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?