I Emission spectra of different materials

[Charles Link]

The spectral radiance function $L(\lambda, T )$ is the one that is most commonly presented in the literature. If you find that $M(\lambda, T)$ is the function that you prefer to work with, in presentations and/or for your own use, all you need to do is multiply all of the $L(\lambda, T)$ results by $\pi$. $\\$ For your second question above, the discussion concerns the case where surface $A_2$ encloses surface $A_1$. $\\$ And for your 3rd comment, I'm glad that you find the Planck spectral function of much interest. The Planck function IMO is one of the more important successes of Quantum Mechanics, and it adds a very important detail to the much simpler radiant emittance $M=\sigma T^4$. Knowing the spectral content is important in determining how much visible light that a source is generating. :) $\\$ In studying the general behavior of the Planck function, that the area under the curve increases dramatically with temperature and is $\frac{\sigma T^4}{\pi}$, be sure to also learn Wien's law (which can be derived by taking the derivative of $L(\lambda, T)$ w.r.t. to $\lambda$ and setting it equal to zero), that $\lambda_{max}T= 2.898 E+6$ nm K, where $\lambda_{max}$ is the wavelength where $L(\lambda,T )$ has its peak. At $T=6000$ K, (the approximate temperature of the outside surface of the sun), the peak of $L(\lambda, T)$ occurs at $\lambda_{max}=500$ nm (approximately).$\\$ By Wien's law, the peak goes to shorter wavelengths as the temperature increases. In general, almost exactly 25% of the energy always lies to the left of the peak. (An in-depth analysis with a fair amount of computing shows the fraction to the left of the peak is not exactly .25, but more precisely .25005...). In any case this fraction is very nearly 25%, independent of temperature. $\\$ Another result that is of interest is that the value of $L(\lambda,T)$ at its peak, $L(\lambda_{max}, T )$, is found to be proportional to the 5th power of the temperature.

 

[Charles Link]

@JohnnyGui Please read post #201, but there is one somewhat simple additional calculation that I think you would find worthwhile involving blackbodies: $\\$ If you treat the sun as a blackbody at temperature $T_s=6000$ K, and compute the total power radiated using the radius of the sun $R_s=7.0 E+8$ m, you can compute the approximate average temperature of the Earth $T_e$ if you assume the Earth to be a blackbody that absorbs all of the energy from the sun that it intercepts, and assuming it radiates as a blackbody (at the temperature $T_e$ that we are calculating) over its entire surface, and that a dynamic equilibrium has been reached so the radiated power of the Earth is the same as the power that it receives from the sun. The radius of the Earth is $R_e=6.4 E+6$ m, and the distance from the sun to the Earth $s_{se}=1.50 E+11$ m. $\\$ The calculation simply uses $M=\sigma T^4$ and does not use the Planck function as part of this computation. $\\$ Perhaps one of the trickier, but still simple, parts of the calculation is to observe that the Earth as seen from the sun appears to be a circle of area $\pi R_e^2$, so that the fraction of power from the sun that it intercepts is $\frac{\pi R_e^2}{4 \pi s_{se}^2}$. Anyway, I thought you might find this calculation of interest.

 

[JohnnyGui]

If you treat the sun as a blackbody at temperature Ts=6000Ts=6000 T_s=6000 K, and compute the total power radiated using the radius of the sun Rs=7.0E+8Rs=7.0E+8 R_s=7.0 E+8 m, you can compute the approximate average temperature of the Earth TeTe T_e if you assume the Earth to be a blackbody that absorbs all of the energy from the sun that it intercepts

Thanks for the exercise! I read that first sentence of your post and decided to reason the calculation by myself before reading the rest of it. I'm happy to say that I reasoned that the receiving surface of the Earth can indeed be estimated as $\pi R_E^2$ and that it will radiate from a surface of $4 \pi R_E^2$.
I'd think that the power that the Earth would receive from the sun is equal to:
$$T_{sun}^4 \cdot \sigma \cdot R_{sun}^2 \cdot \frac{R_e^2 \cdot \pi}{1AU^2} = P_{incident}$$
Where AU is the astronomical unit in metres. This gives me a $P_{incident}$ of around $2.07048 \cdot 10^{17} W$.

I'm not sure about the next part but here it goes. If there's a thermal equilibrium then I'd reason that $2.07048 \cdot 10^{17} = T_e^4 \cdot \sigma \cdot 4\pi R_e^2$. This gives $T_E = 290.22K$. If this is correct, then I'll be appreciating the world's greenhouse effect a lot more from now on.

By Wien's law, the peak goes to shorter wavelengths as the temperature increases. In general, almost exactly 25% of the energy always lies to the left of the peak. (An in-depth analysis with a fair amount of computing shows the fraction to the left of the peak is not exactly .25, but more precisely .25005...). In any case this fraction is very nearly 25%, independent of temperature. \\ Another result that is of interest is that the value of L(λ,T)L(λ,T) L(\lambda,T) at its peak, L(λmax,T)L(λmax,T) L(\lambda_{max}, T ) , is found to be proportional to the 5th power of the temperature.

I actually didn't know that it has more or less constant energy distribution like that. Quite interesting as all of your explanations!

The spectral radiance function L(λ,T)L(λ,T) L(\lambda, T ) is the one that is most commonly presented in the literature. If you find that M(λ,T)M(λ,T) M(\lambda, T) is the function that you prefer to work with, in presentations and/or for your own use, all you need to do is multiply all of the L(λ,T)L(λ,T) L(\lambda, T) results by ππ \pi .

Apologies, I've noticed you've explained this several times to me but I didn't pay attention to that. However, I'd think that to be able to draw a $L(\lambda, T)$ in the first place, one must have derived this from a $M (\lambda, T)$ curve since it's quite hard to measure the spectral curve coming from 1 $dA$ at one steradian. So I'd reason that the $L(\lambda,T)$ must be deduced from a $M(\lambda, T)$ curve and not the other way around, right?

PS: I'll be asking something in a following post about the example used to derive Planck's function, since this post is quite long for now ;). I'll go read your PM.

 

[Charles Link]

In general $L=\frac{M}{\pi}$. The same holds for $L(\lambda, T)=\frac{M(\lambda, T)}{\pi}$. $\\$ And yes, very good, I believe you computed $T_e$ correctly ! (It might interest you that the factor $\sigma R_e^2$ actually cancels on both sides of the equation, so that you can do much algebraic reduction before finally plugging in the numbers).

 

[JohnnyGui]

And yes, very good, I believe you computed TeTe T_e correctly ! (It might interest you that the factor σR2eσRe2 \sigma R_e^2 actually cancels on both sides of the equation, so that you can do much algebraic reduction before finally plugging in the numbers).

Wow, I rushed too fast there. I noticed the whole formula can be reduced to $\frac{T_s^4 \cdot R_s^2}{1 AU^2} = 4T_e$.

Regarding Planck's function, I noticed in the wiki that an example of a hollow cube containing EM waves is used for the derivation. Very simply put, it says that such a cube in thermal equilibrium would contain standing EM waves propagating in differerent directions inside the cube, as long as the distance between the endpoints that they propogate in can fit an integer amount of $0.5 \lambda$ of the concerning standing wave.
Furthermore, it says that each standing wave in the cube has an amount of states depending on the frequency. The energy of each state and that the energy that each state determines the likelihood of that state being present in the cube or not.
What boggles my mind is, how can an example of a hollow cube with standing waves inside be used as an analogue for a radiating material/atom/electron? Does a radiating source also contain standing waves when it's in thermal equilibrium, such that its spectral curve only contain wavelengths of which half of their wavelengths can fit an integer amount of times in the electrons of that source?

 

[Charles Link]

To answer your question, they drill a small hole in the hollow cube and examine what emerges. Inside of the cavity, in a particle description, there are N photons bouncing around in the volume $V$ (with a distribution of wavelengths), and each has velocity $c$. There is a result from chemistry in studiyng the effusion of gases that the number of particles emerging per unit area per unit time from the hole in the box will be $R=\frac{N \bar{v}}{4 V}$ where $\bar{v}$ is the average speed of the particles. In this case $\bar{v}=c$. $\\$ One other thing that is used in the derivation is the emissivity=absorption law. If we put a small hole in the hollow box and send a light beam into the hole, it will bounce around a lot (if the walls are somewhat reflective inside), but if the hole is small enough, the beam will never find its way back out. Therefore, we conclude for a small aperture with a hollow box that this thing behaves like a blackbody in whatever radiates out the hole with emissivity $\epsilon=1$, independent of the material of the walls inside of our box. $\\$ Meanwhile, the energy density inside of the box and number of photons is computed by counting modes, just as you mentioned. There is one other factor in this calculation, and that is the average occupancy number of a mode of photon energy $E_p$. That is given by the Bose factor: $\bar{n}=\frac{1}{\exp(\frac{E_p}{k_b T})-1}$. There is also a factor of 2 for polarization. The combination of knowing the mode density= how many photon modes there are per energy interval or wavelength interval, along with how many photons on the average are in each mode allows us to compute the total number of photons as well as their spectral distribution. This tells what the energy and photon count are inside the volume $V$. The effusion formula $R=\frac{Nc}{4V}$ is then used to calculate how many photons come out of the hole per unit area per unit time, and this number is multiplied by the energy of each photon $E_p=\frac{hc}{\lambda}$ to compute the spectral energy that comes out of the hole per unit area per unit time. In this method of calculation, we actually computed $M(\lambda,T)$ since we computed everything that got out over the hemisphere with our $R=\frac{N \bar{v}}{4V}$ formula. $\\$ (And one minor correction: Your $T_e$ above should read $T_e^4$).

