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Looks like you have a good handle on it. I'd enjoy seeing a graph of the complete solution to the differential equation as a function of time for a couple of cases.
Charles Link said:Looks like you have a good handle on it. I'd enjoy seeing a graph of the complete solution to the differential equation as a function of time for a couple of cases.
Charles Link said:I am no expert in differential equations, but I think it is likely that the exact and final TT T will not be reached in a finite time. In any case the function T(t)T(t) T(t) could be found by numerical methods (again of which I am no expert), and I think it might be the case that T(t)T(t) T(t) approaches its final value asymptotically but never quite gets there.
Charles Link said:To answer question (2): In general the emissivity at a given wavelength for an opaque material is equal to the absorbance A=1−RA=1−R A=1-R where RR R is the reflectivity. This can get complicated by surfaces having reflectivities that depend on the angle of incidence, etc., so it is more applicable to diffuse reflectors that absorb a certain fraction of the incident light. In any case, if A=1A=1 A=1 for all wavelengths and all angles, then the emissivity ϵ=1ϵ=1 \epsilon=1 .
The emissivity affects how much energy is radiated. With the filament, you supply electrical power to raise the temperature. If you supply zero electrical power, the filament will be in thermal equilibrium with the walls of the room=which, regardless of their emissivity, the room, because it is enclosed will look like a blackbody insofar as how much thermal energy (in the form of electromagnetic waves) is in the air (i.e. being radiated and/or reflected off of the walls). The tungsten filament will be in equilibrium even with emissivity of .2. At equilibrium with zero input power, it will radiate at the same rate that it absorbs energy and be at the same temperature as the surface temperature of the walls of the room. Even if the room is in a total vacuum (no air), it will still have the electromagnetic wave energy density characteristic of what occurs inside any enclosure at temperature ## T ##. ## \\ ## And if you drill a small hole of area ## A ## in the wall, the energy coming out of that aperture will be exactly like that of a blackbody of area ## A ## at the same temperature as the surface temperature inside the walls of the room (We're assuming the light bulb is turned off.) . Meanwhile, if a light beam from the outside goes into the aperture, it bounces around inside the room and gets absorbed and doesn't find its way back out=thereby the aperture can be considered to have absorbtivity=1. (I don't want to use the letter ## A ## again, because I just used it for area.) That's why the aperture is said to have an emissivity of 1. (If the walls are so hot that they are red hot, the aperture would appear to be reddish in color and look exactly like the walls.) (We're assuming the walls inside the box are hot, but are insulated on the outside so that they are cool. The only thing that will look red from the outside is the aperture of area ## A ##. Anyway, that's a quick introduction to some of the details behind the science of "blackbodies".JohnnyGui said:Thanks. What surprises me is that the temperature in the SB-formula is determinede by the amount absorbance of wavelengths, since it has an emissivity factor in it. What if the rise in temperature of a material is caused by something else than the amount of absorbed wavelengths? Why would that then need an emissivity factor?
Thanks for the detailed explanation.Charles Link said:The emissivity affects how much energy is radiated. With the filament, you supply electrical power to raise the temperature. If you supply zero electrical power, the filament will be in thermal equilibrium with the walls of the room=which, regardless of their emissivity, the room, because it is enclosed will look like a blackbody insofar as how much thermal energy (in the form of electromagnetic waves) is in the air (i.e. being radiated and/or reflected off of the walls). The tungsten filament will be in equilibrium even with emissivity of .2. At equilibrium with zero input power, it will radiate at the same rate that it absorbs energy and be at the same temperature as the surface temperature of the walls of the room. Even if the room is in a total vacuum (no air), it will still have the electromagnetic wave energy density characteristic of what occurs inside any enclosure at temperature ## T ##. ## \\ ## And if you drill a small hole of area ## A ## in the wall, the energy coming out of that aperture will be exactly like that of a blackbody of area ## A ## at the same temperature as the surface temperature inside the walls of the room (We're assuming the light bulb is turned off.) . Meanwhile, if a light beam from the outside goes into the aperture, it bounces around inside the room and gets absorbed and doesn't find its way back out=thereby the aperture can be considered to have absorbtivity=1. (I don't want to use the letter ## A ## again, because I just used it for area.) That's why the aperture is said to have an emissivity of 1. (If the walls are so hot that they are red hot, the aperture would appear to be reddish in color and look exactly like the walls.) (We're assuming the walls inside the box are hot, but are insulated on the outside so that they are cool. The only thing that will look red from the outside is the aperture of area ## A ##. Anyway, that's a quick introduction to some of the details behind the science of "blackbodies".
