Why is ##\text{flux} = \pi\text{intensity}##

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In summary, based on the diagram and the equation, the flux at a surface of uniform brightness is simply ##\pi B##.
  • #1
yucheng
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Homework Statement
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Relevant Equations
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1632904653656.png

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?

SmartSelect_20210929-175549_Samsung Notes.jpg

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.
SmartSelect_20210929-175558_Samsung Notes.jpg
 
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  • #2
yucheng said:
r is not the hypothenuse of the triangle
Looks like the hypotenuse to me.
You must have the wrong triangle.
The R is from the sphere's centre to where the tangent from P touches the sphere. It makes a right angle to that tangent, not to r.
 
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  • #3
haruspex said:
Looks like the hypotenuse to me.
You must have the wrong triangle.
The R is from the sphere's centre to where the tangent from P touches the sphere. It makes a right angle to that tangent, not to r.
Oops, wrong triangle indeed. So when ##r \to R##? So it does become like figure 3 (In the second thumbnail), albeit a different orientation. Ah I see it. Thanks! But the interpretation of ##F = \pi B##?
 
  • #4
yucheng said:
But the interpretation of ##F = \pi B##?
I'm not sure why it doesn't turn out to be ##F = 2\pi B##. On reaching the sphere, the sphere occupies one half of what can be seen from P, so that should be a solid angle of ##2\pi##.
Need to think about it some more.
 
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  • #5
haruspex said:
I'm not sure why it doesn't turn out to be ##F = 2\pi B##. On reaching the sphere, the sphere occupies one half of what can be seen from P, so that should be a solid angle of ##2\pi##.
Need to think about it some more.
Let ##B_v## be the intensity (##\text{flux} \; \text{rad}^{-1} Hz^{-1}##)

However, the amount of radiation emitted is not isotropic. It depends on the angle from the normal of the surface (or projected surface).

$$F_v = \int B_v \cos{\theta} \, d\Omega =\int^{2\pi}_{0} \int^{ \pi}_{0} B_v \cos {\theta} \sin{\theta} \, d\theta \, d\phi = \pi B_v $$

I guess this makes sense, that F is not ##4\pi B##, nor is it ##2\pi B##
 

Related to Why is ##\text{flux} = \pi\text{intensity}##

1. What is flux and why is it important in science?

Flux is a measure of the flow of a physical quantity through a given area. It is important in science because it helps us understand the transfer of energy, mass, or particles in various systems. Flux is used in many fields of science, including physics, chemistry, and engineering.

2. How is flux related to intensity?

Flux and intensity are directly related to each other. Intensity is a measure of the amount of energy or particles passing through a given area per unit time, while flux is the total amount of energy or particles passing through that area. The relationship between flux and intensity is represented by the equation flux = intensity x area.

3. Why is the value of flux often represented by ##\pi## times the intensity?

The value of flux is often represented by ##\pi## times the intensity because it takes into account the direction of the flow. Flux is a vector quantity, meaning it has both magnitude and direction. By multiplying the intensity by ##\pi##, we are accounting for the direction of the flow and getting a more accurate measure of the total flux.

4. Can flux be negative?

Yes, flux can be negative. This occurs when the direction of the flow is opposite to the direction of the chosen area. In this case, the value of flux will be negative, indicating that the flow is in the opposite direction of the chosen area.

5. How is flux calculated in real-world situations?

In real-world situations, flux is calculated by measuring the intensity of the flow and the area through which it is passing. The intensity can be measured using various instruments such as sensors or detectors, while the area can be measured using geometric calculations or physical measurements. Once these values are obtained, they can be plugged into the equation flux = intensity x area to calculate the total flux.

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