Emission spectra of different materials

[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.

 

[JohnnyGui]

Yes. The radiance LL L is the surface "brightness.

Let's say an observer has moved in the x and y-axis (2 dimensional) to keep the same distance $R$ to the source. Regarding the projected area, is this schematic correct?

If this is correct, then I have 2 questions:

1. How is $x = cos(*) \cdot R$ while my schematic shows that it's $x = sin(*) \cdot R$? Perhaps I considered $x$ as your $z$ coordinate?

2. For some reason my instinct says that the size of a length when viewed from an angle should be $B$ in the below picture instead of $A$. Don't know if there is any way to explain why it's $A$ and not $B$. EDIT: Hold on, I have a hunch it's the same thing, but it looks as if the projected lengths are different.

Btw; The extra info you always add to your posts is very interesting to read. I'm not ignoring them, it's just that I'm trying to understand this first :)

 

[Charles Link]

From a previous post (#88) where I discussed the projected area of the source, your y here is that z, and y is the dimension into the paper. The source will appear to be shrunk in the x-direction so that, instead of being circular with radius $r$, it will be appear to be elliptical, and in the x direction it will appear to have a half-width of $a=rcos(\theta)$. (In the y direction it will be unaffected and appear to have a half -width of $b=r$, which is exactly what it has.) $\\$ Incidentally $z=Rcos(\theta)$ and $\frac{\sqrt{x^2+y^2}}{R}=sin(\theta)$, so you do have it correct in your figure that $x=Rsin(\theta)$, but that is really irrelevant in computing the apparent dimensions of the source as seen from angle $\theta$. $\\$ As previously mentioned, the location $R>>r$, so that the viewing angle $\theta$ is constant across the entire source. $\\$ Suggestion: Take a small circular object (like a penny) and view it from an angle off-axis. Does it look elliptical in appearance?

 

[JohnnyGui]

From a previous post (#88) where I discussed the projected area of the source, your y here is that z, and y is the dimension into the paper. The source will appear to be shrunk in the x-direction so that, instead of being circular with radius $r$, it will be appear to be elliptical, and in the x direction it will appear to have a half-width of $a=rcos(\theta)$. (In the y direction it will be unaffected and appear to have a half -width of $b=r$, which is exactly what it has.) $\\$ Incidentally $z=Rcos(\theta)$ and $\frac{\sqrt{x^2+y^2}}{R}=sin(\theta)$, so you do have it correct in your figure that $x=Rsin(\theta)$, but that is really irrelevant in computing the apparent dimensions of the source as seen from angle $\theta$. $\\$ As previously mentioned, the location $R>>r$, so that the viewing angle $\theta$ is constant across the entire source. $\\$ Suggestion: Take a small circular object (like a penny) and view it from an angle off-axis. Does it look elliptical in appearance?

Thanks, I understand now the influence of viewing angle on how the surface of the radiating source appears to be smaller, but the following stumped me:

If a blackbody with a true surface of A is sending an X amount of energy to an observer who views the blackbody at an angle, and the observer sees the blackbody having a smaller apparent surface from which he receives that X amount of energy. Shouldn't he think that it radiates MORE energy per surface unit and thus also per steradian than when he's observing it from the normal?

 

[Charles Link]

Thanks, I understand now the influence of viewing angle on how the surface of the radiating source appears to be smaller, but the following stumped me:

If a blackbody with a true surface of A is sending an X amount of energy to an observer who views the blackbody at an angle, and the observer sees the blackbody having a smaller apparent surface from which he receives that X amount of energy. Shouldn't he think that it radiates MORE energy per surface unit and thus also per steradian than when he's observing it from the normal?

