fredag 11 januari 2013

Why Measuring Temperature is Easier than Radiance

The recent posts show that in using remote infrared sensing, it is easier to measure temperature than radiance, because the detector records temperature rather than radiance. To see this we recall that a detector in principle is a blackbody in radiative contact with an object to be measured with the following energy balance:
  • dE/dt + Output = Input  
where dE/dt is the rate of change of the internal energy E of the detector with a certain relation between  the energy and the temperature T (for example E ~ T^2, page 64), Output is the (net) outgoing radiance from the detector and Input is the (net) incoming radiance from the object. The detector at lower temperature than the object reacts by heating up until in radiative equilibrium with dE/dt = 0 and Output = Input, it assumes the same temperature as the object. The detector thus records an increase in temperature from its known instrument background temperature and can thus report the temperature of the object, by remote infrared thermal sensing based on radiative equilibrium.

To measure radiance, would require measuring the dynamic response of the detector by reading dE/dt and then reconstructing Input from dE/dt, E and Output, which is much more difficult. Remote infrared sensing involving a sensor of the form described (e.g. thermopile),  reports temperature. Radiance is then constructed from temperature using some guess work and calibration. In the absence of true direct measurements of radiance, this construction has come to define radiance in a circular way.

To record a lower temperature than the instrument background temperature is also difficult since it requires effective shielding of the detector.

PS Recall How to Fool Yourself with a Pyrgeometer.

PS2 There is a connection to the option market with the value of an option constructed by the Black-Scholes formula, which has created an artificial fictional market leading to the present crisis in the world economy. Replace Black-Scholes with an artificial incorrect version of Planck, and you have climate alarmism adding to the crisis.

32 kommentarer:

  1. dE/dt + output = Input,

    is not the correct equation.

    Take a bolometer for instance. It works by connecting the absorbing part of the detector with a heat sink that is held at a low temperature, it is more than safe to ignore any losses from the detector.

    For a bolometer the correct equation read

    C*d(DT)/dt + G*DT = P,

    C is the heat capacity of the receiving part of the detector, G is the conductance to a heat sink at temperature T0, P is the absorbed incoming power and DT is the temperature difference between the absorbing part and the heat sink.

    In steady state

    DT = P/G

    which is a good starting point if you want to analyze the performance further. You have a direct relation between absorbed power and temperature difference.

    SvaraRadera
  2. There are several possible models. Is the conductance G constant in your model? How is G determined?

    SvaraRadera
  3. It's not my model.

    Look it up in any text on designing radiometry circuits, or an text about observational astronomy or something.

    Never the less, it kind of obliterates your thoughts about how remote sensing of radiance is done in reality.

    SvaraRadera
  4. This is a trivial model that can model anything, which no commercial
    company could dare to try to sell. It is like modeling
    the world with straight line through the origin and claim that
    the slope can be determined by calibration.

    SvaraRadera
  5. But seriously, come on. Do you not have more important things to do then flogging a stone cold dead horse?

    But if you want to pursue, as a starting point you could look at

    Jones (1953), J. Opt. Soc. America 43, 1,

    it discusses the general theory and realizations into usable circuits.

    Treatment of the different sources of noise can be found in

    Mather (1982), Appl. Optics, 21, 1125.

    SvaraRadera
  6. Yes, about the linearity.

    By designing your circuit in a suitable manner you get an operating point and a load line that sets the operational range for the intended application. Very basic, you should know this.

    SvaraRadera
  7. This is very dangerous, since if the dynamics is not taken into account, you cannot distinguish a weak source (low emissivity) from a strong source (emissivity =1) having the same temperature.

    SvaraRadera
  8. You measure a time averaged quantity...

    SvaraRadera
  9. What are you de facto measuring? Could it be DT?

    SvaraRadera
  10. You measure the applied amount of electric energy that is needed to heat a resistor (you heat it with a DC current) so that it's resistance matches the resistance that is caused by the increased temperature in the bolometer. This increasing temperature is the result of applied power to the bolometer.

    If I remember correctly. You could measure DT to if you like I guess.

    SvaraRadera
  11. As as you measure DT at stationary state, it is temperature you are measuring, and connecting radiance to temperature requires emmisivity which is unknown in general. To guess radiance from temperature is guess work.

    SvaraRadera
  12. It needs to be explained why one cannot simply identify the term C*d(DT)/dt as dE/dt, and G*DT as output from the detector while it comes to temperature balance with the object--thus making the two equations under discussion the same.

    SvaraRadera
  13. Your brain seems to be in a locked mode... Give it a day or two.

    In the meantime, do you see why your own model is useless in making qualitative statements about the detector and the incoming radiation?

    SvaraRadera
  14. If the aim is to measure temperature, no model is needed, since the temperature is what is registered. If the aim is to measure radiance then a model is needed, and then my model as a wave model with small damping is much better than the trivial model suggested which can model anything. But as said measuring radiance is more difficult.

    SvaraRadera
  15. What are you measuring if you shine laser light on the bolometer?

    SvaraRadera
  16. That is an interesting question

    SvaraRadera
  17. Have you come up with any idea of what you are measuring if you shine laser light on a bolometer yet?

    SvaraRadera
  18. Why wont you answer?

    What do you think is measured if the light collected in the detector is in a very narrow band of a few wavelengths?

    SvaraRadera
  19. You can only measure temperature, or net heat transfer, but not gross input because you don't know the net output. Gross input/output are fictional quantities connected to an incorrect reading of Planck's radiation law as expressing two-way heat transfer, while the correct reading is one-way from warm to cold as net heat transfer.

    SvaraRadera
  20. Well, assume the detector is really cold, you deposit energy in it with a narrow band laser.

    What are you measuring?

    SvaraRadera
  21. We are speaking about black or grey body radiation, not lasers. Please stick to the subject.

    SvaraRadera
  22. On the contrary, it must be really important for the discussion.

    If you could measure spectral radiance, say from a nearly monochromatic source, then you could of course measure radiance by appropriate filtering.

    So, what are you measuring with a narrow band laser? The conditions are set up so that all energy loss from the detector is negligible.

    SvaraRadera
  23. This is the same discussion as an earlier one about heating chickens in microwave ovens, see e g

    http://claesjohnson.blogspot.se/2011/07/sky-dragon-strikes-back.html

    The chicken will in stationary state assume the temperature of blackbody with the radiance of the input. By checking the state of the chicken you can determine that temperature. If the input temp is too low the chicken will stay raw.

    SvaraRadera
  24. Exactly what is input temperature?

    SvaraRadera
  25. It is the temperature of a blackbody emitting into a surrounding at 0 K the same radiance (in W/m2)
    as that of the microwave oven.

    SvaraRadera
  26. Are you saying that a laser has a black body temperature, or what?

    It makes no sense.

    Why is it not possible to talk about power [J/s] straight away? Are you against local conservation of energy?

    SvaraRadera
  27. You have not read the thread on the microwave oven. Understanding the proof of Planck's law is essential. You are welcome back with questions when you have digested the proof.

    SvaraRadera
  28. Well, a proof is less then a half page and very straight forward...

    So, what are the bloggpost suppose to say about measuring the intensity of laser light?

    SvaraRadera
  29. Read and digest my proof and then return.

    SvaraRadera
  30. Where can I find your proof?

    SvaraRadera
  31. In my book Computational Blackbody Radiation which you can download from this blog.

    SvaraRadera
  32. The title of the book is Mathematical Physics of Blackbody Radiation available under Upcoming Books By CJ.

    SvaraRadera