I Emission spectra of different materials

[sophiecentaur]

The answer is yes. Incidentally, at my workplace, using a thermal imager to test this concept , we once put a gold-coated plate in an electrical frying pan. (gold has reflectivity nearly 1 and thereby emissivity near zero throughout the infrared). The gold plate appeared very cold=(room temperature, caused by reflected energy from the walls of the room), while another object, (with emissivity close to 1), appeared to be nearly at the temperature of the frying pan. As cold as it looked according to the thermal imager, you couldn't touch the gold plate=it was hot !

An interesting experiment and the results need careful analysis before reaching a conclusion, I think. Very large thermal gradient at surface of the gold, I think - same as if it were covered with a good 'insulator'.

 

[Charles Link]

An interesting experiment and the results need careful analysis before reaching a conclusion, I think. Very large thermal gradient at surface of the gold, I think - same as if it were covered with a good 'insulator'.

The experiment was quite conclusive. (Incidentally gold would have a fairly high thermal conductivity, and it was kept in the frying pan for quite a long period of time. The material below the gold coating was also a metal.)

 

[sophiecentaur]

The experiment was quite conclusive.

Not intuitive though. The apparent paradox would need explaining and any misconceptions dealt with.

 

[Charles Link]

Not intuitive though. The apparent paradox would need explaining and any misconceptions dealt with.

The emissivity of the gold coating is nearly zero=this one is not paradoxical. $M=\epsilon \sigma T^4$ . The emissivity factor $\epsilon$ is very small for gold.

 

[sophiecentaur]

The emissivity of the gold coating is nearly zero=this one is not paradoxical. $M=\epsilon \sigma T^4$ . The emissivity factor $\epsilon$ is very small for gold.

I could ask you. If there is no apparent paradox , why was the experiment conducted at all? Are you sure that everyone would have predicted that result?
Don't get me wrong, it was a lovely experiment to do. You were lucky to have had the means, motivation and opportunity to do it and, post hoc, the result makes sense but . . . .

 

[JohnnyGui]

The answer is yes. Incidentally, at my workplace, using a thermal imager to test this concept , we once put a gold-coated plate in an electrical frying pan. (gold has reflectivity nearly 1 and thereby emissivity near zero throughout the infrared). The gold plate appeared very cold=(room temperature, caused by reflected energy from the walls of the room), while another object, (with emissivity close to 1), appeared to be nearly at the temperature of the frying pan. As cold as it looked according to the thermal imager, you couldn't touch the gold plate=it was hot !

Sorry, but I have a hard time understanding the link between emissivity and the amount of reflection. I don't see how a low emissivity automatically means having highly reflective property. Can't a material fully absorb any wavelength but NOT emit all of them at the same energy amount that it absorbed them with afterwards? (lattice interaction for example)

 

[Charles Link]

Sorry, but I have a hard time understanding the link between emissivity and the amount of reflection. I don't see how a low emissivity automatically means having highly reflective property. Can't a material fully absorb any wavelength but NOT emit all of them at the same energy amount that it absorbed them with afterwards? (lattice interaction for example)

It's called Kirchhoff's law (the relation between emissivity and reflectivity.)

 

[sophiecentaur]

I don't see how a low emissivity automatically means having highly reflective property.

If the two were not equivalent, you could build an engine that would get hotter and hotter for free! Another Maxwell's Demon. Verboten, I'm afraid.

 

[JohnnyGui]

If the two were not equivalent, you could build an engine that would get hotter and hotter for free! Another Maxwell's Demon. Verboten, I'm afraid.

How about a material that has a higher absorptivity than emissivity because it loses the absorbed energy in some other form than radiation?

 

[sophiecentaur]

How about a material that has a higher absorptivity than emissivity because it loses the absorbed energy in some other form than radiation?

This situation occurs in many cases - for instance when the object is being warmed up by incident radiation. That will happen when its temperature is lower than the surroundings. The heat it radiates will (of course) be less than the radiant heat it absorbs, but it's not the emissivity and absorptivity that are different - it's the temperatures.

