coherent light from stars
John C
Coherency is required for interferometry, isn't it?
-jc
Douglas B Sweetser
John C wrote:
Stars are point sources, so they are spatially coherent. For dual slit
experiments, a flashlight shown on the two slits will not create an
interference pattern. If the flashlight first travels through one slit
before reaching the two slits, then that light will be spatially coherent
and thus show an interference pattern.
doug
quaternions.com
Uncle Al
John C wrote:
>
> Could someone explain to this novice why star light is coherent?
> Coherency is required for interferometry, isn't it?
One can argue a point source is a de facto coherent source. Starlight
as such is neither intrinsically coherent nor monochromatic.
Coherence (would you like that in time or space?) can give you a nice
clean set of flat interferometric fringes that contain almost no
information. Interferometry of messy light is rich with information:
Split a clean coherent beam and then recombine with a generous helping
of garble in one leg, get a hologram at the distal intersection.
A large optic, transmission or reflection, is a Fourier filter. Area
gives you intensity, span gives you phase information. Of the two,
the span and its phase information are the important stuff - wherein
the information resides. Each Keck telescope in Hawaii is
respectible. Ganged into an interferometer the pair together have
awesome resolving power. Ditto farms of radio telescopes.
The object of astronomic inteferometry is to disentangle the rich
garble into information. Similarly, you can do continuous wave NMR
and get a nice spectrum, or go pulsed Fourier transform and kick
butt. One would not guess that smashing a case of leaded crystal wine
glasses is the best and fastest way to find out how each one rings
(and how it affects it neighbors' rings, and vice-versa), but it is if
you are clever.
--
Uncle Al
http://www.mazepath.com/uncleal/eotvos.htm
(Do something naughty to physics)
Robert Kolker
John C wrote:
> Could someone explain to this novice why star light is coherent?
> Coherency is required for interferometry, isn't it?
Is sunlight coherent?
Bob Kolker
Student
John C wrote:
> Could someone explain to this novice why star light is coherent?
> Coherency is required for interferometry, isn't it?
I am not fully satisfied by the replies you got to your question.
Though I know little of the matter, it seems to me that star interferometry
calls for Hanbury Brown-Twiss type of experiments, i.e., measure of
*second* order coherence, which has nothing at all to do with first order
coherence (though Uncle Al makes interesting claims in that regard).
1st order coherence measures field-field correlations. This is the usual
measure by which a field can exhibit interference and displays fringes. HBT
experiment measure intensity-intensity correlations. There the photon
statistics is measured. If the light is 2nd order coherent, there are no
photons correlations. If the light is thermal (as is the case of a star),
the photons show bunching effect: the probability increase to measure one
when you already measured another. Of course, if the light comes from two
distinct thermal sources, no correlations either.
I would not qualify light from a star as coherent by any standard. Just
some statistical effect in higher coherence order is measured.
Comments welcome.
Douglas B Sweetser
Bob Kolker wrote:
> Is sunlight coherent?
No. The Sun is not spatially coherent because the disc in the sky takes up
a clearly visible area. If I were a better physicist, I am sure there is a
way to calculate how thin a source must be to be spatially coherent, but I
do not know it.
Sunlight bounces off atoms in the upper atmosphere with a 4th power
dependence on wavelength, which is why it is blue (was Einstein really the
first to state this, or is that an Einstein hyperbola?). The scattering
process polarizes the light 90 degrees away from the source. Bees use the
polarization of scattering to tell each other about good pollen sources.
The darkness of the sky also gets altered when wearing polarized sunglasses
and twisting your head.
doug
quaternions.com
[Moderator's note: Einstein hyperbola, or hyperbole? - jb]
Andrew Resnick
Sunlight, at a wavelength of 555 nm (the peak wavelength) has a
transverse coherence length of approximately 0.06 mm.
--
Andrew Resnick, Ph. D.
National Center for Microgravity Research
NASA Glenn Research Center
Andrew Resnick
> Bob Kolker wrote:
>> Is sunlight coherent?
> No. The Sun is not spatially coherent because the disc in the sky
> takes up a clearly visible area. If I were a better physicist, I am
> sure there is a way to calculate how thin a source must be to be
> spatially coherent, but I do not know it.
