# I Is the Sun a black body?

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1. Aug 27, 2018

### YoungPhysicist

Black bodies are objects that don't reflect radiation.
But If I shoot light beams to the sun, it doesn't reflect it , so does that mean the sun is a black body(or at least very close to a theoretical one)?

2. Aug 27, 2018

The sun is too far away to do reflectivity measurements of its surface, and there isn't a source around with enough light to measure the reflectivity of the sun. For the planets in the solar system, as well as the earth's moon, the light from the sun can be used to measure their reflectivity throughout the spectrum radiated by the sun. $\\$ I do believe the surface of the sun may be close to a blackbody, both in the visible as well as the infrared, because of the plasma nature of its surface that makes for good electromagnetic absorption, but I don't know of any way presently that can be used to make a reflectivity measurement.

3. Aug 28, 2018

### Staff: Mentor

Can't we use temperature to predict luminous intensity, then compare that to actual to find emissivity?

See:

4. Aug 28, 2018

@Young physicist I think the curve that @russ_watters presents above would also give reason to believe that the surface of the sun is "approximately" a blackbody without any spectral reflectivity measurements. It would be rather unlikely that the surface would have a nearly constant emissivity/reflectivity independent of wavelength for such a wide spectral range unless that emissivity was quite near 1.0.

5. Aug 28, 2018

### Orodruin

Staff Emeritus
6. Aug 28, 2018

I think @Young physicist might need some explanation of what @Orodruin just presented: I believe it is a spectral measurement of the approximately $T=2.73^o$ K blackbody-like background radiation of deep space.

7. Aug 28, 2018

### Staff: Mentor

Anyway, the one I posted seems just a touch off point. The total area [average irradiance] appears to have been purposely set equal, but the sun's graph looks clearly shifted to the right, indicating the temperatures don't match. I suggested the opposite: match the temperatures, then compare the areas to find the emissivity.

Googling, I find values around 0.985.

8. Aug 28, 2018

### Staff: Mentor

Also an inside joke. Ignore.

9. Aug 28, 2018

### Orodruin

Staff Emeritus
It is indeed. It is the FIRAS measurement of the cosmic microwave background, which is as close as anything we know to being a blackbody spectrum. It is a remnant of the early hot universe which became transparent to radiation at around 3000 K. Due to cosmological redshift, the radiation is now at around 2.73 K.

(Also, as usual, I cannot help but mentioning my pet peeves ... There is no $^\circ$ in K. It is just Kelvin, not degrees Kelvin. ... And the LaTeX for $^\circ$ is ^\circ)

Seems pretty close to blackbody to me.

10. Aug 28, 2018

### davenn

thankyou, one of mine too

saved me the effort