 

[JohnnyGui]

This situation occurs in many cases - for instance when the object is being warmed up by incident radiation. That will happen when its temperature is lower than the surroundings. The heat it radiates will (of course) be less than the radiant heat it absorbs, but it's not the emissivity and absorptivity that are different - it's the temperatures.

But isn't it possible that there's thermal equilibrium (same temperatures) because a part of the absorbed energy is lost through another energy form at the same time as the radiation, while the surroundings receives both that radiation and that other form of energy?
So "Absorbed Q = Radiation loss + other form of energy loss".

 

[Charles Link]

But isn't it possible that there's thermal equilibrium (same temperatures) because a part of the absorbed energy is lost through another energy form at the same time as the radiation, while the surroundings receives both that radiation and that other form of energy?
So "Absorbed Q = Radiation loss + other form of energy loss".

At thermal equilibrium, the radiation loss is equal to the radiation gain. If there is another method of acquiring heat at thermal equilibrium, that mechanism also has a balance/equality between energy input and energy output: $\\$ e.g. gas molecules at temperature T that make contact with the surface. If some of the collisions are inelastic, so that some of the gas molecules lose energy to the surface, in the same number of collisions, some gas molecules acquire energy as they collide with the surface. The energy must balance in thermal equilibrium.

 

[JohnnyGui]

At thermal equilibrium, the radiation loss is equal to the radiation gain. If there is another method of acquiring heat at thermal equilibrium, that mechanism also has a balance/equality between energy input and energy output: $\\$ e.g. gas molecules at temperature T that make contact with the surface. If some of the collisions are inelastic, so that some of the gas molecules lose energy to the surface, in the same number of collisions, some gas molecules acquire energy as they collide with the surface. The energy must balance in thermal equilibrium.

Got it. This might be a too farfetched scenario but what if the material receives radiation from the surroundings, loses it part in the form of radiation and part in another energy form as I said, and then the surroundings receive both forms and turn it all into radiation energy again so that the material solely receives radiation energy but loses it radiation + some other form. Will there be a thermal equilibrium with different absorptivity and emissivity of the material in this case? Not sure if this is even possible.

 

[Charles Link]

Got it. This might be a too farfetched scenario but what if the material receives radiation from the surroundings, loses it part in the form of radiation and part in another energy form as I said, and then the surroundings receive both forms and turn it all into radiation energy again so that the material solely receives radiation energy but loses it radiation + some other form. Will there be a thermal equilibrium with different absorptivity and emissivity of the material in this case? Not sure if this is even possible.

The emissivity being equal to the absorbtivity is not fixed in stone because there are materials that are partially transmissive. In general though, thermal equilibrium requires a balance, which for opaque materials normally implies absorbtivity=emissivity. $\\$ Editing... On a very related topic, it might interest you to read up on the "spectral distribution" of the energy that is radiated according to the Stefan-Boltzman law=the "spectral distribution" is the Planck blackbody function, $M(\lambda, T)$, and the total area under the curve of the Planck function vs. wavelength gives $M= \sigma T^4$.

 

[JohnnyGui]

The emissivity being equal to the absorbtivity is not fixed in stone because there are materials that are partially transmissive. In general though, thermal equilibrium requires a balance, which for opaque materials normally implies absorbtivity=emissivity

Can't a material be partially transmissive and still be in thermal equilibrium so that its absorptivity = emissivity? For example, 2 materials A and B in a vacuum enclosed room all at the same temperature, material A being highly emissive and thus absorptive and B losing radiation energy through reflection and transmission. In order for A to not change in temperature it needs to emit equal radiation energy as it absorbs. In order for B to not change in temperature it needs to also emit (less) radiation energy as it absorbs. So that for material B: $E_{in} - E_{reflection} - E_{transmission} = E_{absorption} = E_{emission}$.

Editing... On a very related topic, it might interest you to read up on the "spectral distribution" of the energy that is radiated according to the Stefan-Boltzman law=the "spectral distribution" is the Planck blackbody function, M(λ,T)M(λ,T) M(\lambda, T) , and the total area under the curve of the Planck function vs. wavelength gives M=σT4M=σT4 M= \sigma T^4 .