You are wrong. The sun is not infinitely large, so there is some
spatial coherence. The formula is A = l^2/w, where A is the coherence
area (taken to be a circle), l the mean wavelength of the source, and w
is the angular size of the source. The sun subtends approximately 16
minutes of arc, so the solid angle is about 6.8 *10^-5 steradians. Using
a narrowband filter at 500 nm give a coherence area of 3.7*10^-3 mm^2,
which is a circle of radius 0.03 mm
> Sunlight bounces off atoms in the upper atmosphere with a 4th power
> dependence on wavelength, which is why it is blue (was Einstein really
> the first to state this, or is that an Einstein hyperbola?). The
> scattering process polarizes the light 90 degrees away from the
> source. Bees use the polarization of scattering to tell each other
> about good pollen sources. The darkness of the sky also gets altered
> when wearing polarized sunglasses and twisting your head.
No again. It's called "Rayleigh scattering".
Andrew Resnick
In <b8mkc1$ikm$1...@panther.uwo.ca> Student wrote:
> John C wrote:
>> Could someone explain to this novice why star light is coherent?
>> Coherency is required for interferometry, isn't it?
> I am not fully satisfied by the replies you got to your question.
> Though I know little of the matter, it seems to me that star
> interferometry calls for Hanbury Brown-Twiss type of experiments, i.e.,
> measure of *second* order coherence, which has nothing at all to do
> with first order coherence (though Uncle Al makes interesting claims
> in that regard).
Existing stellar interferometers use first-order correlations. No photon
counting required. Purely classical.
<snip>
> I would not qualify light from a star as coherent by any standard.
> Just some statistical effect in higher coherence order is measured.
>
> Comments welcome.
Eh? Could you please expand those final two sentences? You seem to claim
that starlight is not coherent because only statstical correlations are
measured. But that's the definition of coherence!
Student
> Existing stellar interferometers use first-order correlations. No
> photon counting required. Purely classical.
Maybe the next time I encounter this topic somewhere I would like to
hear from you again telling that. Unfortunatly I don't think I'll
enter a vast litterature right now just for that purpose. Allow me
please to delay this discussion.
>> I would not qualify light from a star as coherent by any standard.
>> Just some statistical effect in higher coherence order is measured.
> Eh? Could you please expand those final two sentences? You seem to
> claim that starlight is not coherent because only statstical
> correlations are measured. But that's the definition of coherence!
Yes, statistical correlations (which define 2nd order coherence) are
measured, but they are *not* met. To be sure, if you don't have
correlations, you are coherent, if you have correlations, you are not
coherent. So why I think starlight is not first order, neither second
order coherent is that it is thermal light which does make good
fringes over appreciable optical path delay (so it's not 1st order)
and it exhibits photon bunching, i.e., it has photon-correlations (so
it's not 2nd order either). I don't think I contradict here the
previous statement and I think we can probably agree on this since
this is standard definition of coherence to be found everywhere. When
I say "not 1st order coherent" I do not mean, of course, that it has
no coherence whatsoever (its coherent degree is less than 1, that's
what I mean, now of course it is certainly > 0 so it is "a little" 1st
order coherent, as everything else).
That is the part I know for sure. Now I still do not know what people
do with coherence and stars. Hanbury Brown and Twiss made
interferometry with stars, and it was 2nd order, it was designed for
this purpose, and it allowed to measure star radius or such things. I
do not know the details. However when I hear "coherence" and "stars"
in the same sentence I would believe that this is related to 2nd order
coherence, through this HBT setup. Also, HBT is not quantum by nature,
you can perform the experiment on classical fields (as one emanating
from a star! the very historical application).
Best Regards.
BllFs6
Hi
maybe this will clear things up a bit...
Yes, the Hanbury Brown-Twiss interferometry DID use somekinda funky secondary
coherence which I cant even pretend to explain...
Funny, many "experts" at the time just KNEW it wouldnt work even IN
theory...much less practice...
They did this way back in the 60's I believe
Measured the apparent diameters of 80 or so nearby/large stars...which was
pretty useful to astronomers at the time..
But there was NO imaging involved....just big crude seachlight like mirrors,
which served as light buckets that feed into photon counters....
It was intensity/time interferometry....
They even published a nice little book on the whole theory and the story of its
construction and use. The did ALOT with just a little...any major univ library
will probably have it the astronomy section.
