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EDG
2009-Feb-09, 01:28 AM
So, we all know (I hope) that we shouldn't stare at our own sun for very long or we'll damage (or destroy) our vision.

But how does this damage scale when we're talking about other stars? For example, if we orbited Sirius (luminosity of 25 Sols) at 1 AU, we'd be looking at a star that's 25 times brighter in the sky, right? So wouldn't that mean that if we even glimpsed at it, we'd fry our eyes? (and yes, I know we'd have other problems if Earth orbited Sirius at that distance ;) ).

I guess I'm looking for some kind of rough, qualitative relationship between the luminosity of the star, the amount of time we look at it with normal human eyes, and the amount of damage done to the eyes after that time.

I ask because it seems that nobody really considers this sort of thing in SF stories.

Hornblower
2009-Feb-09, 01:33 AM
So, we all know (I hope) that we shouldn't stare at our own sun for very long or we'll damage (or destroy) our vision.

But how does this damage scale when we're talking about other stars? For example, if we orbited Sirius (luminosity of 25 Sols) at 1 AU, we'd be looking at a star that's 25 times brighter in the sky, right? So wouldn't that mean that if we even glimpsed at it, we'd fry our eyes? (and yes, I know we'd have other problems if Earth orbited Sirius at that distance ;) ).

I guess I'm looking for some kind of rough, qualitative relationship between the luminosity of the star, the amount of time we look at it with normal human eyes, and the amount of damage done to the eyes after that time.

I ask because it seems that nobody really considers this sort of thing in SF stories.Yes, we would be fried at that range. Sirius is about 25 times the Sun's luminosity in visible light and considerably more than that in ultraviolet, because of its higher temperature. Our eyes likely would be toast in a split second.

Even at Jupiter's distance, with the visible light down to the intensity we get from the Sun, the UV would be much stronger.

Ronald Brak
2009-Feb-09, 01:41 AM
Well, our eyes are adapted so that if we accidently glance at the sun they won't be damaged. However, I do think that for anything much brighter than the sun human eyes would be in big trouble, especially when you consider that larger, brighter stars put out a higher portion of their energy as ultraviolet. Even though a habitable planet around Sirius would orbit further out than earth, I think just glimpsing Sirius would cause eye damage. Of course a habitable planet would need some sort of thick atmosphere to protect against ultraviolet, so that might mitigate the problem, but even if your skin was safe, Sirius would appear smaller than our sun at a habitable distance and would be an intense source of light and UV very very roughly the apparent size of a lady beetle on the finger tip of an outstretched hand. I think that looking at such an intense, point like source would cause real problems for our eyes.

EDG
2009-Feb-09, 01:50 AM
So we'd need really good sunglasses all the time, right :). So we'd need to be at a distance of 5 AU before Sirius' apparent luminosity would be the same as Sol's, right? (5 times further = 25 times dimmer).

I'm half wondering if even ambient light would be a problem. If Sirius is 25 times more luminous than Sol, could the light reflected even by rocky ground on a planet orbiting at 1 AU damage our eyesight? (I'm pretty sure that Siriuslight reflected by water or a wet road surface would be pretty fatal to our vision - it's bad enough on Earth with our own sun!).

And I know UV would be a huge problem, but I'm just considering the visible light for now.

RalofTyr
2009-Feb-09, 06:13 AM
Not to mention we'd probably be set on fire from the heat of Sirius at 1 AU.

EDG
2009-Feb-09, 08:11 AM
So obviously UV damages eyes, but what if the star emitted more IR radiation (i.e. if it was an orange or red giant)?

Ronald Brak
2009-Feb-09, 08:38 AM
So obviously UV damages eyes, but what if the star emitted more IR radiation (i.e. if it was an orange or red giant)?

As long as it isn't intense enough to burn your skin your eyes should be okay.

eburacum45
2009-Feb-09, 12:25 PM
If we assume that we are standing on a planet that receives the same total amount of radiation as the Earth does, then we know that we are in the effective habitable zone and we can make some direct comparisons. First off, let's imagine a planet around a red dwarf in that situation. We would be quite close to the star, and it would look larger in the sky than our own sun, and somewhat yellower or even pinker. But the star would be pumping out lots of infra-red which we couldn't see. I think that excess of infra-red might be dangerous, because we assume that the fact that the star looks slightly dimmer than our own star in visible light means it is safe to look at, which probably isn't the case.

More obviously dangerous would be the situation of a star orbiting a hotter star in the habitable zone. Once again the star looks dimmer, as more of the light is now emitted in UV, and the star looks smaller because it is quite a bit further away. We would be very ill advised to look at such a star with the naked eye from such a planet, despite the fact that it would look dimmer than the Sun, because of the excess UV. But there is a possibility that a planet in such a location might have a denser ozone layer, which might mitigate the danger somewhat.

Or perhaps we could engineer a thicker ozone layer on purpose...

grant hutchison
2009-Feb-09, 01:19 PM
Hot stars shed less light in their habitable zones than does the sun, because their luminous efficacy is lower. But the surface brightness of their discs is higher, so they will deliver a greater density of luminous energy to a spot on the retina if you look directly at the disc.
Although UV and IR are potentially dangerous, recall that your eye doesn't focus those wavelengths effectively, and that your eye lenses are opaque to UV that is close to the well-focused (visible) range.

So I'm thinking we want to know what level of surface brightness causes retinal damage before you can blink. We know that atomic bomb air blasts hit about 1010 cd.m-2, and we know that people outside Hiroshima and Nagasaki who happened to be looking in the wrong direction sustained retinal injuries.
That surface brightness corresponds to a black body of about 9000K: so we're talking about stars hotter than spectral class A2.
If the surface brightness is high enough to cause eye injury faster than the blink reflex, you're going to get injured by looking at the disc from any distance short of the point at which your eye diffraction-limits. Once you're far enough away for the disc to fall below an apparent diameter of a few arc-minutes, its brightness on your retina will start to decline towards safe limits.

Grant Hutchison

EDG
2009-Feb-09, 04:48 PM
Remind me, what was "luminous efficacy" again? I remember discussing it with you a long time ago but it's slipped my mind.

BigDon
2009-Feb-09, 05:50 PM
My father was flash blinded for about 45 minutes by a 15 megaton explosion he had his back too.

The little half inch by 2 inch stainless steel serial number tag on a piece of radio equipment reflected enough of the light to do so. Through tightly closed eyelids no less.

Nick Theodorakis
2009-Feb-09, 05:57 PM
So obviously UV damages eyes, but what if the star emitted more IR radiation (i.e. if it was an orange or red giant)?

We discussed this briefly in one of the IR vision threads. Near IR, to a first approximation, has essentially the same ability to damage the retina as visible light. The difference is that we have reflexes that cause us to avoid looking into very bright visible light but not so for near IR.

Nick

EDG
2009-Feb-09, 06:31 PM
My father was flash blinded for about 45 minutes by a 15 megaton explosion he had his back too.

The little half inch by 2 inch stainless steel serial number tag on a piece of radio equipment reflected enough of the light to do so. Through tightly closed eyelids no less.

Wow! :eek:

grant hutchison
2009-Feb-09, 08:17 PM
Remind me, what was "luminous efficacy" again? I remember discussing it with you a long time ago but it's slipped my mind.Lumens per watt. A measure of how much light you get for your energy. For black bodies, luminous efficacy peaks close to the Sun's effective temperature. Obviously, you can do better with a non-black body spectrum. Fluorescent lights better than incandescents, for instance. And everyone loves those green laser pointers, which have higher luminous efficacy than the old red ones.

Grant Hutchison

Celestial Mechanic
2009-Feb-09, 08:31 PM
OK, OK, so I'll only look at Sirius until I need glasses! :lol: ;)

mugaliens
2009-Feb-09, 08:52 PM
So wouldn't that mean that if we even glimpsed at it, we'd fry our eyes?

Hey! Just talking about it hurts my eyes!

George
2009-Feb-10, 12:05 AM
So we'd need really good sunglasses all the time, right :). No, for Sirius you need some very sirius glasses. ;)


So we'd need to be at a distance of 5 AU before Sirius' apparent luminosity would be the same as Sol's, right? (5 times further = 25 times dimmer). Yes. However, Sirius is 1.7x the radius of the Sun, so its surface brightness is only about 8.5 times that of the Sun. That is not likely enough to damage the retina during the time it takes to blink (~ 300 miliseconds).

The Sun is more than 1000 times brighter (surface brightness) than we can handle. Actual retina damage is possible around 107 cd/m2, and the Sun's luminance is a little more than 109 cd/m2.


I'm half wondering if even ambient light would be a problem. If Sirius is 25 times more luminous than Sol, could the light reflected even by rocky ground on a planet orbiting at 1 AU damage our eyesight? (I'm pretty sure that Siriuslight reflected by water or a wet road surface would be pretty fatal to our vision - it's bad enough on Earth with our own sun!). Yes, glare would a problem, no doubt. But illuminated surface objects might not be so bad,... especially at night (sorry, I had to say it :( ).