This is quite interesting to know. Does that also mean that the deriviation of $\sigma \cdot T^4$ gives the spectral curve function of the blackbody, $M(\lambda, T)$?

Two more questions (sorry for bombarding you guys with all this):

1. If a material is highly reflective and thus poorly emissive, does that mean that it would cool down much slower than a blackbody when they're both at the same temperature, both equally conductive/convective and both sitting in an environment with the same lower temperature?

2. I keep on reading that if you polish a surface it would make it more highly reflective and thus be less emissive. However, doesn't polishing a surface merely means that radiation would reflect in a more uniform direction? Before polishing, the surface was still highly reflective as after but it was just scattering the radiation more diffusely in all directions. So shouldn't the amount of reflected energy be the same before and after polishing?

 

[Charles Link]

To just quickly answer one of the questions, the derivation of the Planck blackbody function $M(\lambda,T)$ is somewhat complicated=it took both Einstein and Planck along with Bose to get all of the details. It is also a rather complex integration (non-elementary), but it can be shown to give the precise result $\int\limits_{0}^{+\infty} M(\lambda,T) \, d \lambda=\sigma T^4$. It also gives the expession for $\sigma=\frac{\pi^2}{60} \frac{k_b^4}{\hbar^3 c^2}=5.6697 \, E-8 \, watts/(m^2 (degK)^4)$. I believe the Stefan-Boltzmann law came quite a number of years before they figured out the Planck blackbody function, and the Stefan-Boltzmann law is not used in the derivation.

 

[Charles Link]

Can't a material be partially transmissive and still be in thermal equilibrium so that its absorptivity = emissivity? For example, 2 materials A and B in a vacuum enclosed room all at the same temperature, material A being highly emissive and thus absorptive and B losing radiation energy through reflection and transmission. In order for A to not change in temperature it needs to emit equal radiation energy as it absorbs. In order for B to not change in temperature it needs to also emit (less) radiation energy as it absorbs. So that for material B: $E_{in} - E_{reflection} - E_{transmission} = E_{absorption} = E_{emission}$.
This is quite interesting to know. Does that also mean that the deriviation of $\sigma \cdot T^4$ gives the spectral curve function of the blackbody, $M(\lambda, T)$?

Two more questions (sorry for bombarding you guys with all this):

1. If a material is highly reflective and thus poorly emissive, does that mean that it would cool down much slower than a blackbody when they're both at the same temperature, both equally conductive/convective and both sitting in an environment with the same lower temperature?

2. I keep on reading that if you polish a surface it would make it more highly reflective and thus be less emissive. However, doesn't polishing a surface merely means that radiation would reflect in a more uniform direction? Before polishing, the surface was still highly reflective as after but it was just scattering the radiation more diffusely in all directions. So shouldn't the amount of reflected energy be the same before and after polishing?

Your conclusion in (1) I believe is completely correct. For (2), you can examine a diffuse surface on a microscopic scale: I picture it like a bunch of mountains that will have the light reflect back and forth between them before reflecting back out, so that the multiple bounces will increase the absorption.

 

[JohnnyGui]

you can examine a diffuse surface on a microscopic scale: I picture it like a bunch of mountains that will have the light reflect back and forth between them before reflecting back out, so that the multiple bounces will increase the absorption.

Ah, I kind of had this idea in my head but this verified it.

but it can be shown to give the precise result +∞∫0M(λ,T)dλ=σT4∫0+∞M(λ,T)dλ=σT4 \int\limits_{0}^{+\infty} M(\lambda,T) \, d \lambda=\sigma T^4 .

I had a small question regarding the units. Since $\sigma \cdot T^4$ doesn't contain the sufrace $A$ in it, doesn't that mean that this formula gives the total radiation intensity of the whole spectrum? I'm asking this because for some reason, this link says the following under paragraph 7.2.4:

"It has been shown that the irradiation field in an isothermal cavity is equal to $E_b= \sigma \cdot T^4$. Moreover, the irradiation was same for all planes of any orientation within the cavity. It may then be shown that the intensity of the blackbody radiation $I_b$ is uniform. Thus, blackbody radiation is defined as $E_b = \pi \cdot I_b$"

I find this remarkable because $E_b$ is already the intensity since its units are per unit area.