 

[JohnnyGui]

To answer your question, they drill a small hole in the hollow cube and examine what emerges. Inside of the cavity, in a particle description, there are N photons bouncing around in the volume $V$ (with a distribution of wavelengths), and each has velocity $c$. There is a result from chemistry in studiyng the effusion of gases that the number of particles emerging per unit area per unit time from the hole in the box will be $R=\frac{N \bar{v}}{4 V}$ where $\bar{v}$ is the average speed of the particles. In this case $\bar{v}=c$. $\\$ One other thing that is used in the derivation is the emissivity=absorption law. If we put a small hole in the hollow box and send a light beam into the hole, it will bounce around a lot (if the walls are somewhat reflective inside), but if the hole is small enough, the beam will never find its way back out. Therefore, we conclude for a small aperture with a hollow box that this thing behaves like a blackbody in whatever radiates out the hole with emissivity $\epsilon=1$, independent of the material of the walls inside of our box. $\\$ Meanwhile, the energy density inside of the box and number of photons is computed by counting modes, just as you mentioned. There is one other factor in this calculation, and that is the average occupancy number of a mode of photon energy $E_p$. That is given by the Bose factor: $\bar{n}=\frac{1}{\exp(\frac{E_p}{k_b T})-1}$. There is also a factor of 2 for polarization. The combination of knowing the mode density= how many photon modes there are per energy interval or wavelength interval, along with how many photons on the average are in each mode allows us to compute the total number of photons as well as their spectral distribution. This tells what the energy and photon count are inside the volume $V$. The effusion formula $R=\frac{Nc}{4V}$ is then used to calculate how many photons come out of the hole per unit area per unit time, and this number is multiplied by the energy of each photon $E_p=\frac{hc}{\lambda}$ to compute the spectral energy that comes out of the hole per unit area per unit time. In this method of calculation, we actually computed $M(\lambda,T)$ since we computed everything that got out over the hemisphere with our $R=\frac{N \bar{v}}{4V}$ formula. $\\$ (And one minor correction: Your $T_e$ above should read $T_e^4$).

Thanks for the detailed explanation! What I noticed is that the amount of modes that a certain frequency can have is equal to $\frac{8\pi f^2}{c^3}$ where $f$ is the frequency. This formula is explained by saying that the higher a frequency is, the more “ways” it has to fit itself in a hollow cube so that it has more modes. I find it hard to bring this analogue to a particle, because in “what” exactly does a frequency fit itself into have more modes in case of a radiating particle? In the particle itself?

Furthermore, I deduced 2 conclusions from Planck’s function but I’m not sure if these are correct:

1. Looking at a spectrum curve, you’ll see on the right side that as the frequency gets higher, it will contain less and less energy. This is because, with each higher frequency, the average amount of energy per mode declines stronger than the increasing amount of possible modes per frequency. On the left side you’ll see that as frequency gets lower, it will contain less energy as well. This is because, as frequency gets lower, each mode will reach the maximum fixed energy of $kT$ (photons with lower $hf$ energy fit better in an energy of $kT$). Since the amount of possible modes decreases with lower frequencies while each mode has a more or less fixed energy of $kT$, the net result is that as frequency gets lower so will its energy.

2. I have come to a rather interesting conclusion but it might be a bit too farfetched and even nonsense. The fact that the average calculated energy per mode using Planck’s equation is based on how many integer times an energy of $hf$ fits in $kT$, means that our calculated $kT$ must be accurate in some way. However, when $hf$ is approximately the same as $kT$, then probability would come into play regarding a mode being present or not. Can this emergence of probability be caused by a theory that the way we calculate $kT$ is not accurate enough? Such that when $hf$ approaches our (inacurrate) calculated value of $kT$, it would seem that there is a probability because in reality it’s all about $hf$ reaching a much more accurate energy value than our calculated $kT$?

 

[Charles Link]

The way the modes are counted is given in this post where the application is actually for particles in a gas. Boltzmann vs Maxwell distribution? $\\$ The reason why the energy in the modes stays within reasonable bounds in the ultraviolet (at short wavelength and high frequency) is due to the Bose factor $\bar{n}_s=\frac{1}{\exp(E_s/(k_bT))-1}$. (If it was simply due to the number of modes, the result would diverge=this important factor from quantum mechanics solved the problem that is known as the ultraviolet catastrophe. The mode counting by itself , without this factor, did not work).

 

[JohnnyGui]

The reason why the energy in the modes stays within reasonable bounds in the ultraviolet (at short wavelength and high frequency) is due to the Bose factor ¯ns=1exp(Es/(kbT))−1n¯s=1exp⁡(Es/(kbT))−1 \bar{n}_s=\frac{1}{\exp(E_s/(k_bT))-1} . (If it was simply due to the number of modes, the result would diverge=this important factor from quantum mechanics solved the problem that is known as the ultraviolet catastrophe. The mode counting by itself , without this factor, did not work).

Thanks for the link, I'll go read it. Yes, I indeed realized what the culprit of the ultraviolet catastrophe was. What I meant in my first statement in post #207 is explaining why the Planck function makes the frequencies have less and less energy when going left or right on the spectrum curve. Each direction having its own reason. I was wondering if the explanation in my statement is correct.

I'm very curious as well about my 2nd statement regarding the root cause of emergence of probability.

 

[Charles Link]

Thanks for the link, I'll go read it. Yes, I indeed realized what the culprit of the ultraviolet catastrophe was. What I meant in my first statement in post #207 is explaining why the Planck function makes the frequencies have less and less energy when going left or right on the spectrum curve. Each direction having its own reason. I was wondering if the explanation in my statement is correct.

I'm very curious as well about my 2nd statement regarding the root cause of emergence of probability.

@JohnnyGui Your assessments are correct. Probability does come into play to some degree. There is only an average number of photons $\bar{n}_s=\frac{1}{\exp(hf/(k_b T))-1}$ and that number can range to several photons, even many photons for $k_bT>> hf$, or it can be a small decimal number (much less than 1 photon on the average) for $k_bT<< hf$. There are normally enough modes to integrate that average value estimates are quite good, but even for thermal sources, if you do a photon count over a very short time interval with a measurement by a photodiode (particularly in detecting photons of shorter wavelengths), you can get fluctuations in the incident photon count that is caused by the statistical nature of the photons that are emitted by the thermal source. I believe it follows a binomial type statistics, and if you measure $N$ photons in your measurement, the standard deviation $\sigma$ essentially is $\sigma=\sqrt{N}$. For $N=1.0 \, E+20$, that makes $\Delta N=1.0 \, E+10$, (one part in 10,000,000,000), but for $N=1000$, this makes $\Delta N=30$ (approximately) which is 1 part in 30. $\\$ For an additional detail on this, the $N$ photons are the result of successes of $N'$ binomial trials, with mean $N=pN'$ and $\sigma=\sqrt{N'pq }=\sqrt{N'p}=\sqrt{N}$ with $q$ assumed to be very close to 1, and $p=1-q <<1$. And additional note: The statistical standard deviation $\sigma$ is not to be confused with the Stefan-Boltzmann constant $\sigma$. The same Greek letter is used, but there is no relation otherwise. $\\$ And additional comment: You ask in question 2, is the probability caused by the inaccuracy in our computation of $k_b$ and/or $T$ ? The answer is no=the $k_b T$ product can be known to about 1 part in 1,000 but the statistical count of photons can readily vary by 1 part in 30. This is not caused by inaccuracies in $k_b T$. Meanwhile, the Planck function as computed appears to be quite exact. It accurately predicts the Stefan-Boltzmann constant to be $\sigma=\frac{\pi^2}{60} \frac{k_b^4}{\bar{h}^3 c^2}=5.6703 \, E-8$ watts/(m^2 k^4), and the spectral shape has long been confirmed by experiment.

 

[JohnnyGui]

Your assessments are correct. Probability does come into play to some degree. There is only an average number of photons ¯ns=1exp(hf/(kbT))−1n¯s=1exp⁡(hf/(kbT))−1 \bar{n}_s=\frac{1}{\exp(hf/(k_b T))-1} and that number can range to several photons, even many photons for kbT>>hfkbT>>hf k_bT>> hf , or it can be a small decimal number (much less than 1 photon on the average) for kbT<<hfkbT<<hf k_bTNN N photons in your measurement, the standard deviation σσ \sigma essentially is σ=√Nσ=N \sigma=\sqrt{N} . For N=1.0E+20N=1.0E+20 N=1.0 \, E+20 , that makes ΔN=1.0E+10ΔN=1.0E+10 \Delta N=1.0 \, E+10 , (one part in 10,000,000,000), but for N=1000N=1000 N=1000 , this makes ΔN=30ΔN=30 \Delta N=30 (approximately) which is 1 part in 30.

