I Emission spectra of different materials

[Charles Link]

Your curve of post #140 that has watts on the y-axis can not be a $A(\lambda)$ type curve. It has to be mislabeled with "watts". The reason is that a $A(\lambda)$ curve must increase or stay the same in power from left to right. $A (\lambda)$ can never drop back down, since there is no contribution from $\Phi(\lambda)$ that will give a negative result. In mathematical terms, $A(\lambda)$ is a monotonically increasing function. $\\$ Assuming you did have both the $A(\lambda)$ curve and the $\Phi(\lambda)$ curve, $A(\lambda_2)-A(\lambda_1)=\int\limits_{\lambda_1}^{\lambda_2} \Phi(\lambda) \, d\lambda$. That's why in probability theory, the tabulated $A(\lambda)$ has proven to be quite useful. Spectroscopists simply have never bothered to implement it though.

 

[Charles Link]

@JohnnyGui Please see also the edited part of post #151.

 

[JohnnyGui]

Your curve of post #140 that has watts on the y-axis can not be a $A(\lambda)$ type curve. It has to be mislabeled with "watts". The reason is that a $A(\lambda)$ curve must increase or stay the same in power from left to right. $A (\lambda)$ can never drop back down, since there is no contribution from $\Phi(\lambda)$ that will give a negative result. In mathematical terms, $A(\lambda)$ is a monotonically increasing function. $\\$ Assuming you did have both the $A(\lambda)$ curve and the $\Phi(\lambda)$ curve, $A(\lambda_2)-A(\lambda_1)=\int\limits_{\lambda_1}^{\lambda_2} \Phi(\lambda) \, d\lambda$.

Ah, this makes sense to me now. $A(\lambda)$ is an "accumulating" energy curve, just like distance cannot decrease over time if there's a velocity over time.

But, can't there be a graph of some sort that shows the amount of absolute Watts for each wavelength individually??

JohnnyGui Please see also the edited part of post #151.

Reading it now.

 

[Charles Link]

Ah, this makes sense to me now. $A(\lambda)$ is an "accumulating" energy curve, just like distance cannot decrease over time if there's a velocity over time.

But, can't there be a graph of some sort that shows the amount of absolute Watts for each wavelength individually??

The wavelengths are normally continuous. A discrete type spectrum could be used to model a source consisting of laser lines that could be each said to be a single discrete wavelength, but otherwise, most spectra=e.g., incadescent lamps and blackbodies, and other sources, the spectrum is normally continuous. Even a laser, under very high resolution, actually has a finite wavelength range and could be said to be continuous, and some laser lines are more monochromatic than others. e.g. laser diodes often have a wider spread of wavelengths. $\\$ One example: A 2 mwatt HeNe laser at $\lambda=632.8$ nm. You could say it is a single wavelength putting out 2 mwatts of power. Other than lasers, and perhaps a couple other sources such as sodium or mercury arc lamps, and other such sources that result from atomic transitions, the spectral curves you see of sources will, in general, be continuous. In these couple of exceptions, (the same thing happens with the probability curves as well, e.g. if the variable only takes on integer values), you could model them discretely at the selected wavelengths.

 

[JohnnyGui]

The wavelengths are normally continuous. A discrete type spectrum could be used to model two or 3 laser lines that could be each said to be a single discrete wavelength, but otherwise, most spectra=e.g., incadescent lamps and blackbodies, and other sources, the spectrum is normally continuous. Even a laser, under very high resolution, actually has a finite wavelength range and could be said to be continuous, and some laser lines are more monochromatic than others. One example: A 2 mwatt HeNe laser at $\lambda=632.8$ nm. You could say it is a single wavelength putting out 2 mwatts of power. Other than lasers, the spectral curves you see of sources will, in general, be continous.

This is the answer to my problem in a previous post of mine in which I was wondering to what extent one would have to go to sum up the energies of each wavelength individually (energy of each wavelength with wavelengths differing in steps of 0.001 or 0.0000000..001, etc.).
I ended up concluding that this would give me an infinite energy sum. I was basically reinventing an analogue of the ultraviolet catastrophe all over again XD

It boggles my mind though that there doesn't seem to be discrete step at even the highest resolution. I know it's wrong but it either makes me conclude that there's an infinite amount of photons, each with 1 exact specific wavelength at infinitesimally small discrete steps or that each photon consists of a very small range of wavelengths. I'm not sure how I should accept a continuous spectrum without talking about infinite amount of energy or photons.