My father was flash blinded for about 45 minutes by a 15 megaton explosion he had his back too. I would guess it was painful, too. I accidentally got "arc eye" from intentionally looking away from our welder, but looked at a reflective wall instead. A dozen or more seconds of this was all it took for me to be in ugly pain later that evening when it hit me. The real pain came when the ER nurse elected to flush my eyes out.

BigDon
2009-Feb-10, 12:47 AM
George, the "Castle Bravo" shot was intended to be a 2 to 3 megatonner according to my dad. Wiki says different

http://en.wikipedia.org/wiki/Castle_Bravo

Here's the pertinent quote though:


Cause of high yield
The yield of 15 Megatons was two and a half times what was expected. The cause of the high yield was a laboratory error made by designers of the device at Los Alamos National Laboratory.

It was expected that lithium-6 isotope would absorb a neutron from the fissioning plutonium, emit an alpha particle and tritium in the process, of which the latter would then fuse with deuterium (which was already present in the LiD) and increase the yield in a predicted manner.

The designers missed the fact that when the lithium-7 isotope (which was considered basically inert) is bombarded with high-energy neutrons, it absorbs a neutron then decomposes to form an alpha particle, another neutron, and a tritium nucleus. This means that much more tritium was produced than expected, and the extra tritium in fusion with deuterium (as well as the extra neutron from lithium-7 decomposition) produced many more neutrons than expected and induced more fission of the uranium tamper, increasing yield.

This resultant extra fuel (both lithium-6 and lithium-7) contributed greatly to the fusion reactions and neutron production, and in this manner greatly increased the device's yield. The test used lithium with a high percentage of lithium-7 only because lithium-6 was (at the time) scarce and expensive; the later Castle Union test used almost pure lithium-6. Had more lithium-6 been available, the usability of the common lithium-7 might not have been discovered.

Of the total 15-megaton yield, 10 megatons were from fission of the natural uranium tamper.

Ronald Brak
2009-Feb-10, 01:07 AM
We discussed this briefly in one of the IR vision threads. Near IR, to a first approximation, has essentially the same ability to damage the retina as visible light. The difference is that we have reflexes that cause us to avoid looking into very bright visible light but not so for near IR.

I believe the earth's atmosphere is fairly transparent to near IR, so I guess that below a certain size a red dwarf would be dangerous to human eyes. I was thinking that a red dwarf would be bright enough to still make us look away, the way I look away when I accidently look at a bright incandescant bulb, but often at sunset you can watch the sun sink below the horizon without looking away and a red dwarf would be much dimmer, so people on a planet around a red dwarf might easily get eyefulls of near IR. Looks like I was wrong about it not being a problem.

BigDon
2009-Feb-10, 02:02 AM
Remember those brush fires we had a couple of months back here in California? (And God be with those poor Australians). The fires were more than a hundred miles from me so the smoke was very high and just gave the illusion of being a light overcast.

Except for the effect it had on the Sun.

I know, George is going to say DON'T!

To the naked eye the Sun looked like a slightly pink pearl without any hint at all of the discomfort normal for such foolishness. I only mention George because I immediately thought him and what he would think of the rather cool phenomena.

Plus a foot tall, mental bobble-head of George materialized over my right shoulder and said, "It's going on four seconds Dingweed! You give up liking to look at pretty ladies?"

(I guess George was a bit testy that day.)

George
2009-Feb-10, 02:30 PM
Remember those brush fires we had a couple of months back here in California? (And God be with those poor Australians). The fires were more than a hundred miles from me so the smoke was very high and just gave the illusion of being a light overcast.

Except for the effect it had on the Sun.

I know, George is going to say DON'T! :) I wasn't going to, but I had forgotten about IR until:


But the star would be pumping out lots of infra-red which we couldn't see. I think that excess of infra-red might be dangerous...
This must be taken seriously.

Look at this Solar Radiation Spectrum (http://en.wikipedia.org/wiki/File:Solar_Spectrum.png). The red portion is the reduced amount of radiation seen at sea level due to 1 or, perhaps 1.5, atmospheres of scattering and absorption. It also shows the IR portion of the band. Notice that the IR portions around 1000nm and at 1250nm are not attenuated that much by our atmosphere. Since the eye allows light in the range of ~380nm to 1400nm to reach the retina, then IR in these regions will heat the retina. This can be damaging, according to this NASA eye safety (http://eclipse.gsfc.nasa.gov/SEhelp/safety2.html) article.

On the horizon, the portion of the spectrum that is attenuated by about 20% due to one atmosphere is reduced by about 600x when the Sun is on a clear horizon. However, if we pick an IR region that is reduced by only, say, 5% by one atmosphere, then the air mass along a clear horizon may only reduce this IR region by only about 5x, or 20% of its original value, which is, thus, much more intense than the visible region when we view the Sun while it is on the horizon.

I'm kicking myself for forgetting this fact that IR is so much less attenuated by our atmosphere compared to visible light attenuation. When the Sun has been low on a dusty horizon, I have even used binoculars for a quick look for sunspots. This was wrong!

In your case, it could be that the smoke particles attenuated the IR band since scattering could be more likely with these larger particles, assuming the particles were numerous enough and of the size smaller than 1000nm, which is possible. But, we can't tell, at least I can't, by looking at the Sun whether IR is at a safe level or not.


To the naked eye the Sun looked like a slightly pink pearl without any hint at all of the discomfort normal for such foolishness. I only mention George because I immediately thought him and what he would think of the rather cool phenomena. There are a few reports where the Sun and Moon have appeared bluish due to smoke particles. This occurs when the particles are too large for blue light to scatter, yet these particles are small enough to allow red light to scatter, thus the red end is scattered more than the blue end and the Sun or Moon and the halo around it will appear somewhat blue. This is the explanation for the blue halo around the Sun as seen from Mars by the rovers.


Plus a foot tall, mental bobble-head of George materialized over my right shoulder and said, "It's going on four seconds Dingweed! You give up liking to look at pretty ladies?" I wondered where that thing went. Try not to break its nose, its neck has limited spring action due to its age. :)

tracer
2009-Feb-11, 02:23 AM
Not to mention we'd probably be set on fire from the heat of Sirius at 1 AU.

Assuming the Earth could even retain an oxygen atmosphere at that temperature in the first place -- can't catch fire without an oxidizer!

RalofTyr
2009-Feb-11, 05:38 AM
Assuming the Earth could even retain an oxygen atmosphere at that temperature in the first place -- can't catch fire without an oxidizer!

I was assuming he was in a space suit.

EDG
2009-Feb-11, 06:23 AM
Actually, just assume you're a pair of eyes :). I'm not assuming anything about atmospheres or anything else, just interested in the effects of the different luminosities on human vision.

eburacum45
2009-Feb-11, 08:07 AM
George, the "Castle Bravo" shot was intended to be a 2 to 3 megatonner according to my dad. Your dad saw Castle Bravo? That was a particularly unusual test. No wonder he had problems with unexpected brightness.

eburacum45
2009-Feb-11, 08:36 AM
So for a planet orbiting an F Class star in the habitable zone, the visible sunlight would be slightly dimmer and the disk of the star would be smaller than the Sun as seen from Earth. But the surface brightness of the star would be brighter, so that smaller disk would be intrinsically brighter per square radian (or per square degree if you prefer). You certainly don't want to look at it.

If I understand Grant's post correctly, you can however protect yourself against an accidental glimpse of an F-class star by blinking; you can't do that for a brighter A-class star, any more than you can protect your eyes against a glimpse of an arc-welder's torch by blinking.

Another effect of living around a hotter star would be increased flourescence because of the increased UV. White buildings and clothing might look brighter, depending on the pigments used, and flourescent clothing would positively glow, making anyone wearing a hi-vis jacket visible at a greater distance.

EDG
2009-Feb-11, 04:52 PM
On a related note, Bolometric Corrections are used to convert bolometric magnitudes to visual magnitudes, right?

I note from the table here:
http://www.peripatus.gen.nz/Astronomy/SteMag.html

that an A0 V star's bolometric correction is only -0.15... that doesn't sound like a lot to me. Doesn't that mean that it wouldn't emit that much more UV than Sol? And an F0 V's bolometric magnitude is almost the same as its visual magnitude (correction is only -0.01). Or am I misunderstanding these tables? Of course you'd still have to worry about the increased visual magnitude anyway.

And does anyone have a link to a more complete table of bolometric corrections than that?

Also, I know the sun's visual magnitude (from Earth) is about -26.7 and its absolute visual magnitude is about 4.83. According to that website its bolometric correction would be -0.128, so does that mean that its bolometric (absolute) magnitude is 4.83-0.128 = 4.702?

George
2009-Feb-11, 06:20 PM
If I understand Grant's post correctly, you can however protect yourself against an accidental glimpse of an F-class star by blinking; you can't do that for a brighter A-class star, any more than you can protect your eyes against a glimpse of an arc-welder's torch by blinking.

A class A star at 9,000K produces roughly 13,000 w/m^2 (9000 to 16000 w/m^2 per nm) in the 200 to 315nm UV range. According to this OSHA Safety Guide (http://www-nehc.med.navy.mil/downloads/ih/uvdoc2.doc) the maximum exposure for 300ms (best blink time) is 10,000 w/m^2 (Table 2).

grant hutchison
2009-Feb-11, 10:01 PM
On a related note, Bolometric Corrections are used to convert bolometric magnitudes to visual magnitudes, right? Correct.