 

[Charles Link]

Ah, I kind of had this idea in my head but this verified it.
I had a small question regarding the units. Since $\sigma \cdot T^4$ doesn't contain the sufrace $A$ in it, doesn't that mean that this formula gives the total radiation intensity over whole spectrum?
I'm asking this because for some reason, this link says the following under paragraph 7.2.4:

"It has been shown that the irradiation field in an isothermal cavity is equal to $E_b= \sigma \cdot T^4$. Moreover, the irradiation was same for all planes of any orientation within the cavity. It may then be shown that the intensity of the blackbody radiation $I_b$ is uniform. Thus, blackbody radiation is defined as $E_b = \pi \cdot I_b$"

I find this remarkable because $E_b$ is already the intensity since its units are per unit area.

The source from the blackbody aperture radiates with a $cos(\theta)$ fall-off in intensity from-on axis ($\theta$ is the polar angle), because the aperture appears to change in size/area as it is viewed at an angle. Thereby the effective solid angle of a hemisphere for this radiator thereby turns out to be $\pi$ rather than $2 \pi$. Meanwhile, solid angle (steradian) is defined as $\Omega=\frac{A}{r^2}$. Because of this, the steradian is actually dimensionless, and intensity $I$ (watts/steradian) and power $P$ (watts) essentially have the same units. The topic is tricky enough that you will likely find that some articles give much more thorough and better explanations of some of the finer details than others. $\\$ Meanwhile the $M$ that I used, ($M=\sigma T^4$ ), is radiant emittance per unit area off of the surface: Power $P=M A$. You may also see another unit called radiance, ($L$ ), used, where $I=L A$ and $M=L \pi$. (I'm trying to keep it as simple as i can.) Notice also that $P=I_o \pi$ where $I_o$ is the on-axis intensity, and $I (\theta)=I_o cos(\theta)$. Suggestion is to compute $P=\int \int I(\theta) \, d \Omega$ where $d \Omega=sin(\theta) \, d \theta \, d \phi$ for a hemisphere. You should be able to show that $P=I_o \pi$. $\\$ One more unit you will likely see in the literature is the irradiance $E$ defined as power/area. $E=\frac{I}{r^2}$ (watts/m^2) for a point source that radiates spherically. One question people often have for this is how do the units work? If $I$ is in watts/steradian, where did the "steradian" go? ( $E$ is in watts/m^2 without any steradian.) And the answer is that the steradian is dimensionless and can be pulled out or put in, depending upon the quantity that is represented.

 

[JohnnyGui]

Thank you so much for this detailed explanation. I just noticed that I was confusing the amount of energy emitted from a unit area with the amount of energy received per unit area. The latter being the intensity and a function of distance. Which would need a $cos$ function since the distance from a source is dependent on the angle from the zenith.

Meanwhile the MM M that I used, (M=σT4M=σT4 M=\sigma T^4 ), is radiant emittance per unit area off of the surface: Power P=MAP=MA P=M A . You may also see another unit called radiance, (LL L ), used, where I=LAI=LA I=L A and M=LπM=Lπ M=L \pi . (I'm trying to keep it as simple as i can.) Notice also that P=IoπP=Ioπ P=I_o \pi where IoIo I_o is the on-axis intensity, and I(θ)=Iocos(θ)I(θ)=Iocos(θ) I (\theta)=I_o cos(\theta) . Suggestion is to compute P=∫∫I(θ)dΩP=∫∫I(θ)dΩ P=\int \int I(\theta) \, d \Omega where dΩ=sin(θ)dθdϕdΩ=sin(θ)dθdϕ d \Omega=sin(\theta) \, d \theta \, d \phi for a hemisphere. You should be able to show that P=IoπP=Ioπ P=I_o \pi

I'm taking this step by step. Reading this explanation, I pictured the following:

Let's say this is a hemisphere of an amount of radiant emittance $M$ of 1 unit area. Are you saying that the radiance $L$ is the amount of energy in the blue cone ($Ω$) that has a solid angle that fits $\pi$ times in the whole hemisphere? Such that the solid angle of 1 $L$ is 1 radian i.e. 57.3 degrees?