 

[Charles Link]

This is the answer to my problem in a previous post of mine in which I was wondering to what extent one would have to go to sum up the energies of each wavelength individually (energy of each wavelength with wavelengths differing in steps of 0.001 or 0.0000000..001, etc.).
I ended up concluding that this would give me an infinite energy sum. I was basically reinventing an analogue of the ultraviolet catastrophe all over again XD

It boggles my mind though that there doesn't seem to be discrete step at even the highest resolution. I know it's wrong but it either makes me conclude that there's an infinite amount of photons, each with 1 exact specific wavelength at infinitesimally small discrete steps or that each photon consists of a very small range of wavelengths. I'm not sure how I should reason this continuous spectrum without talking about infinite amount of energy or photons.

It would give you many, many points to sum, but remember, you multiply each point by $\Delta \lambda$. Regardless of the increased resolution, you still get the same area under the curve. In calculus, you actually take the limit as $\Delta x$ goes to zero. Suggestion for you: Graph $y=x^2$ from $x=0$ to $x=1$ and integrate it numerically at resolution $\Delta x=.1, .01, .001, \,and \, .0001$. Compare each to the exact calculus answer of 1/3. Even the $\Delta x=.1$ should get you reasonably close.

 

[Drakkith]

It boggles my mind though that there doesn't seem to be discrete step at even the highest resolution. I know it's wrong but it either makes me conclude that there's an infinite amount of photons, each with 1 exact specific wavelength at infinitesimally small discrete steps or that each photon consists of a very small range of wavelengths. I'm not sure how I should accept a continuous spectrum without talking about infinite amount of energy or photons.

If I remember correctly, there is some uncertainty in the energy/frequency of a photon, so it doesn't really even have a single frequency until you measure it. Or, another way of looking at it, is that the EM wave doesn't have a set amount of photons of specific frequencies. It will give you some spread across the number of photons and the frequency of each photon such that this always adds up to the total energy of the EM wave.

 

[Charles Link]

If I remember correctly, there is some uncertainty in the energy/frequency of a photon, so it doesn't really even have a single frequency until you measure it. Or, another way of looking at it, is that the EM wave doesn't have a set amount of photons of specific frequencies. It will give you some spread across the number of photons and the frequency of each photon such that this always adds up to the total energy of the EM wave.

In the case of a monochromatic source, such as a laser, you could actually compute the number of photons (but because of uncertainty principles the number is never exact), each with energy $E_p=\frac{hc}{\lambda}$. If you have so many milliwatts of laser power, that would mean so many photons per second, but the numbers are enormous, and would be on the order of Avogadro's number. It would be like a chemist wanting to count atoms instead of measuring things in grams.

 

[Drakkith]

In the case of a monochromatic source, such as a laser, you could actually compute the number of photons (but because of uncertainty principles the number is never exact), each with energy $E_p=\frac{hc}{\lambda}$.

Ah, but didn't you just say in post #154 that lasers are not perfectly monochromatic? I'm not aware of any perfectly monochromatic sources.

 

[Charles Link]

Ah, but didn't you just say in post #154 that lasers are not perfectly monochromatic? I'm not aware of any perfectly monochromatic sources.

Back to the OP's @JohnnyGui question: In general, you work with a continuous spectrum and compute the area under the spectral curve for the wavelength interval of interest. There are exceptions, but in general, this is how it is done. A similar thing applies to r-f (radio frequency)=radio waves. In some cases, the sources are essentially power generated at one frequency, in which case, it is not necessary to use integral calculus and/or computational techniques to compute the area under the curve. If you have a radio station at 98.6 MHz, at very high resolution there may be (there of course is) some spectral structure, but basically this is so and so many watts at 98.6 Mhz. A similar thing applies to some electromagnetic sources in the visible region. Most of them that you will encounter though, will be continuous even at medium resolution.

 

[JohnnyGui]

It would give you many, many points to sum, but remember, you multiply each point by ΔλΔλ \Delta \lambda . Regardless of the increased resolution, you still get the same area under the curve. In calculus, you actually take the limit as ΔxΔx \Delta x goes to zero. Suggestion for you: Graph y=x2y=x2 y=x^2 from x=0x=0 x=0 to x=1x=1 x=1 and integrate it numerically at resolution Δx=.1,.01,.001,and.0001Δx=.1,.01,.001,and.0001 \Delta x=.1, .01, .001, \,and \, .0001 . Compare each to the exact calculus answer of 1/3. Even the Δx=.1Δx=.1 \Delta x=.1 should get you reasonably close.

So it is possible to calculate the energy from a graph that shows the absolute energy in Watts for each individual wavelenght, by integrating with $Δ\lambda$?

I seem to be able to accept this calculation if it's about mathematical formulas. But as soon as I accept the theory that there are individual physical particles (photons), each having a specific different wavelength in a continuous spectrum, the multiplication of each point with $Δ\lambda$ gets thrown out of the window for me (how can a range of $Δ\lambda$ be used for 1 specific amount of energy while in that $Δ\lambda$ there are photons with different energies?)