I note from the table here:
http://www.peripatus.gen.nz/Astronomy/SteMag.html

that an A0 V star's bolometric correction is only -0.15... that doesn't sound like a lot to me. Doesn't that mean that it wouldn't emit that much more UV than Sol?Nah. Remember that the bolometric magnitude includes both IR and UV. What you're seeing there is a shift from predominantly IR (for the sun) to predominantly UV (for the A0V), without much change in the proportion of visible light.
Integrating the black body curve for 5800K, we get about 37% of the energy in the visible range, 51% at longer wavelengths (predominantly IR), and 12% at shorter wavelengths (predominantly UV). For an A0V at 9520K, the corresponding figures are 34% visible, 21% IR and 45% UV. So you have almost four times the UV for a very small shift in bolometric correction.


Also, I know the sun's visual magnitude (from Earth) is about -26.7 and its absolute visual magnitude is about 4.83. According to that website its bolometric correction would be -0.128, so does that mean that its bolometric (absolute) magnitude is 4.83-0.128 = 4.702?The standard figures for the Sun (see, for instance, Allen's Astrophysical Quantities, 4th Ed.) are MV = +4.82, Mbol = +4.74. So BC = -0.08.

Grant Hutchison

JustAFriend
2009-Feb-12, 02:37 PM
I ask because it seems that nobody really considers this sort of thing in SF stories.

A) Because unless you HAVE to colonize a particular planet for some specific reason, humans are going to look for planets that are similar in conditions to the Earth. There are lots of planets out there and there's no reason not to be comfortable....

And they HAVE dealt with this in SF. Here's one example:
"The Mutant" (http://www.youtube.com/watch?v=fUnZq6pGdHs), a 1964 episode of "The Outer Limits"

grant hutchison
2009-Feb-12, 04:22 PM
A sun too bright to glance at (CY Aquarii) was also used as a significant plot element in Niven's 1968 short story Grendel. But such thoughtfulness by SF writers does seem to be rare.

Grant Hutchison

eburacum45
2009-Feb-12, 06:02 PM
In a recent story I gave the human colonists special genetic adaptations to protect against the slightly bluer light of the local star
http://www.voicesoa.net/fall-from-grace3/

...because of the fierce short wavelength radiation from the local star, a class F5 dwarf somewhat hotter than the norm, these people have purplish black skin and heavy brows.

BigDon
2009-Feb-12, 07:04 PM
Eburacum, what name do you write under, if I may ask?

grant hutchison
2009-Feb-12, 07:18 PM
Eburacum, what name do you write under, if I may ask?Don, check the bottom of the story eburacum45 linked to. :)

Grant Hutchison

mahesh
2009-Feb-12, 08:45 PM
With regards to EDG's OP...
Where life / eyes evolved in circumstances like in the habitable zones near Sirius, or any other life sustaining star(s), wouldn't natural evolution (Earthly kind) do the necessary and ensure that visual acuity is not compromised?

All things being equal, Mother Nature follows the same patterns, taking the best lines of least resistance! I would have thought.

oh by the way, thanks Grant!

EDG
2009-Feb-12, 10:44 PM
With regards to EDG's OP...
Where life / eyes evolved in circumstances like in the habitable zones near Sirius, or any other life sustaining star(s), wouldn't natural evolution (Earthly kind) do the necessary and ensure that visual acuity is not compromised?

Probably. But I was thinking of specifically human eyes here. :)

EDG
2009-Feb-13, 07:32 PM
Nah. Remember that the bolometric magnitude includes both IR and UV. What you're seeing there is a shift from predominantly IR (for the sun) to predominantly UV (for the A0V), without much change in the proportion of visible light.
Integrating the black body curve for 5800K, we get about 37% of the energy in the visible range, 51% at longer wavelengths (predominantly IR), and 12% at shorter wavelengths (predominantly UV). For an A0V at 9520K, the corresponding figures are 34% visible, 21% IR and 45% UV. So you have almost four times the UV for a very small shift in bolometric correction.

grant - how do you calculate those percentages for each temperature? I'd be interested to know how to do that.

George
2009-Feb-13, 09:13 PM
grant - how do you calculate those percentages for each temperature? I'd be interested to know how to do that. I would also be interested in how he does it. It is the integral across the range of desired frequencies or wavelengths...

http://scienceworld.wolfram.com/physics/pimg258.gif

or by wavelength..

[see subsequent post]


However, you can use a simple spreadsheet layout. I have two control cells: initial wavelength; and increment. Then I have about 100 rows that compute the value for each incremented wavelength.

To find the percentages across any band, simply sum the desired range, then divide by the total. You must find the initial wavelength that yields a low wattage result, then increment as necessary so that the final (after 100 or whatever) result is also a low wattage value. [My results were within 1% or 2% of Grant's.]

You may want to convert to photon flux if you are comparing these results for what people see since the energy requirements are nearly double that for blue light compared to red.

grant hutchison
2009-Feb-15, 12:48 PM
grant - how do you calculate those percentages for each temperature? I'd be interested to know how to do that.Same technique as George describes, essentially.
I use the equation for radiant exitance as a function of wavelength, and let MathCAD do the numerical integration for me.

Grant Hutchison

EDG
2009-Feb-16, 01:56 AM
George - your wavelength equation isn't showing up, I just see a "physicsforums.com" hotlink graphic instead.

George
2009-Feb-16, 01:39 PM
George - your wavelength equation isn't showing up, I just see a "physicsforums.com" hotlink graphic instead.Odd, it seemed to work when I first posted it.

Here is the wavelength version.

[Corrected version:]

(2πhc2/λ5)*(1/(ehc/λkT – 1))

EDG
2009-Feb-16, 04:42 PM
Thanks... er, what are all the terms in it?

George
2009-Feb-16, 06:41 PM
This site on blackbody (http://en.wikipedia.org/wiki/Black_body) radiation should explain them for you.

Sanbeko
2009-Feb-17, 04:50 PM
Wouldn't a planet orbiting an F star have a higher ozone concentration to help offset the UV radiation? Maybe the planet has a slightly higher oxygen concentration, so it creates more ozone?
Hey, maybe this planet has an atmosphere and complex life. So even though we would rather be on a planet orbiting a star like our Sun, we'd need to go where the action was, no?

Fun video, btw.

Sanbeko

EDG
2009-Feb-20, 06:51 AM
Ok, let me see if I've done this right...

I've attached a graph for a star with temperature of 7625K (should be an A8 IV star). I used the wavelength equation that George gave, and made an excel sheet that calculated the value for each wavelength. I started at (log 0) wavelength and incremented the wavelengths in units of (log -0.1), so it went 0, -0.1, -0.2, -0.3 etc.

One question: Am I calculating "the amount of energy per unit surface area per unit time per unit solid angle emitted in the wavelength range between w and w+dw by a black body at temperature T"? So this is... Joules per square metre per second per steradian?

The graph shows the log wavelength on the x-axis, and the total energy on the y-axis (so shorter wavelength is to the left). George/Grant - does that graph look about right for this temperature? I think these BB graphs are supposed to be a bit skewed on one side, and this does look skewed, so that may be promising?

If that is right, then I suppose I just divide it up according to the chart on http://en.wikipedia.org/wiki/Electromagnetic_spectrum and find out what proportion of the energy is in each "bin" that I've divided it into?

grant hutchison
2009-Feb-20, 11:46 AM
One question: Am I calculating "the amount of energy per unit surface area per unit time per unit solid angle emitted in the wavelength range between w and w+dw by a black body at temperature T"? So this is... Joules per square metre per second per steradian?No, you're calculating radiant exitance, which is just W.m-2. The units in the radiant exitance wavelength equation are W.m-2.m-1, so integrating the area under the curve gives you W.m-2.
Edit: Mind you, George's equations seem to be missing some factors and adding others in a way I don't understand. The radiant exitance formulae for wavelength and frequency are given here (http://www.astro.uvic.ca/~tatum/stellatm/atm2.pdf) (200KB pdf) in the form I'm used to seeing (equations 2.6.1 and 2.6.3).
Edit again: Ah. George gave you numbers for spectral energy density, which has units of energy per volume (per unit wavelength or unit frequency). That may explain why your graph looks odd. You actually need the radiant exitance equations as given in my link.


The graph shows the log wavelength on the x-axis, and the total energy on the y-axis (so shorter wavelength is to the left). George/Grant - does that graph look about right for this temperature? I think these BB graphs are supposed to be a bit skewed on one side, and this does look skewed, so that may be promising?It doesn't look right to me. What are your units? You look as if you've got a peak in the submillimetre range, which isn't appropriate for the temperature you're working with. The shape is also too symmetrical to my eye, though I confess I'm used to lin-lin or log-log plots, so the log-lin may just be confusing me.


If that is right, then I suppose I just divide it up according to the chart on http://en.wikipedia.org/wiki/Electromagnetic_spectrum and find out what proportion of the energy is in each "bin" that I've divided it intoBe sure that when you do your numerical integration, it's with respect to a linear wavelength axis: integrating the area under the log curve will completely skew your result.

