I Emission spectra of different materials

[Charles Link]

Great. So if the first integration over $dA_1$ for $E$ isn't needed, then the first integration formula would collapse to $\frac{I_0 \cdot cos(\theta)^4}{R^2} = E$. Now for the second integration this would mean, according to post #110:
$$\int \frac{I_0 \cdot cos(\theta)^4}{R^2} \cdot dA_2 = P_{A_2} = T^4_2 \cdot \sigma \cdot A_2$$
This means that $E = T^4_2 \cdot \sigma$ and $P_{incident} = T^4_2 \cdot \sigma \cdot A_1$.
Correct?

100 "likes". You got it correct. One minor correction or two though: Change the $I_o$ to an $L _2$, and instead of a $P_{A_2}$ on the right side, call it $E$. And leave off the last $A_2$. (Note: $L_2=\frac {\sigma T_2^4}{\pi}$).

 

[JohnnyGui]

100 "likes". You got it correct. One minor correction or two though: Change the $I_o$ to an $L _2$, and instead of a $P_{A_2}$ on the right side, call it $E$. And leave off the last $A_2$. (Note: $L_2=\frac {\sigma T_2^4}{\pi}$).

As much as I'm glad that I got it mostly correct, after typing the previous post I realized something XD.

If I have calculated that $\frac{I_0 \cdot cos(\theta)^4}{R^2} = E$ for the whole surface $A_1$ from one $dA_2$, doesn't this mean that $\int E \cdot dA_2$ would give me the $E$ for surface $A_1$ but from the whole surface $A_2$? This would lead me to conclude that $E = T^4_2 \cdot \sigma \cdot A_2$ and that $P_{incidence} = T^4_2 \cdot \sigma \cdot A_2 \cdot A_1$.
In post #110 the $E$ and $dA$ in the integration were directed at 1 surface so that multiplying $E$ with $dA$ would indeed give the total $P_{incidence}$ for that surface. But in this case the $E$ is for $A_1$ while the $dA_2$ is obviously for $A_2$ so here I'm calculating the total $E$ on $A_1$ from the whole surface $A_2$..

What is it that I'm reasoning wrong here?

 

[Charles Link]

As much as I'm glad that I got it mostly correct, after typing the previous post I realized something XD.

If I have calculated that $\frac{I_0 \cdot cos(\theta)^4}{R^2} = E$ for the whole surface $A_1$ from one $dA_2$, doesn't this mean that $\int E \cdot dA_2$ would give me the $E$ for surface $A_1$ but from the whole surface $A_2$? This would lead me to conclude that $E = T^4_2 \cdot \sigma \cdot A_2$ and that $P_{incidence} = T^4_2 \cdot \sigma \cdot A_2 \cdot A_1$.

What is it that I'm reasoning wrong here?

This is readily corrected: Your elemental source on surface $A_2$ is $dI_2=L_2 \, dA_2 \, cos(\theta)$, and $dE_{perpendicular \, to \, path}=\frac{dI}{s^2}=\frac{dI}{R^2} cos^2(\theta)$. Finally $dE=dE_{perpendicular \, to \, path} cos(\theta)$, (where $dE$ refers to irradiance onto surface $A_1$), so that $\\$ $E=\int \frac{ L_2 cos^4(\theta)}{R^2} \, dA_2$. $\\$ (To make things more clear, I re-derived the $cos^4(\theta)$ term here.) $\\$ Note that $L_2=\frac{ \sigma T_2^4}{\pi}$. $\\$ (Note the distance $s=\frac{R}{cos(\theta)}$ in the inverse square law. Note also the $cos(\theta)$ dependence for the elemental source $dI_2$. Comes from $I_o=LA$, and $I(\theta)=I_o cos(\theta)$). $\\$ Upon evaluating this integral, the result will cancel the $\pi$ in the denominator of the expression for $L_2$ giving $E=\sigma T_2 ^4$.