So perhaps this theory about each photon having a specific wavelength is wrong, as @Drakkith pointed out?

 

[Charles Link]

Perhaps one way to illustrate the concept of the discrete case vs. the continuous case is to take a plastic ruler. We could make a discrete graph that assigned 10 grams at each marking=at 1", at 2", at 3",...and then count them up and we would find on a 12" ruler that we had 120 grams of plastic. If we asked how much plastic is at the 6" mark, the answer would be 10 grams. $\\$ The continuous case would assign a density $\delta (x)=10$ grams/inch, independent of x. If we want to know how much plastic is within .25" of the 6" mark, it would be $\delta (6") \, .25=2.5$ grams. The continuous case more accurately represents the make-up of the ruler. Can we say there are 10 grams at the 6" mark? Or do we say there are 10 grams per inch at the 6" mark? In a discrete representation, we could say there are 10 grams at the 6" mark, but the way it is presented in a spectal measurement, (and the spectrometer acceptance window $\Delta \lambda$ is often adjustable when a spectral scan is performed), is to say that there are 10 grams per inch at the 6" mark. If the spectrometer (measuring our ruler) uses a width of $\Delta x=.25$ inches, the measurement would record a mass of 2.5 grams, but in processing the data that would be taken into account, and the experimenter would say we had a density of 2.5 grams/.25"=10 grams/inch at x=6". $\\$ If we counted individual photons (but really impossible to count that way=there are too many of them), we would actually be doing a discrete representation of the spectrum, and we would need to assign bins to the individual wavelengths in nanometers, like we did with the ruler in the discrete case. If the wavelength was 635.63 nm, it would go in the 636 nm bin, etc. Instead though, the spectrum can be sampled in a spectrometer run with arbitrary resolution $\Delta \lambda$. Sometimes the spectrometer may use $\Delta \lambda=1$ nm, but if another $\Delta \lambda$ is used e.g. $\Delta \lambda =.25$ nm, it is still the $\Phi(\lambda)$ in watts/nm that is presented. $\\$Note: In the prism type of spectrometer, you can adjust the width of the slit over which you are sampling the spectrum. (e.g. You can take a sample over a narrow part of the blue region, $\lambda=450$ nm (approximately)You might have the slit adjusted so that $\Delta \lambda=10$ nm ). The light comes out of the prism over a continuous spread of angles with the colors separated into an angular spread. Diffraction gratings are often also used in spectrometers, and the spreading that occurs is similar.

 

[JohnnyGui]

Perhaps one way to illustrate the concept of the discrete case vs. the continuous case is to take a plastic ruler. We could make a discrete graph that assigned 10 grams at each marking=at 1", at 2", at 3",...and then count them up and we would find on a 12" ruler that we had 120 grams of plastic. If we asked how much plastic is at the 6" mark, the answer would be 10 grams. $\\$ The continuous case would assign a density $\delta (x)=10$ grams/inch, independent of x. If we want to know how much plastic is within .25" of the 6" mark, it would be $\delta (6") \, .25=2.5$ grams. The continuous case more accurately represents the make-up of the ruler. Can we say there are 10 grams at the 6" mark? Or do we say there are 10 grams per inch at the 6" mark? In a discrete representation, we could say there are 10 grams at the 6" mark, but the way it is presented in a spectal measurement, (and the spectrometer acceptance window $\Delta \lambda$ is often adjustable when a spectral scan is performed), is to say that there are 10 grams per inch at the 6" mark. If the spectrometer (measuring our ruler) uses a width of $\Delta x=.25$ inches, the measurement would record a mass of 2.5 grams, but in processing the data that would be taken into account, and the experimenter would say we had a density of 2.5 grams/.25"=10 grams/inch at x=6". $\\$ If we counted individual photons (but really impossible to count that way=there are too many of them), we would actually be doing a discrete representation of the spectrum, and we would need to assign bins to the individual wavelengths in nanometers, like we did with the ruler in the discrete case. If the wavelength was 635.63 nm, it would go in the 636 nm bin, etc. Instead though, the spectrum can be sampled in a spectrometer run with arbitrary resolution $\Delta \lambda$. Sometimes the spectrometer may use $\Delta \lambda=1$ nm, but if another $\Delta \lambda$ is used e.g. $\Delta \lambda =.25$ nm, it is still the $\Phi(\lambda)$ in watts/nm that is presented. $\\$Note: In the prism type of spectrometer, you can adjust the width of the slit over which you are sampling the spectrum. (e.g. You can take a sample over a narrow part of the blue region, $\lambda=450$ nm (approximately)You might have the slit adjusted so that $\Delta \lambda=10$ nm ). The light comes out of the prism over a continuous spread of angles with the colors separated into an angular spread. Diffraction gratings are often also used in spectrometers, and the spreading that occurs is similar.

This helped me understand it a bit better. So basically you assign photons that have infinitesimally small difference in wavelength to 1 bin if you want to calculate the total energy. However, regarding the number of photons, this doesn't remove the fact that each photon has a infinitesimally different wavelength right? How would one explain that there's a finite energy in a seemingly infinite amount of photons? I think I'm delving into quantum physics here.

 

[Charles Link]

This helped me understand it a bit better. So basically you assign photons that have infinitesimally small difference in wavelength to 1 bin if you want to calculate the total energy. However, regarding the number of photons, this doesn't remove the fact that each photon has a infinitesimally different wavelength right? How would one explain that there's a finite energy in a seemingly infinite amount of photons? I think I'm delving into quantum physics here.

The photon count isn't infinite. The energy of each ( a result from quantum mechanics) is $E_p=\frac{hc}{\lambda}$. You can count them approximately, e.g. by saying you have $n= 2.3546 E+20$ photons per second. If you do the math, you will find a reasonable number of watts. Planck's constant $h=6.626 \, E-34$ joule-sec, speed of light $c=3.0 E+8$ m/sec and let wavelength $\lambda=550 E-9$ m (550 nm). $\\$ (1 watt=1 joule/sec, and power $P=n E_p$ where $n$ is the number of photons per unit time). Normally this calculation is done in reverse: You know $P$ and you know $E_p$ so that you can compute $n$.

 

[JohnnyGui]

The photon count isn't infinite. The energy of each ( a result from quantum mechanics) is $E_p=\frac{hc}{\lambda}$. You can count them approximately, e.g. by saying you have $n= 2.3546 E+20$ photons per second. If you do the math, you will find a reasonable number of watts. Planck's constant $h=6.626 \, E-34$ joule-sec, speed of light $c=3.0 E+8$ m/sec and let wavelength $\lambda=550 E-9$ m (550 nm). $\\$ (1 watt=1 joule/sec, and $P=n E_p$ where $n$ is the number of photons per unit time). Normally this calculation is done in reverse: You know $P$ and you know $E_p$ so that you can compute $n$.

Can I say that if a laser emits a wavelength of 550nm, that we'd calculate that the total energy of 1 photon is $E_p=\frac{hc}{550E-9}$, but since that emitted wavelength is in reality actually continuous (i.e. a range around 550nm that we've put in a bin of 550nm), that amount of energy per photon is actually spread among a number of photons around that wavelength that together sum up that amount of energy $E_p$ that we would think is the energy of 1 photon at exactly 550nm?

 

[Charles Link]

Can I say that if a laser emits a wavelength of 550nm, that we'd calculate that the total energy of 1 photon is $E_p=\frac{hc}{550E-9}$, but since that emitted wavelength is in reality actually continuous (i.e. a range around 550nm that we've put in a bin of 550nm), that amount of energy per photon is actually spread among a number of photons around that wavelength that together sum up that amount of energy $E_p$ that we would think is the energy of 1 photon at exactly 550nm?

If the spectrometer is set to a slit width of $\Delta \lambda =1$ nm and is set at $\lambda=550$ nm, and we measured 2 watts, the conclusion would be that $\Phi(\lambda)=2$ watts/nm at $\lambda=550$ nm (in the interval $\Delta \lambda=1$ nm from $\lambda=549.5$ to $550.5$ and zero outside of it ). A higher resolution spectral scan might show that we actually have 20 watts/nm in a spectral line that is only $\Delta \lambda=.1$ nm wide, perhaps centered at 550.2 nm . Alternatively, if we used a spectrometer that had $\Delta \lambda=10$ nm, we would measure 2 watts from 545 nm to 555 nm, but all we could say is that $\Phi(\lambda)=.2$ watts/nm from 545 nm to 555 nm. Sometimes, the results that get presented for $\Phi(\lambda)$ depend upon the resolution of the spectrometer during the measurement, but in any case, the total measured power $P=\int \Phi(\lambda) \, d \lambda$ should be the same in all cases. $\\$ Editing: A specific example is the sodium doublet from a sodium arc lamp. At low resolution, it is a single spectral line (bright spot) at $\lambda=589$ nm. A higher resolution spectrum will show it actually consists of two spectral lines, one at $\lambda=589.0$ nm and the other at $\lambda=589.6$ nm. $\\$ Meanwhile, to answer your question, if you put the constants in the numerator, the single photon energy calculation is quite accurate. This is the energy of a single photon, and it is not spread out around other photons. Changing the wavelength from 550 nm to 551 nm won't change the photon energy appreciably. The calculation is a good one for computing e.g. if a given photon can cause an electronic transition in an atom to occur, like being able to excite the electron in a hydrogen atom from the ground state to an excited state. In some semiconductors, the material is transparent to longer wavelength photons because the single photon doesn't carry sufficient energy to cause an electronic transition in the semiconductor that would cause the material to absorb it. Meanwhile shorter wavelengths get absorbed and the material can be used as a long (wavelength) pass filter.