Thanks. I noticed something else; if $\frac{1}{\exp(hf/(k_b T))-1}$ gives the average number of photons per mode of a particular frequency $f$ and it follows a binomial statistic, doesn't that mean that $\frac{1}{\exp(hf/(k_b T))-1} = np$ in which $n$ is the number of times that mode has been tested?
If so, wouldn't the standard deviation of each mode then be:
$$\sigma_{mode} = \sqrt{\frac{1}{\exp(hf/(k_b T))-1} \cdot q}$$
If there are $\frac{8\pi \cdot f^2}{c^3}$ modes per frequency, then the total amount of photons in that frequency $N_f$ would therefore be:
$$\frac{\sigma_{mode}^2}{q} \cdot \frac{8\pi \cdot f^2}{c^3} = N_f$$
Notice that I'm defining $N_f$ here as the number of photons of one particular frequency $f$ instead of the number of photons of a whole spectrum. If I'm defining $N_f$ that way, do these formulas make sense?

And additional comment: You ask in question 2, is the probability caused by the inaccuracy in our computation of kbkb k_b and/or TT T ? The answer is no=the kbTkbT k_b T product can be known to about 1 part in 1,000 but the statistical count of photons can readily vary by 1 part in 30. This is not caused by inaccuracies in kbTkbT k_b T . Meanwhile, the Planck function as computed appears to be quite exact. It accurately predicts the Stefan-Boltzmann constant to be σ=π260k4b¯h3c2=5.6703E−8σ=π260kb4h¯3c2=5.6703E−8 \sigma=\frac{\pi^2}{60} \frac{k_b^4}{\bar{h}^3 c^2}=5.6703 \, E-8 watts/(m^2 k^4), and the spectral shape has long been confirmed by experiment.

Thanks for the correction. If it's all about probability and about the average number of photons per mode, is there a chance that a certain wavelength of $hf << kT$ would not be emitted at all, even if that chance is very low?

I believe it follows a binomial type statistics, and if you measure NN N photons in your measurement, the standard deviation σσ \sigma essentially is σ=√Nσ=N \sigma=\sqrt{N} . For N=1.0E+20N=1.0E+20 N=1.0 \, E+20 , that makes ΔN=1.0E+10ΔN=1.0E+10 \Delta N=1.0 \, E+10 , (one part in 10,000,000,000), but for N=1000N=1000 N=1000 , this makes ΔN=30ΔN=30 \Delta N=30 (approximately) which is 1 part in 30.

I'm sorry but I have trouble finding out how you deduced $ΔN$. Didn't you mean $E - 10$ instead of $E +10$?

 

[Charles Link]

I have only done very limited statistical analyses of photons. For me, I have limited expertise with this topic. The $q$ in your formula is assumed to be nearly 1, if I'm not mistaken. Meanwhile for a photon count of $N=1.0 E+20$, the $\Delta N=1.0 E+10$, but that also means $N=1.0+/-.0000000001 \, E+20$, so I can see where you questioned the result. There is likely to be some good papers on this topic in the literature, but I haven't researched it in depth. When considering many modes, you need the equation $\sigma^2_{total}=\sigma^2_1+\sigma^2_2+\sigma^2_3+...$ so that $\sigma_{total}=\sqrt{N}$. Except in the case of a laser, usually the photons that you measure cover quite a number of modes, and even lasers in general are not single mode. In general, even the highest resolution spectrometers are unable to isolate on a single photon mode. The $\Delta \lambda$, even at high resolution, covers quite a large number of modes.

 

[JohnnyGui]

I have only done very limited statistical analyses of photons. For me, I have limited expertise with this topic. The $q$ in your formula is assumed to be nearly 1, if I'm not mistaken. Meanwhile for a photon count of $N=1.0 E+20$, the $\Delta N=1.0 E+10$, but that also means $N=1.0+/-.0000000001 \, E+20$, so I can see where you questioned the result. There is likely to be some good papers on this topic in the literature, but I haven't researched it in depth. When considering many modes, you need the equation $\sigma^2_{total}=\sigma^2_1+\sigma^2_2+\sigma^2_3+...$ so that $\sigma_{total}=\sqrt{N}$. Except in the case of a laser, usually the photons that you measure cover quite a number of modes, and even lasers in general are not single mode. In general, even the highest resolution spectrometers are unable to isolate on a single photon mode. The $\Delta \lambda$, even at high resolution, covers quite a large number of modes.

Not very good at statistics myself. I see that my deduced formulas do agree with your sigma summation. So I take it that since it's about probability, there is a chance that any frequency, regardless of how much hf is w.r.t. kT, not be emitted at all? (the chances merely differ)

In the meantime I have delved into emissivity yet again and there are things that I find a bit confusing.

I keep reading that emissivity doesn't have anything to do with color. A scenario: if a white object and black object are exposed to an equal amount of radiation energy, the black object absorbs more and emits more than the white one. Since the received power is less for the white one, it will therefore have a thermal equilibrium at a cooler T than the black one, even though they're exposed to an equal amount of radiation.
Here's what I deduce from this; a lower absorptivity leads to a lower T which therefore explains why a white object emits less energy. But the thing about emissivity is that even at the same T it should emit less energy.
Since the absorptivity of the white one is less, and e = a, this should mean that if you put the white and black object a non-radiating environment (dark room) at the same T, the white one should still emit less energy than the black object.
However, I read that both objects at the same T in a dark room would emit an equal amount of energy. But this means that they have the same emissivity and since e = a, the white object should then absorb an equal amount of energy as the black one when they're sitting in a radiating environment, which is not the case.
Where's the culprit in my reasoning here?

 

[Charles Link]

Not very good at statistics myself. I see that my deduced formulas do agree with your sigma summation. So I take it that since it's about probability, there is a chance that any frequency, regardless of how much hf is w.r.t. kT, not be emitted at all? (the chances merely differ)

In the meantime I have delved into emissivity yet again and there are things that I find a bit confusing.

I keep reading that emissivity doesn't have anything to do with color. A scenario: if a white object and black object are exposed to an equal amount of radiation energy, the black object absorbs more and emits more than the white one. Since the received power is less for the white one, it will therefore have a thermal equilibrium at a cooler T than the black one, even though they're exposed to an equal amount of radiation.
Here's what I deduce from this; a lower absorptivity leads to a lower T which therefore explains why a white object emits less energy. But the thing about emissivity is that even at the same T it should emit less energy.
Since the absorptivity of the white one is less, and e = a, this should mean that if you put the white and black object a non-radiating environment (dark room) at the same T, the white one should still emit less energy than the black object.
However, I read that both objects at the same T in a dark room would emit an equal amount of energy. But this means that they have the same emissivity and since e = a, the white object should then absorb an equal amount of energy as the black one when they're sitting in a radiating environment, which is not the case.
Where's the culprit in my reasoning here?

With the black object and white object, the emissivity can be wavelength dependent=we know the emissivity of the black object in the visible region of the spectrum is nearly 1.0, and the emissivity of the white object in the visible region of the spectrum is nearly zero, but the emissivity in the various portions of the infrared can be totally different. $\\$ It may be worth mentioning at this point that there are actually two types of reflectivities of surfaces. A surface with high reflectivity (close to 1.0) can either have a specular type reflectance where angle of incidence=angle of reflectance, or it can have a diffuse reflectance that scatters the light. A mirror is an example of specular reflectance, and a white sheet of paper is diffuse reflectance. A smooth sample of silver gives specular reflectance, but if you roughen the surface it can be somewhat diffuse but will still be part specular. $\\$ Additional item, you may wonder where gold and copper get their color from: The reason is they are highly reflective at the longer visible wavelengths (yellow, orange, and red regions of the visible), but much less reflective for the shorter visible wavelengths (violet, blue , and green regions of the visible).

 

[JohnnyGui]

With the black object and white object, the emissivity can be wavelength dependent=we know the emissivity of the black object in the visible region of the spectrum is nearly 1.0, and the emissivity of the white object in the visible region of the spectrum is nearly zero, but the emissivity in the various portions of the infrared can be totally different. $\\$ It may be worth mentioning at this point that there are actually two types of reflectivities of surfaces. A surface with high reflectivity (close to 1.0) can either have a specular type reflectance where angle of incidence=angle of reflectance, or it can have a diffuse reflectance that scatters the light. A mirror is an example of specular reflectance, and a white sheet of paper is diffuse reflectance. A smooth sample of silver gives specular reflectance, but if you roughen the surface it can be somewhat diffuse but will still be part specular. $\\$ Additional item, you may wonder where gold and copper get their color from: The reason is they are highly reflective at the longer visible wavelengths (yellow, orange, and red regions of the visible), but much less reflective for the shorter visible wavelengths (violet, blue , and green regions of the visible).