Todays optical/near IR interferometry is a whole other ballpark....they combine
the optical IMAGES from 2 or more seperate scopes....they must actively
correct, real time, the incoming wave fronts for each scope to a fraction of a
wavelength of light, (quite a feat for a mirror many meters across).
Then they have to combine the 2 or more beams and keep all the images stable
while maintianing the optical path difference to a fraction of wavelength of
light....AMAZING
take care
Bill
Douglas B Sweetser
>> Sunlight bounces off atoms in the upper atmosphere with a 4th power
>> dependence on wavelength, which is why it is blue (was Einstein really
>> the first to state this, or is that an Einstein hyperbol[e]?). The
>> scattering process polarizes the light 90 degrees away from the
>> source. Bees use the polarization of scattering to tell each other
>> about good pollen sources. The darkness of the sky also gets altered
>> when wearing polarized sunglasses and twisting your head.
> No again. It's called "Rayleigh scattering".
I have read a few sources ("Subtle is the Lord..." pg 100-103 and the
"Encyclopedia of Physics"). John Tyndall was an experimentalist who in
1869 first suggested the sky was blue due to scattering of light off of
water droplets and dust. Rayleigh did the theoretical work, first
published in 1871 that should the fourth power dependenc of scattering on
the wavelenght, so (6500/4500)^4 ~ 4.3/1.
Near its critical temperature, a gas becomes opaque. The scientist Marian
Smoluchowski studied this issue experimentally and theoretically. He was
able to determine an equation for the mean square particle number
fluctuations, a number that blows up at the critical temperature.
Rayleigh believed that inhomogeneities in the air might explain some of
the things known as Tyndall effects. Smoluchowski thought his critical
opalescence might be the key, but this was a qualitative argument. In 1910
Einstein did the quantitative work, deriving an equation for scattering in
a weakly inhomogeneous nonabsorptive medium. In a particular limit, the
equation becomes Rayleigh scattering. So Einstein did contribute a
refinement to the issue of why the sky is blue.
Thanks for the formula and calculation that for the Sun, the spatial
coherence works out to be a circle with a radius of 0.03 mm. That is a
small number. How would that effect the double slit experimental setup?
doug
quaternions.com
Andrew Resnick
> Hello Andrew:
>
<snip>
>
> Thanks for the formula and calculation that for the Sun, the spatial
> coherence works out to be a circle with a radius of 0.03 mm. That is
> a small number. How would that effect the double slit experimental
> setup?
I'm not sure what you mean here, but if you use sunlight (through a
narrowband filter) to illuminate an object and look at it under high
magnification, you will see speckle. The speckle is from spatial
coherence. So I imagine if your slits (pinholes might be better) were
less than one speckle diameter apart, you could generate interference
fringes in the far field.
Andrew Resnick
In <3ebab04e$1...@news.sentex.net> Student wrote:
> Andrew Resnick wrote:
>
<snip>
>
>> Eh? Could you please expand those final two sentences? You seem to
>> claim that starlight is not coherent because only statstical
>> correlations are measured. But that's the definition of coherence!
>
> Yes, statistical correlations (which define 2nd order coherence) are
> measured, but they are *not* met. To be sure, if you don't have
> correlations, you are coherent, if you have correlations, you are not
> coherent. So why I think starlight is not first order, neither second
> order coherent is that it is thermal light which does make good
> fringes over appreciable optical path delay (so it's not 1st order)
> and it exhibits photon bunching, i.e., it has photon-correlations (so
> it's not 2nd order either). I don't think I contradict here the
> previous statement and I think we can probably agree on this since
> this is standard definition of coherence to be found everywhere. When
> I say "not 1st order coherent" I do not mean, of course, that it has
> no coherence whatsoever (its coherent degree is less than 1, that's
> what I mean, now of course it is certainly > 0 so it is "a little" 1st
> order coherent, as everything else).
>
This paragraph is very confusing to me. Let's see if I can parse it
correctly:
First, the claim that '...if you don't have correlations, you are
coherent, if you have correlations, you are not coherent.'
