Mixing
Peaks different in wavelength and frequency space
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According to high-school physics, light frequency and wavelength are inversely proportional, and their product is the speed of light. But when discussing spectral power, somehow sunlight peaks differently in these two spaces:
http://wtamu.edu/~cbaird/sq/2013/07/03/what-is-the-color-of-the-sun/
How is this possible?
electromagnetic-radiation
asked 2 hours ago
Cindy Almighty
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up vote
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down vote
favorite
According to high-school physics, light frequency and wavelength are inversely proportional, and their product is the speed of light. But when discussing spectral power, somehow sunlight peaks differently in these two spaces:
http://wtamu.edu/~cbaird/sq/2013/07/03/what-is-the-color-of-the-sun/
How is this possible?
electromagnetic-radiation
asked 2 hours ago
Cindy Almighty
1112
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
According to high-school physics, light frequency and wavelength are inversely proportional, and their product is the speed of light. But when discussing spectral power, somehow sunlight peaks differently in these two spaces:
http://wtamu.edu/~cbaird/sq/2013/07/03/what-is-the-color-of-the-sun/
How is this possible?
electromagnetic-radiation
asked 2 hours ago
Cindy Almighty
1112
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
According to high-school physics, light frequency and wavelength are inversely proportional, and their product is the speed of light. But when discussing spectral power, somehow sunlight peaks differently in these two spaces:
http://wtamu.edu/~cbaird/sq/2013/07/03/what-is-the-color-of-the-sun/
How is this possible?
electromagnetic-radiation
electromagnetic-radiation
asked 2 hours ago
Cindy Almighty
1112
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Cindy Almighty
1112
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Cindy Almighty
1112
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Cindy Almighty
1112
asked 2 hours ago
Cindy Almighty
1112
1112
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Cindy Almighty is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Cindy Almighty 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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add a comment |Â
3 Answers
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2
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The units of the y-axis quantity spectral irradience, $rm W , m^-2 , nm^-1$ and $rm W , m^-2 , THz^-1$, differ in a subtle way.
They both are in terms of power $rm W$ per unit area $rm m^-2$ but the interval over which the power per unit area is measured is different, $rm nm^-1$ and $rm THz^-1$.
Noting that a frequency interval of $1,rm THz$ corresponds to a wavelength interval of $300,000,rm nm$ accounts for the difference in the spectral power graphs that you are referring to.
answered 1 hour ago
Farcher
45.6k33388
add a comment |Â
up vote
1
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This is all about the difference between "a quantity Q" and "the density of that quantity per unit X" where X is some other quantity. A good way to get a feel for this is to convert statements about density to statements about amount. For example, if $u$ is "the spectral energy density per unit wavelength range" then
(The amount of energy in the range $lambda$ to $lambda + dlambda$) = $|u , dlambda|$.
Now let $rho$ be "the spectral energy density per unit frequency range". Then
(The amount of energy in the range $nu$ to $nu + dnu$) = $|rho , dnu|$.
Next we need to realise that these two amounts of energy are the same. For, when the wavelength is $lambda$ the frequency is $nu$, and when the wavelength is $lambda + dlambda$ the frequency is $nu + dnu$, so the two energies we just referred to are one and the same energy, in the sense that they are two ways of referring to the same "chunk" out of the complete spectrum. Thus we find
$|u , dlambda| = |rho , dnu|$
and therefore
$ u = rho, |dnu/dlambda| = rho , c / lambda^2$
That's the algebra, but I hope I have also given a feel for what is going on. It does take some getting used to.
If you now take some distribution $rho$ and multiply it by the function $1/lambda^2$ you can see that any local peak in the distribution is going to be moved a bit: the long wavelength part gets reduced and the short wavelength part gets increased, so the peak moves to shorter wavelength. It is not because of any change in the relationship between frequency and wavelength, but because of the difference in "per unit frequency change" as compared to "per unit wavelength change". A similar shift will apply to a study of quantities involving other things related by an inverse, such as time and frequency, resistivity and conductivity, etc.
answered 48 mins ago
Andrew Steane
2005
add a comment |Â
up vote
1
down vote
It's an interesting question!