Ah, this explains it. So if I expose a white and black object to radiation that consists of the same infrared/visible light energy ratio as what the white object would emit in a dark room, both would have the same temperature under that radiation?

 

[Charles Link]

Ah, this explains it. So if I expose a white and black object to radiation that consists of the same infrared/visible light energy ratio as what the white object would emit in a dark room, both would have the same temperature under that radiation?

The black object will absorb more visible light, but they could have very similar emissivities in the infrared. The visible light will heat up the black object, but with an incadescent lamp, it is unknown how much infrared it is absorbing. For the white object, it absorbs little visible, but its infrared absorption is also unknown. Likewise the emissivity(=absorbtion) in the infrared is unknown. At room temperature, the Planck function peaks at around wavelength $\lambda=$10,000 nm and has almost zero visible output ($L(\lambda, 300K)$ is nearly zero for the visible wavelengths), so the emissivity values in the far infrared region is what matters for how much they radiate and how quickly they cool down after heating them with a lamp, as well as thermal conduction from the air.

 

[JohnnyGui]

The black object will absorb more visible light, but they could have very similar emissivities in the infrared. The visible light will heat up the black object, but with an incadescent lamp, it is unknown how much infrared it is absorbing. For the white object, it absorbs little visible, but its infrared absorption is also unknown. Likewise the emissivity(=absorbtion) in the infrared is unknown. At room temperature, the Planck function peaks at around wavelength $\lambda=$10,000 nm and has almost zero visible output ($L(\lambda, 300K)$ is nearly zero for the visible wavelengths), so the emissivity values in the far infrared region is what matters for how much they radiate and how quickly they cool down after heating them with a lamp, as well as thermal conduction from the air.

Makes sense to me. So if for example the white object has a much higher emissivity in the infrared than the black object, and they're both sitting in a dark room of mostly infrared radiation, the white object would be hotter than the black one?

 

[Charles Link]

Makes sense to me. So if for example the white object has a much higher emissivity in the infrared than the black object, and they're both sitting in a dark room of mostly infrared radiation, the white object would be hotter than the black one?

Yes, that is correct. What you are referring to is where there is an infrared source present whose temperature is much higher than the ambient and surrounding walls. Yes, then the white object with high infrared absorbance and emissivity will get heated by this infrared source.

 

[JohnnyGui]

Yes, that is correct. What you are referring to is where there is an infrared source present whose temperature is much higher than the ambient and surrounding walls. Yes, then the white object with high infrared absorbance and emissivity will get heated by this infrared source.

Apologies for being stubborn but what if there isn't an infrared source but merely walls that emit (a larger wavelength) infrared radiation? Are you implying that in that case the result would be different?

 

[Charles Link]

Apologies for being stubborn but what if there isn't an infrared source but merely walls that emit (a larger wavelength) infrared radiation? Are you implying that in that case the result would be different?

Regardless of the emissivity of the walls, if the objects inside are not large enough to alter the temperature of the walls, thermal equilibrium will occur and eventually everything will be at the same temperature as the walls.

 

[JohnnyGui]

Regardless of the emissivity of the walls, if the objects inside are not large enough to alter the temperature of the walls, thermal equilibrium will occur and eventually everything will be at the same temperature as the walls.

I indeed expected this, but when I think this through something collides.
If both objects only interact through radiation with the environment and one of the objects is highly reflective for infrared radiation (that the walls emit), how does it reach the same temperature while the energy that the highly reflective object receives is less than the object that fully absorbs the infrared radiation? Since the e = a, the highly reflective object would emit also less radiation, which leads to a thermal equilibrium at a lower temperature. I'm noticing this is the explanation in case there's an infrared source, but how is this explanation wrong in case of just walls?

 

[Charles Link]

I indeed expected this, but when I think this through something collides.
If both objects only interact through radiation with the environment and one of the objects is highly reflective for infrared radiation (that the walls emit), how does it reach the same temperature while the energy that the highly reflective object receives is less than the object that fully absorbs the infrared radiation? Since the e = a, the highly reflective object would emit also less radiation, which leads to a thermal equilibrium at a lower temperature. I'm noticing this is the explanation in case there's an infrared source, but how is this explanation wrong in case of just walls?

If both objects start out at temperature $T$ cooler than the walls, (so that $T < T_o$ initially), and it is simply walls at temperature $T_o$ around them, a higher emissivity object will absorb energy faster, but remember the differential equation is $c (\frac{dT}{dt})=-A \epsilon \sigma( T^4-T_o^4)$ where $c$ is the heat capacity (see post # 31). Equilibrium means $\frac{dT}{dt}=0$, and the result is $T=T_o$ independent of the emissivity. At equilibrium, regardless of the emissivity, the material radiates exactly at the same rate that it absorbs.

 

[JohnnyGui]

If both objects start out at temperature $T$ cooler than the walls, (so that $T < T_o$ initially), and it is simply walls at temperature $T_o$ around them, a higher emissivity object will absorb energy faster, but remember the differential equation is $c (\frac{dT}{dt})=-A \epsilon \sigma( T^4-T_o^4)$ where $c$ is the heat capacity (see post # 31). Equilibrium means $\frac{dT}{dt}=0$, and the result is $T=T_o$ independent of the emissivity. At equilibrium, regardless of the emissivity, the material radiates exactly at the same rate that it absorbs.

After reading this and trying to understand it, it actually makes perfect sense. Because the emissivitiy factor is in both the energy input and the energy output, it merely has influence on the time until equilibrium is reached but doesn't change the equilibrium temperature itself.

However, I noticed that I can't accept both ways. I now don't understand again how a black object is warmer than a white object when they're both under solar radiation XD. I understand that the solar radiation may contain wavelenghts that the white object has very low emissivities for, but if it absorbs and emits the same low amount, shouldn't it have the same equilibrium temperature as the black one? After all, we concluded that a different emissivity merely influences the time until equilibrium is reached.

 

[Charles Link]

After reading this and trying to understand it, it actually makes perfect sense. Because the emissivitiy factor is in both the energy input and the energy output, it merely has influence on the time until equilibrium is reached but doesn't change the equilibrium temperature itself.

However, I noticed that I can't accept both ways. I now don't understand again how a black object is warmer than a white object when they're both under solar radiation XD. I understand that the solar radiation may contain wavelenghts that the white object has very low emissivities for, but if it absorbs and emits the same amount, shouldn't it have the same equilibrium temperature as the black one? After all, we concluded that a different emissivity merely influences the time until equilibrium is reached.

The differential equation we have is simply for equilibrium with the walls of the room. We included a source $P_{in}$ in the differential equation of the electrical kind=(see post #31), but you could make $P_{in}$ be a source of irradiance $E_s$ onto our samples, each of area $A$. The power absorbed $P_{in}=\epsilon \, E_s \, A$. And that equation from post #31 is $\frac{dT}{dt}=\frac{P_{in}-\epsilon (\sigma (T^4-T_{ambient}^4))A}{Cm}$. For dynamic equilibrium, $\frac{dT}{dt}=0$. This equation as it is written is incomplete, and it assumes the emissivity to be independent of wavelength. $\\$ If the emissivity $\epsilon$ is completely independent of wavelength, then the analysis shows, the dynamic equilibrium temperature that is reached is independent of the emissivity, because the emissivity constant $\epsilon$ factors out of the terms in the numerator... $\\$ A more detailed equation would contain a spectral integral over $\lambda$ with the Planck function, and include an emissivity that has $\epsilon=\epsilon(\lambda)$. That equation would also include a source with a spectral irradiance function $E_s(\lambda)$. $\\$ $\frac{dT}{dt}=\frac{ A \, \int \epsilon (\lambda) E_s(\lambda) \, d \lambda \, - A \int \epsilon (\lambda) \sigma (M(\lambda,T)-M(\lambda,T_{ambient}) ) \, d \lambda}{Cm}$ is that equation in its spectral form. ($M(\lambda, T)$ is the Planck blackbody function in the radiant emittance form). The integrals are over all wavelengths $\lambda =0$ to $+\infty$. $\\$ The $\epsilon (\lambda)$ function will play a significant role in making the dynamic equilibrium of the black object (with high emissivity in the visible region of the spectrum) to have a higher dynamic equilibrium temperature $T$ because the first integral in the numerator will be larger for the black object, while the second term in the numerator would be nearly the same for both objects if the emissivity in the infrared region is the same for both objects. $\\$ For the white object, (which may have a high emissivity in the infrared), its emissivity in the visible region is small, so it may behave as if the source $E_s$ was very small, because it absorbs very little from that source. The dynamic equilibrium temperature $T$ for the white object will be very nearly equal to $T_{ambient}$. (This basically tells you the advantage to wearing reflective (e.g. white and light-colored) clothing when in the bright sunshine).