An interesting experiment and the results need careful analysis before reaching a conclusion, I think. Very large thermal gradient at surface of the gold, I think - same as if it were covered with a good 'insulator'.Charles Link said:The answer is yes. Incidentally, at my workplace, using a thermal imager to test this concept , we once put a gold-coated plate in an electrical frying pan. (gold has reflectivity nearly 1 and thereby emissivity near zero throughout the infrared). The gold plate appeared very cold=(room temperature, caused by reflected energy from the walls of the room), while another object, (with emissivity close to 1), appeared to be nearly at the temperature of the frying pan. As cold as it looked according to the thermal imager, you couldn't touch the gold plate=it was hot !
The experiment was quite conclusive. (Incidentally gold would have a fairly high thermal conductivity, and it was kept in the frying pan for quite a long period of time. The material below the gold coating was also a metal.)sophiecentaur said:An interesting experiment and the results need careful analysis before reaching a conclusion, I think. Very large thermal gradient at surface of the gold, I think - same as if it were covered with a good 'insulator'.
Not intuitive though. The apparent paradox would need explaining and any misconceptions dealt with.Charles Link said:The experiment was quite conclusive.
The emissivity of the gold coating is nearly zero=this one is not paradoxical. ## M=\epsilon \sigma T^4 ## . The emissivity factor ## \epsilon ## is very small for gold.sophiecentaur said:Not intuitive though. The apparent paradox would need explaining and any misconceptions dealt with.
I could ask you. If there is no apparent paradox , why was the experiment conducted at all? Are you sure that everyone would have predicted that result?Charles Link said:The emissivity of the gold coating is nearly zero=this one is not paradoxical. ## M=\epsilon \sigma T^4 ## . The emissivity factor ## \epsilon ## is very small for gold.
Sorry, but I have a hard time understanding the link between emissivity and the amount of reflection. I don't see how a low emissivity automatically means having highly reflective property. Can't a material fully absorb any wavelength but NOT emit all of them at the same energy amount that it absorbed them with afterwards? (lattice interaction for example)Charles Link said:The answer is yes. Incidentally, at my workplace, using a thermal imager to test this concept , we once put a gold-coated plate in an electrical frying pan. (gold has reflectivity nearly 1 and thereby emissivity near zero throughout the infrared). The gold plate appeared very cold=(room temperature, caused by reflected energy from the walls of the room), while another object, (with emissivity close to 1), appeared to be nearly at the temperature of the frying pan. As cold as it looked according to the thermal imager, you couldn't touch the gold plate=it was hot !
It's called Kirchhoff's law (the relation between emissivity and reflectivity.)JohnnyGui said:Sorry, but I have a hard time understanding the link between emissivity and the amount of reflection. I don't see how a low emissivity automatically means having highly reflective property. Can't a material fully absorb any wavelength but NOT emit all of them at the same energy amount that it absorbed them with afterwards? (lattice interaction for example)
If the two were not equivalent, you could build an engine that would get hotter and hotter for free! Another Maxwell's Demon. Verboten, I'm afraid.JohnnyGui said:I don't see how a low emissivity automatically means having highly reflective property.
sophiecentaur said:If the two were not equivalent, you could build an engine that would get hotter and hotter for free! Another Maxwell's Demon. Verboten, I'm afraid.