 

[Charles Link]

The reason for the quantity $L$, which is known as the radiance, also referred to as the "brightness", is that it is a quantity indicative of how "bright" the surface appears (in an intensive sense, independent of size or area), and this remains constant (for a blackbody or other radiator with constant emissivity) independent of viewing angle. $\\$ The units of $L$ are watts/(m^2 sr), and it has a couple of simple formulas, like $I_o$ (on axis) $=LA,$ and $I(\theta)$ (off axis) $=LA_{effective}$ where $A_{effective}=Acos(\theta)$. It is somewhat difficult to explain this quantity in detail, but the name "brightness" is a very good description of what it represents. $\\$ [Perhaps there is a somewhat simple way to explain this: Qualitatively, if you say something is a "bright" white, such as a white t-shirt in the sun, it wouldn't matter if the t-shirt is a small or an extra-large, it is the material itself that has the "brightness" feature. Meanwhile, if you put the t-shirt in the shade, the "brightness" level would decrease. That's what this radiance quantity $L$ represents.]... $\\$ $\\$ ...Back to some formulas: You can also write irradiance $E=\frac{I}{s^2}$ (watts/m^2), so that $E=\frac{LA_{eff}}{s^2}$. $\\$ $\\$ In a good deal of the literature, the Planck blackbody function is written as $L(\lambda, T)=\frac{2hc^2}{\lambda^5(exp^{hc/(\lambda k_b T)}-1)}$ without the factor of $\pi$ in the numerator, ( i.e. $M(\lambda, T)$ has $2 \pi h c^2$ in the numerator because $M(\lambda, T)=L(\lambda,T) \pi$ ). $\\$ The result is $\int\limits_{0}^{+\infty} L(\lambda,T) \, d \lambda=\frac{\sigma T^4}{\pi}$. $\\$ A very interesting as well as important part of the Planck blackbody function is that integrating the Planck function actually provides the theoretical expression for $\sigma$ in terms of the other constants: $\sigma=\frac{\pi^2}{60} \frac{ k_b^4}{\hbar^3 c^2}$. This lends much credence to its accuracy. (The integral is non-elementary, but some textbooks such as F.Reif's Statistical Physics show the complete evaluation of the integral.) $\\$ Additional item is from the Planck function, you can also derive Wien's law: $\lambda_{max}{T}=2898$ micron degK , where $\lambda_{max}$ is the wavelength where the peak of the spectrum (Planck blackbody function) occurs. (By taking $\frac{\partial{L(\lambda,T)}}{\partial{\lambda}}=0$, you can derive Wien's law. )

 

[Charles Link]

@JohnnyGui Be sure to see the edited additions at the bottom of post #81=a couple important features of the Planck blackbody function.

 

[JohnnyGui]

