Emission spectra of different materials

[Charles Link]

I think I have found an alternative way to show why $\Phi(\lambda)$ is not the slope of a spectrum curve, although I'm not sure it's the correct way to explain this. Consider the following spectrum curve that shows many watts each wavelength is emitted from a source.
View attachment 205862
Say I'd want to calculate the energy from the left side of this curve, starting from 400nm, up to the peak wavelength (around 560 nm) that is emitted at 680 Watt as shown.
If $\Phi(\lambda)$ is the slope (i.e. derivative) of this curve in Watts/nm, then integrating $\Phi(\lambda)$ would only give the difference in Watts between 400 nm and 560nm, which is 680 Watts. Correct?

So, this means that $\Phi(\lambda)$ must be the function of the curve itself since integrating the function of the curve itself would give the area beneath it.
Is this reasoning correct?

Let $A(\lambda)=\int\limits_{0}^{\lambda} \Phi(\lambda ') \, d \lambda '$, then $\frac{d A(\lambda)}{d \lambda}=\Phi(\lambda)$. $\frac{d A(\lambda)}{d \lambda }$ is of course the slope of $A(\lambda)$ vs. $\lambda$ at any point. In spectroscopy, (unlike probability theory where the integral (area under the curve) of the Gaussian distribution is often tabulated), the function $A(\lambda )$ simply isn't in widespread use. Editing: I should qualify the last statement: Many times the integration of $\Phi(\lambda)$ is performed between two wavelengths to get $P=\int\limits_{\lambda_1}^{\lambda_2} \Phi(\lambda) \, d \lambda$, but this is done without ever tabulating $A(\lambda)$. The power $P=A(\lambda_2)-A(\lambda_1)$, but $\Phi(\lambda)$ is normally tabulated, and I think I can say I have never seen a table of the function $A(\lambda)$ in any spectroscopic publication. Mathematically, it could be done this way, but it simply isn't.

 

[JohnnyGui]

Let $A(\lambda)=\int\limits_{0}^{\lambda} \Phi(\lambda ') \, d \lambda '$, then $\frac{d A(\lambda)}{d \lambda}=\Phi(\lambda)$. $\frac{d A(\lambda)}{d \lambda }$ is of course the slope of $A(\lambda)$ vs. $\lambda$ at any point. In spectroscopy, (unlike probability theory where the integral (area under the curve) of the Gaussian distribution is often tabulated), the function $A(\lambda )$ simply isn't in widespread use. Editing: I should qualify the last statement: Many times the integration of $\Phi(\lambda)$ is performed between two wavelengths to get $P=\int\limits_{\lambda_1}^{\lambda_2}$, but this is done without ever tabulating $A(\lambda)$. The above $P=A(\lambda_2)-A(\lambda_1)$, but $\Phi(\lambda)$ is normally tabulated, and I think I can say I have never seen a table of the function $A(\lambda)$ in any spectroscopic publication.

So if I understand correctly $A(\lambda)$ is the total energy in a chosen wavelength range of the spectrum curve shown in my previous post. And $A(\lambda)$ has the derivatie $\Phi(\lambda)$ which is the function of the spectrum curve? (apologies if I misunderstood your post)

 

[Charles Link]

So if I understand correctly $A(\lambda)$ is the total energy in a specific wavelength range of the spectrum curve shown in my previous post that has the derivatie $\Phi(\lambda)$ which is the function of the spectrum curve? (apologies if I misunderstood your post)

Starting at the left end of the curve, yes, $A (\lambda)$ is the area under the curve up to (and including) wavelength $\lambda$. It is analogous mathematics that distance $s(t)=\int\limits_{0}^{t} v(t') \, dt'$ and $\frac{d s(t)}{dt}=v(t)$. If you have a graph of the velocity $v$ vs. time $t$, the area under the curve is the distance traveled. Meanwhile on a graph of $s$ vs. $t$, $v=\frac{ds}{dt}$ is the slope of that graph at any point.

 

[JohnnyGui]

Starting at the left end of the curve, yes, $A (\lambda)$ is the area under the curve up to (and including) wavelength $\lambda$.

Great, because this leads me to the root of my problem XD. If $\Phi(\lambda)$ is the function of the spectrum curve, then how does it have units of Watts per nanometre? Shouldn't it give only Watts for a chosen wavelength? The per nanometre part kind of implies that it's a derivative of the spectrum curve while it isn't.

 

[Charles Link]

Starting at the left end of the curve, yes, $A (\lambda)$ is the area under the curve up to (and including) wavelength $\lambda$.

