Why is short-wave infrared portion of electromagnetic spectrum so sensitive to fire?
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
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
add a comment |
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
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
1
This sounds like a question more on-topic at the Earth Science Stack Exchange than here.
– PolyGeo♦
Mar 22 at 8:34
2
@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.
– radouxju
Mar 22 at 8:48
add a comment |
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
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
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
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
edited Mar 22 at 8:35
PolyGeo♦
53.8k1781245
53.8k1781245
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
asked Mar 22 at 7:57
Robert BuckleyRobert Buckley
4,8701256115
4,8701256115
1
This sounds like a question more on-topic at the Earth Science Stack Exchange than here.
– PolyGeo♦
Mar 22 at 8:34
2
@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.
– radouxju
Mar 22 at 8:48
add a comment |
1
This sounds like a question more on-topic at the Earth Science Stack Exchange than here.
– PolyGeo♦
Mar 22 at 8:34
2
@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.
– radouxju
Mar 22 at 8:48
1
1
This sounds like a question more on-topic at the Earth Science Stack Exchange than here.
– PolyGeo♦
Mar 22 at 8:34
This sounds like a question more on-topic at the Earth Science Stack Exchange than here.
– PolyGeo♦
Mar 22 at 8:34
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
Mar 22 at 8:48
@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.
– radouxju
Mar 22 at 8:48
add a comment |
2 Answers
2
active
oldest
votes
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.
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
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
Mar 22 at 22:24
add a comment |
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 fact is that 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
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
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
Mar 22 at 22:24
add a comment |
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.
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
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
Mar 22 at 22:24
add a comment |
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.
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
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.
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
edited Mar 22 at 8:49
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
answered Mar 22 at 8:36
radouxjuradouxju
41.3k144122
41.3k144122
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
Mar 22 at 22:24
add a comment |
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
Mar 22 at 22:24
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
Mar 22 at 22:24
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
Mar 22 at 22:24
add a comment |
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 fact is that 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
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
add a comment |
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 fact is that 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
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
add a comment |
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 fact is that 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
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
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 fact is that 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
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
edited Mar 25 at 13:43
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
answered Mar 22 at 14:54
gustavovelascohgustavovelascoh
1514
1514
add a comment |
add a comment |
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1
This sounds like a question more on-topic at the Earth Science Stack Exchange than here.
– PolyGeo♦
Mar 22 at 8:34
2
@PolyGeo this is a question about spectral signature, so it also belongs here as long as this site hosts remote sensing questions.
– radouxju
Mar 22 at 8:48