The blackbody will have more energy per steradian and thereby higher intensity $I(\theta)$ when viewed from the normal because intensity $I(\theta)=LA_{effective}$. Perhaps one way of looking at the concept of radiance or brightness $L$ is to bring along your own aperture when you measure the brightness, and have that aperture be smaller than the area of the blackbody source. Place it in front of the source and view the appearance. It is not necessary to focus on the source which may be a couple inches behind the aperture. Instead. you focus on your own aperture. Sources that have the same brightness $L$ will be indistinguishable. You also will not be able to tell whether you are viewing the blackbody from on-axis or from off-axis. (Be sure and have the normal to your own aperture always pointing at you so that its area doesn't change (and always keep the distance between you and your own aperture constant.) $\\$ Alternatively, instead of viewing the source by focusing on your own aperture, you could measure the power reaching a detector a specified distance $R$ from your own aperture. Again, you could not distinguish whether you are observing from on-axis or off-axis. The power received will be the same. (Again, always have the normal to your own aperture pointing in the direction of the detector.)

 

[JohnnyGui]

Alternatively, instead of viewing the source by focusing on your own aperture, you could measure the power reaching a detector a specified distance RR R from your own aperture. Again, you could not distinguish whether you are observing from on-axis or off-axis. The power received will be the same. (Again, always have the normal to your own aperture pointing in the direction of the detector.)

Exactly, so the power will be the same from all angles. So, let's say the observer is watching the source form an angle and measures the power, he'd say that the intensity is $P$ / $A_{effective}$. Now, the observer moves towards the normal of the surface (he doesn't know he does, but he just moves in that direction) and then looks at the source. He'll still measure the same power (as you said) but he'll also see a larger surface $A$ since he's on-axis. He'll therefore calculate a smaller intensity since $P$ is the same but the surface is larger from the on-axis view.

What am I missing here?

 

[Charles Link]

Exactly, so the power will be the same from all angles. So, let's say the observer is watching the source form an angle and measures the power, he'd say that the intensity is $P$ / $A_{effective}$. Now, the observer moves towards the normal of the surface (he doesn't know he does, but he just moves in that direction) and then looks at the source. He'll still measure the same power (as you said) but he'll also see a larger surface $A$ since he's on-axis. He'll therefore calculate a smaller intensity since $P$ is the same but the surface is larger from the on-axis view.

What am I missing here?

This is why you bring your own aperture=the area stays the same. It blocks the parts of the source whose line of sight is outside your own aperture. Right now the purpose is to measure the brightness $L$ and nothing else. To measure the intensity $I$ you need to be able to see the entire source. $\\$ (Intensity is perhaps a poor name for the quantity, because it does not refer to the intensity of the surface. Intensity $I$ refers to how much power is radiated from the whole surface per unit solid angle. It really is not "intensity" in the literal sense. It also does not refer to how intense it feels, or how much reaches a detector. That instead is given by the irradiance $E$, which is sometimes very loosely(and incorrectly) referred to as the intensity.) $\\$ To summarize something quickly, there are four radiometric quantities: $L$, $I$ , $M$, and $E$, and each of these is well defined, and represents something very specific. $I$ is called intensity, (perhaps a poor name for it, but that's what they call it.) $M$ refers to power coming off of a surface per unit area, and $E$ refers to the power per unit area across a surface or onto a surface. Of the four, $E$ and $M$ are the most closely related. (Occasionally, in some of the older literature these four quantities are referred to with the letters $N$, $J$, $W$, and $H$ respectively. In any case, the letters $L$, $I$, $M$, and $E$ are now quite standard for these quantities.) $\\$ Additional comment: These quantities are all quite useful in doing calculations such as the one with the lamp filament in the OP. It is good to use the standard terminology in the calculations, so it makes for easy reading by others.

 

[JohnnyGui]

Thanks, I think I'm starting to get it now. You've probably been trying to say this the whole time but after reading some sources, I now understand that it's about the solid angle in the observer's sphere himself, not in the sphere of the emitting source. After all, $L$ is about what the observer preceives.

Please bear with me here as I try to punch this in my head. Here's the observer again who has just move in the $x$ direction again w.r.t. the surface $A$.

As I see it, the $cos(\theta) \cdot A$ is approximately the same as its projection on the observer's sphere only in the case if the observer is very far away ($R$ is high) or if the surface $A$ is very small.