Great, because this leads me to the root of my problem XD. If $\Phi(\lambda)$ is the function of the spectrum curve, then how does it have units of Watts per nanometre? Shouldn't it give only Watts for a chosen wavelength? The per nanometre part kind of implies that it's a derivative of the spectrum curve while it isn't.

Please read the edited version of my previous post. $\\$ To answer the nanometer question, if the wavelength is in nanometers, then $\Phi(\lambda)$ has units of watts per nanometer. Over a 50 nm interval, say from 550-600 nm, it will integrate correctly if you take the height of the curve at 575 nm (assuming it is a uniform height) and multiply by 50. Numerically, you could do the following $P=\Phi(550) \, 5 \, nm+\Phi(555) \, 5 \, nm +\Phi(560) \, 5 \,nm +... +\Phi(595) \, 5 \, nm$. Alternatively, at higher resolution $P=\Phi(550) \, 1 \, nm +\Phi(551) \, 1 \, nm+... +\Phi(599) \, 1 \, nm$. You'll basically get the same answer both ways with the higher resolution possibly providing increased accuracy. $\\$ Meanwhile, yes, $\Phi(\lambda)$ is the derivative of $A(\lambda)$, but they don't call $A(\lambda)$ the spectrum curve (they call $\Phi(\lambda)$ the spectral curve) =they normally don't give $A(\lambda)$ any name=again, mathematically, they could have gone the route you are asking about, but it simply has never been presented that way. $\\$ Additional item of interest is the Planck blackbody spectral function $L(\lambda, T)$ back in post # 81. You may find it of interest that this function also has units of watts/.../per unit wavelength. If a graph is displayed with wavelength in nm, then it has units of watts/.../nm. Other than the extra units of watts/(m^2 sr nm), this function is very much like the function $\Phi(\lambda)$.

 

[JohnnyGui]

Please read the edited version of my previous post. $\\$ To answer the nanometer question, if the wavelength is in nanometers, then $\Phi(\lambda)$ has units of watts per nanometer. Over a 50 nm interval, say from 550-600 nm, it will integrate correctly if you take the height of the curve at 575 nm (assuming it is a uniform height) and multiply by 50. Numerically, you could do the following $P=\Phi(550) \, 5 \, nm+\Phi(555) \, 5 \, nm +\Phi(560) \, 5 \,nm +... +\Phi(595) \, 5 \, nm$. Alternatively, at higher resolution $P=\Phi(550) \, 1 \, nm +\Phi(551) \, 1 \, nm+... +\Phi(599) \, 1 \, nm$. You'll basically get the same answer both ways with the higher resolution possibly providing increased accuracy. $\\$ Meanwhile, yes, $\Phi(\lambda)$ is the derivative of $A(\lambda)$, but they don't call $A(\lambda)$ the spectrum curve (they call $\Phi(\lambda)$ the spectral curve) =they normally don't give $A(\lambda)$ any name=again, mathematically, they could have gone the route you are asking about, but it simply has never been presented that way. $\\$ Additional item of interest is the Planck spectral function $L(\lambda, T)$ back in post # ... You may find it of interest that this function also has units of watts/.../per unit wavelength. If a graph is displayed with wavelength in nm, then it has units of watts/.../nm.

I think your analogy with the velocity-time diagram made me find the culprit. My spectrum graph shown in post #140 has a y-axis in Watts, and not in Watts/nanometre. Hence if you integrate a function that has Watts/nm as units ($\Phi(\lambda)$) you'll merely get the difference in Watts between the 2 chosen wavelengths in my Watts vs $\lambda$ spectrum curve in post #140.

So with your mentioned example of integrating $\Phi(\lambda)$ between $550nm-600nm$, in a Watts vs $\lambda$ graph this would give give the following answer:

However, in a graph of Watts/nm vs $\lambda$, integrating $\Phi(\lambda)$ would give the area under that Watts/nm vs $\lambda$ graph.

So my question is; if I'm using a Watts vs $\lambda$ graph (like in my post #140) instead of a Watts/nm vs $\lambda$ graph, and I want to calculate the energy within a specific wavelength range using that graph, shouldn't I integrate a function other than $\Phi(\lambda)$ to get the area under that Watts vs $\lambda$ graph? Is that the step by step numerical calculation that you were talking about in your post #145?