In a good deal of the literature, the Planck blackbody function is written as L(λ,T)=2hc2λ5(exphc/(λkbT)−1)L(λ,T)=2hc2λ5(exphc/(λkbT)−1) L(\lambda, T)=\frac{2hc^2}{\lambda^5(exp^{hc/(\lambda k_b T)}-1)} without the factor of ππ \pi in the numerator, ( i.e. M(λ,T)M(λ,T) M(\lambda, T) has 2πhc22πhc2 2 \pi h c^2 in the numerator because M(λ,T)=L(λ,T)πM(λ,T)=L(λ,T)π M(\lambda, T)=L(\lambda,T) \pi ). \\ The result is +∞∫0L(λ,T)dλ=σT4π∫0+∞L(λ,T)dλ=σT4π\int\limits_{0}^{+\infty} L(\lambda,T) \, d \lambda=\frac{\sigma T^4}{\pi} . \\ A very interesting as well as important part of the Planck blackbody function is that integrating the Planck function actually provides the theoretical expression for σσ \sigma in terms of the other constants: σ=π260k4bℏ3c2σ=π260kb4ℏ3c2 \sigma=\frac{\pi^2}{60} \frac{ k_b^4}{\hbar^3 c^2} . This lends much credence to its accuracy. (The integral is non-elementary, but some textbooks such as F.Reif's Statistical Physics show the complete evaluation of the integral.) \\ Additional item is from the Planck function, you can also derive Wien's law: λmaxT=2898λmaxT=2898 \lambda_{max}{T}=2898 micron degK , where λmaxλmax \lambda_{max} is the wavelength where the peak of the spectrum (Planck blackbody function) occurs. (By taking ∂L(λ,T)∂λ=0∂L(λ,T)∂λ=0 \frac{\partial{L(\lambda,T)}}{\partial{\lambda}}=0 , you can derive Wien's law. )

This is really interesting to know. And actually all makes sense to me. I indeed expected the derivatie of $\sigma \cdot T^4$ to be Planck's black body function, as I stated in my post #75. Glad I predicted this correctly.

Your explanation on deriving Wien's law is also intuitive. I think that writing down an expression to solve $\frac{dL (\lambda, T)}{d\lambda} = 0$ at a specific $T_1$ to get $\lambda_{max1}$ and then divide this expression by the same expression but for another temperature $T_2$ to get $\lambda_{max2}$ would eventually show that:
$$\frac{T_2}{T_1} = \frac{\lambda_{max1}}{\lambda_{max2}}$$
Cross multiplying these fractions together should give 2989/2989 = 1.

I still need a bit of googling to fully understand the relation between $M$ and $L$. What I got so far from your post #79 is the following:
- The hemisphere has a solid angle of $2\pi$ but since the $M$ differs with angle, the effective solid angle is $\pi$ (really interested how this is deduced).
- Therefore, the radiance $L$, which is a quantity per 1 unit solid angle, is deduced by divding $M$ by $\pi$ instead of $2\pi$

Are these statements correct so far?

 

[Charles Link]

You got it reasonably accurate, but a couple of corrections: $\\$ 1) $\int\limits_{0}^{+\infty} M(\lambda,T)=\sigma T^4$, but you can't take a derivative of $\sigma T^4$ to get $M(\lambda, T)$. (If the upper limit on the integral was $\lambda$ then you could take a derivative to get $M(\lambda, T)$, but in this case you can't.) $\\$ 2) To derive Wien's law, you take $\frac{\partial{L(\lambda, T)}}{\partial{\lambda}}=0$ and solve for $\lambda=\lambda_{max}$ where this occurs. Given $x=hc/(\lambda k_b T)$, you will get $e^{-x}=1-x/5$ upon taking the above partial derivative. The equation has two solutions: $x=0$ which is extraneous(essentially tells you the slope approaches zero as you go to very long wavelengths), and $x=4.96$ or thereabouts. The $x=4.96$ solution gives you Wien's law. $\\$ Additional item: To get the factor of $\pi$ as opposed to $2 \pi$ for a hemisphere, see the 2nd to the last sentence of post #79="Suggest you compute... " . Please give the computation a try=I think you will be pleased with the result.

 

[JohnnyGui]

You got it reasonably accurate, but a couple of corrections: $\\$ 1) $\int\limits_{0}^{+\infty} M(\lambda,T)=\sigma T^4$, but you can't take a derivative of $\sigma T^4$ to get $M(\lambda, T)$. (If the upper limit on the integral was $\lambda$ then you could take a derivative to get $M(\lambda, T)$, but in this case you can't.) $\\$ 2) To derive Wien's law, you take $\frac{\partial{L(\lambda, T)}}{\partial{\lambda}}=0$ and solve for $\lambda=\lambda_{max}$ where this occurs. Given $x=hc/(\lambda k_b T)$, you will get $e^{-x}=1-x/5$ upon taking the above partial derivative. The equation has two solutions: $x=0$ which is extraneous(essentially tells you the slope approaches zero as you go to very long wavelengths), and $x=4.96$ or thereabouts. The $x=4.96$ solution gives you Wien's law. $\\$ Additional item: To get the factor of $\pi$ as opposed to $2 \pi$ for a hemisphere, see the 2nd to the last sentence of post #79="Suggest you compute... " . Please give the computation a try=I think you will be pleased with the result.