This situation occurs in many cases - for instance when the object is being warmed up by incident radiation. That will happen when its temperature is lower than the surroundings. The heat it radiates will (of course) be less than the radiant heat it absorbs, but it's not the emissivity and absorptivity that are different - it's the temperatures.JohnnyGui said:How about a material that has a higher absorptivity than emissivity because it loses the absorbed energy in some other form than radiation?
sophiecentaur said:This situation occurs in many cases - for instance when the object is being warmed up by incident radiation. That will happen when its temperature is lower than the surroundings. The heat it radiates will (of course) be less than the radiant heat it absorbs, but it's not the emissivity and absorptivity that are different - it's the temperatures.
At thermal equilibrium, the radiation loss is equal to the radiation gain. If there is another method of acquiring heat at thermal equilibrium, that mechanism also has a balance/equality between energy input and energy output: ## \\ ## e.g. gas molecules at temperature T that make contact with the surface. If some of the collisions are inelastic, so that some of the gas molecules lose energy to the surface, in the same number of collisions, some gas molecules acquire energy as they collide with the surface. The energy must balance in thermal equilibrium.JohnnyGui said:But isn't it possible that there's thermal equilibrium (same temperatures) because a part of the absorbed energy is lost through another energy form at the same time as the radiation, while the surroundings receives both that radiation and that other form of energy?
So "Absorbed Q = Radiation loss + other form of energy loss".
Charles Link said:At thermal equilibrium, the radiation loss is equal to the radiation gain. If there is another method of acquiring heat at thermal equilibrium, that mechanism also has a balance/equality between energy input and energy output: ## \\ ## e.g. gas molecules at temperature T that make contact with the surface. If some of the collisions are inelastic, so that some of the gas molecules lose energy to the surface, in the same number of collisions, some gas molecules acquire energy as they collide with the surface. The energy must balance in thermal equilibrium.
The emissivity being equal to the absorbtivity is not fixed in stone because there are materials that are partially transmissive. In general though, thermal equilibrium requires a balance, which for opaque materials normally implies absorbtivity=emissivity. ## \\ ## Editing... On a very related topic, it might interest you to read up on the "spectral distribution" of the energy that is radiated according to the Stefan-Boltzman law=the "spectral distribution" is the Planck blackbody function, ## M(\lambda, T) ##, and the total area under the curve of the Planck function vs. wavelength gives ## M= \sigma T^4 ##.JohnnyGui said:Got it. This might be a too farfetched scenario but what if the material receives radiation from the surroundings, loses it part in the form of radiation and part in another energy form as I said, and then the surroundings receive both forms and turn it all into radiation energy again so that the material solely receives radiation energy but loses it radiation + some other form. Will there be a thermal equilibrium with different absorptivity and emissivity of the material in this case? Not sure if this is even possible.
Charles Link said:The emissivity being equal to the absorbtivity is not fixed in stone because there are materials that are partially transmissive. In general though, thermal equilibrium requires a balance, which for opaque materials normally implies absorbtivity=emissivity
Charles Link said:Editing... On a very related topic, it might interest you to read up on the "spectral distribution" of the energy that is radiated according to the Stefan-Boltzman law=the "spectral distribution" is the Planck blackbody function, M(λ,T)M(λ,T) M(\lambda, T) , and the total area under the curve of the Planck function vs. wavelength gives M=σT4M=σT4 M= \sigma T^4 .
Your conclusion in (1) I believe is completely correct. For (2), you can examine a diffuse surface on a microscopic scale: I picture it like a bunch of mountains that will have the light reflect back and forth between them before reflecting back out, so that the multiple bounces will increase the absorption.JohnnyGui said:Can't a material be partially transmissive and still be in thermal equilibrium so that its absorptivity = emissivity? For example, 2 materials A and B in a vacuum enclosed room all at the same temperature, material A being highly emissive and thus absorptive and B losing radiation energy through reflection and transmission. In order for A to not change in temperature it needs to emit equal radiation energy as it absorbs. In order for B to not change in temperature it needs to also emit (less) radiation energy as it absorbs. So that for material B: ##E_{in} - E_{reflection} - E_{transmission} = E_{absorption} = E_{emission}##.