 

[Charles Link]

@JohnnyGui Once we have that one solved, an alternative enclosure shape could be used where the "ceiling" is a hemisphere in shape. The same result will be found to occur that $E=\sigma T_2 ^4$. In general this result will be found to be the case for an enclosure of any shape. $\\$ And the reason for this result is that $E=\int L_2 \, cos(\theta) \, d \Omega$ where $d \Omega$ is the solid angle subtended by the source on surface $A_2$ as seen from $A_1$. The solid angle integral just depends upon the hemisphere being completely enclosed=the shape is immaterial.

 

[JohnnyGui]

@JohnnyGui Once we have that one solved, an alternative enclosure shape could be used where the "ceiling" is a hemisphere in shape. The same result will be found to occur that $E=\sigma T_2 ^4$. In general this result will be found to be the case for an enclosure of any shape. $\\$ And the reason for this result is that $E=\int L_2 \, cos(\theta) \, d \Omega$ where $d \Omega$ is the solid angle subtended by the source on surface $A_2$ as seen from $A_1$. The solid angle integral just depends upon the hemisphere being completely enclosed=the shape is immaterial.

Ah, so that's what was corrected in your post #181

I was falsely assuming that the integration $\int E \cdot dA_2$ would sum every $E$ from each $dA_2$ while this actually would give the total energy from $E$ that surface 1 receives from each $dA_2$ but on $A_2$ itself. So you've corrected the integration so that integrating over $dA_2$ would give $E$.

Your corrected integration makes sense. So actually my deduced integration:
$$\int I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R^2} \cdot dA_2$$
Does not really give $P_{incident}$ coming from the whole surface $A_2$ because I'm multiplying by $dA_2$.
Instead, as an alternative to your method, I'd have to numerically add multiple times $I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R^2}$ for each $dA_2$ until I have calculated $P_{incident}$ on $A_1$ coming from the whole $A_2$ and then divide that by $A_1$ to get $E$?

 

[Charles Link]

Ah, so that's what was corrected in your post #181

I was falsely assuming that the integration $\int E \cdot dA_2$ would sum every $E$ from each $dA_2$ while this actually would give the total energy from $E$ that surface 1 receives from each $dA_2$ but on $A_2$ itself. So you've corrected the integration so that integrating over $dA_2$ would give $E$.

Your corrected integration makes sense. So actually my deduced integration:
$$\int I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R^2} \cdot dA_2$$
Does not really give $P_{incident}$ coming from the whole surface $A_2$ because I'm multiplying by $dA_2$.
Instead, as an alternative to your method, I'd have to numerically add multiple times $I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R^2}$ for each $dA_2$ until I have calculated $P_{incident}$ on $A_1$ coming from the whole $A_2$ and then divide that by $A_1$ to get $E$?

You could compute $P_{incident}=\int E \, dA_1$ for whatever you pick $A_1$ to be, e.g. a small circle, etc. But here we have a simplification: Since $A_2$ is uniform in brightness $L_2$ and infinite in extent, $E$ is independent of position on $A_1$, so that $P=\int E \, dA_1=E \int dA_1=EA_1$. We only need to compute $E$ (by doing an integral over $dA_2$). $\\$ It is a much more difficult calculation when $A_2$ is finite in size=e.g. a large or medium size circle and $A_1$ is also a large or medium size circle. Then the calculation for $P_{incident}$ onto $A_1$ involves what you mentioned with your diagram in post #115. Your diagram in post #115 should also show a bunch of rays coming from the middle parts of the source as well, but you have the right idea.

 

[Charles Link]

And take a minute to look at what the result $E=\sigma T_2^4$ is telling us: The irradiance on $A_1$ from $A_2$ would be the same if we butted surface $A_2$ right up against $A_1$. (The irradiance $E$ onto surface $A_1$ is the same as the radiant emittance $M=\sigma T_2^4$ that is leaving surface $A_2$). It doesn't matter how far away surface $A_2$ is from $A_1$ because it's like looking at a uniform blue sky: If you painted a uniformly lit blue ceiling over your head with no contrast whatsoever, no cracks, no other markings or light fixtures, you could not tell how far away it is: It is the same effect here. You could even make the dome a hemisphere, but if it were uniformly illuminated, you could not even tell its shape: There would be no objects present to focus on to give you distance information.