If your graph is plotted as a function of wavelength,the solar irradiance is maximum in the visible and if it were plotted as a function of frequency the maximum is,interestingly,in the near infrared.
How is this possible?
This happens because the relationship between wavelength and frequency is not linear.
$ vert d nuvert = vert c/ lambda^2 dlambda vert$
Reference:
http://www.oceanopticsbook.info/view/light_and_radiometry/level_2/a_common_misconception#fig:photons-wave-freq
answered 22 mins ago
ayc
32417
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The units of the y-axis quantity spectral irradience, $rm W , m^-2 , nm^-1$ and $rm W , m^-2 , THz^-1$, differ in a subtle way.
They both are in terms of power $rm W$ per unit area $rm m^-2$ but the interval over which the power per unit area is measured is different, $rm nm^-1$ and $rm THz^-1$.
Noting that a frequency interval of $1,rm THz$ corresponds to a wavelength interval of $300,000,rm nm$ accounts for the difference in the spectral power graphs that you are referring to.
answered 1 hour ago
Farcher
45.6k33388
add a comment |Â
up vote
2
down vote
The units of the y-axis quantity spectral irradience, $rm W , m^-2 , nm^-1$ and $rm W , m^-2 , THz^-1$, differ in a subtle way.
They both are in terms of power $rm W$ per unit area $rm m^-2$ but the interval over which the power per unit area is measured is different, $rm nm^-1$ and $rm THz^-1$.
Noting that a frequency interval of $1,rm THz$ corresponds to a wavelength interval of $300,000,rm nm$ accounts for the difference in the spectral power graphs that you are referring to.
answered 1 hour ago
Farcher
45.6k33388
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The units of the y-axis quantity spectral irradience, $rm W , m^-2 , nm^-1$ and $rm W , m^-2 , THz^-1$, differ in a subtle way.
They both are in terms of power $rm W$ per unit area $rm m^-2$ but the interval over which the power per unit area is measured is different, $rm nm^-1$ and $rm THz^-1$.
Noting that a frequency interval of $1,rm THz$ corresponds to a wavelength interval of $300,000,rm nm$ accounts for the difference in the spectral power graphs that you are referring to.
answered 1 hour ago
Farcher
45.6k33388
The units of the y-axis quantity spectral irradience, $rm W , m^-2 , nm^-1$ and $rm W , m^-2 , THz^-1$, differ in a subtle way.
They both are in terms of power $rm W$ per unit area $rm m^-2$ but the interval over which the power per unit area is measured is different, $rm nm^-1$ and $rm THz^-1$.
Noting that a frequency interval of $1,rm THz$ corresponds to a wavelength interval of $300,000,rm nm$ accounts for the difference in the spectral power graphs that you are referring to.
answered 1 hour ago
Farcher
45.6k33388
answered 1 hour ago
Farcher
45.6k33388
answered 1 hour ago
Farcher
45.6k33388
answered 1 hour ago
Farcher
45.6k33388
45.6k33388
add a comment |Â
add a comment |Â
up vote
1
down vote
This is all about the difference between "a quantity Q" and "the density of that quantity per unit X" where X is some other quantity. A good way to get a feel for this is to convert statements about density to statements about amount. For example, if $u$ is "the spectral energy density per unit wavelength range" then
(The amount of energy in the range $lambda$ to $lambda + dlambda$) = $|u , dlambda|$.
Now let $rho$ be "the spectral energy density per unit frequency range". Then
(The amount of energy in the range $nu$ to $nu + dnu$) = $|rho , dnu|$.
Next we need to realise that these two amounts of energy are the same. For, when the wavelength is $lambda$ the frequency is $nu$, and when the wavelength is $lambda + dlambda$ the frequency is $nu + dnu$, so the two energies we just referred to are one and the same energy, in the sense that they are two ways of referring to the same "chunk" out of the complete spectrum. Thus we find
$|u , dlambda| = |rho , dnu|$
and therefore
$ u = rho, |dnu/dlambda| = rho , c / lambda^2$
That's the algebra, but I hope I have also given a feel for what is going on. It does take some getting used to.
