Undergrad Is the Sun a Perfect Black Body? Investigating the Reflection of Radiation

[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)?

 

[Charles Link]

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.

 

[russ_watters]

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.

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

See:
https://upload.wikimedia.org/wikipedia/commons/0/0d/EffectiveTemperature_300dpi_e.png

 

[Charles Link]

@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.

 

[Orodruin]

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

See:
https://upload.wikimedia.org/wikipedia/commons/0/0d/EffectiveTemperature_300dpi_e.png

View attachment 229943

In other words, pretty close, but not as close as

 

[Charles Link]

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.

 

[russ_watters]

Please, no...

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.

 

[russ_watters]

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.

Also an inside joke. Ignore.

 

[Orodruin]

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.

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)

Googling, I find values around 0.985.

Seems pretty close to blackbody to me.

 

[davenn]

(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)

thankyou, one of mine too

saved me the effort