Thanks for the feedback. Will try the integration, but I noticed something.

You said that the effective solid angle of the hemisphere is $\pi$. But doesn't a hemisphere always have a solid angle of $\pi$ anyway? Or did you mean that the effective area of the hemisphere contains $\pi$ steradians instead of $2 \pi$?. Such that the effective solid angle of the whole hemisphere is $\pi$ steradians × 0.572 radians (solid angle of 1 steradian) = 1,797 radians?

 

[Charles Link]

Thanks for the feedback. Will try the integration, but I noticed something.

You said that the effective solid angle of the hemisphere is $\pi$. But doesn't a hemisphere always have a solid angle of $\pi$ anyway? Or did you mean that the effective area of the hemisphere contains $\pi$ steradians instead of $2 \pi$?. Such that the effective solid angle of the whole hemisphere is $\pi$ steradians × 0.572 radians (solid angle of 1 steradian) = 1,797 radians?

A sphere has solid angle $\Omega=$(surface area)$/R^2=4 \pi R^2/R^2=4 \pi$ steradians, so that a hemisphere has $2 \pi$ steradians. A flat surface that radiates as $I(\theta)=I_o cos(\theta)$ puts out power $P=\int I(\theta) \, d \Omega=I_o \Omega_{effective}$ where $I_o$ is on-axis intensity and $\Omega_{effective}=\pi$ steradians. Do the integration and I think you will see that the $\Omega_{effective}$ is a useful concept.

 

[JohnnyGui]

A sphere has solid angle $\Omega=$(surface area)$/R^2=4 \pi R^2/R^2=4 \pi$ steradians, so that a hemisphere has $2 \pi$ steradians. A flat surface that radiates as $I(\theta)=I_o cos(\theta)$ puts out power $P=\int I(\theta) \, d \Omega=I_o \Omega_{effective}$ where $I_o$ is on-axis intensity and $\Omega_{effective}=\pi$ steradians. Do the integration and I think you will see that the $\Omega_{effective}$ is a useful concept.

Sorry, but I'm still at the part where I'm trying to understand why radiance decreases when deviating from the on-axis of the hemisphere in the first place. I have a feeling I'm missing something very obvious.
Here's a 2 dimensional schematic of what I think is the cause:

All surfaces $S, S2$ and $S3$ have the same length. They all have a surface that, when put on the radiation arc (orange), has a solid angle of 1 steradian. Surface $S$ is sitting on the radiation arc and thus has a solid angle of 1 steradian. However, when deviating from the radiation arc, like $S2$ and $S3$ shown in the picture, it shows that the more it goes away from the on-axis, the smaller the angle of radiation that it receives.

Is this the cause why radiance decreases when deviating from the on-axis or is it something else? Sorry for bothering you with all this but if this is incorrect, perhaps it's more handy for me to see in a schematic what causes the decrease in radiance when deviating from the on-axis.

 

[Charles Link]

Sorry, but I'm still at the part where I'm trying to understand why radiance decreases when deviating from the on-axis of the hemisphere in the first place.
Here's a 2 dimensional schematic of what I think is the cause:
View attachment 204285
All surfaces $S, S2$ and $S3$ have the same length. They all have a surface that, when put on the radiation arc (orange), has a solid angle of 1 steradian. Surface $S$ is sitting on the radiation arc and thus has a solid angle of 1 steradian. However, when deviating from the radiation arc, like $S2$ and $S3$ shown in the picture, it shows that the more it goes away from the on-axis, the smaller the angle of radiation that it receives.

Is this the cause why radiance decreases when deviating from the on-axis or is it something else? Sorry for bothering you with all this but if this is incorrect, perhaps it's more handy for me to see in a schematic what causes the decrease in radiance when deviating from the on-axis.