 

[JohnnyGui]

The differential equation we have is simply for equilibrium with the walls of the room. We included a source $P_{in}$ in the differential equation of the electrical kind=(see post #31), but you could make $P_{in}$ be a source of irradiance $E_s$ onto our samples, each of area $A$. The power absorbed $P_{in}=\epsilon \, E_s \, A$. And that equation from post #31 is $\frac{dT}{dt}=\frac{P_{in}-\epsilon (\sigma (T^4-T_{ambient}^4))A}{Cm}$. For dynamic equilibrium, $\frac{dT}{dt}=0$. This equation as it is written is incomplete, and it assumes the emissivity to be independent of wavelength. $\\$ If the emissivity $\epsilon$ is completely independent of wavelength, then the analysis shows, the dynamic equilibrium temperature that is reached is independent of the emissivity, because the emissivity constant $\epsilon$ factors out of the terms in the numerator... $\\$ A more detailed equation would contain a spectral integral over $\lambda$ with the Planck function, and include an emissivity that has $\epsilon=\epsilon(\lambda)$. That equation would also include a source with a spectral irradiance function $E_s(\lambda)$. $\\$ $\frac{dT}{dt}=\frac{ A \, \int \epsilon (\lambda) E_s(\lambda) \, d \lambda \, - A \int \epsilon (\lambda) \sigma (M(\lambda,T)-M(\lambda,T_{ambient}) ) \, d \lambda}{Cm}$ is that equation in its spectral form. ($M(\lambda, T)$ is the Planck blackbody function in the radiant emittance form). The integrals are over all wavelengths $\lambda =0$ to $+\infty$. $\\$ The $\epsilon (\lambda)$ function will play a significant role in making the dynamic equilibrium of the black object (with high emissivity in the visible region of the spectrum) to have a higher dynamic equilibrium temperature $T$ because the first integral in the numerator will be larger for the black object, while the second term in the numerator would be nearly the same for both objects if the emissivity in the infrared region is the same for both objects. $\\$ For the white object, (which may have a high emissivity in the infrared), its emissivity in the visible region is small, so it may behave as if the source $E_s$ was very small, because it absorbs very little from that source. The dynamic equilibrium temperature $T$ for the white object will be very nearly equal to $T_{ambient}$. (This basically tells you the advantage to wearing reflective (e.g. white and light-colored) clothing when in the bright sunshine).

Thanks a lot, I think I got it. Just to make sure I get this:

1. The emissivity is dependent on the wavelength
2. Therefore, the emissivity in the first integral of your formula is different from the emissivity of the second integral?
3. This is because the object is receiving in the first integral a certain range of wavelengths (sun's spectrum), while it's emitting a different range of wavelengths in the second integral (based on the object's temperature)?

Are these correct?

 

[Charles Link]

Thanks a lot, I think I got it. Just to make sure I get this:

1. The emissivity is dependent on the wavelength
2. Therefore, the emissivity in the first integral of your formula is different from the emissivity of the second integral?
3. This is because the object is receiving in the first integral a certain range of wavelengths (sun's spectrum), while it's emitting a different range of wavelengths in the second integral (based on the object's temperature)?

Are these correct?

@JohnnyGui Almost completely correct. Your second statement regarding the emissivity is incorrect though. The emissivity function $\epsilon(\lambda)$ is the same for both integral terms, but it is different for the two objects. (We do the calculation for each object, using $\epsilon_{black}(\lambda)$ for one, and $\epsilon_{white}(\lambda)$ for the other.) And yes, in (3), you really got it correct: $E_s(\lambda)$ typically will include a large contribution in the visible region of the spectrum (for the sun at approximately $T=6000$ k, about 1/2 of the energy is in the visible, and for an incadescent source, about 15% is in the visible region, while $M(\lambda, T)$ and $M(\lambda,T_{ambient})$ terms have almost nothing in the visible if $T<500 K$. Thereby, you actually got the reason correct in statement (3). $\\$ Additional comment about the sun and wearing light -colored clothing=you still can get very significant absorption in the infrared region from the sun even with the lightest clothing. I don't know whether they typically make any special types that are reflective in the infrared, but I believe for the astronauts who went to the moon, their spacesuits were also designed to be reflective in the infrared. $\\$ Meanwhile, there are sources $E_s(\lambda)$ whose output is almost entirely in the visible region. For fluorescent lamps, I believe at least 80% of the output is in the visible, and for some of the LED sources, I believe that number may be nearly 100%.

 

[JohnnyGui]

Your second statement regarding the emissivity is incorrect though. The emissivity function ϵ(λ)ϵ(λ) \epsilon(\lambda) is the same for both integral terms, but it is different for the two objects

Ah, I get what you're saying. I think I worded it poorly. How about if I say the following:

If we take an infinitesimally small range from the spectrum of the sun for which an object has approximately the same emissivity $\epsilon(d\lambda_{ab})$, then the object would absorb energy from that small range according to $A\sigma \cdot \epsilon(d\lambda_{ab}) \cdot M(\lambda, T_{sun}) \cdot d\lambda$. Let's say the object rises in temperature because of this absorbed energy but emits a very small different range of larger wavelengths for which it has an approximate fixed emissivity $\epsilon(d\lambda_{emit})$ but different from the $\epsilon(d\lambda_{ab})$. In equilibrium this should mean that:
$$A\sigma \cdot \epsilon(d\lambda_{ab}) \cdot M(\lambda, T_{sun}) \cdot d\lambda = A\sigma \cdot \epsilon(d\lambda_{emit}) \cdot (M(\lambda, T) - M(\lambda, T_{ambient})) \cdot d\lambda$$
Since $A\sigma \cdot (M(\lambda, T) - M(\lambda, T_{ambient})) \cdot d\lambda$ at the larger wavelength range must be lower than $A\sigma \cdot M(\lambda, T_{sun}) \cdot d\lambda$ which is at the smaller wavelength range, this must mean that $\epsilon(d\lambda_{ab}) < \epsilon(d\lambda_{emit})$,

Please correct me if this doesn't make sense.

 

[Charles Link]

Ah, I get what you're saying. I think I worded it poorly. How about if I say the following:

If we take an infinitesimally small range from the spectrum of the sun for which an object has approximately the same emissivity $\epsilon(d\lambda_{ab})$, then the object would absorb energy from that small range according to $A\sigma \cdot \epsilon(d\lambda_{ab}) \cdot M(\lambda, T_{sun}) \cdot d\lambda$. Let's say the object rises in temperature because of this absorbed energy but emits a very small different range of larger wavelengths for which it has an approximate fixed emissivity $\epsilon(d\lambda_{emit})$ but different from the $\epsilon(d\lambda_{ab})$. In equilibrium this should mean that:
$$A\sigma \cdot \epsilon(d\lambda_{ab}) \cdot M(\lambda, T_{sun}) \cdot d\lambda = A\sigma \cdot \epsilon(d\lambda_{emit}) \cdot M(\lambda, T_{object}) \cdot d\lambda$$
Since $A\sigma \cdot M(\lambda, T_{object}) \cdot d\lambda$ at the larger wavelength range must be lower than $A\sigma \cdot M(\lambda, T_{sun}) \cdot d\lambda$ which is at the smaller wavelength range, this must mean that $\epsilon(d\lambda_{ab}) < \epsilon(d\lambda_{emit})$,

Please correct me if this doesn't make sense.

Would recommend you do a much simpler case: Let $\epsilon_{black}=1$ independent of wavelength, and let $\epsilon_{white}=0$ for $\lambda< 750$ nm and $\epsilon_{white}=1$ for $\lambda >750$ nm. If you look at the general shapes of these $M(\lambda, T)$ blackbody functions, e.g. study the graph in the "link" https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/ [select the linear graph option on both axes as opposed to the log (logarithmic) graph] for temperatures near $T=300 K$, (close to room temperature), you will see there is pretty much zero output in the visible region ( $380 \, nm < \lambda<750 \, nm$) with nearly 100% of its output in the infrared. Just a quick qualitative look at these functions, along with the function $\epsilon(\lambda)$ should give you some idea of the whether each term in the integral is significant or not. Remember the integral is the area under the curve, and if we make the product function $\epsilon(\lambda) M(\lambda, T)$ somewhat simple, the integral is easier to evaluate. For a simple case, you could also have $E_s(\lambda)$ be a source which has output only in the visible region of the spectrum. $\\$ Additional note: If you let the source $E_s(\lambda)$ be the sun, you still need to scale it to its irradiance $E_s$ value: $E_s(\lambda)=\frac{L(\lambda,6000 \, K)A_s}{s^2}$. For the sun, $\frac{A_s}{s^2}=7.0 E-5$ (approximately). $\\$ Also note that T(degrees K)=T(Centigrade)+273.