This is quite interesting to know. Does that also mean that the deriviation of ##\sigma \cdot T^4## gives the spectral curve function of the blackbody, ##M(\lambda, T)##?
Two more questions (sorry for bombarding you guys with all this):
1. If a material is highly reflective and thus poorly emissive, does that mean that it would cool down much slower than a blackbody when they're both at the same temperature, both equally conductive/convective and both sitting in an environment with the same lower temperature?
2. I keep on reading that if you polish a surface it would make it more highly reflective and thus be less emissive. However, doesn't polishing a surface merely means that radiation would reflect in a more uniform direction? Before polishing, the surface was still highly reflective as after but it was just scattering the radiation more diffusely in all directions. So shouldn't the amount of reflected energy be the same before and after polishing?
Charles Link said:you can examine a diffuse surface on a microscopic scale: I picture it like a bunch of mountains that will have the light reflect back and forth between them before reflecting back out, so that the multiple bounces will increase the absorption.
Charles Link said:but it can be shown to give the precise result +∞∫0M(λ,T)dλ=σT4∫0+∞M(λ,T)dλ=σT4 \int\limits_{0}^{+\infty} M(\lambda,T) \, d \lambda=\sigma T^4 .
The source from the blackbody aperture radiates with a ## cos(\theta) ## fall-off in intensity from-on axis (## \theta ## is the polar angle), because the aperture appears to change in size/area as it is viewed at an angle. Thereby the effective solid angle of a hemisphere for this radiator thereby turns out to be ## \pi ## rather than ## 2 \pi ##. Meanwhile, solid angle (steradian) is defined as ## \Omega=\frac{A}{r^2} ##. Because of this, the steradian is actually dimensionless, and intensity ## I ## (watts/steradian) and power ## P ## (watts) essentially have the same units. The topic is tricky enough that you will likely find that some articles give much more thorough and better explanations of some of the finer details than others. ## \\ ## Meanwhile the ## M ## that I used, (## M=\sigma T^4 ## ), is radiant emittance per unit area off of the surface: Power ## P=M A ##. You may also see another unit called radiance, (## L ## ), used, where ## I=L A ## and ## M=L \pi ##. (I'm trying to keep it as simple as i can.) Notice also that ## P=I_o \pi ## where ## I_o ## is the on-axis intensity, and ## I (\theta)=I_o cos(\theta) ##. Suggestion is to compute ## P=\int \int I(\theta) \, d \Omega ## where ## d \Omega=sin(\theta) \, d \theta \, d \phi ## for a hemisphere. You should be able to show that ## P=I_o \pi ##. ## \\ ## One more unit you will likely see in the literature is the irradiance ## E ## defined as power/area. ## E=\frac{I}{r^2} ## (watts/m^2) for a point source that radiates spherically. One question people often have for this is how do the units work? If ## I ## is in watts/steradian, where did the "steradian" go? ( ## E ## is in watts/m^2 without any steradian.) And the answer is that the steradian is dimensionless and can be pulled out or put in, depending upon the quantity that is represented.JohnnyGui said:Ah, I kind of had this idea in my head but this verified it.
I had a small question regarding the units. Since ##\sigma \cdot T^4## doesn't contain the sufrace ##A## in it, doesn't that mean that this formula gives the total radiation intensity over whole spectrum?
I'm asking this because for some reason, this link says the following under paragraph 7.2.4:
"It has been shown that the irradiation field in an isothermal cavity is equal to ##E_b= \sigma \cdot T^4##. Moreover, the irradiation was same for all planes of any orientation within the cavity. It may then be shown that the intensity of the blackbody radiation ##I_b## is uniform. Thus, blackbody radiation is defined as ##E_b = \pi \cdot I_b##"
I find this remarkable because ##E_b## is already the intensity since its units are per unit area.
