What are the RGB values that correctly represent a 5800 K white surface on a calibrated 6500 K monitor?

Consider a high-quality monitor calibrated at the standard parameters: 6500 K, 2.2 gamma, 120 cd/m^2. Calibration is accomplished with a LaCie hardware sensor + its software, and it's quite accurate.

I intend to take a picture of the Sun through a telescope, using a safe, dedicated solar filter (full aperture Baader solar film for telescopes). The Sun's temperature is 5800 K. The filter is "white", quite decent actually, but I'm sure it's spectrum is not 100% flat - rigorously speaking it cannot be. Also, the camera may capture some infrared and so on, and further alter the color of the solar surface.

I want to process the resulting image so that, on the calibrated 6500 K monitor, the Sun's color is represented as close to original as possible. I expect the result to look like a soft creamy white.

Basically, that boils down to representing a 5800 K "white" on a 6500 K monitor. How do I do that?

I could load the image and tweak the tint settings (white balance) in software until the RGB triads on the solar disk fall in the required range, but I don't know what that range is. Sounds like there should be a formula for it somewhere ("given T1 the temperature of the monitor, then T2 white is represented when xR + yG = zB" or something like that, I'm just making stuff up).

Another approach: it would be nice if there was an app that could just generate "white" at any temperature, given that the monitor is calibrated at a certain color temperature. Then I could compare the generated white with the Sun's image, and make adjustments. But I'm now aware of any such app.

Any suggestions?

I do most of my raw file processing in Lightroom, I can use GIMP for additional color channel tricks. I'm not a photography expert, obviously, but I can follow directions. :)

Thanks!

The answer is: sRGB = (255, 241, 234).

The details of the calculation:

I calculated the spectrum of a blackbody at 5800 K using the Planck’s formula, then multiplied by the CIE color-matching functions of the standard 2 degrees observer and integrated over the wavelengths to get the (X, Y, Z) color. I then divided by X+Y+Z to get the chromaticity:

(x, y) = (0.3260, 0.3354)


multiplying (x, y, 1-x-y) by the XYZ to sRGB matrix, and dividing by the greatest component (R) yields:

(R, G, B) = (1, 0.8794, 0.8267)


I then gamma-encoded, multiplied by 255 and rounded to the nearest integer and got:

(R’, G’, B’) = (255, 241, 234)


Caveat: My answer is in the sRGB color space, which is almost, but not quite 6500 K with 2.2 gamma. BTW, “6500 K with 2.2 gamma” is not a color space specification: you also need the chromaticities of the primaries to get a fully-specified color space.

• Whoa! Jaw dropped to the floor. That's exactly what I was asking. Thanks! BTW, at (255, 241, 234) I think it would look like white with a slight golden hue, which makes sense. May 22, 2012 at 0:00
• This is an excellent answer. I have three questions: Sep 2, 2015 at 16:23
• "integrated over the wavelengths to get the (X, Y, Z) color. I then divided by X+Y+Z to get the chromaticity:" How did you go from a 3 vector to a 2 vector by scalar division? (Where did Z go?) Sep 2, 2015 at 16:24
• "I then gamma-encoded" Does this mean you raised R, G, and B to the power gamma, like [this]? What value of gamma did you use? There seem to be many options. Sep 2, 2015 at 16:24
• @kdbanman: No, I mean I transformed the linear RGB values to the sRGB non-linear representation, as per the equations (1.2) of the document you referenced. This is close to, although not exactly, a power law with exponent 1/2.2. Sep 2, 2015 at 20:18

Are you looking to change the color of the sun in your photographs, or simply represent the color that is there accurately? The two are very different tasks. The former would probably require a lot of work, and I'm not sure it would actually be accurate. The latter is actually already taken care of for you with ICM and ICC profiles.

It should also be noted that "white" is a highly subjective thing. The "white" of your monitor would, technically, be too blue for a "true white", given that 6500k models daylight, not sunlight. The white of the sun as imaged directly, without the interference of an atmosphere or any filtration, is probably more accurately modeled at 5785 K in the photosphere on a normalized basis, but it can fluctuate between around 4000 K and 6000 K depending on location and time (sunspots tend to be cooler). There is also the Chromosphere, above the photosphere, which ranges from about 6000 K to tens of thousands of degrees Kelvin until you hit the Corona, which spikes into the millions of degrees. When you image the sun without a filter, the only time your actually photographing the photosphere would be through sunspots, otherwise the white point of the sun can fluctuate wildly over its surface. With a filter, your ultimate white point will be affected by its design and the wavelengths it actually is designed to pass through, so again nailing down an exact white point is probably going to be a tough thing to start with. A neutral, true white to the human eye is probably in the realm of 5500 K, however that actually changes depending on whether you are observing an emitter or a reflector.