 

[JohnnyGui]

Would recommend you do a much simpler case: Let $\epsilon_{black}=1$ independent of wavelength, and let $\epsilon_{white}=0$ for $\lambda< 750$ nm and $\epsilon_{white}=1$ for $\lambda >750$ nm. If you look at the general shapes of these $M(\lambda, T)$ blackbody functions, e.g. study the graph in the "link" https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/ [select the linear graph option on both axes as opposed to the log (logarithmic) graph] for temperatures near $T=300 K$, (close to room temperature), you will see there is pretty much zero output in the visible region ( $380 \, nm < \lambda<750 \, nm$) with nearly 100% of its output in the infrared. Just a quick qualitative look at these functions, along with the function $\epsilon(\lambda)$ should give you some idea of the whether each term in the integral is significant or not. Remember the integral is the area under the curve, and if we make the product function $\epsilon(\lambda) M(\lambda, T)$ somewhat simple, the integral is easier to evaluate. For a simple case, you could also have $E_s(\lambda)$ be a source which has output only in the visible region of the spectrum. $\\$ Additional note: If you let the source $E_s(\lambda)$ be the sun, you still need to scale it to its irradiance $E_s$ value: $E_s(\lambda)=\frac{L(\lambda,6000 \, K)A_s}{s^2}$. For the sun, $\frac{A_s}{s^2}=7.0 E-5$ (approximately). $\\$ Also note that T(degrees K)=T(Centigrade)+273.

One important question for me before I delve into the link and play with this. If emissivity of an object changes with wavelength, and the emitted energy of wavelengths changes with temperature $T$ of the object based on that change in emissivity, couldn't one write $\epsilon$ as a function of $T$ as well?

Here's what I mean:
So instead of integrating $M(\lambda, T) \cdot \epsilon(\lambda) \cdot A \cdot d\lambda$ up to infinity to calculate the total radiating energy of an object at $T$, couldn't one just measure the total emitted $M$ that the object emits at its $T$ and divide that by $\sigma \cdot T^4$?
If one keeps measuring the $M$ at different $T$'s and divide each of that measured energy by the corresponding $\sigma \cdot T^4$, one would have drawn a function of $\epsilon$ of the object over the change in $T$. In this case, the total energy that the object emits can be written as $T^4 \cdot \sigma \cdot \epsilon(T)$ where now the emissivitiy is a function of $T$.

Now that we know how $\epsilon$ changes over $T$, we can then solve for:
$$T^4_{sun} \cdot A\sigma \cdot \epsilon(T_{sun})) = A\sigma \cdot (T^4 \cdot \epsilon(T) - T^4_{ambient} \cdot \epsilon(T_{ambient}))$$
All of those $\epsilon$'s are the emissivities of the object but at different temperatures. They "correct" for the change in energy radiation/absorption by the object based on its temperature instead of its absorbed/emitted wavelengths.

Is it possible to do it this way as well, as an alternative?

 

[Charles Link]

The emissivity of an object is normally not a function of temperature. Fortunately it doesn't get that mathematically complicated. The case we are doing is somewhat difficult in that the emissivity is wavelength-dependent, and we have simplified that somewhat by just using a white object (or a black object) in the visible and assuming they both have emissivity equal to 1.0 in the infrared. $\\$ For the tungsten filament of an incadescent bulb, its resistance increases with temperature, and perhaps its emissivity might also change with temperature, but it would take a very lengthy experiment to determine something of this sort. There was some data that I saw in a google that showed emissivity for the filament as a function of wavelength. I don't recall if they had a couple of different graphs for a couple of different temperatures. $\\$ The emissivity normally would vary more with wavelength than it does with temperature. The way this experiment would be done is to use a spectrometer at a selected wavelength with a given $\Delta \lambda$ determined by the spectrometer slit width, and measure the output in radiance $L$ (which is $L= \epsilon (\lambda) \, L(\lambda,T) \, (\Delta \lambda)$ , and then divide by $L(\lambda,T) \, (\Delta \lambda)$, where $L(\lambda, T )$ is the Planck blackbody (radiance) function. The experimentalist might take 100 different wavelength points to get the emissivity $\epsilon (\lambda)$ as a function of $\lambda$. The exercise could be repeated for a different temperature $T$, to have the emissivity graphs for two different temperatures. $\\$ The experiment where the total power radiated per unit area is divided by $\sigma T^4$ and doing this for several different temperatures would not be necessary or useful, because it can not be assumed that the emissivity is independent of wavelength. For a very crude emissivity measurement, you could measure the total power radiated and divide by $\sigma T^4$, but if the $\epsilon$ showed any change for measurements at two different temperatures, the cause of the effect could be that the emissivity is a function of wavelength.

 

[JohnnyGui]

For a very crude emissivity measurement, you could measure the total power radiated and divide by σT4σT4 \sigma T^4 , but if the ϵϵ \epsilon showed any change for measurements at two different temperatures, the cause of the effect could be that the emissivity is a function of wavelength.

But it's indeed my intention to take the change of emissivity as a function of wavelength into account through the temperature change.

If I understand correctly, the fact that an object could have a different equilibrium $T$ than another object under the same (hot) radiation is because, as its temperature rises to reach the energy equilibrium as $P_{in}$, its radiating energy does not necessarily increase proportionally to $T^4$ because its specific emissivity and absorptivity is wavelength dependent and the spectrum curve is distributed differently over the wavelengths as temperature rises.

So by measuring the $M$'s of the object at different $T$'s and dividing by the corresponding $\sigma T^4$'s, you would be taking the wavelength dependency of emissivity also into account (along with the influence of $T$ change itself)

The whole reason I'm doing this is so that I can write the equation as the full radiating energy $P_{in} = A \sigma(T^4 \cdot \epsilon(T) - A \sigma(T^4 \cdot \epsilon(T_{ambient}))$ that contain the specific $\epsilon$'s for different temperatures and wavelength dependency at those temperatures to calculate the end equilibrium $T$ of the object, instead of using the integration of $M(\lambda,T)$. The $\epsilon(T)$ for a white object might be different from the $\epsilon (T)$ of the black object but this is done through measurements.

Come to think of it, this concludes that $A \int \epsilon (\lambda) \sigma (M(\lambda,T) \cdot d\lambda$ is equal to $A \sigma(T^4 \cdot \epsilon(T))$ and $A \int \epsilon(\lambda) \sigma M(\lambda,T_{ambient}) d \lambda = A \sigma(T^4 \cdot \epsilon(T_{ambient}))$. Not sure if this is correct.

Apologies for my stubbornness but I don't understand why it's not possible to do it this way.

 

[Charles Link]

But it's indeed my intention to take the change of emissivity as a function of wavelength into account through the temperature change.

If I understand correctly, the fact that an object could have a different equilibrium $T$ than another object under the same (hot) radiation is because, as its temperature rises to reach the energy equilibrium as $P_{in}$, its radiating energy does not necessarily increase proportionally to $T^4$ because its specific emissivity and absorptivity is wavelength dependent and the spectrum curve is distributed differently over the wavelengths as temperature rises.

So by measuring the $M$'s of the object at different $T$'s and dividing by the corresponding $\sigma T^4$'s, you would be taking the wavelength dependency of emissivity also into account (along with the influence of $T$ change itself)

The whole reason I'm doing this is so that I can write the equation as the full radiating energy $P_{in} = A \sigma(T^4 \cdot \epsilon(T) - T^4_{ambient} \cdot \epsilon (T_{ambient})$ that contain the specific $\epsilon$'s for different temperatures and wavelength dependency at those temperatures to calculate the end equilibrium $T$ of the object, instead of using the integration of $M(\lambda,T)$. The $\epsilon(T)$ for a white object might be different from the $\epsilon (T)$ of the black object but this is done through measurements.

Apologies for my stubbornness but I don't understand why it's not possible to do it this way.

If I understand what you are proposing, it is to use an effective emissivity as a function of temperature $\epsilon_{effective}(T)$. That could work ok if the incident energy always has a blackbody type spectrum whose temperature is known. This method would not give information on the spectral output of the sample=it would simply give the total power per unit area $M =\epsilon_{effective}(T) \, \sigma \, T^4$. Thereby, I can see why you might think it could be useful, but it just would not be used in enough calculations.

 

[JohnnyGui]

If I understand what you are proposing, it is to use an effective emissivity as a function of temperature $\epsilon_{effective}(T)$. That could work ok if the incident energy always has a blackbody type spectrum whose temperature is known. This method would not give information on the spectral output of the sample=it would simply give the total power per unit area $M =\epsilon_{effective}(T) \, \sigma \, T^4$. Thereby, I can see why you might think it could be useful, but it just would not be used in enough calculations.

I think that's indeed what I mean. I've edited my above post to include something after you replied, but I'm not sure if it's correct. Because my proposition of using $e_{effective}$ as a function of temperature implies that:
$$A \int \epsilon (\lambda) \sigma M(\lambda,T) \cdot d\lambda = A \sigma(T^4 \cdot \epsilon_{eff}(T))$$
And that:
$$A \int \epsilon(\lambda) \sigma M(\lambda,T_{ambient}) d \lambda = A \sigma(T^4 \cdot \epsilon_{eff}(T_{ambient}))$$
Not sure if this makes sense or not.