 

[JohnnyGui]

You could compute Pincident=∫EdA1Pincident=∫EdA1 P_{incident}=\int E \, dA_1 for whatever you pick A1A1 A_1 to be, e.g. a small circle, etc. But here we have a simplification: Since A2A2 A_2 is uniform in brightness L2L2 L_2 and infinite in extent, EE E is independent of position on A1A1 A_1 , so that P=∫EdA1=E∫dA1=EA1P=∫EdA1=E∫dA1=EA1 P=\int E \, dA_1=E \int dA_1=EA_1 . We only need to compute EE E (by doing an integral over dA2dA2 dA_2 ).

Yes, I indeed understand that we don't need to integrate $E$ over $dA_1$ and that I can use $P = E \cdot A_1$. What I meant is an alternative how to compute $E$ from the whole surface $A_2$ instead of doing an integral over $dA_2$. (this is merely to understand the mathematical definition of $E$)

Say we have calculated $P_{incidence}$ on the whole surface $A_1$ but from only one $dA_2$. To get the total $E$, can I calculate every $P_{incident}$ from every $dA_2$ from $A_2$ by using $I_0 \cdot cos(\theta_{dA_2})^4 \cdot \frac{A_1}{R^2}$ (each $dA_2$ having its own $\theta$, doing this calculation $\frac{A_2}{dA_2}$ times), add all those $P_{incident}$'s together to get $P_{incident}$ based on the whole surface $A_2$ and then divide that by $A_1$ and still get the answer $\sigma T_2^4$?

I know this is a very time-consuming way but I'm merely asking this to see if I understand how $E$ is defined mathematically.

And take a minute to look at what the result E=σT42E=σT24 E=\sigma T_2^4 is telling us: The irradiance on A1A1 A_1 from A2A2 A_2 would be the same if we butted surface A2A2 A_2 right up against A1A1 A_1 . (The irradiance EE E onto surface A1A1 A_1 is the same as the radiant emittance M=σT42M=σT24 M=\sigma T_2^4 that is leaving surface A2A2 A_2 ). It doesn't matter how far away surface A2A2 A_2 is from A1A1 A_1 because it's like looking at a uniform blue sky: If you painted a uniformly lit blue ceiling over your head with no contrast whatsoever, no cracks, no other markings or light fixtures, you could not tell how far away it is: It is the same effect here. You could even make the dome a hemisphere, but if it were uniformly illuminated, you could not even tell its shape: There would be no objects present to focus on to give you distance information.

This is actually very interesting to know! Can I explain this also the following way:
The received $P_{incidence}$ is constant ($\sigma T_2^4 \cdot A_1$) and independent from the distance between $A_1$ to $A_2$ because if I move $A_1$ closer to $A_2$, the received energy from the sides of $A_2$ decreases because of the increased angle $\theta$ but gets exactly compensated by an increased received energy caused by the decreased distance between $A_1$ and $A_2$?

 

[Charles Link]

The answer to your question is yes, in principle it could be computed this way. The problem that this method has is that the $dP_{incident}$ would be found to be a function of $r$ and $\phi$ , (i.e. when using polar coordinates=you could also say the result for $dP_{incident}$ would be a function of (x,y)=corresponding to the position of your elemental source on $A_2$ ). As a result, the calculation for $P_{incident}$ when you are integrating over $dA_2$ could get extremely complicated. If you were to go this route by selecting some shape for $A_1$, e.g. a circle, you would necessarily get the result that $P_{incident}=A_1 \sigma T_2^4$, but it's possible the integral expression when integrating over $dA_1$ to compute $dP_{incident}$ from the elemental source $dA_2$ at location (x,y) would become unmanageable. (If you did get an answer for $dP_{incident}$ (by integrating $E$ over $dA_1$)as a function of (x,y) of the position of $dA_2$ on surface $A_2$, you could then integrate over $dA_2$ to compute $P_{incident}=A_1 \sigma T_2^4$, but it's quite possible the mathematics to do these two steps would be next to impossible). It is a much easier route to pick one single point on $A_1$, and compute the $E$ by integrating over $dA_2$ first. $\\$ And even though you apparently missed a Latex symbol in your second comment, I was able to read it, and yes, it is correct. This $E=\sigma T_2^4$ is a very useful result that we previously used in your differential equation in post #31, where $T_2=T_{ambient}$.