If you now take some distribution $rho$ and multiply it by the function $1/lambda^2$ you can see that any local peak in the distribution is going to be moved a bit: the long wavelength part gets reduced and the short wavelength part gets increased, so the peak moves to shorter wavelength. It is not because of any change in the relationship between frequency and wavelength, but because of the difference in "per unit frequency change" as compared to "per unit wavelength change". A similar shift will apply to a study of quantities involving other things related by an inverse, such as time and frequency, resistivity and conductivity, etc.
answered 48 mins ago
Andrew Steane
2005
add a comment |Â
up vote
1
down vote
This is all about the difference between "a quantity Q" and "the density of that quantity per unit X" where X is some other quantity. A good way to get a feel for this is to convert statements about density to statements about amount. For example, if $u$ is "the spectral energy density per unit wavelength range" then
(The amount of energy in the range $lambda$ to $lambda + dlambda$) = $|u , dlambda|$.
Now let $rho$ be "the spectral energy density per unit frequency range". Then
(The amount of energy in the range $nu$ to $nu + dnu$) = $|rho , dnu|$.
Next we need to realise that these two amounts of energy are the same. For, when the wavelength is $lambda$ the frequency is $nu$, and when the wavelength is $lambda + dlambda$ the frequency is $nu + dnu$, so the two energies we just referred to are one and the same energy, in the sense that they are two ways of referring to the same "chunk" out of the complete spectrum. Thus we find
$|u , dlambda| = |rho , dnu|$
and therefore
$ u = rho, |dnu/dlambda| = rho , c / lambda^2$
That's the algebra, but I hope I have also given a feel for what is going on. It does take some getting used to.
If you now take some distribution $rho$ and multiply it by the function $1/lambda^2$ you can see that any local peak in the distribution is going to be moved a bit: the long wavelength part gets reduced and the short wavelength part gets increased, so the peak moves to shorter wavelength. It is not because of any change in the relationship between frequency and wavelength, but because of the difference in "per unit frequency change" as compared to "per unit wavelength change". A similar shift will apply to a study of quantities involving other things related by an inverse, such as time and frequency, resistivity and conductivity, etc.
answered 48 mins ago
Andrew Steane
2005
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is all about the difference between "a quantity Q" and "the density of that quantity per unit X" where X is some other quantity. A good way to get a feel for this is to convert statements about density to statements about amount. For example, if $u$ is "the spectral energy density per unit wavelength range" then
(The amount of energy in the range $lambda$ to $lambda + dlambda$) = $|u , dlambda|$.
Now let $rho$ be "the spectral energy density per unit frequency range". Then
(The amount of energy in the range $nu$ to $nu + dnu$) = $|rho , dnu|$.
Next we need to realise that these two amounts of energy are the same. For, when the wavelength is $lambda$ the frequency is $nu$, and when the wavelength is $lambda + dlambda$ the frequency is $nu + dnu$, so the two energies we just referred to are one and the same energy, in the sense that they are two ways of referring to the same "chunk" out of the complete spectrum. Thus we find
$|u , dlambda| = |rho , dnu|$
and therefore
$ u = rho, |dnu/dlambda| = rho , c / lambda^2$
That's the algebra, but I hope I have also given a feel for what is going on. It does take some getting used to.
If you now take some distribution $rho$ and multiply it by the function $1/lambda^2$ you can see that any local peak in the distribution is going to be moved a bit: the long wavelength part gets reduced and the short wavelength part gets increased, so the peak moves to shorter wavelength. It is not because of any change in the relationship between frequency and wavelength, but because of the difference in "per unit frequency change" as compared to "per unit wavelength change". A similar shift will apply to a study of quantities involving other things related by an inverse, such as time and frequency, resistivity and conductivity, etc.
answered 48 mins ago
Andrew Steane
2005
This is all about the difference between "a quantity Q" and "the density of that quantity per unit X" where X is some other quantity. A good way to get a feel for this is to convert statements about density to statements about amount. For example, if $u$ is "the spectral energy density per unit wavelength range" then
(The amount of energy in the range $lambda$ to $lambda + dlambda$) = $|u , dlambda|$.