 

[Charles Link]

I think that's indeed what I mean. I've edited my above post to include something after you replied, but I'm not sure if it's correct. Because my proposition of using $e_{effective}$ as a function of temperature implies that:
$$A \int \epsilon (\lambda) \sigma (M(\lambda,T) \cdot d\lambda = A \sigma(T^4 \cdot \epsilon_{eff}(T))$$
And that:
$$A \int \epsilon(\lambda) \sigma M(\lambda,T_{ambient}) d \lambda = A \sigma(T^4 \cdot \epsilon_{eff}(T_{ambient}))$$
Not sure if this makes sense or not.

@JohnnyGui Very good. You have it figured out correctly. The $\epsilon(\lambda)$ function works well enough in the integrals that $\epsilon_{eff}(T)$ is not required.

 

[JohnnyGui]

@JohnnyGui Very good. You have it figured out correctly. The $\epsilon(\lambda)$ function works well enough in the integrals that $\epsilon_{eff}(T)$ is not required.

I'm very curious to prove this equation. So far I was able to find a graph showing the total emissivity of tungsten over the temperature https://www.researchgate.net/figure/273005745_fig4_Fig-4-Measured-spectrum-average-emissivity-of-tungsten-against-temperature-and-curve.
From that graph, I've picked the total emissivity of 0.15 at the temperature of approximately 1265K.

I don't know if there's a known formula for $\epsilon(\lambda)$ of tungsten so that I can fill in the equation:
$$\int \epsilon (\lambda) \sigma (M(\lambda,1265K) \cdot d\lambda = \sigma(1265K^4 \cdot 0.15)$$
And then do the integration with $\lambda$ as $x$.

 

[Charles Link]

I'm very curious to prove this equation. So far I was able to find a graph showing the total emissivity of tungsten over the temperature https://www.researchgate.net/figure/273005745_fig4_Fig-4-Measured-spectrum-average-emissivity-of-tungsten-against-temperature-and-curve.
From that graph, I've picked the total emissivity of 0.15 at the temperature of approximately 1265K.

I don't know if there's a known formula for $\epsilon(\lambda)$ of tungsten so that I can fill in the equation:
$$\int \epsilon (\lambda) \sigma (M(\lambda,1265K) \cdot d\lambda = \sigma(1265K^4 \cdot 0.15)$$
And then do the integration with $\lambda$ as $x$.

Any data for $\epsilon(\lambda)$ will of course be subject to experimental error. An additional suggestion for you, which I had previously mentioned is to numerically integrate the Planck function for any temperature that you pick, ( $T>500 \, K$ works better or you may need to extend the upper limit past $\lambda=20000 \, nm$ to get good accuracy), and show that $\int\limits_{0}^{+\infty} L(\lambda, T) \, d \lambda=\frac{\sigma T^4}{\pi}$. An integration interval $\Delta \lambda=10 \, nm$ should give reasonably good accuracy=+/- 1% or thereabouts. See also post #192 for additional details.

 

[JohnnyGui]

Any data for $\epsilon(\lambda)$ will of course be subject to experimental error. An additional suggestion for you, which I had previously mentioned is to numerically integrate the Planck function for any temperature that you pick, ( $T>500 \, K$ works better or you may need to extend the upper limit past $\lambda=20000 \, nm$ to get good accuracy), and show that $\int\limits_{0}^{+\infty} L(\lambda, T) \, d \lambda=\frac{\sigma T^4}{\pi}$. An integration interval $\Delta \lambda=10 \, nm$ should give reasonably good accuracy=+/- 1% or thereabouts. See also post #192 for additional details.

Yes. I was curious though if the effective $\epsilon$ of 0.15 would give a reasonable estimation on the total emissive power $M$ at 1265K for tungsten, just as one would use the integration with $\epsilon(\lambda)$. Best way I think is to find a graph that shows the total $M$ or $L$ over the temperature and see if it approximately equals $0.15 \cdot \sigma \cdot 1265^4$. Can't find any so far but I'll keep looking.

If tungsten should behave as a black body at that temperature, then I 'm surprised that the graph says that it would have an effective emissivity of 0.15 at that temperature instead of 1.

 

[Charles Link]

The answer is that apparently even when tungsten is so hot that it glows almost white in color, it still has a fairly high reflectivity. The spectral shape of the output of hot tungsten is nearly that of a blackbody, but with an emissivity that is apparently around $\epsilon=.15$.

 

[JohnnyGui]

The answer is that apparently even when tungsten is so hot that it glows almost white in color, it still has a fairly high reflectivity. The spectral shape of the output of hot tungsten is nearly that of a blackbody, but with an emissivity that is apparently around $\epsilon=.15$.

You mean the spectral shape at 1265K that includes also the reflected energy? What if the reflected energy is merely room temperature radiation?

Or do you mean the spectral shape of only its radiated energy having the same energy ratios among the emitted wavelengths as of a blackbody but not the same absolute wavelength energies amounts?
I can't see otherwise how a material with emissivity of 0.15 would emit the same spectral shape of that of a blackbody.

 

[Charles Link]

You mean the spectral shape at 1265K that includes also the reflected energy? What if the reflected energy is merely room temperature radiation?

Or do you mean the spectral shape of only its radiated energy having the same energy ratios among the emitted wavelengths as of a blackbody but not the same absolute wavelength energies amounts?
I can't see otherwise how a material with emissivity of 0.15 would emit the same spectral shape of that of a blackbody.

The reflectivity is apparently $\rho=1-\epsilon=.85$ and is nearly independent of wavelength. Otherwise, the radiated spectrum $M(\lambda,T)=\epsilon(\lambda) \, M_{bb}(\lambda,T)$ would not be that of a blackbody shape. i.e. if $\epsilon(\lambda)$ had hills and valleys, it would alter the shape.

 

[JohnnyGui]

The reflectivity is apparently $\rho=1-\epsilon=.85$ and is nearly independent of wavelength. Otherwise, the radiated spectrum $M(\lambda,T)=\epsilon(\lambda) \, M_{bb}(\lambda,T)$ would not be that of a blackbody shape. i.e. if $\epsilon(\lambda)$ had hills and valleys, it would alter the shape.

Ah, so it has the same spectral shape as a blackbody but at lower energy levels for each wavelength because the energy level of each wavelength is multiplied by a fixed constant $\epsilon(\lambda)$)?

 

[Charles Link]

Ah, so it has the same spectral shape as a blackbody but at lower energy levels for each wavelength because the energy level of each wavelength is multiplied by a fixed constant $\epsilon(\lambda)$)?

Yes. I don't know myself why it is that way, but that's apparently how it is. One time, I did measure the spectrum of an incadescent lamp, and it did have very much a $T=2500 \, K$ blackbody shape. The area of the filament was unknown, so that I was unable to compute the emissivity.

 

[JohnnyGui]

Yes. I don't know myself why it is that way, but that's apparently how it is. One time, I did measure the spectrum of an incadescent lamp, and it did have very much a $T=2500 \, K$ blackbody shape. The area of the filament was unknown, so that I was unable to compute the emissivity.

But integrating the curve would yield a lower total $M$ right compared to integrating a spectral curve of a BB at the same temp, right?

There's something else I forgot to ask regarding equilibrium.
Can I say the following:
If an object at equilibrium has the same temperature as the walls $T=T_o$, this means that the object has the same approximate emissivity (i.e. absorption) for the radiation that the walls emit at $T_o$ as the emissivity that it had at its initial ambient temperature $T_{ambient}$.
For emissivity not to change too much from $T_{ambient}$ to the equilibrium $T=T_o$, they must not differ too much from each other initially.

 

[Charles Link]

In previous posts, when writing the Planck function, I used $L(\lambda, T)$ and/or $M(\lambda, T)$, but it really would have been better to write it as $L_{bb}(\lambda, T)$ and/or $M_{bb}(\lambda, T)$ to distinguish it from a greybody that has $M(\lambda,T)=\epsilon(\lambda) \, M_{bb}(\lambda, T)$. $\\$ Meanwhile, an important concept is that in an enclosed space, at temperature $T= T_{ambient}$ where everything is at thermal equilibrium, the emissivity does not matter. This is assuming no other sources $E_s(\lambda )$ are present that are not at thermal equilibrium. At thermal equilibrium in an enclosed space, everything will be at $T=T_{ambient}$. Basically $E_{incident}(\lambda)$ as well as what is emerging off of the wall is the same at thermal equilibrium, independent of the emissivity $\epsilon(\lambda)$. $E_{incident}(\lambda)$ will in fact be equal to $M_{bb}(\lambda,T_{ambient})$ at thermal equilibrium regardless of $\epsilon(\lambda)$ of that portion of the wall. And as previously mentioned, this result is also completely independent of the shape of the enclosure.

 

[JohnnyGui]

Apologies if I missed your point. I worded it poorly as well.

What I meant is if an object is exposed to a hot radiation source like the sun, the object won't necessarily reach the equilibrium temperature that one would expect according to $E_s = \sigma \cdot T^4$ because the object may have a total different absorptivity/emissivity for the wavelengths from the hot source that it's being exposed to. Therefore it can be at equilibrium at a lower $T$ for example. Just like we discussed and what I said in my statement #3 in post #225.