 

[JohnnyGui]

Thanks for your verifications. I noticed after typing my second comment a furtther detail; that the energy is constant because the received energy decreases by a factor of $cos(\theta)^2$ because of the smaller projected $A_1$ and because of the Lambertian law, but also increases because of the distance decreases by a factor of $cos(\theta)$ such that the energy increases by the square of that.

So if the received energy is equal to $\sigma T_2^4 \cdot A_1$ and the emissivity and absorptivity stay the same even when $A_1$ and $A_2$ have different temperatures, let's say the surface $A_1$ has a low emissivity and is highly reflective and I have a detector that measures the energy coming off from $A_1$, can I say that the detector would measure a false energy of:
$$\sigma T_1^4 \cdot A_1 \cdot \epsilon + \sigma T_2^4 \cdot A_1 \cdot (1 - \epsilon)$$
The part with $(1 - \epsilon)$ is the energy that gets reflected towards the detector. When not taking the integration with $cos(\theta)^4$ into account for simplicity's sake, is this correct?

 

[Charles Link]

Thanks for your verifications. I noticed after typing my second comment a furtther detail; that the energy is constant because the received energy decreases by a factor of $cos(\theta)^2$ because of the smaller projected $A_1$ and because of the Lambertian law, but also increases because of the distance decreases by a factor of $cos(\theta)$ such that the energy increases by the square of that.

So if the received energy is equal to $\sigma T_2^4 \cdot A_1$ and the emissivity and absorptivity stay the same even when $A_1$ and $A_2$ have different temperatures, let's say the surface $A_1$ has a low emissivity and is highly reflective and I have a detector that measures the energy coming off from $A_1$, can I say that the detector would measure a false energy of:
$$\sigma T_1^4 \cdot A_1 \cdot \epsilon + \sigma T_2^4 \cdot A_1 \cdot (1 - \epsilon)$$
The part with $(1 - \epsilon)$ is the energy that gets reflected towards the detector. When not taking the integration with $cos(\theta)^4$ into account for simplicity's sake, is this correct?

Yes, that would be the total power that appears to come off of surface $A_1$. An interesting case is when $\epsilon=0$ (100% reflective). In that case the surface looks like it is at temperature $T_2$.

 

[Charles Link]