That being said, when an area of $cos(\theta) \cdot A$ is on the observer's sphere, it has a certain 3D angle $\alpha$ in the observer's sphere. From what I know, you can calculate $\alpha$ by rearranging the following formula: $cos(\theta) \cdot A = 2 \pi \cdot R^2 (1 - cos(\frac{\alpha}{2}))$

That aside, the surface $cos(\theta) \cdot A$ is part of the surface of the observer's sphere. The amount of steradians that this surface $cos(\theta) \cdot A$ contains is calculated by $\frac{cos(\theta) \cdot A}{R^2}$.
So to get the amount of energy per steradian in the observer's sphere, I'd have to divide the energy that the observer preceives by that factor. This would eventually give the amount of energy that the observer perceives per steradian (but not necessarly per steradian per m²). So:
$$\frac{E_{observed} \cdot R^2}{cos(\theta) \cdot A}=L$$
Two questions:
1. Does this all make sense? If not, where exactly did I go wrong?
2. I'm probably having a huge blackout at the moment because I'm not sure how $E_{observerd}$ is calculated first nor what quantity this is. It's the energy that the source is emitting to the eye of the observer but with a certain spatial angle of the observer's sphere himself, not of the sphere of the source. How is this energy calculated?

(Intensity is perhaps a poor name for the quantity, because it does not refer to the intensity of the surface. Intensity II I refers to how much power is radiated from the whole surface per unit solid angle. It really is not "intensity" in the literal sense. It also does not refer to how intense it feels, or how much reaches a detector. That instead is given by the irradiance EE E , which is sometimes very loosely(and incorrectly) referred to as the intensity.)

I totally agree, and that is what confused me in the first place. So the energy that a radiating surface is putting out per steradian of its own hemisphere is the Radiant Intensity, right? However, isn't Irradiance the amount of energy that the observer perceives per $m^2$ of his own surface and not per steradian of his own sphere?

 

[Charles Link]

@JohnnyGui I think you are starting to get a good handle on it. If I'm not mistaken, the brightness $L$ can be computed either from looking from the surface and how it radiates (per square m^2 per steradian), or alternatively, from the observers' reference and computing the solid angle that the source subtends,( using the irradiance $E$ at the observer), so that your calculation is correct. Keep up the good work !
$\\$ Just an additional comment: The irradiance $E_{observed}$ is often measured with a photodetector. Typically the response is linear, and the voltage from the photodetector (usually coupled to a current to voltage amplifier) is proportional to the irradiance. Typically, a photodetector can be calibrated using a calibration source of known irradiance $E_{cal}$ at a specific distance. Then $\frac{E_{observed}}{E_{cal}}=\frac{V_{observed}}{V_{cal}}$ where $V_{cal}$ is the voltage that the photodector gave for the calibration source. This allows you to compute $E_{observed}$ from the voltage $V_{observed}$ of the photodetector.

 

[JohnnyGui]

Glad I'm on the right track.

Just an additional comment: The irradiance EobservedEobserved E_{observed} is often measured with a photodetector.

But isn't irradiance the amount of energy the observer receives but per m² of the radiating surface of $cos(\theta) \cdot A$? Since my mentioned $E_{observed}$ is the energy received by the observer by the whole surface $cos(\theta) \cdot A$, I think the relationship should be:
$$\frac{E_{observed}}{cos(\theta) \cdot A} = Irradiance$$

 

[Charles Link]

Glad I'm on the right track.
But isn't irradiance the amount of energy the observer receives but per m² of the radiating surface of $cos(\theta) \cdot A$? Since my mentioned $E_{observed}$ is the energy received by the observer by the whole surface $cos(\theta) \cdot A$, I think the relationship should be:
$$\frac{E_{observed}}{cos(\theta) \cdot A} = Irradiance$$

When the letter $E$ is used in radiometrics, like $E_{observed}$, it stands for irradiance (watts/m^2). (In other contexts it can represent an electric field, or it can represent energy, but in the present context, it represents irradiance (watts/m^2)). The letter $P$ is used for power (watts). The brightness $L=E_{observed}/\frac{Acos(\theta)}{R^2} =E_{observed}R^2/(A cos(\theta))$, where $E_{observed}=P_{observed}/A_{detector}$, but normally you can do a measurement of $E_{observed}$ as described in my previous post, and you don't need to know $A_{detector}$.