Then the claim that '...thermal light which does make good fringes over
appreciable optical path delay...and it exhibits photon bunching'
And finally, '...this is standard definition of coherence to be found
everywhere'
I think that is the essence of the paragraph above. Assuming I read it
properly, I respond thusly:
As to the first point, that is exactly backwards. The degree of
coherence is actually defined in terms of the two-point two-time
correlation:
G(P1,P2,s) = <E(P1,t+s),E*(P2,t)>, where <> means time average, E is the
electric field and E* the complex conjugate. Thus, if G is a delta
function (no field correlations) then the degree of coherence is zero.
This is first-order coherence; higher-order coherence terms are
calculated straightforwardly from higher-order correlations.
As to the second point, it appears you are mixing spatial coherence with
spectral coherence. Spatial coherence (the kind needed for Young's
double slit type interference) requires a non-zero transverse coherence
length (or coherence area). Spectral coherence (the kind needed for
Michaelson/Twyman-Green interferometer-type interference) requires a non-
zero coherence length. Broad-band thermal radation has a vanishingly
small coherence length, but could have a very large coherence area, if
the radiation originates from a sufficiently small source. I am aware
of some recent results which claim that certain thermal radiation
exhibits photon bunching, but I don't know much more than that. Stellar
interferometers have exceeingly tight control over the path length
differences- but they are not Twyman-Green interferometers, they are
Young's interferometers.
Finally, for the third point, let's just state the definition of the
mutual coherence function, which is G(P1, P2, s) and contains both
temporal (spectral) and spatial coherence parts.
Or am I misunderstanding your claims?
p.ki...@imperial.ac.uk
Andrew Resnick <andy.r...@nospam.grc.nasadotgov> wrote:
> Douglas B Sweetser wrote:
>> Sunlight bounces off atoms in the upper atmosphere with a 4th power
>> dependence on wavelength, which is why it is blue (was Einstein really
>> the first to state this, or is that an Einstein hyperbola?). The
>> scattering process polarizes the light 90 degrees away from the
>> source. Bees use the polarization of scattering to tell each other
>> about good pollen sources. The darkness of the sky also gets altered
>> when wearing polarized sunglasses and twisting your head.
> No again. It's called "Rayleigh scattering".
I'm not sure what you are saying "no" to here. If I wear
polarized sunglasses, and twist my head, I see the brightness
of the sky change. This surely means that the light from
the sky has, in some (most) places, at least a partial degree of
polarization.
--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714
Imperial College London, Dr.Paul...@physics.org
SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/
Public Key: http://www.qols.ph.ic.ac.uk/~kinsle/key/work-key-2002a
Andrew Resnick
wrote:
>
> Andrew Resnick <andy.r...@nospam.grc.nasadotgov> wrote:
>> Douglas B Sweetser wrote:
>>> Sunlight bounces off atoms in the upper atmosphere with a 4th power
>>> dependence on wavelength, which is why it is blue (was Einstein
>>> really the first to state this, or is that an Einstein hyperbola?).
>>> The scattering process polarizes the light 90 degrees away from the
>>> source. Bees use the polarization of scattering to tell each other
>>> about good pollen sources. The darkness of the sky also gets
>>> altered when wearing polarized sunglasses and twisting your head.
>
>> No again. It's called "Rayleigh scattering".
>
> I'm not sure what you are saying "no" to here. If I wear
> polarized sunglasses, and twist my head, I see the brightness
> of the sky change. This surely means that the light from
> the sky has, in some (most) places, at least a partial degree of
> polarization.
I should have clipped the above paragraph earier. Atmospheric
scattering, and specifically the 4th-power wavelength dependance, is
referred to as "Rayleigh scattering" rather than something Einstein came
up with.
Light from the sky is polarized, that is correct. Two nice references
are:
Sekera, Z., "Light Scattering in the Atmosphere and Polarization of Sky
Light" and Chandrasekhar, S, and Elbert, D. D., "The Illumination and
Polarization of the Sunlit Sky on Rayleigh Scattering", both reprinted
in Selected Papers on Applications of Polarized Light (Spie Milestone
Series, Vol. MS 57)
by Bruce H. Billings, Brian J. Thompson (Editor)
There are several "neutral points" in the sky (called the Babinet,
Brewster, and Arago) that trace out well-defined paths with day and date.
And, of course, the Haidiger's brush phenomenon. I can't see it, BTW.