One other thing we could look at that you might find of interest would be to take the Planck blackbody spectral function, (see post #81 for its functional form), for the incadescent source you had in the early parts of this thread. If you assume the emissivity $\epsilon$ is independent of wavelength, (it's precise value isn't fussy; $\epsilon=.3$ would work ok), and assign a temperature $T$ (degrees kelvin) to the filament, you can then take the ratio of the integral of the Planck function integrated over the visible wavelengths (wavelength $\lambda=$380 nm to 750 nm) vs. the integral over the whole spectrum, and this will give you the fraction of the radiated light that is in the visible region of the spectrum. I believe this number will be something like $r=.15$ so that these light bulbs basically have an efficiency of about 15%. Most of the output of an incadescent lamp is in the infrared region of the spectrum.$\\$ A numerical integration should work ok, and an integration for the denominator of wavelengths $\lambda$ from 0 to 20,000 nm would be ok, (instead of $+\infty$ for the upper limit on the integral). An increment of $\Delta \lambda =10$ nm would give reasonably good accuracy. $\\$ (It is also of interest that the exact answer for the integral $L=\int\limits_{0}^{+\infty} L(\lambda,T) \, d \lambda$ for the function $L(\lambda,T)$ in post #81 is $L=\frac{\sigma T^4}{\pi}$ where $\sigma=(\frac{\pi^2}{60})(\frac{k_b^4}{\hbar^3 c^2})$. Plugging in these constants will indeed give the value for $\sigma=5.67 E-8$ watts/(m^2 K^4) ). $\\$ A couple of helpful items in this integration: $\\$ 1) $L(\lambda,T)$ is undefined at $\lambda=0$ but for $L(0,T)$ you use the value of $L(\lambda,T)$ as $\lambda \rightarrow 0$ which is $L(0,T)=0$. $\\$ 2) For the function $L(\lambda,T)$, you can use constants $c_1=2hc^2=1.19E+20$ Watts/(m^2 nm) (I believe I computed it correctly=it actually gets a unit of nm^5 in the numerator next to watts, but that cancels the nm^5 units in $\lambda^5$ ), and $c_2=\frac{hc}{k_b} =1.438 E+7$ nm K. $\\$ [Use $\lambda$ measured in nanometers in the formula for $L(\lambda,T)=\frac{c_1}{\lambda^5 (e^{\frac{c_2}{\lambda T}}-1)} \,$]. $\\$ Additional note: A google of the subject shows there are a couple of programs you can find on-line that will perform the above mathematics for you.

 

[JohnnyGui]

Yes, that would be the total power that appears to come off of surface A1A1A_1 . An interesting case is when ϵ=0ϵ=0 \epsilon=0 (100% reflective). In that case the surface looks like it is at temperature T2T2 T_2 .

Yes, that makes sense to me. Two more things if you don't mind:
1. Does that mean that a detector that measures the radiated power from an object with $\epsilon < 1$ will always overestimate that radiated power because the material is reflecting energy in addition to its temperature dependent radiation? Even when the surroundings are colder than the object?

2. If I have a detector that can estimate the temperature of an object of surface $A_1$, temperature $T_1$ and a low emissivity $\epsilon$ in a surrounding with temperature $T_2$, and the detector is set to an emissivity of 1, will that mean that the detector will calculate a (false) observed temperature of the object $T_o$ according to:
$$(T_1^4 \cdot \epsilon + T_2^4 \cdot (1-\epsilon))^{\frac{1}{4}} = T_o$$
(Again ignoring the $cos(\theta)^4$ dependence for simplicity's sake)

 

[Charles Link]

Yes, that makes sense to me. Two more things if you don't mind:
1. Does that mean that a detector that measures the radiated power from an object with $\epsilon < 1$ will always overestimate that radiated power because the material is reflecting energy in addition to its temperature dependent radiation? Even when the surroundings are colder than the object?

2. If I have a detector that can estimate the temperature of an object of surface $A_1$, temperature $T_1$ and a low emissivity $\epsilon$ in a surrounding with temperature $T_2$, and the detector is set to an emissivity of 1, will that mean that the detector will calculate a (false) observed temperature of the object $T_o$ according to:
$$(T_1^4 \cdot \epsilon + T_2^4 \cdot (1-\epsilon))^{\frac{1}{4}} = T_o$$
(Again ignoring the $cos(\theta)^4$ dependence for simplicity's sake)

If the detector emissivity is set for $\epsilon=1$ , in general, the temperature reading it gives will be too low if the object's emissivity is less than 1, and the error can be quite significant in cases where the emissivity is very low, such as in reflective materials. In general, using an infrared camera to measure temperatures can be subject to considerable errors unless the object of interest has a well known emissivity and/or the effects from the background are small enough that they don't significantly distort the results. $\\$ And please take a look at the spectral computation of post #192. I think you should find it somewhat easy to follow since we looked at the concept of spectral intensity functions in posts #140-#160 and thereabouts. And if you know how to use a computer spreadsheet, you can write the formula for the spectral intensity in one of the columns of the spreadsheet and copy it for all the different wavelengths that you have in the first column. You can then do the sum/integral yourself by summing the column and multiplying the sum by $\Delta \lambda$.