Grant Hutchison

grant hutchison
2009-Feb-20, 12:37 PM
More information on the various forms of the Planck equation:

Radiant exitance is the total number of watts per square metre a surface radiates. The spectral radiant exitance forms of Planck's equation (wavelength or frequency) feature a multiplicative constant of 2π.

Radiance is a measure of surface brightness, measured in watts per square metre per steradian. The spectral radiance forms of Planck's equation (wavelength or frequency) feature a multiplicative constant of 2.

Energy density is a measure of the energy content of a photon "gas", measured in joules per cubic metre. The spectral energy density forms of Planck's equation (wavelength or frequency) are the radiance formulae multiplied by 4π/c (or the exitance formulae multiplied by 4/c).

We're primarily interested in radiant exitance, here, so that's the appropriate formula.

Grant Hutchison

EDG
2009-Feb-20, 04:34 PM
OK, thanks... I'll have a look at the link properly tonight, though a quick skim reveals a LOT of equations... am I basically doing what the Example is about in there?

(the units in the graph were SI - metres on the x-axis(when "de-logarithm" it), and whatever the SI units are for the equation on the y-axis).

George
2009-Feb-20, 05:23 PM
One question: Am I calculating "the amount of energy per unit surface area per unit time per unit solid angle emitted in the wavelength range between w and w+dw by a black body at temperature T"? So this is... Joules per square metre per second per steradian?
The Planck distribution equation I gave, as Grant noted, does require some extra effort.

Using the equation [post 41], you will find you need to multiply by 10-6 to convert to watts/m2. This value is the unit area energy production at the surface of the object (ie. a star's "surface"), and this value is per unit wavelength and the energy values are in a derivative form. Thus, when you integrate across the spectrum you will obtain the actual energy production per unit area.

Since this radiant energy diminishes as to the inverse square, simply square the ratio of the distances (Sun's radius/1 AU), 4.65x10-3, to convert to the level we see at Earth (above our atmosphere).

This method is likely correct, but Grant's review would be nice to have.

grant hutchison
2009-Feb-20, 05:41 PM
Using the equation [post 41], you will find you need to multiply by 10-6 to convert to watts/m2. This value is the unit area energy production at the surface of the object (ie. a star's "surface"), and this value is per unit wavelength and the energy values are in a derivative form. Thus, when you integrate across the spectrumyou will obtain the actual energy production per unit area.George, it's not immediately evident why you would need to insert a factor of a million, unless you are using inappropriate units elsewhere.
As I pointed out above, your equations are for spectral energy density (although you are missing a factor of 4π in the equation for the frequency distribution), and so won't give radiant exitance directly. They'll be fine for comparing the power in various parts of the spectrum (since the differences between the various formulae involve only constants), but they shouldn't be used to obtain values for exitance or irradiance. If you check the dimensional analysis for your equations, you'll see that this is so.

For wavelength we have these units:
hc/λ5 => J.s.m.s-1/m5 = J.m-3.m-1

For frequency we have these units:
hν3/c3 => J.s.s-3/m3.s-3 = J.m-3.Hz-1

Grant Hutchison

grant hutchison
2009-Feb-20, 05:46 PM
I've attached a graph for a star with temperature of 7625K ... For reference, here (http://www.ghutchison.pwp.blueyonder.co.uk/blackbody.jpg) is the curve MathCAD plots for Planck's spectral radiant exitance formula at 7625K. Wavelength in metres on the X axis, radiant exitance in watts per square metre on the Y axis.

Grant Hutchison

EDG
2009-Feb-20, 06:50 PM
Ok, useful stuff... will see if I can figure it out tonight. Thanks!

George
2009-Feb-20, 07:34 PM
George, it's not immediately evident why you would need to insert a factor of a million, unless you are using inappropriate units elsewhere.
The wavlength would have to be in meters in order to avoid conversion. I like to use nanometers. [I actually have two conversions since I convert to sq. meters in lieu of sq. cm.]


For wavelength we have these units:
hc/λ5 => J.s.m.s-1/m5 = J.m-3.m-1
Oops, thanks, I have corrected my carlessly posted equation in post 41. This is the appropriate equation that should give Joules/sec. (watts) per unit volume. [I failed to square "c", and 8 pi was for another equation. This exercise now reminds me just how much bumbling I did to do to get to where EDG wants to be. I may only be complicating things for EDG, though there is no simple place I have found that allows an amateur like me to grab a simple equation and run with it. Had you not delayed your membership here ;), perhaps I would have found the high road, though back then I was a bit too shy to ask, no doubt.]

I then assume that the energy produced within this unit volume is the same energy that must exit at the surface, ie. per unit area.

EDG
2009-Feb-20, 10:15 PM
Well that might explain why my graph looked so different :). I'll try this all out tonight...

grant hutchison
2009-Feb-20, 10:38 PM
Oops, thanks, I have corrected my carlessly posted equation in post 41. This is the appropriate equation that should give Joules/sec. (watts) per unit volume.Since the idea of the spectral curve is to produce a quantity per unit wavelength, you actually have watts per unit area, per unit wavelength. The integration with respect to wavelength then yields the desired watts per unit area.


Had you not delayed your membership here ;), perhaps I would have found the high road, though back then I was a bit too shy to ask, no doubt.I'm having difficulty understanding my area of responsibility in that hypothetical.
But I'm sorry for whatever it was.
Or pleased, if I should be pleased.

Grant Hutchison

George
2009-Feb-21, 09:39 PM
Since the idea of the spectral curve is to produce a quantity per unit wavelength, you actually have watts per unit area, per unit wavelength. The integration with respect to wavelength then yields the desired watts per unit area.
I have seen this stated in a derivative form, though perhaps only once, and recall being a bit struck by such an expression as I do not remember seeing a y-axis stated as a derivative, but then I realized that because integration calculates the area under a curve that the objective of the integration across the whole spectrum is to obtain the total wattage per unit area. I suspect the deriviative form may be the correct way to state it because stating it as wattager per unit area per wavelength doesn't seem to work, at least for me. Say the values across a 1000 nm portion of the spectrum were all 1000 watts (per unit area). There would be some who would think that one would need to multiply the 1000 watts (per unit area per nm) by the 1000 nm in order to obtain total watts per unit area. Thus, they would incorrectly get 1 megawatts per unit area, and not the 1000 watts as the sp. irr. plot is intending to represent.

I'm elaborating mainly to get your view on this, and it is historicaly important for me because when I first began considering whether or not to engage in the "Quest for the color of Sun", I found the derivative form stated in a sp. irr. graph and I suddenly felt like a calf staring at a new gate; an event that reminded me why I elected not to attempt to be an astrophysicist. But, I grit my teeth and gave it some thought, and soon the integration angle popped-up.


I'm having difficulty understanding my area of responsibility in that hypothetical. But I'm sorry for whatever it was. Or pleased, if I should be pleased. It was a compliment to your generous and expressive talents in a facetious sort of way. :)

grant hutchison
2009-Feb-21, 10:09 PM
Say the values across a 1000 nm portion of the spectrum were all 1000 watts (per unit area). There would be some who would think that one would need to multiply the 1000 watts (per unit area per nm) by the 1000 nm in order to obtain total watts per unit area. Thus, they would incorrectly get 1 megawatts per unit area, and not the 1000 watts as the sp. irr. plot is intending to represent.But this neglects the fact that the spectral irradiance plot Y axis is marked off in watts per unit area per unit wavelength, because that's what the Planck formula provides. So if the axis is marked in watts per unit area per nanometer, the user is correct to multiply by 1000 to get the area under the horizontal line, in watts per unit area. If the axis were marked off in watts per unit area per metre, then the user would be correct to multiply by 10-6 to get the area under the horizontal line (since 1000nm = 10-6m).

Grant Hutchison

George
2009-Feb-22, 02:02 AM
Ah, we are still on the conversion factor, and rightfully so.


If the axis were marked off in watts per unit area per metre, then the user would be correct to multiply by 10-6 to get the area under the horizontal line (since 1000nm = 10-6m). Well, after stating how I like to work in nanometers, I went back to the spreadsheet and, as you have determined, it is in meters after all. [I show it as nanometers in another column and assumed I was using nanometer, but I used meters for the equation which is consistent with the Planck equation terms, duh!]

Has my bumbling messed you up too much, EDG?

If you want to use a spreadsheet approach, as I do, use the corrected wavelength equation I gave and multiply by the square of the distance ratio (Sun's radius/1 AU), as previously stated, and you will have watts per sq. meter per meter of wavelength at 1 AU. Convert to the units of wavelength you prefer. The 1 million division, of course, will give you microns.

EDG
2009-Feb-22, 02:13 AM
I'll see if I can look at it again tomorrow :) - had to set up my new computer and reinstall everything for the past few days.

Though why do I need to involve the distance? The proportion of UV/IR/Visible etc would be the same at any distance wouldn't it?