Charles Link said:Meanwhile the MM M that I used, (M=σT4M=σT4 M=\sigma T^4 ), is radiant emittance per unit area off of the surface: Power P=MAP=MA P=M A . You may also see another unit called radiance, (LL L ), used, where I=LAI=LA I=L A and M=LπM=Lπ M=L \pi . (I'm trying to keep it as simple as i can.) Notice also that P=IoπP=Ioπ P=I_o \pi where IoIo I_o is the on-axis intensity, and I(θ)=Iocos(θ)I(θ)=Iocos(θ) I (\theta)=I_o cos(\theta) . Suggestion is to compute P=∫∫I(θ)dΩP=∫∫I(θ)dΩ P=\int \int I(\theta) \, d \Omega where dΩ=sin(θ)dθdϕdΩ=sin(θ)dθdϕ d \Omega=sin(\theta) \, d \theta \, d \phi for a hemisphere. You should be able to show that P=IoπP=Ioπ P=I_o \pi
Charles Link said:In a good deal of the literature, the Planck blackbody function is written as L(λ,T)=2hc2λ5(exphc/(λkbT)−1)L(λ,T)=2hc2λ5(exphc/(λkbT)−1) L(\lambda, T)=\frac{2hc^2}{\lambda^5(exp^{hc/(\lambda k_b T)}-1)} without the factor of ππ \pi in the numerator, ( i.e. M(λ,T)M(λ,T) M(\lambda, T) has 2πhc22πhc2 2 \pi h c^2 in the numerator because M(λ,T)=L(λ,T)πM(λ,T)=L(λ,T)π M(\lambda, T)=L(\lambda,T) \pi ). \\ The result is +∞∫0L(λ,T)dλ=σT4π∫0+∞L(λ,T)dλ=σT4π\int\limits_{0}^{+\infty} L(\lambda,T) \, d \lambda=\frac{\sigma T^4}{\pi} . \\ A very interesting as well as important part of the Planck blackbody function is that integrating the Planck function actually provides the theoretical expression for σσ \sigma in terms of the other constants: σ=π260k4bℏ3c2σ=π260kb4ℏ3c2 \sigma=\frac{\pi^2}{60} \frac{ k_b^4}{\hbar^3 c^2} . This lends much credence to its accuracy. (The integral is non-elementary, but some textbooks such as F.Reif's Statistical Physics show the complete evaluation of the integral.) \\ Additional item is from the Planck function, you can also derive Wien's law: λmaxT=2898λmaxT=2898 \lambda_{max}{T}=2898 micron degK , where λmaxλmax \lambda_{max} is the wavelength where the peak of the spectrum (Planck blackbody function) occurs. (By taking ∂L(λ,T)∂λ=0∂L(λ,T)∂λ=0 \frac{\partial{L(\lambda,T)}}{\partial{\lambda}}=0 , you can derive Wien's law. )
Charles Link said:You got it reasonably accurate, but a couple of corrections: ## \\ ## 1) ## \int\limits_{0}^{+\infty} M(\lambda,T)=\sigma T^4 ##, but you can't take a derivative of ## \sigma T^4 ## to get ## M(\lambda, T) ##. (If the upper limit on the integral was ## \lambda ## then you could take a derivative to get ## M(\lambda, T) ##, but in this case you can't.) ## \\ ## 2) To derive Wien's law, you take ## \frac{\partial{L(\lambda, T)}}{\partial{\lambda}}=0 ## and solve for ## \lambda=\lambda_{max} ## where this occurs. Given ## x=hc/(\lambda k_b T) ##, you will get ## e^{-x}=1-x/5 ## upon taking the above partial derivative. The equation has two solutions: ## x=0 ## which is extraneous(essentially tells you the slope approaches zero as you go to very long wavelengths), and ## x=4.96 ## or thereabouts. The ## x=4.96 ## solution gives you Wien's law. ## \\ ## Additional item: To get the factor of ## \pi ## as opposed to ## 2 \pi ## for a hemisphere, see the 2nd to the last sentence of post #79="Suggest you compute... " . Please give the computation a try=I think you will be pleased with the result.
