Discussion:
What does it mean in astrophysics for X-rays to be reflected?
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Thomas Womack
2019-01-14 18:14:36 UTC
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Various articles about black holes talk about X-rays emitted near the
event horizon 'reflecting off the accretion disc'.

What kind of material is it that can *reflect* X-rays? I've worked in
X-ray crystallography, and we needed grazing incidence off very
precisely figured monocrystalline silicon to get something that
reflected X-rays at 12.7keV (selenium K line); astrophysical X-rays
seem to be more at iron K which is about half that energy, but still
generally-occuring materials either absorb or transmit them.

Is this in fact more like the process around a nuclear detonation,
where things absorb X-rays and are themselves heated to X-ray-emitting
temperatures?

Tom

[[Mod. note -- Yes, thermal re-emission is one possibility.
Compton scattering is another possibility. As you note, coherent
reflection seems unlikely.
-- jt]]
Martin Hardcastle
2019-01-15 19:59:51 UTC
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Post by Thomas Womack
Is this in fact more like the process around a nuclear detonation,
where things absorb X-rays and are themselves heated to X-ray-emitting
temperatures?
The emission lines people talk about are X-ray fluorescence from
(relatively) cold material.

Martin
--
Martin Hardcastle
School of Physics, Astronomy and Mathematics, University of Hertfordshire, UK
Please replace the xxx.xxx.xxx in the header with herts.ac.uk to mail me
Eric Flesch
2019-01-16 03:15:59 UTC
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Post by Thomas Womack
generally-occuring materials either absorb or transmit them.
[[Mod. note -- ... As you note, coherent reflection seems unlikely.
Does not radiation reflection *always* consist of absorption &
re-transmission? By "coherent reflection", JT seems to imply
preservation of the original photons. But photons don't "bounce" in
the rubber-ball sense, right?

I'm reminded of transferring money in the banking system -- it's not
the same dollar that moves around, money is "fungible" in the sense
that a dollar has no individual identity as such. I am suspecting
that photons are fungible in the same way, but that that element is
not built into the model of light as we know it.

[[Mod. note -- My apologies for being unclear. What I was trying to
get at with the phrase "coherent reflection" (which in hindsight was
a poor choice of words on my part) was "reflecting like a beam of
optical light from a mirror, with angle-of-reflection =
angle-of-incidence".

As to whether elastic scattering of any sort yields the "same" photon,
I suspect that you're right and that photons don't have an individual
identify. In fact, I suspect that "the same photon" isn't even a
meaningful concept in quantum optics. But my knowledge of quantum
optics is alas very small, so I can't speak with any authority on
this....
-- jt]]
Steve Willner
2019-02-05 15:50:44 UTC
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Post by Eric Flesch
As to whether elastic scattering of any sort yields the "same" photon,
I suspect that you're right and that photons don't have an individual
identify.
Doesn't this have to be true? Wouldn't the Planck law have a
different form if photons were distinguishable? When I studied
thermodynamics, there were four cases: particles could be
distinguishable or indistinguishable, and they could or could not
occupy the same state. (Distinguishable particles that can occupy
the same state are rare.)
--
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