 

[JohnnyGui]

When the letter $E$ is used in radiometrics, like $E_{observed}$, it stands for irradiance (watts/m^2). (In other contexts it can represent an electric field, or it can represent energy, but in the present context, it represents irradiance (watts/m^2)). The letter $P$ is used for power (watts). The brightness $L=E_{observed}/\frac{Acos(\theta)}{R^2} =E_{observed}R^2/(A cos(\theta))$, where $E_{observed}=P_{observed}/A_{detector}$, but normally you can do a measurement of $E_{observed}$ as described in my previous post, and you don't need to know $A_{detector}$.

I noticed something but I'm not sure if this is correct. If the observer is a point, then one cannot speak of irradiance since that would require a receiving surface, not a point. One can only speak of the received energy per unit solid angle in the sphere of the point observer.
Furthermore, to speak of the amount of energy the emitted surface emits per m² of that emitting surface according to that point observer (received energy / ($cos(\theta) \cdot A$) is also not possible because this amount differs with distance from the emitting source.

However, if you consider the receiver as having a surface as well, just like the emitting source, then one would get the following scenario:

I think that in this case, one cannot speak of any quantity that has a solid angle unit in it (for example like radiant intensity) because they're not points. Except if you try to extrapolate the lines further up until they intersect. But to know the radius of that circle, one would have to know the angle at which the rays pass by the sides of the receiving surface. Since the detector you mentioned has a surface, what it does is measure the amount of energy received, by using the proportionality with the calibrated voltage like you said, and divide that amount of energy by it's own detector surface to get the irradiance in W/m². So only if the receiver is considered a surface can one speak of irradiance.

Does my reasoning make any sense?

 

[Charles Link]

I noticed something but I'm not sure if this is correct. If the observer is a point, then one cannot speak of irradiance since that would require a receiving surface, not a point. One can only speak of the received energy per unit solid angle in the sphere of the point observer.
Furthermore, to speak of the amount of energy the emitted surface emits per m² of that emitting surface according to that point observer (received energy / ($cos(\theta) \cdot A$) is also not possible because this amount differs with distance from the emitting source.

However, if you consider the receiver as having a surface as well, just like the emitting source, then one would get the following scenario:
View attachment 204547

I think that in this case, one cannot speak of any quantity that has a solid angle unit in it (for example like radiant intensity) because they're not points. Except if you try to extrapolate the lines further up until they intersect. But to know the radius of that circle, one would have to know the angle at which the rays pass by the sides of the receiving surface. Since the detector you mentioned has a surface, what it does is measure the amount of energy received, by using the proportionality with the calibrated voltage like you said, and divide that amount of energy by it's own detector surface to get the irradiance in W/m². So only if the receiver is considered a surface can one speak of irradiance.

Does my reasoning make any sense?

I think you are starting to get a good idea of all of the concepts, but it might be good to practice with a couple of calculations. Here is one that I think you could solve: $\\$ A source consists of a flat circular blackbody surface of temperature $T=1000 K$ that has area $A=.001 m^2$ and is observed at a location $\theta=60$ degrees off-axis and at a distance $R=1.0$ m from the blackbody. The receiver/detector has area $A_{detector}= .0001$ m^2, and its normal points at the center of the blackbody. Compute the radiant emittance (per unit area ) $M$ (watts/m^2) of the source (over the whole spectrum), the radiance $L$ (watts/(m^2 sr)) of the source, the intensity $I$ (watts/sr) of the source, both on-axis, and at angle $\theta=60$ degrees, and the irradiance $E$ (watts/m^2) at the detector. Finally compute the power $P_d$ incident on the detector. Also compute the total power $P$ radiated by the source. If you can successfully compute these quantities, you have a good start at understanding the different terms that arise in this type of calculation.