Student
Andrew Resnick wrote:
I would say there is misunderstanding on the second point since reading
your comments I find myself in full agreement with them. I speak of
spectral coherence only. When refering to optical path, I mean that which
you adjust in a Michelson by displacing a mirror. Also I do not know what
astronomers really do with HBT experiment on stars. However, not certain
thermal sources exhibit photon bunching but all of them. Probably this is a
criteria to tell apart if two streams of photons come from the same star:
they don't if they are not correlated (not bunched in the case of thermal
radiation). I do not want to stress that point further since, may be apart
from terminology, we seem to agree. (probably if you think we don't I am
faulty since a thermal light can be 1st order coherent, but not stars which
I believe have broad spectrum)
The first point is more serious, though. Here we disagree completly.
> As to the first point, that is exactly backwards. The degree of
> coherence is actually defined in terms of the two-point two-time
> correlation:
>
> G(P1,P2,s) = <E(P1,t+s),E*(P2,t)>
should be normalized by 1/(<E(P1,t+s)E*(P1,t+s)><E(P2,t)E*(P2,t)>) and
absolute value taken. If there are no correlations, i.e., if
<E(P1,t+s)E*(P2,t)>=<E(P1,t+s)E*(P1,t+s)><E(P2,t)E*(P2,t)>, then obviously
you have a degree of 1 (in absolute value).
With your delta, take P1=P2, then (with proper normalization), the
corelation time which is the integral of |G(t)|dt, i.e., the integral of a
constant (once you killed the delta with < >), is infinite. That's
coherence indeed!
That's definition of coherence: fields separate under < >. Field exhibit no
correlations. Thermal is bunched, photons are antibunched, etc... If you
are not convinced, check Glauber's PRL articles, or Loudon, or Mandel and
Wolf, or indeed, any reference.
Andrew Resnick
> The first point is more serious, though. Here we disagree completly.
>
>> As to the first point, that is exactly backwards. The degree of
>> coherence is actually defined in terms of the two-point two-time
>> correlation:
>>
>> G(P1,P2,s) = <E(P1,t+s),E*(P2,t)>
>
> should be normalized by 1/(<E(P1,t+s)E*(P1,t+s)><E(P2,t)E*(P2,t)>) and
> absolute value taken. If there are no correlations, i.e., if
> <E(P1,t+s)E*(P2,t)>=<E(P1,t+s)E*(P1,t+s)><E(P2,t)E*(P2,t)>, then
> obviously you have a degree of 1 (in absolute value).
>
> With your delta, take P1=P2, then (with proper normalization), the
> corelation time which is the integral of |G(t)|dt, i.e., the integral
> of a constant (once you killed the delta with < >), is infinite.
> That's coherence indeed!
I don't see it. The delta function only removes the spatial part of the
average, not the temporal part of the average. I should point out that
you tried to define the complex degree of coherence rather than the
cross correlation fucntion, and the normalization is really a factor of
1/(<I(P1,t)>^0.5<I(P2,t)>^0.5) (Mandel and Wolf, p. 163)... The cross-
spectral density has a similar definition on page 171. But we don't
need to be so detailed right now.
In any case, following Mandel and Wolf, we define the coherence time as (
cursing ASCII as we go....)
(Ds)^2 = Int(s^2|G(P1,P2,s)|^2)/Int(|G(P1,P2,s)|^2), where Ds =
coherence time.
Let's set |G(P1,P2,s)|^2 = d(P1-P2)exp(-(as)^2) for the sake of argument.
d is a delta function, so I kept the P1=P2 statement above. The temporal
part is a useful way to define bandwidth, because the Fourier transform
of a Gaussian function is a Gaussian function. Then, the coherence time
is given by:
(Ds)^2 = k/a, where k is some numerical constant.
So the coherence time is inversely proportional to the (spectral)
bandwidth. If s = 0, then the coherence time is indeed infinite.
> That's definition of coherence: fields separate under < >. Field
> exhibit no correlations. Thermal is bunched, photons are antibunched,
> etc... If you are not convinced, check Glauber's PRL articles, or
> Loudon, or Mandel and Wolf, or indeed, any reference.
Those are good references, and another good one is "Elements of Optical
Coherence theory", by Arvind Marathay. Marchand and Wolf's 1974 paper
is included in an appendix.