George
2009-Feb-22, 02:22 AM
Yes, that is true regarding proportions, but I thought our eyes view of other stars would be at 1 AU. Besides, might as well give you the whole enchilada and not just the cheese. :)

EDG
2009-Feb-22, 07:33 AM
I'm guessing that one could extract the visual luminosity too from all this - if for example 40% of the total luminosity of a 1000 Sol star is in the visible part of the spectrum then that means its visual luminosity is 400 Sols, right? (1 Sol = 1 Solar Luminosity).

timb
2009-Feb-22, 09:38 AM
I'm guessing that one could extract the visual luminosity too from all this - if for example 40% of the total luminosity of a 1000 Sol star is in the visible part of the spectrum then that means its visual luminosity is 400 Sols, right? (1 Sol = 1 Solar Luminosity).

No. Presumably you want to know how its visual luminosity compares with the Sun's visual luminosity, in which case you have to divide its absolute output in the visible spectrum by the Sun's absolute output in the visible spectrum.

grant hutchison
2009-Feb-22, 01:01 PM
And because you can't convert watts to lumens directly. Different parts of the visible spectrum produce different levels of luminous sensation: you get a peak of luminous efficiency at 683 lm.W-1 in the middle of the range, and a couple of tails in which luminous efficiency is very poor. So very blue stars or very red stars will look less bright for the same energy output in the visible range.

Grant Hutchison

neilzero
2009-Feb-22, 02:44 PM
All good answers, except I think the light from a red dwarf would be red, not pink or orange. From a Class k star possibly pink or orange. Eye damage from the coolest red dwarf = class m is likely as there is a lot of near infrared, but very little visible light.

grant hutchison
2009-Feb-22, 02:53 PM
All good answers, except I think the light from a red dwarf would be red, not pink or orange.Depends on the viewing conditions.
The spectrum for most of the M stars defines a shade of orange. As a bright illuminant, that will nevertheless look white (which is why the filament of an incandescent lightbulb looks white). As a distant point source, it can also look red, if it is dim enough for us to lose blue and green sensation while retaining the dominant red stimulus (which is why Antares looks red).

Grant Hutchison

George
2009-Feb-22, 09:23 PM
As a distant point source, it can also look red, if it is dim enough for us to lose blue and green sensation while retaining the dominant red stimulus (which is why Antares looks red). Antares, I suspect, is too bright for it to serve as an example. Also, if the energy distribution is close to a black body, I wonder if red ever becomes the dominant color. Using Antares, 36% of the energy is in the red end (to 700nm, though some can see further). Yet 46% of the energy is in the green+yellow+orange. This should shift the net result to more of an orange color, which happens to be the color that I see it, or perhaps redish-orange. [I could wait till its on the horizon. ;)]

Actually, the proper proportions should be in photon flux and not simply energy distribution. Thus, although red gains a further advantage (41% flux), the G-Y-O portion is still greater at 44.5%.

Further, as you stated, the eye sees in the green range best, so the product of spectral irradiance (compensated for photon flux) and spectral sensitivity (slightly favoring green) determines the color seen, assuming the flux is still in the photopic portion of our vision, as you pointed out.

timb
2009-Feb-22, 09:24 PM
And because you can't convert watts to lumens directly. Different parts of the visible spectrum produce different levels of luminous sensation: you get a peak of luminous efficiency at 683 lm.W-1 in the middle of the range, and a couple of tails in which luminous efficiency is very poor. So very blue stars or very red stars will look less bright for the same energy output in the visible range.


I was avoiding that complication. How good an approximation would it be to compare outputs just at that one frequency in the middle of the range?

grant hutchison
2009-Feb-22, 10:19 PM
I was avoiding that complication. How good an approximation would it be to compare outputs just at that one frequency in the middle of the range?I've seen it done: Stimets and Sheldon use a delta function at the peak of the luminous efficiency curve, to simplify their calculations in The celestial view from a relativistic starship (JBIS 1981; 34: 83-99), for instance.
It's always been my intention to check the difference between that approximation and a proper luminous efficiency curve, but I haven't got around to it, so far.

Grant Hutchison

grant hutchison
2009-Feb-22, 10:32 PM
Antares, I suspect, is too bright for it to serve as an example. Also, if the energy distribution is close to a black body, I wonder if red ever becomes the dominant color. Using Antares, 36% of the energy is in the red end (to 700nm, though some can see further). Yet 46% of the energy is in the green+yellow+orange. This should shift the net result to more of an orange color, which happens to be the color that I see it, or perhaps redish-orange.Red for me, which was why I chose it as an example. :)
It's not at a good declination for my northerly latitude, but I've seen it high in the sky from equatorial regions often enough to know that its apparent colour to my eye doesn't match its spectrum.
I find Antares a good match for Mars when Mars is about the same magnitude, but Mars changes from Antares-red to a more mellow orange when it's at its opposition brightest: again, green and blue cones become active with the brighter source, and serve to desaturate and moderate the red signal.

Grant Hutchison

grant hutchison
2009-Feb-22, 11:24 PM
Further, as you stated, the eye sees in the green range best, so the product of spectral irradiance (compensated for photon flux) and spectral sensitivity (slightly favoring green) determines the color seen, assuming the flux is still in the photopic portion of our vision, as you pointed out.But I'm trying, above, to distinguish between nice, bright photopic vision (by which we might hope to see the "true" colour of the stellar spectrum) and the loss of colour detail that comes in the lower mesopic range (the sort of thing we're using when looking at the brighter stars in Earth's sky).
The first thing lost in colour-matching tests which explore the mesopic range is the green sensation. Greens are classified as blues, and we stop being able to tell orange from yellow from red. Hence the redness of "red" stars in mesopic vision: there's more red energy, and less green sensitivity.
I'm guessing you and I differ in our mesopic threshold for green loss, or we differ in how our neural wiring interprets that unusually isolated red sensation.

Grant Hutchison

George
2009-Feb-22, 11:29 PM
Red for me, which was why I chose it as an example. :) Ah yes, you're limited to about 7 or 8 degrees altitude for Antares. That should redden it.


It's not at a good declination for my northerly latitude, but I've seen it high in the sky from equatorial regions often enough to know that its apparent colour to my eye doesn't match its spectrum.
So how red do you recall it? Here is the famous Crayola (http://en.wikipedia.org/wiki/List_of_Crayola_crayon_colors)color list. Is it "Awesome Red" (#7) or more of their "Red Orange" (#107), or even their, uh oh... "Fun in the Sun Red" (#48)? [What is wrong with people and their idea of the Sun?.] I see Antares more as an "Orange" (#84). [But often not so orange.]

The image below shows just how orange, even yellow-orange, it appears from my location. [Obviously, I wasn't holding it all that steady. ;)]

grant hutchison
2009-Feb-23, 12:49 AM
So how red do you recall it? Here is the famous Crayola (http://en.wikipedia.org/wiki/List_of_Crayola_crayon_colors)color list.Can't be that famous: I'd never heard of it. ;)


Is it "Awesome Red" (#7) or more of their "Red Orange" (#107), or even their, uh oh... "Fun in the Sun Red" (#48)?More red as in #106, "Red".


The image below shows just how orange, even yellow-orange, it appears from my location. [Obviously, I wasn't holding it all that steady. ;)]Interesting that you see it the same colour as (some part of) that smear. The idea of defocussing a long exposure and smearing it was pioneered by (I believe) David Malin, as a way of showing us the black-body colours rather than the colours we actually see. His view of Orion (http://zuserver2.star.ucl.ac.uk/~idh/apod/ap961202.html), for instance, produces an orange Betelgeuse which is a long way from what I see.

Grant Hutchison

George
2009-Feb-23, 04:27 AM
Can't be that famous: I'd never heard of it. ;) I suppose I only wish it so since it is the more colorful approach, and I enjoy puns more than I should.


More red as in #106, "Red". That's surprising.


Interesting that you see it the same colour as (some part of) that smear. The idea of defocussing a long exposure and smearing it was pioneered by (I believe) David Malin, as a way of showing us the black-body colours rather than the colours we actually see. His view of Orion (http://zuserver2.star.ucl.ac.uk/~idh/apod/ap961202.html), for instance, produces an orange Betelgeuse which is a long way from what I see. Yes, he's somewhat.... well known in astrophotography. :)

I like this one of the Southern Cross (http://antwrp.gsfc.nasa.gov/apod/ap040708.html) by Stefan Seip. I used this technique to help determine the color of Solar twins (G2 yellow dwarfs *cough*) such as 18 Sco. It is called progressive defocusing.

In uploading 18 Sco, I found another Antares shot taken from Kitt Peak.

grant hutchison
2009-Feb-23, 11:49 AM
... I enjoy puns more than I should.Indeed, you enjoy puns more than I thought was possible.:)


That's surprising.Colour perception is pretty variable at these light levels. But if you check through naked-eye astronomy references, you'll fairly often find Antares described with phrases like "the reddest naked-eye star", "deep red", "unusual metallic red": my perception is not that uncommon. Our own BA calls it "rusty red", from which I assume he sees more than a hint of orange, as you do.

Grant Hutchison

neilzero
2009-Feb-23, 01:09 PM
I learned a lot from these posts. Are there never stars with a color temperature of about 1000 K, radiating mostly in far infrared? These would have only a small amount of radiation in near infrared, and almost none in visible red, but could still appear moderately bright if you were only a few thousand kilometers above the apparent surface of the star. Would typical human eyes still see in black and white? How hot would a black body get at an altitude of 10,000 kilometers assuming no other heat source? Would trace elements in the photosphere, besides 89% hydrogen and 10% helium change the answers, significantly? Would 10,000 kilometers altitude be inside the corona? Perhaps a star that cool would not have a significant corrona, if there had been no recent solar flares? Neil

George
2009-Feb-23, 01:30 PM
Indeed, you enjoy puns more than I thought was possible. I would have thought you enjoyed the paranomasian Shakespeare. ;) If I tell you that many of mine are malaprops, would you believe me? Admittedly, many of those are done accidentally on purpose. I do try, however, to toss in a few oxymorons to avoid being too monochromatic. Ok, most are in the low energy end of the spectrum, but itís hard to get to the other end and not appear too blue. [/corn]


Colour perception is pretty variable at these light levels. But if you check through naked-eye astronomy references, you'll fairly often find Antares described with phrases like "the reddest naked-eye star", "deep red", "unusual metallic red": my perception is not that uncommon. This does surprise me a bit even though I have assumed star colors are to be understood on a more subjective level. Iíll try to find a spectrum of it when time allows. It seems to me that if it is Planck-like then it should be at best orange-red, ignoring atmospheric effects.

grant hutchison
2009-Feb-23, 01:59 PM
I would have thought you enjoyed the paranomasian Shakespeare.I do indeed love Shakespeare. So I forgive him his puns. :)


This does surprise me a bit even though I have assumed star colors are to be understood on a more subjective level. Iíll try to find a spectrum of it when time allows. It seems to me that if it is Planck-like then it should be at best orange-red, ignoring atmospheric effects.For a quoted temperature of 3400K, the black-body spectrum is a rather pretty orange, R=255 G=143 B=61.
My impression is that most folk see a redder star than that: which is easily explained by the blue level perhaps dropping below the colour threshold entirely, combined with the onset of green insensitivity in the mesopic range. So it may be that the variation in reported colours is really a symptom of a sort of acquired reversible Daltonism. We're all getting the same deuteranomalous signal at a cortical level, but are invoking different qualia from our usual photopic repertoire in order to explain it.

Grant Hutchison

George
2009-Feb-23, 02:18 PM
Are there never stars with a color temperature of about 1000 K, radiating mostly in far infrared? This temperature would be found in T-class stars. These cool brown dwarfs have molecules in their atmospheres that allow for strong variations in their spectrum. Long, long ago a thread was introduced by a non-paranomasian that nicely addressed some interesting revelation regarding T-class color (http://www.bautforum.com/astronomy/19534-magenta-brown-dwarfs.html) (ie. maroon :D).


These would have only a small amount of radiation in near infrared, and almost none in visible red, but could still appear moderately bright if you were only a few thousand kilometers above the apparent surface of the star. 1000K seems to be the temperature that is just hot enough for an object to glow red.


Would typical human eyes still see in black and white? Objects that have light levels below about 1 candela/m2 may stimulate the rods in the eye, but not the "color" cones.

Hmmm, I wonder if some cool brown dwarfs might appear gray (Y-class?) for this reason?:doh:


How hot would a black body get at an altitude of 10,000 kilometers assuming no other heat source? Interestingly, being close to a star does not improve its surface brightness (brightness per unit area), though the object would appear brighter as a whole.


Would trace elements in the photosphere, besides 89% hydrogen and 10% helium change the answers, significantly? T-class stars have heavy methane and other molecular compound atmospheres, which greatly alters its spectrum.

George
2009-Feb-23, 03:53 PM
I do indeed love Shakespeare. So I forgive him his puns. :) Oh happy dagger! :)



For a quoted temperature of 3400K, the black-body spectrum is a rather pretty orange, R=255 G=143 B=61.
My impression is that most folk see a redder star than that: which is easily explained by the blue level perhaps dropping below the colour threshold entirely, combined with the onset of green insensitivity in the mesopic range. So it may be that the variation in reported colours is really a symptom of a sort of acquired reversible Daltonism. We're all getting the same deuteranomalous signal at a cortical level, but are invoking different qualia from our usual photopic repertoire in order to explain it.
Me thinks a luster so luminous
will afford a flux so numerous
that the photopic range will be
that which we will see
Greens and yellows are bright
Though, to be fair, red is half again in might

EDG
2010-Aug-06, 08:19 AM
*bump*

I'm looking at this again (for the same project funnily enough) and I've gone through the thread and still not sure I follow how to do this graph - I don't think I ever got a handle on it when I initially asked about it either, to be honest. So I'm going to have a go at this qualitatively here, rather than by crunching the numbers.

Here's the situation I'm looking at - I have a planet orbiting an A8 IV star. The star's total luminosity is 46 Sols, and the surface temperature is 7600 K. The planet orbits at a distance of about 10 AU from the star. The star's apparent diameter in the planet's sky is 0į 12' 9" (about 0.2 degrees, just under half the diameter of the sun in Earth's sky).

Obviously the star is going to be putting out a lot more UV than Sol. From what Grant was saying in post #29, qualitatively speaking it sounds like I'd probably get around 35% of the star's luminosity in the visible spectrum, 35% of it in the IR and longer spectrum, and 30% in the UV - so approximately a third in each 'bin'. That should mean a planet at 1 AU should be getting roughly 15.3 times the luminosity of Sol in each of the IR, visible, and UV spectra. Obviously, not a good place to be! (that said I have an asteroid belt in the system between 0.65 and 1.2 AU from the star, and in this SF setting there are people mining the belt so their ships are going to be enduring some seriously harsh conditions!).

But... at 10 AU the star is 100 times dimmer. It's still around apparent magnitude -25.9 from there, but the star's actually about 0.46 times the apparent luminosity of Sol as seen from the planet. But again, there's more UV from this star that there would be from Sol. While the apparent luminosity is just under half that of Sol fom Earth, it's subtending a disc that's just under half the angular diameter of Sol from Earth. So even if we say that there's a total of 0.153 Solar luminosities worth that's actually in visible light at this distance then it's still going to be pretty bright to look at, and the other 0.153 Sols of UV is probably going to be a bit more damaging than on Earth (since at Earth's distance from Sol, Earth is getting about 0.12 Solar luminosities of UV).

So if my logic is right, then at 10 AU we've got a star that's (to the human eye, approximately) almost as visibly bright as Sol and we're receiving about the same amount of UV as Earth does from Sol. So overall at this distance it's about as damaging to look at as Sol is from Earth (the other difference of course is that the light is going to be much bluer than Sol's, since it's a white star rather than a yellow one). Does that sound about right, or have I seriously messed up somewhere? And would the smaller size of the stellar disc mean that looking at the star might actually be more damaging to the retina (since the light's concentrated into a smaller point?).

Romanus
2010-Aug-06, 10:18 AM
I don't think it would be that straightforward; the Wien displacement law gives a star that hot a luminosity peak that's well in the ultraviolet range. That is, even if it was visually as bright as the Sun at your distance, the UV flux would be exponentially higher compared to the Sun, much more than a straightforward extrapolation of the inverse square law will give you. I lack the maths to be more specific, but that's my understanding of the physics.

George
2010-Aug-06, 06:02 PM
Here's the situation I'm looking at - I have a planet orbiting an A8 IV star. The star's total luminosity is 46 Sols, and the surface temperature is 7600 K. The planet orbits at a distance of about 10 AU from the star. The star's apparent diameter in the planet's sky is 0į 12' 9" (about 0.2 degrees, just under half the diameter of the sun in Earth's sky).

Obviously the star is going to be putting out a lot more UV than Sol. From what Grant was saying in post #29, qualitatively speaking it sounds like I'd probably get around 35% of the star's luminosity in the visible spectrum, 35% of it in the IR and longer spectrum, and 30% in the UV - so approximately a third in each 'bin'. Using the spectral exitance (I'm glad Grant cleared that issue up for me) equation for the bulk of the spectrum of a 7600K star, the various unit wavelength values can simply be added and divided into the total to obtain these three percentages. I get: 27%, 46%, 27% for UV+, Visible (400nm to 750nm), IR+ respectively. [There will be some minor variation from this blackbody representation, and I think I am using a broader visible spectrum than either you or Grant.]

That should mean a planet at 1 AU should be getting roughly 15.3 times the luminosity of Sol in each of the IR, visible, and UV spectra. At 1 AU, the UV band for your A8 star will be about 100x stronger than what we see at 1AU from the Sun (46 x 27%/12%), ignoring our atmosphere.

But... at 10 AU the star is 100 times dimmer. It's still around apparent magnitude -25.9 from there, but the star's actually about 0.46 times the apparent luminosity of Sol as seen from the planet. But again, there's more UV from this star that there would be from Sol. While the apparent luminosity is just under half that of Sol fom Earth, it's subtending a disc that's just under half the angular diameter of Sol from Earth. So even if we say that there's a total of 0.153 Solar luminosities worth that's actually in visible light at this distance then it's still going to be pretty bright to look at, and the other 0.153 Sols of UV is probably going to be a bit more damaging than on Earth (since at Earth's distance from Sol, Earth is getting about 0.12 Solar luminosities of UV). As long as the star is close enough to appear as a disk, it will appear about 3.3 times brighter per unit area as the Sun, regardless of your distance from it. [46/14.4 = 3.3, where 14.4x is the disk area ratio for your star and the Sun.]


