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George
2008-Oct-28, 03:20 AM
This one is kinda hairy...

What is the highest temperature an object the size of the Sun could be and still appear dark to the naked eye at 1 AU?

I realize this is very likely going to be just guess work, unless some super computer has messed with this, but I am still curious.

My calculations show that if the object were to magically behave like a blackbody, then a temperature of ~ 605K should do the trick. This is because at that temperature the visible brightness of the Sun drops 32.7 magnitudes down to a 6th magntiude, or ~ 100 quadrillionith (1e13) the wattage of the visible portion of the spectrum (~ 37% of the 1375 w/m^2 total solar wattage hitting the Earth).

Of course, at that temperature, molecules are all over the chart and there is no hope of a blackbody distribution.

So, what variables come into play and what wild guesses are out there that say how much they kick this 605K temperature value around?

I assume these are some of the major hurdles to jump...

1) Percent of visible spectrum that gets converted to non-visible spectrum wavelengths, mainly IR.

2) Surface brightness issues since our object is about 1/2 deg. in arc diameter.

Would 1000K be more likely?

Jeff Root
2008-Oct-28, 11:16 AM
Since the Sun is visible to the unaided eye as an extended object, its size
shouldn't make any difference. Except for the presence of a significant
thickness of atmosphere, it should be just like looking at a hot object right
in front of you. As I said in the thread about non-glowing molten lead,
a temperature of about 800 degrees Fahrenheit (700 kelvins) is needed
to produce a dark red glow visible in dim light. An even lower temperature
is sufficient to produce a glow visible in total darkness. Mugaliens claimed
somewhere here that he can see farther into the infrared than most people.
I'd guess that a minimum blackbody temperature that would be visible under
ideal conditions might be about the figure you came up with, 600 K or so.

-- Jeff, in Minneapolis

grant hutchison
2008-Oct-28, 11:45 AM
Yes, the calculation based on the visibility of point sources will be misleading when you're looking at an object that subtends more than an couple of arc-minutes.
As Jeff says, since luminance doesn't vary with distance, the question reduces to finding the threshold luminance we can detect.
In the situation given, I think we're not looking for an absolute threshold. We're probably dealing with the problem of edge detection: what luminance is required for us to detect the edge of a luminous disc superimposed on the luminance of the night sky?
These thresholds are tricky at low luminances (the nice smooth curves at high luminances go a bit wonky at the bottom end) and I'm remote from my references at present. I'll see what I can dig out later today, if no-one else chimes in with information.

Grant Hutchison

George
2008-Oct-28, 01:50 PM
As I said in the thread about non-glowing molten lead,
a temperature of about 800 degrees Fahrenheit (700 kelvins) is needed
to produce a dark red glow visible in dim light. An even lower temperature
is sufficient to produce a glow visible in total darkness. That seems to come pretty close. I wonder what a predominantly hydrogen/helium atmosphere might do to the spectrum?

Mugaliens claimed
somewhere here that he can see farther into the infrared than most people. It is known that many can. I only used a 400nm to 700nm visible range, though 750nm would have been better. However, at lower temperatures, there is not much difference. [I did extend to 750nm when I was in the higher temp. range as it becomes more significant to the total amount of flux.]

Your statement reminds me that I have a soldering iron with variable tip temp. control to about 800F. I'll have to give it a try tonite. :)

Yes, the calculation based on the visibility of point sources will be misleading when you're looking at an object that subtends more than an couple of arc-minutes. That's what I thought since the concentrated light (used in magnitude values) becomes diluted as it is spread out over a larger area. Or, looking at it another way (accident pun), the flux at any given eye's cone or rod is reduced as the object appears larger (and maintains distance), since the same amount of light is spread over a larger area of the retina.

I assume the rods rule here [over the cones] so the dimmest extended object would appear gray.

We're probably dealing with the problem of edge detection: what luminance is required for us to detect the edge of a luminous disc superimposed on the luminance of the night sky? That's a good point and it is what we do, I think, when we try to see the "dark side of the Moon", at least in the astronomical sense. :)

I am curious about this as it applies to protostars. If, during the earliest stage of stellar formation, some nascent protostars are enshrouded in close orbiting dust and gas, could an accretion disk exist and yet we still might have no visible detection of the nascent star itself? It seems to me that a body massive enough to have formed an accretion disk might already be too hot to not be visible to the naked eye. But, perhaps, not.

mugaliens
2008-Oct-28, 05:42 PM
These thresholds are tricky at low luminances...

A near-black circle against a nearer-black background made darker by having been pinpointed with stars...

George
2008-Oct-29, 04:58 PM
Yes, the calculation based on the visibility of point sources will be misleading when you're looking at an object that subtends more than an couple of arc-minutes.
My questionable math shows about a 7 mag. reduction for the Solar disc would yield the mag per unit square arcminute, or -19.7 for each arcminute of the Solar disc (ignoring the CLV).

If the Sun were to increase in diameter by about 8x, then a 12 mag. reduction would be seen as the square arcminute reduction from -26.7, due only to size (ignoring large mag. reductions due to temp. reduction).

George
2008-Nov-06, 05:32 PM
Your statement reminds me that I have a soldering iron with variable tip temp. control to about 800F. I'll have to give it a try tonite. :)
I could not see it glow in the dark though I turned the dial to the 900F level. [I admit that I did not wait more than about 5 or 6 minutes in the dark for my eyes to dilate.]

BioSci
2008-Nov-06, 08:12 PM
I could not see it glow in the dark though I turned the dial to the 900F level. [I admit that I did not wait more than about 5 or 6 minutes in the dark for my eyes to dilate.]

900F is just a tad low to be visible. Several sources indicate that:

525C = faint glow
700C = dark red
900C = cherry red
etc. :)

George
2008-Nov-06, 10:47 PM
Thanks. My ~1000F guess wasn't all that bad. :)

I am still curious if a protostar's photosphere could be much hotter than this and yet its dusty shroud, assuming it has one, cooler than 1000F and not simply as per the inverse square law would suggest. Thermal absorption by the dust, I suppose, is where I am headed with this.