imaging of plasma jets at the center of black holes:

At 50 million light-years from Earth an object moving at the speed of light would move across the sky at four milli-arcseconds per year. When we consider that an arcsecond is 1/3600 of a degree, then four-thousandths of this may sound like a tiny angle to measure, but such separations are easily resolvable with an instrument like the VLBA. The VLBA has already imaged the base of this jet to within less than about thirty Schwarzschild radii of its supermassive black hole.

What's exactly meant by that?

$\endgroup$
1
  • 3
    $\begingroup$ Note that the jets are not "at the center of black holes". They are very much outside the black holes. $\endgroup$
    – throx
    Jun 28 at 1:24

1 Answer 1

14
$\begingroup$

The Schwarzschild radius is being used as a unit of distance from a black hole, where $$r_s = 2\frac{GM}{c^2}$$ and is approximately 3 km if $M$, the mass of the black hole, is 1 solar mass and would scale up as $M$ increased.

In this case the author is saying that radio imaging using interferometry has been able to trace the jet back to about 30 times this radius from the centre of some supermassive black hole that resides (presumably) at the centre of an active galaxy.

The paragraph in isolation makes no sense because which supermassive black hole isn't mentioned, but I assume it is one at the centre of a galaxy 50 million light years away.

$\endgroup$
2
  • 2
    $\begingroup$ For any future readers not familiar with the Schwarzschild radius, it's the event horizon, the depth into the gravity well where the escape velocity = the speed of light. (For the special case of non-rotating, non-charged black holes, aka Schwarzschild black holes). So time dilation and red shift approach infinity there for outside observers looking at something falling towards a black hole, thus making it a highly-relevant length scale. $\endgroup$ Jun 28 at 3:45
  • 2
    $\begingroup$ M87 is about 50 Mly away. And VLBA at 43 GHz can apparently achieve 30 Schwarzschild radii resolution for that source. $\endgroup$
    – 9769953
    Jun 28 at 8:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .