this post was submitted on 20 Dec 2023
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I thought that the frequency of light was directly inverse to the wavelength by a constant. In other words, I assumed that graphing the frequency of light as a function of wavelength would be a straight inverse line. Because of that, the graphs for the distribution of light from the sun as functions of frequency and wavelength would be exactly the same, but reversed. Yet, this is not what is reported in the linked article. Even more confusing to me is that the different functions peak at different light. When as a function of frequency, the light peaks at infrared. When as a function of wavelength, the light peaks at violet.

What am I misunderstanding? Is the frequency of light not directly proportional to it's wavelength? Or is this something to do with the way we are measuring the light from the Sun?

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[–] [email protected] 9 points 10 months ago (1 children)

The frequency is not directly proportional to the wavelength - it's inversely proportional: https://en.wikipedia.org/wiki/Proportionality_(mathematics)#Inverse_proportionality

Think of this as this: The wavelength is the distance that light travels during one wave i.e. cycle. Light propagates with the speed of light, so the smaller the wavelength, it means the frequency must increase. If the wavelength gets two times lower, the frequency increases two times. If wavelength approaches 0, then frequency starts growing very quickly, approaching infinity.

The plot is not a straight line but a hyperbola.

[–] [email protected] 8 points 10 months ago (2 children)

Interesting. Sparked by your comment, I found this.

What's the maximun frequency of light?

Approximately 2 x 10^43 Hz is the "Planck frequency" (the inverse of the Planck time). At that frequency, individual photons (or any other particle with this much energy) would be black holes (their Compton wavelength would be smaller than their Schwartzchild radius), and so until someone comes up with a theory of light which includes virtual black-holes, photons with a frequency above this cannot be considered sensible.

[–] [email protected] 3 points 10 months ago

wow TIL sth as well I guess

[–] [email protected] 3 points 10 months ago

I think that answer is a touch misleading because it makes it sound like this is a fundamental physical limit, when really it's just the scale where our current theories break down and give nonsense results, so we don't really know what is going on at that scale yet.

[–] [email protected] 6 points 10 months ago (1 children)

There are 2 reasons:

  1. Those two graphs have different scales on the y-axis. One is Irradiance per nanometer of wavelength, and one is Irradiance per terahertz of frequency. Both graph's y-axis are called "spectral irradiance", despite being different things. This causes most of the distortion between the two graphs, and can even change the location of the absolute maximum.

  2. The graphs' x-axis have different units. This causes some distortion too, but wouldn't change the absolute maximum. It would help if they used a log scale in both cases, because wavelength and frequency are inversely related, so then the graphs could just be horizontally flipped.

So, look at the top graph (by wavelength), and see how much power is in that 1000-2000nm area. It's still a lot, just spread out over a large area. It's the same amount of power in the lower graph (by frequency) shoved into the much smaller area from 150THz to 300THz. Since it's in a smaller area on the lower graph, it has more power-per-unit-of-x-axis.

[–] [email protected] 3 points 10 months ago (1 children)

Thank you. I understand most of your comment, and it makes sense. However, I still don't understand how the change of units in the y-axis would cause a different maximum. It seems to me that the y-axis for both use the same formula with their respective x-axes: W/m^2/x.

I'm not in STEM by the way.

[–] [email protected] 3 points 10 months ago (1 children)

It's because the wavelength and frequency are inversely related. When the wavelength is low and the frequency is high, the wavelength is also moving very slowly, compared to the frequency which is moving very quickly. Since the frequency is changing so quickly, the power-per-unit-frequency is lower at higher frequencies, and higher at lower frequencies (at least relative to the power-per-unit-wavelength).

Let me try and use a car analogy:

You're driving home through Wisconsin, and you live on the border between Wisconsin and Minnesota. The mile markers on the road decrease as you go, reaching 0 at the state border, where you happen to live.

The cows along the highway are evenly distributed, so if you count the cows as you drive, but restart your count every mile when you see the mile marker, you will reach the same number of cows every mile.

Now, the frequency is inversely related to the mile number. The frequency in this case refers to your children in the back seat asking, "Are we there yet?" They know damn well how far it is to home, because they can just look at the mile markers. Regardless, their rate of asking increases as the mile markers go down. When you're at mile marker 100, they ask once every 10 minutes. When you're at mile marker 1, they ask 10 times per minute.

If you instead look at the number of cows between "Are we there yet?" asks, then you will find that the cows-per-ask is much different from the cows-per-mile. At high distances (low frequencies), the cows-per-ask is very high, while at low distances (high frequencies), the cows-per-ask is very low.

Now, the article is looking at power-per-unit-frequency, so you'd actually have to measure the rate in change of how often the kids ask "Are we there yet?" And that would give you a little different result. You might need calculus to correctly calculate the derivative of the number of asks. But hopefully this illustrates that you can get different results, by using a different per-thing to measure your value.

[–] [email protected] 1 points 10 months ago

This covers it all well, but I think a simple explanation is that although "W/m^2/x" looks the same on the axes, it's not the same. f=1/w, so one axis is W/m^2/f and one is W/m^2*f. The article makes a big deal out of the differences as if the x axis were the only difference, but they're just very different things being plotted.

[–] [email protected] 4 points 10 months ago* (last edited 10 months ago) (1 children)

In a vacuum c=nu*lamba or the speed of light is equal to the frequency times wavelength. So nu=c/lamba. If you plot 1/x, you don't get a straight inverse line. You get a multiplicative inverse. So not only is the data flipped, but it also has a distortion that will compress portions and stretch others.

As to why the functions peak at different colors, I believe this is due to an oddity in the axis units. Notice how the irradiance is in W/m^2/nm in the first and W/m^2/THz in the second. Are you familiar with histograms? Think of it like binning the power intensity per nm bin and power intensity per THz bin. Since THz and nm are inversely related, the width of the bins is changing when the basis is changed. This leads to another stretching in the data that is less intuitive.

[–] [email protected] 1 points 10 months ago (1 children)

the width of the bins is changing when the basis is changed.

Thank you. Why would they compress/decompress based on how light is measured? I would assume that the x-axis would reflect the same range of light regardless if the light is measured by length or frequency. Why give different ranges of light?

[–] [email protected] 2 points 10 months ago

The x-axis range spans the same region of “photon energy” space in both plots. The data starts at about 280 nm in the first plot, which is 1000 THz (the maximum value in the second plot).

The stretching effect caused by working in different x-axis units is because the units don’t map linearly, but are inversely proportional. A 1 nm wide histogram bin at 1000 nm will contain the histogram counts corresponding to a 0.3 THz wide region at 300 THz in the frequency plot. Another 1 nm wide bin at 200 nm will correspond to a 7.5 THz wide region located at 1500 THz in the frequency plot.

You can get a sense of how this works just by looking at how much space the colorful visible light portion of the spectrum takes up on each plot. In the wavelength plot, by eye I’d say visible light corresponds to about 1/6 the horizontal axis scale. In the frequency plot, it’s more like 1/4.

That normalization is necessary because otherwise exactly how you bin the data would change the vertical scale, even if you used the same units. For example, consider the first plot. Let’s assume the histogram bins are uniformly 1 nm wide. Now imaging rebinning the data into 2 nm wide bins. You would effectively take the contents of 2 bins and combine them into one, so the vertical scale would roughly double. 2 plots would contain the same data but look vastly different in magnitude. But if in both cases you divide by bin width (1 nm or 2 nm, depending) the histogram magnitudes would be equal again. So that’s why the units have to be given in “per nm” or “per THz).

[–] [email protected] 2 points 10 months ago (1 children)

I would assume that the difference comes from the fact that the same intervals of light will correspond to intervals of different length when plotted by frequency vs when plotted by wavelength, thus making one of them appear higher, while the other is just wider

[–] [email protected] 2 points 10 months ago (1 children)

Why would the intervals be different?

[–] [email protected] 2 points 10 months ago (1 children)

What I mean is that, for example, the interval from 1eV to 2eV has the same length as the one from 2eV to 3eV, yet they correspond to the intervals from 1/2 to 1 and from 1/3 to 1/2 (dropping units and constants), which have different lengths.

[–] [email protected] 1 points 10 months ago

oooHHHhhhh! Then, does that explain why the wavelength one has a long skewed right distribution while the frequency one has more of a slope in the other direction if we adjust the scales to match the x-axis on colors?