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Why is short-wave infrared portion of electromagnetic spectrum so sensitive to fire?


Counting animals using near infrared imagery from a droneBest Landsat-5 TM band combination for detecting fire scarsWhy does Landsat 8 panchromatic band NOT include the infrared?Why are Landsat-8 derived NDVI values higher than SAVI values?Why can't get correct RGB img of Landsat 8 using np.dstackWhy Sentinel-2 bands have different spatial resolutionWhy does my rainbow airplane stripes appearing on Sentinel 2 image?Convertion of spectral indices formulas from Landsat to SentinelWhy is satellite imagery data delivered as signed 16 bit integers?InSAR processing: why can't we just compute distances directly from time delays?













3















After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus







share|improve this question



















  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    yesterday







  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    yesterday















3















After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus







share|improve this question



















  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    yesterday







  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    yesterday













3












3








3








After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus







share|improve this question
















After attending a remote sensing workshop and using Copernicus data to monitor natural disaster scenarios, my question could not be fully answered by the instructor. All that was said, was that is in no way associated to thermal IR.



What property does fire have, that makes is visible with shortwave infrared?






remote-sensing copernicus




remote-sensing copernicus






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









PolyGeo♦

53.7k1781244




53.7k1781244










asked yesterday









Robert BuckleyRobert Buckley

4,8601255115




4,8601255115







  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    yesterday







  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    yesterday












  • 1





    This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

    – PolyGeo♦
    yesterday







  • 2





    @PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

    – radouxju
    yesterday







1




1





This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

– PolyGeo♦
yesterday






This sounds like a question more on-topic at the Earth Science Stack Exchange than here.

– PolyGeo♦
yesterday





2




2





@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

– radouxju
yesterday





@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.

– radouxju
yesterday










2 Answers
2






active

oldest

votes


















9














According to Wien's law, the wavelength of the peak emission depends on the temperature. 



wavelength_of_peak = 2.898*10^(-3)/T


where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






share|improve this answer

























  • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

    – MaxW
    yesterday


















4














First, I would reword the question to Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?



Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:




  • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


  • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.




So, the spectrum is not sensitive, the temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






share|improve this answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9














    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






    share|improve this answer

























    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      yesterday















    9














    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






    share|improve this answer

























    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      yesterday













    9












    9








    9







    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.






    share|improve this answer















    According to Wien's law, the wavelength of the peak emission depends on the temperature. 



    wavelength_of_peak = 2.898*10^(-3)/T


    where T is the temperature in degree Kelvin. So the wavelength of the peak emission decrease when the temperature increases.



    For the temparature of the surface of the Earth, you are around 300 K, so the peak is about 10 micrometer (the so called thermal infrared). In case of fire, the temperature is more like 1200 K, so the peak wavelength is close to 2.4, which correspond to the second SWIR band of Landsat.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited yesterday

























    answered yesterday









    radouxjuradouxju

    41.2k144120




    41.2k144120












    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      yesterday

















    • Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

      – MaxW
      yesterday
















    Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

    – MaxW
    yesterday





    Wien's Law is correct so far as it goes, but you should preface that with the assumption that the radiation given off follows blackbody radiation.

    – MaxW
    yesterday













    4














    First, I would reword the question to Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?



    Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:




    • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


    • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.




    So, the spectrum is not sensitive, the temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



    source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






    share|improve this answer








    New contributor




    gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.
























      4














      First, I would reword the question to Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?



      Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:




      • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


      • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.




      So, the spectrum is not sensitive, the temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



      source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






      share|improve this answer








      New contributor




      gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















        4












        4








        4







        First, I would reword the question to Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?



        Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:




        • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


        • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.




        So, the spectrum is not sensitive, the temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



        source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law






        share|improve this answer








        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.










        First, I would reword the question to Why is fire "visible" in the short-wave infrared portion of electromagnetic spectrum ?



        Second, I would like to add my 2 cents to @radouxju answer. These 2 examples could make the explanation clearer:




        • A piece of metal heated by a blow torch first becomes "red hot" as the very longest visible wavelengths appear red, then becomes more orange-red as the temperature is increased, and at very high temperatures would be described as "white hot" as shorter and shorter wavelengths come to predominate the black body emission spectrum. Before it had even reached the red hot temperature, the thermal emission was mainly at longer infrared wavelengths, which are not visible; nevertheless, that radiation could be felt as it warms one's nearby skin.


        • A wood fire at 1500 K puts out peak radiation at about 2000 nm. 98% of its radiation is at wavelengths longer than 1000 nm, and only a tiny proportion at visible wavelengths (390–700 nm). Consequently, a campfire can keep one warm but is a poor source of visible light.




        So, the spectrum is not sensitive, the temperature is related to the electromagnetic radiation emission at a certain wavelength range. A high temperature like the sun (~5700K) will emit a peak around 500nm (green), but it is also emitting in the UV, visible and IR range. On the other side, fire or wood fire (as the example) will have a peak at ~2200nm, having most of the emitted radiation outside the visible range, that is, in the SWIR range.



        source: https://en.wikipedia.org/wiki/Wien%27s_displacement_law







        share|improve this answer








        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|improve this answer



        share|improve this answer






        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered yesterday









        gustavovelascohgustavovelascoh

        1413




        1413




        New contributor




        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        gustavovelascoh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.



























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