A sphere has solid angle ## \Omega= ##(surface area)## /R^2=4 \pi R^2/R^2=4 \pi ## steradians, so that a hemisphere has ## 2 \pi ## steradians. A flat surface that radiates as ## I(\theta)=I_o cos(\theta) ## puts out power ## P=\int I(\theta) \, d \Omega=I_o \Omega_{effective} ## where ## I_o ## is on-axis intensity and ## \Omega_{effective}=\pi ## steradians. Do the integration and I think you will see that the ## \Omega_{effective} ## is a useful concept.JohnnyGui said:Thanks for the feedback. Will try the integration, but I noticed something.
You said that the effective solid angle of the hemisphere is ##\pi##. But doesn't a hemisphere always have a solid angle of ##\pi## anyway? Or did you mean that the effective area of the hemisphere contains ##\pi## steradians instead of ##2 \pi##?. Such that the effective solid angle of the whole hemisphere is ##\pi## steradians × 0.572 radians (solid angle of 1 steradian) = 1,797 radians?
Charles Link said:A sphere has solid angle ## \Omega= ##(surface area)## /R^2=4 \pi R^2/R^2=4 \pi ## steradians, so that a hemisphere has ## 2 \pi ## steradians. A flat surface that radiates as ## I(\theta)=I_o cos(\theta) ## puts out power ## P=\int I(\theta) \, d \Omega=I_o \Omega_{effective} ## where ## I_o ## is on-axis intensity and ## \Omega_{effective}=\pi ## steradians. Do the integration and I think you will see that the ## \Omega_{effective} ## is a useful concept.
The radiance ## L ## is independent of viewing angle and stays constant. When viewed from a distance ## R ##, (## R ## needs to be kept constant here for comparisons), at viewing position at location with angle ## \theta ##, a circular source of area ## A ## will appear to be elliptical with area ## A_{effectiive}= A cos(\theta ) ##. The source will appear to be shrunken in the direction that the viewer is located: e.g. If viewed from a location that has ## y=0 ##, with ## x^2+z^2=R^2 ##, the apparent width in the ## y ## direction will be unaffected, but the apparent width in the x-direction, where ## x=R cos(\theta) ## and ## z=R sin(\theta) ## will be reduced by a factor ## cos(\theta) ##. If the source has radius ## r ##, the width in the y direction will be ## r ##, and the apparent width in the x direction will be ## rcos(\theta) ##. The area will appear to be ## A_{effective}=A cos(\theta) ##. ## \\ ## To precisely quantify the above, if you take the ellipse ## \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ##, essentially it is a flattened circle when ## b=r ##, and ## a=rcos(\theta) ##. If you compute the area of the ellipse, you will find that it is ## \pi r^2 cos(\theta) ##. ## \\ ## Note: In the above, we are assuming ## R ## is much greater than ## r ##, so that the viewing angle ## \theta ## is constant across the source. ## \\ ## Additional item: The intensity ## I=L A_{effective} ##. Notice if ## A_{effective}=A cos(\theta) ##, that ## I(\theta)=LAcos(\theta)=I_o cos(\theta) ##. A source that obeys this ## cos(\theta) ## fall-off in intensity, (which is very much expected), is said to be "Lambertian", named after Johann Heinrich Lambert. ## \\ ## For an additional problem, we could explore what happens as the viewer moves from S to S2 and to S3 in your figure above. You might find it of interest that the irradiance ## E ## per unit area incident on the surface falls off as ## cos^4(\theta) ##.JohnnyGui said:Sorry, but I'm still at the part where I'm trying to understand why radiance decreases when deviating from the on-axis of the hemisphere in the first place.
Here's a 2 dimensional schematic of what I think is the cause:
View attachment 204285
All surfaces ##S, S2## and ##S3## have the same length. They all have a surface that, when put on the radiation arc (orange), has a solid angle of 1 steradian. Surface ##S## is sitting on the radiation arc and thus has a solid angle of 1 steradian. However, when deviating from the radiation arc, like ##S2## and ##S3## shown in the picture, it shows that the more it goes away from the on-axis, the smaller the angle of radiation that it receives.