However, in the case of radiating walls instead of a hot source, where the temperature of the walls $T_o$ is hotter than the initial temperature of the object that is brought in, the object could eventually reach an equilibrium temperature $T$ equal to $T_o$. The fact that they're equal at equilibrium means that, although emissivity is wavelength dependent, as the temperature of the object rises to reach equilibrium, the effective emissivities at all those temperatures that the object passed by weren't enough to emit the same $M$ amount as that it receives from the walls. Such that at equilibrium $T=T_o$ and thus:
$$T^4_o \cdot \sigma \cdot \epsilon_{eff}(T_o) = T^4 \cdot \sigma \cdot \epsilon_{eff}(T)$$
The thing is, I find it to be too coincidental if the effective emissivity of the object at initial temperature (when it was brought in) was way different and that exactly at $T=T_o$ it would have the ideal emissivity that is equal to the received energy from the walls. It is therefore more likely that the emissivity of the object at $T=T_o$ (which is $\epsilon_{eff}(T_o)$) didn't change too much from the emissivity of the object at its initial temperature, when it was brought in.
Since emissivity doesn't change drastically with small wavelength/temperature change, to have a thermal equilibrium like that, the initial temperature of the object must not have been too different from the $T_o$ of the walls or else effective emissivity would have changed too much at equilibrium and one would then have a scenario similar to a hot radiating source like the sun in which the temperatures of the object and the source are different at equilibrium.

Not sure if I'm making any sense here. Please correct me if I'm wrong.

 

[Charles Link]

One correction: It doesn't require an ideal emissivity to reach thermal equilibrium. When the object reaches temperature $T_{ambient}$ it is at equilibrium. The reason is at $T_{ambient}$ in an enclosed cavity, $E_s(\lambda)=M_{bb} (\lambda, T_{ambient})$. Now, writing the power as a spectral density function, $P_{absorbed} (\lambda)=A \, \epsilon (\lambda) \, E_s(\lambda)=A \, \epsilon (\lambda) \, M_{bb} (\lambda,T_{ambient})$, and $P_{reflected} (\lambda)=A \, (1-\epsilon (\lambda) ) M_{bb} (\lambda, T_{ambient})$. The power radiated $P_{radiated} (\lambda)= A \, \epsilon (\lambda) \, M(\lambda, T_{ambient})$, so that the power emerging from the wall is $P_{emerging} (\lambda)=P_{reflected} (\lambda) +P_{radiated} (\lambda)=A \, M_{bb} (\lambda, T_{ambient} )$, completely independent of any emissivities. Notice also that $P_{absorbed} (\lambda)=P_{radiated} (\lambda)$. $\\$ Regardless of the emissivity of a portion of the wall, in an enclosure at thermal equilibrium, the surface looks like a blackbody even if its emissivity is very low. This is only the case at equilibrium, but it is a very common scenario, and this concept is used in deriving the Planck function, as we previously discussed. See post #206.

 

[JohnnyGui]

One correction: It doesn't require an ideal emissivity to reach thermal equilibrium. When the object reaches temperature $T_{ambient}$ it is at equilibrium. The reason is at $T_{ambient}$ in an enclosed cavity, $E_s(\lambda)=M_{bb} (\lambda, T_{ambient})$. Now, writing the power as a spectral density function, $P_{absorbed} (\lambda)=A \, \epsilon (\lambda) \, E_s(\lambda)=A \, \epsilon (\lambda) \, M_{bb} (\lambda,T_{ambient})$, and $P_{reflected} (\lambda)=A \, (1-\epsilon (\lambda) ) M_{bb} (\lambda, T_{ambient})$. The power radiated $P_{radiated} (\lambda)= A \, \epsilon (\lambda) \, M(\lambda, T_{ambient})$, so that the power emerging from the wall is $P_{emerging} (\lambda)=P_{reflected} (\lambda) +P_{radiated} (\lambda)=A \, M_{bb} (\lambda, T_{ambient} )$, completely independent of any emissivities. Notice also that $P_{absorbed} (\lambda)=P_{radiated} (\lambda)$. $\\$ Regardless of the emissivity of a portion of the wall, in an enclosure at thermal equilibrium, the surface looks like a blackbody even if its emissivity is very low. This is only the case at equilibrium, but it is a very common scenario, and this concept is used in deriving the Planck function, as we previously discussed. See post #206.

Makes sense, so an object at its equilibrium temperature $T_{ambient}$ in an enclosed cavity absorbs and emits, as discussed, a power of:
$$P_{absorbed} = A \cdot \int M_{bb}(\lambda,T_{ambient}) \cdot \epsilon(\lambda) = T^4_{ambient} \cdot A\sigma \cdot \epsilon(T_{ambient})$$
This means that the power that is incident on the object $P_{emerging}$ must be
$$P_{emerging} = A \cdot \int M_{bb}(\lambda,T_{ambient}) = T^4_{ambient} \cdot A\sigma$$
This concludes that the walls are at the same temperature as that of the object (both $T_{ambient}$).

Question, is it possible for an object in an enclosed cavity to be at equilibrium while having a different temperature than the walls? As discussed, the object at $T_{ambient}$ receives a $P_{absorbed}= T^4_{ambient} \cdot A\sigma \cdot \epsilon(T_{ambient})$. However, is it possible that the object is receiving that exact amount of $P_{absorbed}$ but caused by a different temperature of the walls plus the object having a different emissivity for the wavelengths that those walls emit at that different temperature? So that:
$$T_{walls}^4 \cdot A\sigma \cdot \epsilon(T_{walls}) = T^4_{ambient} \cdot A\sigma \cdot \epsilon(T_{ambient}) = P_{absorbed}$$
Where $\epsilon(T_{walls})$ and $T_{walls}$ are different from $\epsilon(T_{ambient})$ and $T_{ambient}$ respectively. The equation would be the same as:
$$A \int M_{bb}(\lambda, T_{walls}) \cdot \epsilon(\lambda) = A \int M_{bb}(\lambda, T_{ambient}) \cdot \epsilon(\lambda) = P_{absorbed}$$
Is this possible or not?

 

[Charles Link]

Your very last equation answers your question. If $T_{walls}>T_{ambient}$, then $M_{bb}(\lambda, T_{walls})>M_{bb}(\lambda,T_{ambient})$ for all $\lambda$, and thereby these two integrals could not be equal. A similar statement applies with the inequality reversed for $T_{walls}<T_{ambient}$. The only way the two integrals can be equal is if $T_{walls}=T_{ambient}$.

 

[JohnnyGui]

Your very last equation answers your question. If $T_{walls}>T_{ambient}$, then $M_{bb}(\lambda, T_{walls})>M_{bb}(\lambda,T_{ambient})$ for all $\lambda$, and thereby these two integrals could not be equal. A similar statement applies with the inequality reversed for $T_{walls}<T_{ambient}$. The only way the two integrals can be equal is if $T_{walls}=T_{ambient}$.

I get what you're saying. But the reason I find this possible is because $\epsilon(\lambda)$ differs with wavelength. So looking at the blackbody spectra at different temperatures:

Looking at the red line of $300K$, the emissivity $\epsilon(\lambda)$ of the object could be for example 1 for the wavelengths corresponding to that $300K$ line while it could be very low for the wavelengths corresponding to the $10000K$ line. Since the $10000K$ line has a higher total energy when integrated but is compensated by the low emissivity, there is a possibility that the total absorbed energy from that $10000K$ line at a low emissivity is equal to the total absorbed energy from the $300K$ line at a high emissivity. (I see that the $10000K$ line also covers the wavelengths from the 300K line but emissivity can also change due to a higher temperature itself so that it is lower for those $300K$ wavelengths at $10000$).

Isn't this possible?

 

[Charles Link]

I get what you're saying. But the reason I find this possible is because $\epsilon(\lambda)$ differs with wavelength. So looking at the blackbody spectra at different temperatures:
View attachment 206732
Looking at the red line of $300K$, the emissivity $\epsilon(\lambda)$ of the object could be for example 1 for the wavelengths corresponding to that $300K$ line while it could be very low for the wavelengths corresponding to the $10000K$ line. Since the $10000K$ line has a higher total energy when integrated but is compensated by the low emissivity, there is a possibility that the total absorbed energy from that $10000K$ line at a low emissivity is equal to the total absorbed energy from the $300K$ line at a high emissivity.

Isn't this possible?

The answer is no, it is not possible. In your last equation of post #247, every single term $\epsilon (\lambda) M_{bb} (\lambda, T_{walls})$ of the integral on the left hand side of the equation is greater than the corresponding term $\epsilon(\lambda) M_{bb} (\lambda , T_{ambient})$ on the right hand side of the equation for $T_{walls}>T_{ambient}$. The emissivity $\epsilon (\lambda)$ is the same for both terms. For the same $\lambda$, we don't have two different $\epsilon (\lambda)$.