The radiance $L$ is independent of viewing angle and stays constant. When viewed from a distance $R$, ($R$ needs to be kept constant here for comparisons), at viewing position at location with angle $\theta$, a circular source of area $A$ will appear to be elliptical with area $A_{effectiive}= A cos(\theta )$. The source will appear to be shrunken in the direction that the viewer is located: e.g. If viewed from a location that has $y=0$, with $x^2+z^2=R^2$, the apparent width in the $y$ direction will be unaffected, but the apparent width in the x-direction, where $x=R cos(\theta)$ and $z=R sin(\theta)$ will be reduced by a factor $cos(\theta)$. If the source has radius $r$, the width in the y direction will be $r$, and the apparent width in the x direction will be $rcos(\theta)$. The area will appear to be $A_{effective}=A cos(\theta)$. $\\$ To precisely quantify the above, if you take the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, essentially it is a flattened circle when $b=r$, and $a=rcos(\theta)$. If you compute the area of the ellipse, you will find that it is $\pi r^2 cos(\theta)$. $\\$ Note: In the above, we are assuming $R$ is much greater than $r$, so that the viewing angle $\theta$ is constant across the source. $\\$ Additional item: The intensity $I=L A_{effective}$. Notice if $A_{effective}=A cos(\theta)$, that $I(\theta)=LAcos(\theta)=I_o cos(\theta)$. A source that obeys this $cos(\theta)$ fall-off in intensity, (which is very much expected), is said to be "Lambertian", named after Johann Heinrich Lambert. $\\$ For an additional problem, we could explore what happens as the viewer moves from S to S2 and to S3 in your figure above. You might find it of interest that the irradiance $E$ per unit area incident on the surface falls off as $cos^4(\theta)$.

 

[JohnnyGui]

The radiance LL L is independent of viewing angle and stays constant. When viewed from a distance RR R , (RR R needs to be kept constant here for comparisons), at viewing position at location with angle θθ \theta , a circular source of area AA A will appear to be elliptical with area Aeffectiive=Acos(θ)Aeffectiive=Acos(θ) A_{effectiive}= A cos(\theta ) . The source will appear to be shrunken in the direction that the viewer is located: e.g. If viewed from a location that has y=0y=0 y=0 , with x2+z2=R2x2+z2=R2 x^2+z^2=R^2 , the apparent width in the yy y direction will be unaffected, but the apparent width in the x-direction, where x=Rcos(θ)x=Rcos(θ) x=R cos(\theta) and z=Rsin(θ)z=Rsin(θ) z=R sin(\theta) will be reduced by a factor cos(θ)cos(θ) cos(\theta) . If the source has radius rr r , the width in the y direction will be rr r , and the apparent width in the x direction will be rcos(θ)rcos(θ) rcos(\theta) . The area will appear to be Aeffective=Acos(θ)Aeffective=Acos(θ) A_{effective}=A cos(\theta) .

Ah, so this is all done according to the observer? While in reality, the "objective" radiance is the same at every angle?

 

[Charles Link]

Yes. The radiance $L$ is the surface "brightness. You might find it of interest, with an optical pyrometer to determine the temperature of a blackbody, a wire is heated (so that it glows orange)=and you can adjust the temperature of the wire=higher temperature is brighter orange=typical temperature range for an optical pyrometer is about 800-2000 degrees Centigrade. Anyway, you look through the viewer at your blackbody whose temperature you are measuring. You turn the dial until the heated wire blends in with the blackbody source=the brightness matches, so that the wire and the blackbody are at the "same temperature". (Editing... And this needs to be qualified slightly=see the "Additional question..." at the end). Their brightness level $L$ is matched. You then read the calibrated dial on the optical pyrometer which tells you the temperature of the blackbody. $\\$ Additional question you may have is, is the emissivity of the wire=1.0 (or thereabouts)?, and I think the answer, in general, is no=the dial of the optical pyrometer is calibrated by initially using a thermocouple (or other means) to measure the blackbody surface temperature.