So if my logic is right, then at 10 AU we've got a star that's (to the human eye, approximately) almost as visibly bright as Sol and we're receiving about the same amount of UV as Earth does from Sol. Yes, if it’s 100x as bright at 1AU it will be just 1x at 10AU, assuming my math is ok.

So overall at this distance it's about as damaging to look at as Sol is from Earth (the other difference of course is that the light is going to be much bluer than Sol's, since it's a white star rather than a yellow one). Yellow star!!??%%$#@. What yellow star?

And would the smaller size of the stellar disc mean that looking at the star might actually be more damaging to the retina (since the light's concentrated into a smaller point?). Since distance and area scale equally, there is no unit brightness change. So if it is 3.3x brighter per unit area as a whole and the UV is 2x more than the Sun, then the unit brightness will be about 7x more, which is the increase in unit area brightness acting on the retina. The actual effect will likely be less since the eye moves around a lot and the star appears significantly smaller.
This isn’t all that significant given that both of these stars are more than a 1000x too bright for the eye’s comfort level.

EDG
2010-Aug-06, 07:29 PM
OK, thanks. So I was in the right ballpark, the distribution was just closer to 25%/50%/25% than 33% all around. So at 10AU there shouldn't be any problem with people being blinded by glare reflecting back from shiny objects or anything (at 0.6 to 1 AU, that might be a problem though).

Though about the light... if I go find a 7800K light (I think they're available?) and compare that to a 5000K one, that should (luminosity difference aside) indicate the colour difference reasonably accurately, right?

grant hutchison
2010-Aug-06, 07:47 PM
Though about the light... if I go find a 7800K light (I think they're available?) and compare that to a 5000K one, that should (luminosity difference aside) indicate the colour difference reasonably accurately, right?Not really.
Your perceptual "white point" is readily variable according to the ambient lighting. Your 7800K light will look pretty blue if you view it under old-fashioned incandescent lighting; less blue under modern fluorescents or sunlight; and not blue at all if it's your primary illuminant. You can safely assume that any human on the surface of your planet will see A8 sunlight as white, because that's what our brains do.

Grant Hutchison

EDG
2010-Aug-06, 09:19 PM
Well, I was thinking of trying that in an otherwise dark room, with the 7800k light being the only light source, but point taken ;)

George
2010-Aug-06, 11:15 PM
Your perceptual "white point" is readily variable according to the ambient lighting. Your 7800K light will look pretty blue if you view it under old-fashioned incandescent lighting; less blue under modern fluorescents or sunlight; and not blue at all if it's your primary illuminant. Hmmmm. Wouldn't the 7800K (or 7600K) be the "white point" causing the cooler one to look more yellow than normal? [Attached is the photon flux]

grant hutchison
2010-Aug-06, 11:26 PM
Hmmmm. Wouldn't the 7800K (or 7600K) be the "white point" causing the cooler one to look more yellow than normal? [Attached is the photon flux]It depends which source is the dominant illuminance. A room lit by incandescents will make a small source of higher temperature look blue. A room lit by fluorescents will make a a small incandescent source look yellow.

Grant Hutchison

George
2010-Aug-07, 05:26 PM
It depends which source is the dominant illuminance. A room lit by incandescents will make a small source of higher temperature look blue. A room lit by fluorescents will make a a small incandescent source look yellow. [Darn, I thought I had ya! :)]

Since radiance increases as the 4th power of absolutle temperature, the hotter sources will outshine the lower temperature ones. Even though the peak wavelength will shift beyond the visible range, when temperatures rise the light flux will always increase within the visible range regardless of this peak shift. [Oddly enough, the temperature at which the peak wavelength moves beyond our visible range happens to be about 7600K. But, this assumes ~ 380nm is a reasonable edge, though I don't think there is any offical visible spectrum range established due to the variation in visibility of the human eye.]

However, there is no likely reason to assume both light sources are of equal distance from the observer. Also, scattered light can be at a low intensity but at a much higher temperature, as determined by its spectral distribution. Thus, your point is certainly valid.

mugaliens
2010-Aug-07, 08:46 PM
Though about the light... if I go find a 7800K light (I think they're available?) and compare that to a 5000K one, that should (luminosity difference aside) indicate the colour difference reasonably accurately, right?

Yes, they're available. In up to 10000K (http://www.ahsupply.com/bulbs.htm), if you'd like, though the latter are generally used for marine acquariums, to simulate the mostly blue light which filters through the water.

It's a misnomer to think that the higher the color temperature the bluer the light, as it's an asymptotic scale (http://en.wikipedia.org/wiki/File:PlanckianLocus.png)based on black-body radiation, and there's a practical limit in the blue range, equal to infinate temperature. In theoretical terms, 6500K is considered the center of "daylight" color, and this is the the standarized white for NTSC, PAL, and older versions of ATSC.

In practical terms, when it comes to flourescent bulbs, I find 3500K too yellow, and 5500K too blue, so when I shopped for living room lamps, I picked a couple of two-bulb models, and I use one of each (a 13W 3500K (60W equivalent), and a 26W 5500K (100W equivalent)). While it looks a little funny, the ambient light which spreads throughout the room is very easy on the eyes, and a lot less expensive than what my two 300W halogen lamps produced.

George
2010-Aug-07, 11:04 PM
It's a misnomer to think that the higher the color temperature the bluer the light, as it's an asymptotic scale (http://en.wikipedia.org/wiki/File:PlanckianLocus.png)based on black-body radiation, and there's a practical limit in the blue range, equal to infinate temperature. For a blackbody, the hotter the object the steeper the energy distribution slope from red to blue. Around 10 or 15 million K the slope seems to be about that seen in a deep blue sky distribution, thus I suggested several years ago that if one could see the core of our Sun, appropriately attenuated, then it would be blue.


In practical terms, when it comes to flourescent bulbs, I find 3500K too yellow, and 5500K too blue, so when I shopped for living room lamps, I picked a couple of two-bulb models, and I use one of each (a 13W 3500K (60W equivalent), and a 26W 5500K (100W equivalent)). While it looks a little funny, the ambient light which spreads throughout the room is very easy on the eyes, and a lot less expensive than what my two 300W halogen lamps produced. That's a neat idea.

mugaliens
2010-Aug-08, 05:20 AM
For a blackbody, the hotter the object the steeper the energy distribution slope from red to blue. Around 10 or 15 million K the slope seems to be about that seen in a deep blue sky distribution, thus I suggested several years ago that if one could see the core of our Sun, appropriately attenuated, then it would be blue.

As are blue giants, which are magnitudes hotter than our sun.


That's a neat idea.

Thanks! I wish they'd simply develop multispectral/temperature/color filaments which more accurately reflect the light we receive from both direct sunlight combined with random sunlit-reflected/translit clouds.

One might pick a centerpoint, but that's a serious engineering misnomer, as when it comes to replicating the effects of multispectral light, it is not the summation of its frequencies. Perhaps it's the fact that the various frequencies are coming from a multitude of sources (directions). It's also one of the primary reasons most professional portrait photographers employ more than seven sources of light, and most lighting supervisors for films also employ multiple lighting sources on set, including both directive and reflective. And that's just the energy directed on the set. Then comes the filters...

Back to reality and household living, I really wish someone would make a multispectral compact flourescent bulb that filters the mercury spikes.

If someone knows of such a bulb, please! Links. Thanks.

EDG
2010-Aug-08, 11:33 PM
As are blue giants, which are magnitudes hotter than our sun.

They're an order of magnitude hotter than the sun. Rigel's surface temperature is about 11,000 K (not even double the sun's surface temperature), and the brightest blue supergiants are more like 50,000 K.

WayneFrancis
2010-Aug-09, 12:36 AM
So, we all know (I hope) that we shouldn't stare at our own sun for very long or we'll damage (or destroy) our vision.

But how does this damage scale when we're talking about other stars? For example, if we orbited Sirius (luminosity of 25 Sols) at 1 AU, we'd be looking at a star that's 25 times brighter in the sky, right? So wouldn't that mean that if we even glimpsed at it, we'd fry our eyes? (and yes, I know we'd have other problems if Earth orbited Sirius at that distance ;) ).

I guess I'm looking for some kind of rough, qualitative relationship between the luminosity of the star, the amount of time we look at it with normal human eyes, and the amount of damage done to the eyes after that time.

I ask because it seems that nobody really considers this sort of thing in SF stories.

I think our eyes would be the least of our worries. We would be toast no matter if we had our eyes open or shut.

EDG
2010-Aug-09, 04:58 AM
I think our eyes would be the least of our worries. We would be toast no matter if we had our eyes open or shut.

Yes, I know. I said when someone raised that on the first page of the thread that I wasn't concerned about any other physical effects of being at that distance...

EDG
2010-Aug-09, 06:21 PM
One related question:

The system's asteroid belt is located between 0.65 and 1.2 AU. I'm imagining that the ships used to mine the belt are going to be heavily shielded, possibly having deployable sun-shades to protect themselves from the star. The blackbody temperature at these distances from the star ranges between 900K at the inner edge of the belt to 660K at its outer edge. I'm imagining that they're not going to have windows for example - it'd just be easier to put metal shutters over those and use cameras. Also, is it possible to have camera and window filters that would be able to filter out enough external light to prevent blindness anyway?

