- #1

yucheng

- 232

- 57

- Homework Statement
- N/A

- Relevant Equations
- N/A

**Textbook Derivation**

Flux at an arbitrary distance from a sphere of uniform brightness ##B## (that is, all rays leaving the sphere have the same brightness). Such a sphere is clearly an isotropic source. At ##P##, the specific intensity is ##B## if the ray intersects the sphere and zero otherwise (see Fig. 1.6). Then,

$$

F=\int I \cos \theta d \Omega=B \int_{0}^{2 \pi} d \phi \int_{0}^{\theta_{c}} \sin \theta \cos \theta d \theta

$$

where ##\theta_{c}=\sin ^{-1} R / r## is the angle at which a ray from ##P## is tangent to the sphere. It follows that

$$

F=\pi B\left(1-\cos ^{2} \theta_{c}\right)=\pi B \sin ^{2} \theta_{c}

$$

or

$$

F=\pi B\left(\frac{R}{r}\right)^{2}

$$

Setting ##r=R:##

$$

F=\pi B

$$

That is, the flux at a surface of uniform brightness ##B## is simply ##\pi B##.

**My worries**

1. How is the equation valid when ##r=R##? The author made an approximation that ##\theta_{c}=\sin ^{-1} R / r##, and r is not the hypotenuse of the triangle, hence the approximation is only valid when ##r >> R## or ##\text{leg} \approx hypotenuse##. Rather, r is a leg. In this case, doesn't the approximation fail and vitiate the result that ##F=\pi B##?

2. Actually, which case does ##r=R## refer to? Diagram 2 or 3 below?

3. Is there a physical interpretation/explanation for ##F=\pi B##? The way I would think of it is that ##\pi## represents the patch of the sphere where the point resides (it is 1/4 of the whole sphere) and that this whole patch contributes to the flux at that point.

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