 

[JohnnyGui]

If the spectrometer is set to a slit width of $\Delta \lambda =1$ nm and is set at $\lambda=550$ nm, and we measured 2 watts, the conclusion would be that $\Phi(\lambda)=2$ watts/nm at $\lambda=550$ nm (in the interval $\Delta \lambda=1$ nm from $\lambda=549.5$ to $550.5$ and zero outside of it ). A higher resolution spectral scan might show that we actually have 20 watts/nm in a spectral line that is only $\Delta \lambda=.1$ nm wide, perhaps centered at 550.2 nm . Alternatively, if we used a spectrometer that had $\Delta \lambda=10$ nm, we would measure 2 watts from 545 nm to 555 nm, but all we could say is that $\Phi(\lambda)=.2$ watts/nm from 545 nm to 555 nm. Sometimes, the results that get presented for $\Phi(\lambda)$ depend upon the resolution of the spectrometer during the measurement, but in any case, the total measured power $P=\int \Phi(\lambda) \, d \lambda$ should be the same in all cases. $\\$ Editing: A specific example is the sodium doublet from a sodium arc lamp. At low resolution, it is a single spectral line (bright spot) at $\lambda=589$ nm. A higher resolution spectrum will show it actually consists of two spectral lines, one at $\lambda=589.0$ nm and the other at $\lambda=589.6$ nm. $\\$ Meanwhile, to answer your question, if you put the constants in the numerator, the single photon energy calculation is quite accurate. This is the energy of a single photon, and it is not spread out around other photons. Changing the wavelength from 550 nm to 551 nm won't change the photon energy appreciably. The calculation is a good one for computing e.g. if a given photon can cause an electronic transition in an atom to occur, like being able to excite the electron in a hydrogen atom from the ground state to an excited state. In some semiconductors, the material is transparent to longer wavelength photons because the single photon doesn't carry sufficient energy to cause an electronic transition in the semiconductor that would cause the material to absorb it. Meanwhile shorter wavelengths get absorbed and the material can be used as a long (wavelength) pass filter.

I understand that the energy in a continuous spectrum is finite. I think my problem is that I'm assigning each photon to each wavelength. In case of a continuous spectrum this leads to the (false) idea that there are unlimited photons.
How about if I say that each photon has a certain range of wavelengths of a continuous spectrum (and using its average wavelength for calculating its energy is accurate enough)? Is this correct to say?

 

[Charles Link]

I understand that the energy in a continuous spectrum is finite. I think my problem is that I'm assigning each photon to each wavelength. In case of a continuous spectrum this leads to the (false) idea that there are unlimited photons.
How about if I say that each photon has a certain range of wavelengths of a continuous spectrum (and using its average wavelength for calculating its energy is accurate enough)? Is this correct to say?

Yes. That would work. You could make the comparison to the atoms in a ruler. (The number of photons per second with power P=1 watt is on a similar order of magnitude.) Instead of counting grams of material, you could count atoms. For the positions of the atoms, you would need to group them into bins. Maybe every .001" you would have another bin for a new position, etc. It's a similar thing with the energy. In any case, you are still computing the density=the number of atoms per inch which equates to the number of atoms in the bin of .001" wide divided by .001".

 

[JohnnyGui]

Yes. That would work. You could make the comparison to the atoms in a ruler. (The number of photons per second with power P=1 watt is on a similar order of magnitude.) Instead of counting grams of material, you could count atoms. For the positions of the atoms, you would need to group them into bins. Maybe every .001" you would have another bin for a new position, etc. It's a similar thing with the energy. In any case, you are still computing the density=the number of atoms per inch which equates to the number of atoms in the bin of .001" wide divided by .001".

Great, this would make sense to me. I think my culprit is that I couldn't understand "density" in the case of a wavelength vs energy curve. I was basically saying that even a wavelength range width of 0 would still contain energy (an analogue would be saying that a width of 0 of a ruler would contain mass) because I thought that a wavelength is not a dimension like width but a property of a photon which has a particular energy assigned to it.

 

[JohnnyGui]

@Charles Link : Sorry for jumping to a totally different subject (I was going off-topic after all) but I've been delving into emissivity yet again and something came up to me.