Is this the cause why radiance decreases when deviating from the on-axis or is it something else? Sorry for bothering you with all this but if this is incorrect, perhaps it's more handy for me to see in a schematic what causes the decrease in radiance when deviating from the on-axis.
Charles Link said:The radiance LL L is independent of viewing angle and stays constant. When viewed from a distance RR R , (RR R needs to be kept constant here for comparisons), at viewing position at location with angle θθ \theta , a circular source of area AA A will appear to be elliptical with area Aeffectiive=Acos(θ)Aeffectiive=Acos(θ) A_{effectiive}= A cos(\theta ) . The source will appear to be shrunken in the direction that the viewer is located: e.g. If viewed from a location that has y=0y=0 y=0 , with x2+z2=R2x2+z2=R2 x^2+z^2=R^2 , the apparent width in the yy y direction will be unaffected, but the apparent width in the x-direction, where x=Rcos(θ)x=Rcos(θ) x=R cos(\theta) and z=Rsin(θ)z=Rsin(θ) z=R sin(\theta) will be reduced by a factor cos(θ)cos(θ) cos(\theta) . If the source has radius rr r , the width in the y direction will be rr r , and the apparent width in the x direction will be rcos(θ)rcos(θ) rcos(\theta) . The area will appear to be Aeffective=Acos(θ)Aeffective=Acos(θ) A_{effective}=A cos(\theta) .
Charles Link said:Yes. The radiance LL L is the surface "brightness.
Thanks, I understand now the influence of viewing angle on how the surface of the radiating source appears to be smaller, but the following stumped me:Charles Link said:From a previous post (#88) where I discussed the projected area of the source, your y here is that z, and y is the dimension into the paper. The source will appear to be shrunk in the x-direction so that, instead of being circular with radius ## r ##, it will be appear to be elliptical, and in the x direction it will appear to have a half-width of ## a=rcos(\theta) ##. (In the y direction it will be unaffected and appear to have a half -width of ## b=r ##, which is exactly what it has.) ## \\ ## Incidentally ## z=Rcos(\theta) ## and ## \frac{\sqrt{x^2+y^2}}{R}=sin(\theta) ##, so you do have it correct in your figure that ## x=Rsin(\theta) ##, but that is really irrelevant in computing the apparent dimensions of the source as seen from angle ## \theta ##. ## \\ ## As previously mentioned, the location ## R>>r ##, so that the viewing angle ## \theta ## is constant across the entire source. ## \\ ## Suggestion: Take a small circular object (like a penny) and view it from an angle off-axis. Does it look elliptical in appearance?
The blackbody will have more energy per steradian and thereby higher intensity ## I(\theta) ## when viewed from the normal because intensity ## I(\theta)=LA_{effective} ##. Perhaps one way of looking at the concept of radiance or brightness ## L ## is to bring along your own aperture when you measure the brightness, and have that aperture be smaller than the area of the blackbody source. Place it in front of the source and view the appearance. It is not necessary to focus on the source which may be a couple inches behind the aperture. Instead. you focus on your own aperture. Sources that have the same brightness ## L ## will be indistinguishable. You also will not be able to tell whether you are viewing the blackbody from on-axis or from off-axis. (Be sure and have the normal to your own aperture always pointing at you so that its area doesn't change (and always keep the distance between you and your own aperture constant.) ## \\ ## Alternatively, instead of viewing the source by focusing on your own aperture, you could measure the power reaching a detector a specified distance ## R ## from your own aperture. Again, you could not distinguish whether you are observing from on-axis or off-axis. The power received will be the same. (Again, always have the normal to your own aperture pointing in the direction of the detector.)JohnnyGui said:Thanks, I understand now the influence of viewing angle on how the surface of the radiating source appears to be smaller, but the following stumped me:
If a blackbody with a true surface of A is sending an X amount of energy to an observer who views the blackbody at an angle, and the observer sees the blackbody having a smaller apparent surface from which he receives that X amount of energy. Shouldn't he think that it radiates MORE energy per surface unit and thus also per steradian than when he's observing it from the normal?