 

[JohnnyGui]

I think you are starting to get a good idea of all of the concepts, but it might be good to practice with a couple of calculations. Here is one that I think you could solve: \\ A source consists of a flat circular blackbody surface of temperature T=1000KT=1000K T=1000 K that has area A=.001m2A=.001m2 A=.001 m^2 and is observed at a location θ=60θ=60 \theta=60 degrees off-axis and at a distance R=1.0R=1.0 R=1.0 m from the blackbody. The receiver/detector has area Adetector=.0001Adetector=.0001 A_{detector}= .0001 m^2, and its normal points at the center of the blackbody. Compute the radiant emittance (per unit area ) MM M (watts/m^2) of the source (over the whole spectrum), the radiance LL L (watts/(m^2 sr)) of the source, the intensity II I (watts/sr) of the source, both on-axis, and at angle θ=60θ=60 \theta=60 degrees, and the irradiance EE E (watts/m^2) at the detector. Finally compute the power PdPd P_d incident on the detector. Also compute the total power PP P radiated by the source. If you can successfully compute these quantities, you have a good start at understanding the different terms that arise in this type of calculation.

Thanks for the exercise!

- The radiant emittance would be $M = 1000^4\cdot \sigma = 56703,73 W/m^2$ and the total power $P$ by the source $56703,73 \cdot 0.001 = 56,70 W$.

- The radiance of the source is $L = \frac{M}{\pi}$ and therefore $\frac{56703,73}{\pi} = 18049,36 W/(m^2 \cdot sr)$ and its intensity on-axis would be $I = LA = 18049,36 \cdot 0.001 = 18,05 W/sr$ while the off-axis intensity at $\theta = 60$ would be $18,05 \cdot cos(60) = 10,61 W/sr$.

- Regarding the power $P_d$ incident on the detector, if the source emits 10,61 W per steradian (1² m²) at 60 degrees towards the detector, and the detector has a surface of 0.0001 m², then I'd think that it will receive $10,61 W \cdot 0.0001 = 0.001061 W$.
The irradiance would then be $P_d / A_{detector} = 0.001061 / 0.0001 = 10,61 W/m^2$ which is the same as the radiant intensity since $R= 1$.

Is this correct?

 

[Charles Link]

Thanks for the exercise!

- The radiant emittance would be $M = 1000^4\cdot \sigma = 56703,73 W/m^2$ and the total power $P$ by the source $56703,73 \cdot 0.001 = 56,70 W$.

- The radiance of the source is $L = \frac{M}{\pi}$ and therefore $\frac{56703,73}{\pi} = 18049,36 W/(m^2 \cdot sr)$ and its intensity on-axis would be $I = LA = 18049,36 \cdot 0.001 = 18,05 W/sr$ while the off-axis intensity at $\theta = 60$ would be $18,05 \cdot cos(60) = 10,61 W/sr$.

- Regarding the power $P_d$ incident on the detector, if the source emits 10,61 W per steradian (1² m²) at 60 degrees towards the detector, and the detector has a surface of 0.0001 m², then I'd think that it will receive $10,61 W \cdot 0.0001 = 0.001061 W$.
The irradiance would then be $P_d / A_{detector} = 0.001061 / 0.0001 = 10,61 W/m^2$ which is the same as the radiant intensity since $R= 1$.

Is this correct?

Very good. I have one small correction: cos(60)=.500 (and not .573 ). One other comment is that you should compute the irradiance $E_d$ at the detector before you compute the power onto the detector. $E_d=\frac{I}{R^2}$ and then $P_d=E_d A_d$.

 

[JohnnyGui]

Very good. I have one small correction: cos(60)=.500 (and not .573 ). One other comment is that you should compute the irradiance $E_d$ at the detector before you compute the power onto the detector. $E_d=\frac{I}{R^2}$ and then $P_d=E_d A_d$.

Great. One question though; we have now calculated the radiant intensity of an emitting source because its surface is so small that it can be considered a point source. But how is radiant intensity then calculated when the emitting source has a much larger surface and is therefore emitting multiple "hemispheres" (each dA having 1 hemisphere)? Is the radiant intensity then equal to the $I$ per dA multiplied by $\frac{A}{dA}$?