In the case of an object only interacting with its surroundings through radiation, I now totally understand that absorptivity and emissivity of that object must be equal in the case of thermal equilibrium (radiated energy = absorbed energy). But why do the absoprtivity and emissivity have to differ if the object has a different temperature than its surroundings?
If the surrounding is hotter/colder than the object, the object would still heat up/cool down if the absorptivity is equal to the emissivity. It's the difference in temperature that makes it radiate more/less energy than absorbing it.

For example, if the surroundings is at 300K and an object at 500K with an emissivity of 0.5, that object would radiate an emissive power of $500^4 \cdot \sigma \cdot 0.5$. If the absorptivity is also 0.5, it would absorb a part of $300^4 \cdot \sigma \cdot 0.5$ which is less than the radiating energy. Therefore it cools down.
So why would that object require to have an absorptivity different from its emissivity in this case?

 

[Charles Link]

@Charles Link : Sorry for jumping to a totally different subject (I was going off-topic after all) but I've been delving into emissivity yet again and something came up to me.

In the case of an object only interacting with its surroundings through radiation, I now totally understand that absorptivity and emissivity of that object a must be equal in the case of thermal equilibrium (radiated energy = absorbed energy). But why do the absoprtivity and emissivity have to differ if the object has a different temperature than its surroundings?
If the surrounding is hotter/colder than the object, the object would still heat up/cool down if the absorptivity is equal to the emissivity. It's the difference in temperature that makes it radiate more/less energy than absorbing it.
For example, if the surroundings is at 300K and an object at 500K with an emissivity of 0.5, that object would radiate an emissive power of $500^4 \cdot \sigma \cdot 0.5$. If the absorptivity is also 0.5, it would absorb a part of $300^4 \cdot \sigma \cdot 0.5$ which is still less than the radiated energy.
Why would that object require to have a different absorptivity than emissivity?

It doesn't. The equation for rate of heat leaving the object per unit area $W$ for an object of emissivity $\epsilon$ at temperature $T _1$ with surroundings at temperature $T_2$ looks like this: $W=\epsilon \sigma (T_1^4-T_2^4)$. The emissivity is the same for both temperatures in the equation. In the first part of the equation, the emissivity is a factor in how much is radiated. In the second part, it is a factor in how much is absorbed. The factor itself stays the same.

 

[JohnnyGui]

It doesn't. The equation for rate of heat leaving the object per unit area $W$ for an object of emissivity $\epsilon$ at temperature $T _1$ with surroundings at temperature $T_2$ looks like this: $W=\epsilon \sigma (T_1^4-T_2^4)$. The emissivity is the same for both temperatures in the equation. In the first part of the equation, the emissivity is a factor in how much is radiated. In the second part, it is a factor in how much is absorbed. The factor itself stays the same.

Ah ok, I got confused because Kirchoff's law states that a = e IF there's thermal equilibrium. Doesn't that imply that they differ if there's no thermal equilibrium?

Also, regarding your formula, doesn't that one give the amount of energy that makes an object cool down/warm up per unit time? The amount of energy that it radiates per unit area at a particular moment is still equal to $T_1^4 \cdot \sigma \cdot \epsilon$ right? (Where T is a function of your mentioned formula divided by the heat capacity over time)

 

[Charles Link]

Ah ok, I got confused because Kirchoff's law states that a = e IF there's thermal equilibrium. Doesn't that imply that they differ if there's no thermal equilibrium?

Also, regarding your formula, doesn't that one give the amount of energy that makes an object cool down/warm up? The amount of energy that it radiates per unit area at a particular moment is still equal to $T^4 \cdot \sigma \cdot \epsilon$ right? (Where T is a function of your mentioned formula divided by the heat capacity over time)

Yes, but if the object is in an enclosure where the surrounding walls are at temperature $T_2$, (emissivity is assumed to be equal to 1 for the enclosure, although that isn't even necessary), and the object is small compared to the size of the enclosure, then the power incident on the surface per unit area from the enclosure will be $E=\sigma T_2^4$ and the amount absorbed per unit area per unit time will be $W_2=\epsilon \sigma T_2^4$. (Remember, we previously computed the irradiance onto an entire plane from the source, where we had a $cos^4 (\theta )$ computation. The power incident from the entire plane onto the source is a similar calculation. The enclosure could simply be a ceiling that extends over the entire plane. If this ceiling has emissivity equal to one, then you don't need to worry about the surface on the floor in the plane of the source contributing to complete the enclosure, and you also don't need to have your object be small in size). $\\$ And yes, we previously used the same equation in this thread to estimate the temperature of an incadescent filament, as well as how it would cool down once the current/voltage was removed.

