Radiation from the surface of a star

[jorgeavg]

Hi there,
I´m reading about the theory of radiation and came up with some doubts when I started to calculate the net flux that an observer would "see" coming off from the star. I have provided my answers to my own questions but would like to see what you people think about them.
1. Can it be assumed that each point on the surface of a star radiates in ALL directions with the same intensity and not only radially?
My answer would be "yes", as radiation should propagate in all direccions. However the intensity of radiation that I would see from the different rays coming off the very same point on the star, would depend on my angular position with respect to the normal vector right on the emittter point on the surface of the star.
2. For an observer far away enough from the star, could we say that one entire star hemisphere (surface area 2pi) will contribute to the intensity measured by this far away observer?
I would again answer yes. However, assuming the observer could get close enough without burning, the closer the observer gets to the star, the smaller the spherical sector that would contribute to the intensity the observer would be able to measure. At the limit, if this 2D observer (no height) stands flat on the star, then only the radiation coming off that very same spot would be the only contribution to be measured by the observer. Is this reasoning flawed?
Thanks for your comments and feedback!
Jorge

 

[Drakkith]

1. Can it be assumed that each point on the surface of a star radiates in ALL directions with the same intensity and not only radially?

I believe so.

However the intensity of radiation that I would see from the different rays coming off the very same point on the star, would depend on my angular position with respect to the normal vector right on the emittter point on the surface of the star.

Would it? I'm not familiar enough with the topic to say anything on this.

2. For an observer far away enough from the star, could we say that one entire star hemisphere (surface area 2pi) will contribute to the intensity measured by this far away observer?

You can.

At the limit, if this 2D observer (no height) stands flat on the star, then only the radiation coming off that very same spot would be the only contribution to be measured by the observer. Is this reasoning flawed?

No, I think that's correct.

 

[ogg]

I have no particular expertise in this area, take this post with a grain of salt. There may be some simplifying assumptions which can be made. If there are I wouldn't be aware of them. The obvious flaw in your last answer is your assumption that the "surface" is a 2 dimensional spherical surface. In reality, it has thickness, as well as opacity - or should I say translucency - where the radiation intensity varies with height as well as temperature and density. At 'sufficient' distance, this thickness is negligible, but not close up. Look at it this way, assuming you're superman you lay on your back looking up away from the Sun and fall into it. As you fall the black of space changes getting foggier and brighter until looking up is just as opaque as looking down. Clearly at this point, the radiation is impinging from all sides. Of course, the question is with this scenario as you fall in, how far is the visiblity? That is, how far away can radiation come straight into your eyes without being absorbed or scattered and this obviously depends on density. Oh, also S.A. of sphere is 4πr² so a hemisphere is 2πr²

 

[Ken G]

Yes, the two points you make are correct in the limit of an infinitely sharp and well-defined surface. In practice, that idealization will have its limitations.

 

[mfb]

1. Can it be assumed that each point on the surface of a star radiates in ALL directions with the same intensity and not only radially?

Do you see the sun as single spot, or as some disk (don't look at it without proper protection, however)?

However the intensity of radiation that I would see from the different rays coming off the very same point on the star, would depend on my angular position with respect to the normal vector right on the emittter point on the surface of the star.

It does, but not for the reason you probably expect. If the surface would be a well-defined two-dimensional thing, the surface of the sun would have the same brightness as seen from all directions. In reality it does not, and if you look towards the center you look deeper into the sun (literally), so you see hotter regions. If you look from the side, you only see the colder outer regions. The result is called limb darkening.

 

[jorgeavg]

As you fall the black of space changes getting foggier and brighter until looking up is just as opaque as looking down. Clearly at this point, the radiation is impinging from all sides.

Interesting experiment. I would think the point where the impinging radiation from all sides is the same would have to be at the center of the star. At that point, and assuming the radiation flux is isotropic, the net flux should be zero, and this would happen if radiation coming in one direction is canceled out by radiation coming in from another direction diametrally opposed.

Oh, also S.A. of sphere is 4πr² so a hemisphere is 2πr²

I missed r². Thank you for highlighting it.

If you look from the side, you only see the colder outer regions. The result is called limb darkening.

I looked up this limb darkening online and still have plenty reading to do. However, I suppose that the fact the rim is darker should be taken as an indication that the radiation intensity we receive from its points is not the same we receive from, say, the middle point of the solar disk, which has more material "behind" pumping radiation out. I should not look at the start as radiation only from its surface but also consider its interior, which is contributing to the amount of photons leaving the star.
As I said, I still have a lot of thinking to do. Thank you!

Would it? I'm not familiar enough with the topic to say anything on this.

I think this point would be related to the limb darkening that mfb mentioned in his reply.
Thanks!

Yes, the two points you make are correct in the limit of an infinitely sharp and well-defined surface. In practice, that idealization will have its limitations.

I´m finding that now I have to think too much these surface differentials, their implications and limitations. It recall it was easier back in college. Let me go back to think about it. Thank you.

 

[mfb]

Well, "interior". Less than 0.1% of the solar radius.

I merged your posts. You can edit your posts if you want to add things, or directly include multiple quotes in a single post.