 

[Charles Link]

I think your analogy with the velocity-time diagram made me find the culprit. My spectrum graph shown in post #140 has a y-axis in Watts, and not in Watts/nanometre. Hence if you integrate a function that has Watts/nm as units ($\Phi(\lambda)$) you'll merely get the difference in Watts between the 2 chosen wavelengths in my Watts vs $\lambda$ spectrum curve in post #140.

So with your mentioned example of integrating $\Phi(\lambda)$ between $550nm-600nm$, in a Watts vs $\lambda$ graph this would give give the following answer:
View attachment 205866

However, in a graph of Watts/nm vs $\lambda$, integrating $\Phi(\lambda)$ would give the area under that Watts/nm vs $\lambda$ graph.

So my question is; if I'm using a Watts vs $\lambda$ graph (like in my post #140) instead of a Watts/nm vs $\lambda$ graph, and I want to calculate the energy within a specific wavelength range using that graph, shouldn't I integrate a function other than $\Phi(\lambda)$ to get the area under that Watts vs $\lambda$ graph? Is that the step by step numerical calculation that you were talking about in your post #145?

I do believe you have most of it figured out. Good work ! $\\$ In your post #140, that graph is clearly mislabeled on the y-axis and it should read watts/nm. Sometimes, the per nm designation is understood, but it really belongs in the label on the y-axis. $\\$ Just an additional comment: With these spectral curves, you always compute the area under the curve=you don't take a difference such as $N-X$. In that particular instance, I think you are trying to do something that is incorrect. These spectral curves are $\Phi(\lambda)$ type curves in all cases (e.g. per unit wavelength). They are not $A(\lambda)$ type curves. If you were showing a graph of $A(\lambda)$ vs. $\lambda$ in post 146, then what you did is correct, but again, it simply is never done that way in spectroscopy. $\\$ (And to say it again=it is a common technique in probability theory to tabulate the integrated curve $A(\lambda)$. The two letters they typically use are $F(\lambda)$ (for the integrated), and $f(\lambda)$ for what they call the probability density function, but the spectroscopists do it without the $A(\lambda)$).

 

[JohnnyGui]

I do believe you have most of it figured out. Good work ! $\\$ In your post #140, that graph is clearly mislabeled on the y-axis and it should read watts/nm. Sometimes, the per nm designation is understood, but it really belongs in the label on the y-axis.

This has been bothering me for a long while and I can't believe this was actually the culprit. Thanks a lot for explaining this!

Let's say I'm very stubborn and I want to use the graph in my post #140 with just Watts on the y-axis vs $\lambda$ to calculate the energy between 550nm-600nm. Should I read off the amount of Watts for each wavelength within 550-600 nm and add them all together? Or would this still give an incorrect answer?

 

[Charles Link]

This has been bothering me for a long while and I can't believe this was actually the culprit. Thanks a lot for explaining this!

Let's say I'm very stubborn and I want to use the graph in my post #140 with just Watts on the y-axis vs $\lambda$ to calculate the energy between 550nm-600nm. Should I read off the amount of Watts for each wavelength within 550-600 nm and add them all together? Or would this still give an incorrect answer?

When they make what is really a mistake with their units on the y-axis, you need to question whether the numbers such as 600 and 700 watts/nm are correct, but using those numbers, when integrating this source, the answer that you get it that it has about 70,000 watts of radiated power. (I estimated this by taking 500 watts/nm at the shoulders near the peak multiplied by a width of about $\Delta \lambda=100$ nm, (for wavelength $\lambda$ from 600 nm to 700 nm) and then observing the entire area under the curve might be about 35% more). It general, I would recommend finding a book that gets the units correct, but it still made for a useful diagram. $\\$ Note: In computing these integrals numerically, it is much easier to use a lower resolution , e.g. $\Delta \lambda=10 \, nm$ or even $\Delta \lambda =100 \, nm$. The summation you do is $\sum\limits_{i} \Phi(\lambda_i) \, \Delta \lambda$. (At higher resolution=smaller $\Delta \lambda$, there are more $\Phi(\lambda_i)$ terms in the sum. At low resolution (e.g. $\Delta \lambda=100 \, nm$ ), you might have i=1 to 10. At very high resolution (e.g. $\Delta \lambda=1 \, nm$), you would then have i=1 to 1000, etc.) As I showed in post # 145, you can do it at various resolutions, but there is no reason to use $\Delta \lambda=1 \, nm$. When the spectral curve has little structure, you can get a very accurate answer with a low resolution summation. If you followed the explanation of how I estimated 70,000 watts, I basically computed the integral at a resolution of $\Delta \lambda=100 \, nm$. If you compute it at $\Delta \lambda =1 \, nm$ resolution, it might take you 30 minutes or more by hand ,(of course you can simply use a spreadsheet and let the computer do the work), and most likely you will still get an answer for the area under the entire curve of about 70,000 watts.