 

[JohnnyGui]

If the detector emissivity is set for ϵ=1ϵ=1 \epsilon=1 , in general, the temperature reading it gives will be too low if the object's emissivity is less than 1, and the error can be quite significant in cases where the emissivity is very low, such as in reflective materials. In general, using an infrared camera to measure temperatures can be subject to considerable errors unless the object of interest has a well known emissivity and/or the effects from the background are small enough that they don't significantly distort the results.

But won't the detector only measure a too low temperature of the object if the surroundings is colder than the object itself ($T_1 > T_2$)? Here's how I deduced this from the formula:
If the energy that the detector measures is equal to $\sigma T_1^4 \cdot A_1 \cdot \epsilon + \sigma T_2^4 \cdot A_1 \cdot (1 - \epsilon) = P_{detector}$ and you set the detector to an emissivity of 1, then for the detector to calculate the object's temperature $T_1$, it will divide that $P_{detector}$ by $A_1 \cdot \sigma$ and do it to the 0.25th power so that after simplifying this gives:
$$(T_1^4 \cdot \epsilon + T_2^4 \cdot (1-\epsilon))^{\frac{1}{4}} = T_o$$
To let the detector measure a lower temperature than $T_1^4$ you'd have solve for that formula so that $(T_1^4 \cdot \epsilon + T_2^4 \cdot (1-\epsilon))^{\frac{1}{4}} < T_1$ and therefore $T_1^4 \cdot \epsilon + T_2^4 \cdot (1-\epsilon) < T_1^4$
Rewriting this would give eventually $T_1^4 (1 - \epsilon) > T_2^4 (1 - \epsilon)$ which means that $T_1 > T_2$. So in that case the detector would give a lower temperature of the object than it actually is.

I've also read in articles (see 3rd paragraph here and 2nd page here ) that this condition ($T_1 > T_2$) is necessary to let the detector underestimate the object's temperature. So is my statement concluded with the formula correct?

EDIT: I've corrected the equations.

And please take a look at the spectral computation of post #192. I think you should find it somewhat easy to follow since we looked at the concept of spectral intensity functions in posts #140-#160 and thereabouts.

I've read it but I need just one hint to fully understand this. Doesn't integrating a spectrum curve, up until 20,000nm like you mentioned, initially give $\sigma \cdot T^4$? I'm trying to find out how you've rewritten the integration so that you'd be integrating $L$ over $d\lambda$. I think it's a blackout once again for me, I'm delving into this right now.

 

[Charles Link]

I've read it but I need just one hint to fully understand this. Doesn't integrating a spectrum curve, up until 20,000nm like you mentioned, initially give $\sigma \cdot T^4$? I'm trying to find out how you've rewritten the integration so that you'd be integrating $L$ over $d\lambda$. I think it's a blackout once again for me, I'm delving into this right now.

Your logic concerning a sensor that outputs an object's temperature is correct. In general it will be prone to the type of error that you have pointed out. $\\$ The spectral curve $L(\lambda, T)$ can be integrated between two wavelengths and gives the radiance $L$
contained between those two wavelengths. If you integrate $L(\lambda, T)$ from 0 to $+\infty$, you get $L=\frac{M}{\pi}=\frac{\sigma T^4}{\pi}$. The numerical integration from 0 to $+\infty$ for a hotter object can be cut off around 20,000 nm and you'll account for 99% or more of the total energy. $\\$ This integral result from 0 to $+\infty$ for $L(\lambda, T)$ is in fact exact, but it is a rather advanced integral, and even a student who has had 2 or 3 calculus courses would not be expected to know how to solve it. You can do a numerical integration of it for a selected temperature $T$, (e.g. $T=2500 \, K$) , and you should be able to show pretty good agreement to the exact value of $\frac{\sigma T^4}{\pi}$. (within +/- 1% or better). $\\$ In some literature, you will see the Planck function as $L(\lambda,T)$. You will also find it presented at times as $M(\lambda, T)=\pi L(\lambda, T)$. The $L(\lambda, T)$ is the more common form, but you will find it in both forms. For $M(\lambda,T)$ you have the result $M=\int\limits_{0}^{+\infty} M(\lambda, T) \, d \lambda=\sigma T^4$. $\\$ The functional form for the Planck blackbody spectral function $L(\lambda, T)$ can be derived from some very advanced physics principles, but that is beyond the scope of what we are doing here. We're simply using the results of that derivation. And note only does the result of this derivation give us the Planck spectral function, it also supplies the result that when the Planck function is integrated from 0 to $+\infty$, the result is that $M=\sigma T^4$.