Would the high UV output or solar wind cause any problems? Would the sun-shades cause any problems (e.g. is photon pressure significant when it comes to stationkeeping?)

Can anyone think of any other issues this would cause? One thing I'm not sure about is EVA operations (be it by mining robots or people in spacesuits) carried out in the shade. With the sun hidden, wouldn't temperatures under the shade plummet? I'm guessing it's not a problem, since MESSENGER is under a sunshade and still operates fine. In fact, I found this article (http://pdf.aiaa.org/preview/2010/CDReadyMSO10_2129/PV2010_2137.pdf) that seems to say that MESSENGER had more problems being further from the sun because of excessive cooling behind the sunshade. I guess that the trick is to angle the shade so that the craft is kept around 'room temperature'? It also suggests that the solar panels themselves can get overheated if they're at the wrong angle!

Hornblower
2010-Aug-09, 10:28 PM
One related question:

The system's asteroid belt is located between 0.65 and 1.2 AU. I'm imagining that the ships used to mine the belt are going to be heavily shielded, possibly having deployable sun-shades to protect themselves from the star. The blackbody temperature at these distances from the star ranges between 900K at the inner edge of the belt to 660K at its outer edge. I'm imagining that they're not going to have windows for example - it'd just be easier to put metal shutters over those and use cameras. Also, is it possible to have camera and window filters that would be able to filter out enough external light to prevent blindness anyway?

Would the high UV output or solar wind cause any problems? Would the sun-shades cause any problems (e.g. is photon pressure significant when it comes to stationkeeping?)

Can anyone think of any other issues this would cause? One thing I'm not sure about is EVA operations (be it by mining robots or people in spacesuits) carried out in the shade. With the sun hidden, wouldn't temperatures under the shade plummet? I'm guessing it's not a problem, since MESSENGER is under a sunshade and still operates fine. In fact, I found this article (http://pdf.aiaa.org/preview/2010/CDReadyMSO10_2129/PV2010_2137.pdf) that seems to say that MESSENGER had more problems being further from the sun because of excessive cooling behind the sunshade. I guess that the trick is to angle the shade so that the craft is kept around 'room temperature'? It also suggests that the solar panels themselves can get overheated if they're at the wrong angle!

If the civilization in question can build a spacecraft advanced enough to performing mining on asteroids, I would expect that protecting the ship and crew from the star's radiation would be a relatively trivial engineering exercise. Likewise with stationkeeping.

Protecting your eyes from the light is child's play. Welders do it all the time with helmets equipped with special dark glass.

EDG
2010-Aug-09, 10:42 PM
If the civilization in question can build a spacecraft advanced enough to performing mining on asteroids, I would expect that protecting the ship and crew from the star's radiation would be a relatively trivial engineering exercise. Likewise with stationkeeping.

I'm sure it would be, but that doesn't address my point at all. I'd like to know how such engineering problems might be solved (assuming tech similar to what we have today).


Protecting your eyes from the light is child's play. Welders do it all the time with helmets equipped with special dark glass.

How bright would an arc-welder's torch be compared to about 23 solar luminosities of visible light (at 1 AU) though?

Hornblower
2010-Aug-10, 01:37 AM
I'm sure it would be, but that doesn't address my point at all. I'd like to know how such engineering problems might be solved (assuming tech similar to what we have today).



How bright would an arc-welder's torch be compared to about 23 solar luminosities of visible light (at 1 AU) though?

To protect the ship from direct starlight at that level, I would suggest an umbrellalike shade with slats that could be opened or closed as needed like window blinds. Surely a large manned ship with the capability of supporting mining machinery could handle such an apparatus. This is a far cry from a small, unmanned probe like Messenger that has very limited mechanical and electrical capability.

The task of building and operating mining machinery that could take the heat from the asteroid should be comparable to building and operating a steel mill. There we have heavy equipment that withstands a tremendous amount of heat in normal operation.

If the infrared from the hot asteroid is enough to overheat a crew cabin with a highly reflective exterior, we may need some refrigeration, with a radiator on a boom to serve as a heat sink.

For eye and camera protection, are we talking about looking at the A-type star up close, or at the lighted surface of the asteroid? For the latter, a moderately dark glass will suffice. For the former (why, I don't know) if one piece of the darkest welding glass is not enough, just stack up two of them. A #14 glass is sufficient for looking directly at the Sun.

I don't think the shade would be enough of a sail to be a stationkeeping problem. For a large ship the mass to surface area ratio is enormously larger than that of the dust grains of a comet's tail.

See the following reference for welding glass.
http://www.sizes.com/tools/welding_glasses.htm

EDG
2010-Aug-10, 02:15 AM
The task of building and operating mining machinery that could take the heat from the asteroid should be comparable to building and operating a steel mill. There we have heavy equipment that withstands a tremendous amount of heat in normal operation.

Good point. I don't think that would represent a significant problem anyway.
They could also position the ship's shade over a portion of the asteroid too, and just work in the shaded part.

I'll check out that link later too, thanks.

Nereid
2010-Aug-11, 10:04 AM
One related question:

The system's asteroid belt is located between 0.65 and 1.2 AU. I'm imagining that the ships used to mine the belt are going to be heavily shielded, possibly having deployable sun-shades to protect themselves from the star. The blackbody temperature at these distances from the star ranges between 900K at the inner edge of the belt to 660K at its outer edge. I'm imagining that they're not going to have windows for example - it'd just be easier to put metal shutters over those and use cameras. Also, is it possible to have camera and window filters that would be able to filter out enough external light to prevent blindness anyway?

Would the high UV output or solar wind cause any problems? Would the sun-shades cause any problems (e.g. is photon pressure significant when it comes to stationkeeping?)

Can anyone think of any other issues this would cause? One thing I'm not sure about is EVA operations (be it by mining robots or people in spacesuits) carried out in the shade. With the sun hidden, wouldn't temperatures under the shade plummet? I'm guessing it's not a problem, since MESSENGER is under a sunshade and still operates fine. In fact, I found this article (http://pdf.aiaa.org/preview/2010/CDReadyMSO10_2129/PV2010_2137.pdf) that seems to say that MESSENGER had more problems being further from the sun because of excessive cooling behind the sunshade. I guess that the trick is to angle the shade so that the craft is kept around 'room temperature'? It also suggests that the solar panels themselves can get overheated if they're at the wrong angle!
While the ESA folk do not concern themselves with how humans, or mining robots, might handle this, they have put a lot of effort into working out how a space probe could both survive, and do good science, as close to the Sun as 0.23 au - check out their work by visiting the Solar Orbiter website (http://sci.esa.int/science-e/www/area/index.cfm?fareaid=45).

EDG
2010-Aug-15, 06:20 AM
Looking at that link as well as this one (http://www.astronexus.com/node/3), Shade #14 lets through about 3 millionths of the light, and is used to watch solar eclipses safely. Plugging in the numbers, if we were 1 AU from the star that was 46 times brighter as viewed from Earth (I'm being conservative and saying it has to shield against all the wavelengths), then we'd need Shade #18 to view it without being blinded. I divided the 3 millionths by 46, and plugged that back into the equation in the link I just posted to get the shade number that would be equivalent to watching our 1 solar luminosity sun through a #14 shade.

To be really safe (since it says that IR and UV aren't blocked at the levels produced by that formula) I guess we could go all the way to shade #20, which only lets through about 7 billionths of the light energy. #14 is the highest that is produced for welding, and rarely used, so I don't even know if shades between #18 and #20 are even possible to make (they'd probably be nearly opaque!)!

Hornblower
2010-Aug-15, 10:52 AM
Looking at that link as well as this one (http://www.astronexus.com/node/3), Shade #14 lets through about 3 millionths of the light, and is used to watch solar eclipses safely. Plugging in the numbers, if we were 1 AU from the star that was 46 times brighter as viewed from Earth (I'm being conservative and saying it has to shield against all the wavelengths), then we'd need Shade #18 to view it without being blinded. I divided the 3 millionths by 46, and plugged that back into the equation in the link I just posted to get the shade number that would be equivalent to watching our 1 solar luminosity sun through a #14 shade.

To be really safe (since it says that IR and UV aren't blocked at the levels produced by that formula) I guess we could go all the way to shade #20, which only lets through about 7 billionths of the light energy. #14 is the highest that is produced for welding, and rarely used, so I don't even know if shades between #18 and #20 are even possible to make (they'd probably be nearly opaque!)!You can make it as dark as you wish by stacking lesser ones.

grant hutchison
2010-Aug-15, 11:39 AM
To be really safe (since it says that IR and UV aren't blocked at the levels produced by that formula) I guess we could go all the way to shade #20, which only lets through about 7 billionths of the light energy. #14 is the highest that is produced for welding, and rarely used, so I don't even know if shades between #18 and #20 are even possible to make (they'd probably be nearly opaque!)!"Nearly opaque" is what you need. :)
As Hornblower says, you can just make the material thicker if you want more attentuation.

Grant Hutchison