Now let $rho$ be "the spectral energy density per unit frequency range". Then
(The amount of energy in the range $nu$ to $nu + dnu$) = $|rho , dnu|$.
Next we need to realise that these two amounts of energy are the same. For, when the wavelength is $lambda$ the frequency is $nu$, and when the wavelength is $lambda + dlambda$ the frequency is $nu + dnu$, so the two energies we just referred to are one and the same energy, in the sense that they are two ways of referring to the same "chunk" out of the complete spectrum. Thus we find
$|u , dlambda| = |rho , dnu|$
and therefore
$ u = rho, |dnu/dlambda| = rho , c / lambda^2$
That's the algebra, but I hope I have also given a feel for what is going on. It does take some getting used to.
If you now take some distribution $rho$ and multiply it by the function $1/lambda^2$ you can see that any local peak in the distribution is going to be moved a bit: the long wavelength part gets reduced and the short wavelength part gets increased, so the peak moves to shorter wavelength. It is not because of any change in the relationship between frequency and wavelength, but because of the difference in "per unit frequency change" as compared to "per unit wavelength change". A similar shift will apply to a study of quantities involving other things related by an inverse, such as time and frequency, resistivity and conductivity, etc.
answered 48 mins ago
Andrew Steane
2005
answered 48 mins ago
Andrew Steane
2005
answered 48 mins ago
Andrew Steane
2005
answered 48 mins ago
Andrew Steane
2005
2005
add a comment |Â
add a comment |Â
up vote
1
down vote
It's an interesting question!
If your graph is plotted as a function of wavelength,the solar irradiance is maximum in the visible and if it were plotted as a function of frequency the maximum is,interestingly,in the near infrared.
How is this possible?
This happens because the relationship between wavelength and frequency is not linear.
$ vert d nuvert = vert c/ lambda^2 dlambda vert$
Reference:
http://www.oceanopticsbook.info/view/light_and_radiometry/level_2/a_common_misconception#fig:photons-wave-freq
answered 22 mins ago
ayc
32417
add a comment |Â
up vote
1
down vote
It's an interesting question!
If your graph is plotted as a function of wavelength,the solar irradiance is maximum in the visible and if it were plotted as a function of frequency the maximum is,interestingly,in the near infrared.
How is this possible?
This happens because the relationship between wavelength and frequency is not linear.
$ vert d nuvert = vert c/ lambda^2 dlambda vert$
Reference:
http://www.oceanopticsbook.info/view/light_and_radiometry/level_2/a_common_misconception#fig:photons-wave-freq
answered 22 mins ago
ayc
32417
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's an interesting question!
If your graph is plotted as a function of wavelength,the solar irradiance is maximum in the visible and if it were plotted as a function of frequency the maximum is,interestingly,in the near infrared.
How is this possible?
This happens because the relationship between wavelength and frequency is not linear.
$ vert d nuvert = vert c/ lambda^2 dlambda vert$
Reference:
http://www.oceanopticsbook.info/view/light_and_radiometry/level_2/a_common_misconception#fig:photons-wave-freq
answered 22 mins ago
ayc
32417
It's an interesting question!
If your graph is plotted as a function of wavelength,the solar irradiance is maximum in the visible and if it were plotted as a function of frequency the maximum is,interestingly,in the near infrared.
How is this possible?
This happens because the relationship between wavelength and frequency is not linear.
$ vert d nuvert = vert c/ lambda^2 dlambda vert$
Reference:
http://www.oceanopticsbook.info/view/light_and_radiometry/level_2/a_common_misconception#fig:photons-wave-freq
answered 22 mins ago
ayc
32417
answered 22 mins ago
ayc
32417
answered 22 mins ago
ayc
32417
answered 22 mins ago
ayc
32417
32417
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add a comment |Â
Cindy Almighty is a new contributor. Be nice, and check out our Code of Conduct.
Cindy Almighty is a new contributor. Be nice, and check out our Code of Conduct.
Cindy Almighty is a new contributor. Be nice, and check out our Code of Conduct.
Cindy Almighty is a new contributor. Be nice, and check out our Code of Conduct.
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