Re: Fluorescent Spectrum - Spikes vs Broad Slopes

In <9ibp02dqbnt7kgqtmo1jll96okhdkrav85@xxxxxxx>, Victor Roberts wrote:

Don, it will take me a bit longer to digest and reply to
your reply to my note, but I did find two papers that
address the issue of self-reversal vs. radial position in
high pressure mercury lamps. Each is only two pages long and
both are available for no charge.

The first is found at:

http://www.icpig.uni-greifswald.de/proceedings/data/Baeva_1

The first paragraph after equation 1 on the second page
states: "Since the emission spectra from the outer region
are not self-reversed, the temperature evaluation is
possible only in the central part of the discharge."

The second is by the same primary author and was presented
at the same conference. It can be found at:

http://www.icpig.uni-greifswald.de/proceedings/data/Baeva_2

I had to add .pdf to these. But thanks *very much* for digging these
up!

Figure 2 on page 2 shows the calculated line profiles for
the 546 nm line for different optical path lengths. The
self-reversal of the line is obvious in these calculated
results. From the discussion after equation 2 on page 2, my
understanding is that the s/R axis shows the results of the
calculations along different "cords" of the cylindrical
discharge. Only the longest optical path length, s/R = 2,
passes through the center of the arc. As the line of
observation moves toward the outer edge of the discharge,
the self-reversal decreases. This interpretation is
confirmed by the following statement that starts under
Figure 2 and continues to the top of the 2nd column on page
2: "The self-reversal appears near the discharge axis. The
contribution of the outer part of the discharge is of minor
importance."

I took a look, and that 3D graph Figure 2 in the second one I did find
interesting - and to be what I expected.
Look at the axis of the arc and 546.1 nm is self-reversed, and look at
the fringe of the arc and 546.1 nm is not self-reversed. Question now is
whether the absorption to blame for the self-reversal occurs closer to the
axis of the discharge or in its fringes.

I tried to find some sort of statement as to where the absorption
causing the self-reversal occurs, and have yet to find such. I do agree
that concentration of absorbers is highest at the axis of the arc, but I
still feel that if you "peel away" the outer regions of the discharge
there will not be self-reversal.
So far I did see mentions of the mercury vapor being "optically thick"
at these wavelengths. This makes me think that if you had a sufficiently
narrowband 546.074 nm bandpass filter and looked at the discharge through
that filter, you don't see the central part of the arc well because the
arc is largely opaque at that wavelength - you see the outer regions. Or
outer and somewhat-outer regions.

I did see the statement of "The self-reversal appears near the discharge
axis". Does this mean as I think that the self-reversal only appears when
a line of sight passes through or close to the discharge axis? Or is this
a statement that the absorption causing the reversal actually occurs
there?

I saw the next sentence, "The contribution of the outer of the discharge
is of minor importance". To what - the self-reversal? Total output in
the 546.1 nm emission feature as a whole? It does appear that the
bandwidth of the 546 nm emission feature is narrower as one looks along a
line-of-sight farther from the discharge axis.

Back to Fig. 2 - showing spectral power distribution as a function of
"s/R". It appears to me that the peak flattens due to approaching
self-reversal when s/R is about 1.85 and the "dip" of the self-reversal
begins to occur when s/R is 1.87 or 1.88 or so.
If I am correct at understanding s to be the chord length and R to be
the arc tube radius, then:
s/R is 1.85 when the chord makes closest approach to the axis at about
..38R (38% of the way from the axis to the inner surface of the arc tube),
and s/R is 1.88 when the chord makes closest approach to the axis at .34R.
(I hope I did not screw up my math here.) If I am correct as to where the
absorption that causes the self-absorption is located (as opposed to even
higher absorption at/near the discharge axis being outpaced by much higher
emission there), I would want to blame the region around and just outside
the "radial location" that barely has its emission self-reversed.

That 3-D graph considers s/R as low as 1.7, which I figure to be a chord
whose closest approach to the axis is about 52.6% of the way out!

34% of the way from the arc axis to the arc tube inner surface I would
consider to be "fringe region". In fact, I would think of this particular
mercury discharge to be a "fat" or "wide" one.
Most of the time I look at a mercury vapor lamp, 1/3 of the way from the
center of the discharge to the arc tube wall appears to me as "fringe".
52.6% of the way from the axis of the discharge to the inner surface of
the arc tube normally appears to me to be "outside of the arc" - but that
Figure 2 is showing something a little significant there! Am I
misinterpreting this or is this a "fatter"-than-usual mercury arc?

I went back to Fig. 4 of the first document - temperature profile, with
4 curves - "adopted" and based on 3 visible mercury lines. The "adopted"
one has temperature 1/3 of the way out being about 500 K below that at the
discharge axis, and the curves based on spectral lines averaging roughly
600 K lower 1/3 of the way out than at the discharge axis. This appears
to me usual. I remember "The High Pressure Mercury Vapor Discharge" by W.
Elenbaas showing a typical/"usual" temperature profile in a high pressure
mercury vapor lamp to be usefully close to being a parabola. If discharge
centerline is upper 5,000's K and arc tube inner surface is 1,000 K, then
1/3 of the way from discharge axis to arc tube inner surface along a
parabolic temperature profile would be low-mid 500's K cooler than the
discharge axis.