 

[JohnnyGui]

In some literature, you will see the Planck function as L(λ,T)L(λ,T) L(\lambda,T) . You will also find it presented at times as M(λ,T)=πL(λ,T)M(λ,T)=πL(λ,T) M(\lambda, T)=\pi L(\lambda, T) . The L(λ,T)L(λ,T) L(\lambda, T) is the more common form, but you will find it in both forms. For M(λ,T)M(λ,T) M(\lambda,T) you have the result M=+∞∫0M(λ,T)dλ=σT4M=∫0+∞M(λ,T)dλ=σT4 M=\int\limits_{0}^{+\infty} M(\lambda, T) \, d \lambda=\sigma T^4 .

Ah got it. So integrating a $M(\lambda, T)$ curve and a $L(\lambda, T)$ curve would only give a factor of $\pi$ difference?
One question, if $L(\lambda, T)=\frac{2hc^2}{\lambda^5(exp^{hc/(\lambda k_b T)}-1)}$ then what would be the formula for $M(\lambda, T)$ curve? And what are the unity differences on the y-axis between them?

Your logic concerning a sensor that outputs an object's temperature is correct. In general it will be prone to the type of error that you have pointed out.

I deduced some conclusions when plotting my deduced formula ($\epsilon$ as the $x$ variable) that shows what temperature the detector would give based on $T_1$ and $T_2$. I'd really appreciate a verification if these are correct.

1. If $T_1 > T_2$, and the emissivity of the object is anywhere between $0 < \epsilon < 1$, then the detector would always underestimate the temperature of the object.

2. If $T_1 < T_2$, and the emissivity of the object is anywhere between $0 < \epsilon < 1$, then the detector would always overestimate the temperature of the object.

3. If $\epsilon = 1$ then the detector would always give the correct temperature $T_1$ regardless if $T_1$ is lower or higher than $T_2$.

4. If $\epsilon = 0$ then the detector would always give $T_2$ regardless if $T_2$ is lower or higher than $T_1$.

Are these correct?

 

[Charles Link]

Ah got it. So integrating a $M(\lambda, T)$ curve and a $L(\lambda, T)$ curve would only give a factor of $\pi$ difference?
One question, if $L(\lambda, T)=\frac{2hc^2}{\lambda^5(exp^{hc/(\lambda k_b T)}-1)}$ then what would be the formula for $M(\lambda, T)$ curve? And what are the unity differences on the y-axis between them?
I deduced some conclusions when plotting my deduced formula ($\epsilon$ as the $x$ variable) that shows what temperature the detector would give based on $T_1$ and $T_2$. I'd really appreciate a verification if these are correct.

1. If $T_1 > T_2$, and the emissivity of the object is anywhere between $0 < \epsilon < 1$, then the detector would always underestimate the temperature of the object.

2. If $T_1 < T_2$, and the emissivity of the object is anywhere between $0 < \epsilon < 1$, then the detector would always overestimate the temperature of the object.

3. If $\epsilon = 1$ then the detector would always give the correct temperature $T_1$ regardless if $T_1$ is lower or higher than $T_2$.

4. If $\epsilon = 0$ then the detector would always give $T_2$ regardless if $T_2$ is lower or higher than $T_1$.

Are these correct?