 

[JohnnyGui]

When they make what is really a mistake with their units on the y-axis, you need to question whether the numbers such as 600 and 700 watts/nm are correct, but using those numbers, when integrating this source, the answer that you get it that it has about 70,000 watts of radiated power. (I estimated this by taking 500 watts/nm at the shoulders near the peak multiplied by a width of about Δλ=100Δλ=100 \Delta \lambda=100 nm, (for wavelength λλ \lambda from 600 nm to 700 nm) and then observing the entire area under the curve might be about 35% more). It general, I would recommend finding a book that gets the units correct, but it still made for a useful diagram. \\ Note: In computing these integrals numerically, it is much easier to use a lower resolution , e.g. Δλ=10nmΔλ=10nm \Delta \lambda=10 \, nm or even Δλ=100nmΔλ=100nm \Delta \lambda =100 \, nm . The summation you do is ∑iΦ(λi)Δλ∑iΦ(λi)Δλ \sum\limits_{i} \Phi(\lambda_i) \, \Delta \lambda . (At higher resolution=smaller ΔλΔλ \Delta \lambda , there are more Φ(λi)Φ(λi) \Phi(\lambda_i) terms in the sum. At low resolution (e.g. Δλ=100nmΔλ=100nm \Delta \lambda=100 \, nm ), you might have i=1 to 10. At very high resolution (e.g. Δλ=1nmΔλ=1nm \Delta \lambda=1 \, nm ), you would then have i=1 to 1000, etc.) As I showed in post # 145, you can do it at various resolutions, but there is no reason to use Δλ=1nmΔλ=1nm \Delta \lambda=1 \, nm . When the spectral curve has little structure, you can get a very accurate answer with a low resolution summation. If you followed the explanation of how I estimated 70,000 watts, I basically computed the integral at a resolution of Δλ=100nmΔλ=100nm \Delta \lambda=100 \, nm . If you compute it at Δλ=1nmΔλ=1nm \Delta \lambda =1 \, nm resolution, it might take you 30 minutes or more by hand ,(of course you can simply use a spreadsheet and let the computer do the work), and most likely you will still get an answer for the area under the entire curve of about 70,000 watts.

You mentioned here that this computation has to be done for a graph that has Watts/nm on the y-axis (a $\Phi(\lambda)$ vs $\lambda$ curve) right?
I might have misunderstood you, but I meant calculating the energy if the graph is a $A(\lambda)$ vs $\lambda$ curve (i.e. Watts on the y-axis instead of Watts/nm).

I might be blacking out again but here's what I find colliding:

For a $\Phi(\lambda)$ vs $\lambda$ curve that has Watts/nm on the y-axis, you'd have to integrate $\Phi(\lambda)$ for a certain wavelength range to calculate the energy. Just like integrating a $v$ vs $t$ curve between $t_1-t_2$ would give the distance $s$ traveled between $t_1 - t_2$ which is the distance $s$ at $t_2$ minus distance at $t_1$ in a $s$ vs $t$ curve, integration of $\Phi(\lambda)$ between $\lambda_1 - \lambda_2$ in a $\Phi(\lambda)$ vs $\lambda$ curve would give the energy between $\lambda_1-\lambda_2$ which is the energy at $\lambda_2$ minus the energy of $\lambda_1$ in a $A(\lambda)$ curve, if I understand this correctly.

However, say I now have a $A(\lambda)$ vs $\lambda$ curve which has Watts on the y-axis and I want to calculate the energy from that curve between $\lambda_1-\lambda_2$. In this case, my instinct would say that I'd have to add up every amount of Watts corresponding to each wavelength between $\lambda_1-\lambda_2$ (with a high resolution for examle). Notice that this would NOT give the same answer as when I integrate $\Phi{\lambda}$ in a $\Phi(\lambda)$ curve between $\lambda_1 - \lambda_2$, since that would merely give the difference of the energy between $\lambda_1$ and $\lambda_2$ in a $A(\lambda)$ curve.

You probably have been trying to explain me this the whole time, but I don't really get how I should calculate the amount of energy from a $A(\lambda)$ vs $\lambda$ curve to get the same amount of energy as integrating a $\Phi(\lambda)$ vs $\lambda$ curve.

 

[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).