The arc could be "fatter" than those in 175W H39 lamps if it is more
"optically thick" due to the mercury vapor concentration (per unit arc
length) being a lot greater or power density (per unit arc length) is a
lot greater. Then again, my experience watching mercury lamps warm up, at
normal power and overpowered/underpowered (done that and managed to not
need a fire extinguisher) suggests little variation of arc "fatness" with
pressure but higher power input makes the arc slightly "fatter".

What are the arc tube dimensions? The data in the first document
appears to me partial but reminding me of the D2S lamp, whose volume I
would "estimate" to be about .08 cm^3, and 554 ug of mercury and a
parabolic temperature profile 5700K at centerline and 1000K at inner
surface I figure to get a mercury vapor pressure around 9 atmospheres.
Given the dimensions, I expect "optical thickness" to be just a little
more than that in H39, H33, H34 lamps and the like.
Voltage drop figures make me think more like about 15 atmospheres.
I expect the discharge then to be somewhat more "optically thick" than in
an H39 175W lamp but maybe "same ballpark".
Input power shown is generally around 39 watts. Assuming 15 volts for
cathode and anode falls, I see generally 31 watts into the arc. With arc
length .44 centimeter, that's about 70 watts per centimeter - somewhat
higher than the 31 I see for H39 (175W) and the 35-38 I see for H33
(400W). Alltogether, I see the discharge being somewhat more "optically
thick" than in H39 and H33 lamps but probably "basically same ballpark".

======================================================================

Now another bit:

Should I be correct about the self-reversal being from an "absorbing
layer" around an "emitting core", then:

I did propose the oversimplification of the discharge having two
distinct uniform layers to arrive at some determination that at the maxima
of the emission line outside the central absorption feature, the
brightness (per unit area per unit bandwidth) would be 25% of that of a
blackbody at the "emission core" temperature. That 25% figure would be
higher if the "emitting layer" has a wider bandwidth than that of the
"absorbing layer" but for now I am assuming not in a most-oversimplified
model of self-reversal.
And with Fig. 5 of the first document having the line-center brightness
about 60% of maximum, my proposed "absorbing layer" needs to approach
being a blackbody (at th line-center wavelength) with brightness 15% of
that of a blackbody whose temperature is that of the "emission core".
So now to figure what such a temperature would be...
I pick somewhat out of a hat and seeing the figures that the "emission
core" temperature is 5700 K. Temperature required to get brightness at
546.1 nm 15% of that of a 5700K blackbody is ... Doing a bit of hasty
fudging while using a couple homebrew BASIC programs that I wrote many
years ago... I figure close enough to 4150 K, but give-or-take 25 maybe
50 degrees since I did not write in output being in terms of watts per
square centimeter per nanometer bandwidth, so I had to do some "fudging".
I hope I got within 50 degrees!

Now, along a parabolic temperature profile whose axis is 5700 K and
whose edge is 1000 K, 4150 K occurs at 57.4% of the way from the discharge
axis to the arc tube inner surface. If I understand the s/R correctly,
that would be about 1.64. This is a little outside the range being
considered in that 3-D graph Fig. 2 of the second document. Of course, I
am doing some oversimplification to arrive at this!
Fig. 2 of the second document does make me want to blame the
self-reversal on a radial location range closer to the discharge axis than
that, generally between 33 and 50% of the way from the discharge axis to
the arc tube inner surface. And before seeing this, I was thinking of
"arc fringe" as being even closer to the discharge axis, as in 30% of
the way from the discharge axis to the arc tube inner surface - where the
temperature I now realize is only about 400 K cooler than at the discharge
axis (with a parabolic temperature profile), maybe 600 K cooler for the
most extreme curve in Fig. 4 of the first document.

===========================

A little followup work:

Fig. 5 of the first document I attempted to translate the vertical scale
to percentage of a 5700K blackbody. I did see the units shown include per
steradian, and I wonder whether to consider unit area of the discharge to
be a lambertian emitter or an isotropic one. Sounds to me as I type this
to try lambertian, and if I did not do at any screwups I get the peak
being about 44% of that of a 5700K blackbody. The dip appears to be about
55% of that, or about 24% of that of a 5700K blackbody. This makes me
think that the discharge core has more broadening than my proposed
"fringe" absorbing layer.
In my mentioned oversimplified model of a uniform temperature absorbing
layer around a uniform temperature emitting core, this has the absorbing
layer having a temperature around 4400 K. Assuming 5700K discharge axis
temperature, 1000K arc tube inner surface temp. and parabolic temperature
profile, this occurs at 52.6% of the way out, where s/R would be 1.7.
This is still a little farther out that I had in mind but maybe in the
ballpark. Is 4400K hot enough to make a 1 mm thick or whatever layer of
15 atmosphere pressure or whatever mercury vapor excited enough to
approach being a blackbody at the central wavelength of the 546.1 nm
emission line?

- Don Klipstein (don@xxxxxxxxx)
.

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