 

[JohnnyGui]

Yes, but if the object is in an enclosure where the surrounding walls are at temperature $T_2$, (emissivity is assumed to be equal to 1 for the enclosure, although that isn't even necessary), and the object is small compared to the size of the enclosure, then the power incident on the surface per unit area from the enclosure will be $E=\sigma T_2^4$ and the amount absorbed per unit area per unit time will be $W_2=\epsilon \sigma T_2^4$. (Remember, we previously computed the irradiance onto an entire plane from the source, where we had a $cos^4 (\theta )$ computation. The power incident from the entire plane onto the source is a similar calculation. The enclosure could simply be a ceiling that extends over the entire plane. If this ceiling has emissivity equal to one, then you don't need to worry about the surface on the floor in the plane of the source contributing to complete the enclosure.)

Makes sense, so integrating $I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R}$ for each $dA_2$ of the ceiling towards the object would yield $T_2^4 \cdot \sigma \cdot A_2$?

 

[Charles Link]

Makes sense, so integrating $I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R}$ for each $dA_2$ of the ceiling towards the object would yield $T_2^4 \cdot \sigma \cdot A_2$?

T_2^4 \cdot \sigma \cdot cos(\theta)^4

Yes, performing the integral over the complete $dA_2$ would yield $P_{incident}=\sigma T_2^4 A_1$ or equivalently $E=\sigma T_2^4$. (Please see also my edited additions to post #173. And as I recall, I think we even determined that this differential equation has a closed form solution, even though you used numerical methods to solve it. Let me look up that post #... Yes, see post #53 and post #54. You should now have a more complete understanding of what led us to the differential equation that we used in posts #53 and #54).

 

[JohnnyGui]

Yes, performing the integral over the complete $dA_2$ would yield $P_{incident}=\sigma T_2^4 A_1$ or equivalently $E=\sigma T_2^4$.

Hmm, correct me if I'm wrong but I thought that integrating $I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R}$ over each $dA_2$ until all of the ceiling's surface has been calculated would give the total energy from the whole ceiling so that it gives $P = \sigma T_2^4 \cdot A_2$ instead of $E = \sigma T_2^4$. Furthermore, I'd think that doing a double integration of $I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R}$ for each $dA_1$ as well as $dA_2$ would yield $P_{incident} = \sigma T_2^4A_1$.

You should now have a more complete understanding of what led us to the differential equation that we used in posts #53 and #54).

I have indeed. :)

 

[Charles Link]

Hmm, correct me if I'm wrong but I thought that integrating $I_0 \cdot cos(\theta)^4 \cdot \frac{A_1}{R}$ over each $dA_2$ until all of the ceiling's surface has been calculated would give the total energy from the whole ceiling so that it gives $P = \sigma T_2^4 \cdot A_2$ instead of $E = \sigma T_2^4$. Furthermore, I'd think that doing a double integration of $I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R}$ for each $dA_1$ as well as $dA_2$ would yield $P_incident = \sigma T_2^4A_1$.
I have indeed. :)

For the ceiling, you need to compute how much reaches surface 1. The simplest way is to take a point on surface 1 (at (0,0,h)) and sum (integrate) the $dE$ that results from each $dA_2$ to give you the total $E$ at the point on surface 1. (You can make the center of the $A_2$ surface to be at (0,0,0)). The result is that the irradiance at surface 1 from surface 2 is $E=\sigma T_2^4$. $\\$ To get you started with that computation, the brightness $L_2=\frac{\sigma T_2^4}{\pi}$. The integral is really identical in form to the one with $cos^4(\theta)$ that we previously did to show that $P=L_1 A_1 \pi$ for the power emerging from area $A_1$ of surface 1. (See posts #79,#86,#87, and also especially posts #107 and #110). $\\$ And you are correct, the total power $P_2$ coming off of surface $A_2$ is $P_2=A_2 \sigma T_2^4$, but we are only interested in how much reaches $A_1$. It's a clumsy multiple integral if you let $A_1$ be a large area, particularly if $A_2$ is finite in size. It's much easier to just compute $E$ at a point on surface 1 from the entire $A_2$ that extends to infinity over the whole plane.

 

[JohnnyGui]

For the ceiling, you need to compute how much reaches surface 1. The simplest way is to take a point on surface 1 (at (0,0,h)) and sum (integrate) the $dE$ that results from each $dA_2$ to give you the total $E$ at the point on surface 1. (You can make the center of the $A_2$ surface to be at (0,0,0)). The result is that the irradiance at surface 1 from surface 2 is $E=\sigma T_2^4$. $\\$ To get you started with that computation, the brightness $L_2=\frac{\sigma T_2^4}{\pi}$. The integral is really identical in form to the one with $cos^4(\theta)$ that we previously did to show that $P=L_1 A_1 \pi$ for the power emerging from area $A_1$ of surface 1. (See posts #79,#86,#87, and also especially posts #107 and #110). $\\$ And you are correct, the total power $P_2$ coming off of surface $A_2$ is $P_2=\sigma T^4 \cdot A_2$, but we are only interested in how much reaches $A_1$. It's a clumsy multiple integral if you have $A_1$ be a large area. It's much easier to just compute $E$ at a point on surface 1 from the entire $A_2$ that extends to infinity over the whole plane.