$\cdot$ The relationship between $M(\lambda,T)$ and $L(\lambda ,T)$ is very simple: $M(\lambda, T)=\pi \, L(\lambda, T)$ for each and every $\lambda$. ($M(\lambda ,T)$ has a $2 \pi hc^2$ in the numerator). The result for the complete integrals of each are thereby $\int\limits_{0}^{+\infty} M(\lambda , T) \, d \lambda=\sigma T^4 =\pi \int\limits_{0}^{+\infty} L(\lambda ,T) \, d \lambda$. $\\$ $\cdot$ Your conclusions concerning the temperatures the detector would determine are all correct. $\\$ $\cdot$ For something simple concerning the Planck function, take a look at this website: $\\$ https://astrogeology.usgs.gov/tools/thermal-radiance-calculator/ $\\$ You would actually be able to program this yourself, but you might find it easier to simply use their results. For a useful exercise, you might try programming the Planck function yourself, and comparing your results to theirs.

 

[Charles Link]

@JohnnyGui One recommendation in using the above "link" is to chose the linear graph option. Anyway, I think you might find it useful. :)

 

[JohnnyGui]

The relationship between M(λ,T)M(λ,T) M(\lambda,T) and L(λ,T)L(λ,T) L(\lambda ,T) is very simple: M(λ,T)=πL(λ,T)M(λ,T)=πL(λ,T) M(\lambda, T)=\pi \, L(\lambda, T) for each and every λλ \lambda . (M(λ,T)M(λ,T) M(\lambda ,T) has a 2πhc22πhc2 2 \pi hc^2 in the numerator). The result for the complete integrals of each are thereby +∞∫0M(λ,T)dλ=σT4=π+∞∫0L(λ,T)dλ∫0+∞M(λ,T)dλ=σT4=π∫0+∞L(λ,T)dλ \int\limits_{0}^{+\infty} M(\lambda , T) \, d \lambda=\sigma T^4 =\pi \int\limits_{0}^{+\infty} L(\lambda ,T) \, d \lambda .

Ah of course, was about to ask if $M(\lambda ,T)$ has a $\pi$ in the formula. I noticed that the spectral curves always indeed have $L$ on the y-axis. If you measure the total power of each wavelength from a whole emitting surface, and draw a spectral curve based on that, wouldn't you be drawing $M(\lambda,T)$?

And take a minute to look at what the result E=σT42E=σT24 E=\sigma T_2^4 is telling us: The irradiance on A1A1 A_1 from A2A2 A_2 would be the same if we butted surface A2A2 A_2 right up against A1A1 A_1 . (The irradiance EE E onto surface A1A1 A_1 is the same as the radiant emittance M=σT42M=σT24 M=\sigma T_2^4 that is leaving surface A2A2 A_2 ). It doesn't matter how far away surface A2A2 A_2 is from A1A1 A_1 because it's like looking at a uniform blue sky

I was thinking this through a bit and I noticed something. I understand that if you move surface $A_1$ away or towards the emitting $A_2$ you'd still receive the same power from each $dA_2$ from the surface $A_2$. However, (and correct me if I'm wrong) there's one exception: there is exactly one $dA_2$ that you'll get a different power from if you change the distance of $A_1$ from $A_2$. This is the $dA_2$ that is exactly in front of $A_1$ perpendicularly on it. The surface $A_1$ will receive $I_0$ from that particular $dA_2$ and since there is no $cos(\theta)$ factor that changes with distance from that $dA_2$, the power that $A_1$ receives from that particular $dA_2$ will have a net change of the distance change ratio squared.
So if you increase the distance between $A_1$ and $A_2$, wouldn't $A_1$ have a net decrease in received power because of that perpendicular $dA_2$ instead of a constant $T^4_2 \sigma \cdot A_1$?

For something simple concerning the Planck function, take a look at this website:

I was actually reading about the derivation of that function. It's amazing how one would conclude that (and even more amazing that I actually understand how the derivation is done). One thing though, how does the Planck function take the the emissivity of an emitting object into account, to calculate the power of each wavelength?

Your conclusions concerning the temperatures the detector would determine are all correct.

Thanks for verifying!

 

[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.