Charles Link said:Alternatively, instead of viewing the source by focusing on your own aperture, you could measure the power reaching a detector a specified distance RR R from your own aperture. Again, you could not distinguish whether you are observing from on-axis or off-axis. The power received will be the same. (Again, always have the normal to your own aperture pointing in the direction of the detector.)
This is why you bring your own aperture=the area stays the same. It blocks the parts of the source whose line of sight is outside your own aperture. Right now the purpose is to measure the brightness ## L ## and nothing else. To measure the intensity ## I ## you need to be able to see the entire source. ## \\ ## (Intensity is perhaps a poor name for the quantity, because it does not refer to the intensity of the surface. Intensity ## I ## refers to how much power is radiated from the whole surface per unit solid angle. It really is not "intensity" in the literal sense. It also does not refer to how intense it feels, or how much reaches a detector. That instead is given by the irradiance ## E ##, which is sometimes very loosely(and incorrectly) referred to as the intensity.) ## \\ ## To summarize something quickly, there are four radiometric quantities: ## L ##, ## I ## , ## M ##, and ## E ##, and each of these is well defined, and represents something very specific. ## I ## is called intensity, (perhaps a poor name for it, but that's what they call it.) ## M ## refers to power coming off of a surface per unit area, and ## E ## refers to the power per unit area across a surface or onto a surface. Of the four, ## E ## and ## M ## are the most closely related. (Occasionally, in some of the older literature these four quantities are referred to with the letters ## N ##, ## J ##, ## W ##, and ## H ## respectively. In any case, the letters ## L ##, ## I ##, ## M ##, and ## E ## are now quite standard for these quantities.) ## \\ ## Additional comment: These quantities are all quite useful in doing calculations such as the one with the lamp filament in the OP. It is good to use the standard terminology in the calculations, so it makes for easy reading by others.JohnnyGui said:Exactly, so the power will be the same from all angles. So, let's say the observer is watching the source form an angle and measures the power, he'd say that the intensity is ##P## / ##A_{effective}##. Now, the observer moves towards the normal of the surface (he doesn't know he does, but he just moves in that direction) and then looks at the source. He'll still measure the same power (as you said) but he'll also see a larger surface ##A## since he's on-axis. He'll therefore calculate a smaller intensity since ##P## is the same but the surface is larger from the on-axis view.
What am I missing here?
Charles Link said:(Intensity is perhaps a poor name for the quantity, because it does not refer to the intensity of the surface. Intensity II I refers to how much power is radiated from the whole surface per unit solid angle. It really is not "intensity" in the literal sense. It also does not refer to how intense it feels, or how much reaches a detector. That instead is given by the irradiance EE E , which is sometimes very loosely(and incorrectly) referred to as the intensity.)
Charles Link said:Just an additional comment: The irradiance EobservedEobserved E_{observed} is often measured with a photodetector.
When the letter ## E ## is used in radiometrics, like ## E_{observed} ##, it stands for irradiance (watts/m^2). (In other contexts it can represent an electric field, or it can represent energy, but in the present context, it represents irradiance (watts/m^2)). The letter ## P ## is used for power (watts). The brightness ## L=E_{observed}/\frac{Acos(\theta)}{R^2} =E_{observed}R^2/(A cos(\theta)) ##, where ## E_{observed}=P_{observed}/A_{detector} ##, but normally you can do a measurement of ## E_{observed} ## as described in my previous post, and you don't need to know ## A_{detector} ##.JohnnyGui said:Glad I'm on the right track.
But isn't irradiance the amount of energy the observer receives but per m2 of the radiating surface of ##cos(\theta) \cdot A##? Since my mentioned ##E_{observed}## is the energy received by the observer by the whole surface ##cos(\theta) \cdot A##, I think the relationship should be:
$$\frac{E_{observed}}{cos(\theta) \cdot A} = Irradiance$$