Ah, for some reason I falsely considered $E$ to be $P$ and also didn't notice that you've combined $T_2$ with $A_1$ in the formula. It sounds actually very logical but let's see if I can prove this mathematically.

So regarding the calculation from the older post, if we consider the object a point source emitting radiation at the ceiling $A2$, then this means that (in the 2D plane):
$$\int I_0 \cdot cos(\theta)^4 \cdot \frac{dA_2}{R^2} = P_{object}$$
Now, let's make the object have a surface $A_1$ and receiving radiation from the ceiling $A_2$. This means that each $dA_2$ is emitting radiation on the whole $A_1$ (integrating for each $dA_1$) of energy:
$$\int I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R^2} = P_{incident from each dA_2}$$
Integrating this integration for each $dA_2$ would give the incident energy from the whole ceiling onto $A_1$:
$$\int \int I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R^2} \cdot dA_2 = P_{incident from whole A_2}$$
Since $I_0$ is equal to $dA_2 \cdot T^4_2 \cdot \sigma \cdot \frac{1}{π}$, I'd have to prove the following:
$$\int \int \frac{dA_2 \cdot T^4_2 \cdot \sigma}{π} \cdot cos(\theta)^4 \cdot \frac{dA_1}{R^2} \cdot dA_2 = T^4_2 \cdot A_1 \cdot \sigma$$
Am I making any sense? (I'm aware this is a different approach)

 

[Charles Link]

Ah, for some reason I falsely considered $E$ to be $P$ and also didn't notice that you've combined $T_2$ with $A_1$ in the formula. It sounds actually very logical but let's see if I can prove this mathematically.

So regarding the calculation from the older post, if we consider the object a point source emitting radiation at the ceiling $A2$, then this means that (in the 2D plane):
$$\int I_0 \cdot cos(\theta)^4 \cdot \frac{dA_2}{R^2} = P_{object}$$
Now, let's make the object have a surface $A_1$ and receiving radiation from the ceiling $A_2$. This means that each $dA_2$ is emitting radiation on the whole $A_1$ (integrating for each $dA_1$) of energy:
$$\int I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R^2} = P_{incident from each dA_2}$$
Integrating this integration for each $dA_2$ would give the incident energy from the whole ceiling onto $A_1$:
$$\int \int I_0 \cdot cos(\theta)^4 \cdot \frac{dA_1}{R^2} \cdot dA_2 = P_{incident from whole A_2}$$
Since $I_0$ is equal to $dA_2 \cdot T^4_2 \cdot \sigma \cdot \frac{1}{π}$, I'd have to prove the following:
$$\int \int \frac{dA_2 \cdot T^4_2 \cdot \sigma}{π} \cdot cos(\theta)^4 \cdot \frac{dA_1}{R^2} \cdot dA_2 = T^4_2 \cdot A_1 \cdot \sigma$$
Am I making any sense? (I'm aware this is a different approach)

Yes, it makes perfect sense. But you do have one simplifying fact=since the $A_2$ surface is infinite in extent, every point on $A_1$ will necessarily receive the same irradiance level. Thereby, the integral over $dA_1$ is unnecessary, and you simply get the result that $E=\sigma T_2^4$. It follows immediately that $P_{incident}=A_1 \sigma T_2^4$, since $E$ must be uniform across $A_1$. (If you compute the irradiance $E$ at a different point on $A_1$, since $A_2$ is infinite in extent, you will get the exact same answer.) $\\$ Note: You need to integrate $dE$ rather than $dP$. $\\$ Meanwhile, the integral over $dA_2$ is quite straightforward=see post #110 for the details.

 

[JohnnyGui]

Yes, it makes perfect sense. But you do have one simplifying fact=since the $A_2$ surface is infinite in extent, every point on $A_1$ will necessarily receive the same irradiance level. Thereby, the integral over $dA_1$ is unnecessary, and you simply get the result that $E=\sigma T_2^4$. It follows immediately that $P_{incident}=A_1 \sigma T_2^4$, since $E$ must be uniform across $A_1$. (If you compute the irradiance $E$ at a different point on $A_1$, since $A_2$ is infinite in extent, you will get the exact same answer.) $\\$ Note: You need to integrate $dE$ rather than $dP$. $\\$ Meanwhile, the integral over $dA_2$ is quite straightforward=see post #110 for the details.

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?

 

[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!