Why doesn't sunlight have infinite energy?

  • #1
another_dude
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In school we are taught that sunlight contains all different frequencies of light. Also that each frequency has it's own unique wavelength and energy (per photon). So my question is that if there are infinitely many different wavelengths of light (much like infinitely many numbers in an interval) and each wavelength has it's own non-zero energy, how don't all these different and infinitely many energies add up to infinity?
 
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  • #2
sophiecentaur
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In school we are taught that sunlight contains all different frequencies of light. Also that each frequency has it's own unique wavelength and energy (per photon). So my question is that if there are infinitely many different wavelengths of light (much like infinitely many numbers in an interval) and each wavelength has it's own non-zero energy, how don't all these different and infinitely many energies add up to infinity?
Saying that light has an infinite continuum of wavelengths does not imply that 'each wavelength' has to be present with a finite. amount of energy. In pure Euclidean Geometry a white line can be thought of as consisting of an infinity of points. That doesn't mean that the line would be infinitely bright.
Say you choose to break the spectrum of red light (at the long wavelength end) up into intervals of 10nm. Each interval could have an energy of, say E. If you then choose to break it up into intervals of 1nm, each of those intervals would have an energy of E/10. The total energy would be the same, however small you make your divisions.
 
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  • #3
PeroK
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In school we are taught that sunlight contains all different frequencies of light. Also that each frequency has it's own unique wavelength and energy (per photon). So my question is that if there are infinitely many different wavelengths of light (much like infinitely many numbers in an interval) and each wavelength has it's own non-zero energy, how don't all these different and infinitely many energies add up to infinity?
If you have a continuous spectrum, the total energy would be like an integral, not a sum. It would be like the finite area under a curve. Even though the curve has a height at an infinite number of points, the area under the curve is finite.
 
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  • #4
hilbert2
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In school we are taught that sunlight contains all different frequencies of light. Also that each frequency has it's own unique wavelength and energy (per photon). So my question is that if there are infinitely many different wavelengths of light (much like infinitely many numbers in an interval) and each wavelength has it's own non-zero energy, how don't all these different and infinitely many energies add up to infinity?

It's much the same as a metal sphere having infinitely many points in it and a nonzero density at all points, but still having a finite mass. This kind of things are studied in integral calculus.
 
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  • #5
ZapperZ
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In school we are taught that sunlight contains all different frequencies of light. Also that each frequency has it's own unique wavelength and energy (per photon). So my question is that if there are infinitely many different wavelengths of light (much like infinitely many numbers in an interval) and each wavelength has it's own non-zero energy, how don't all these different and infinitely many energies add up to infinity?

There's a huge error in physics here.

Sunlight does NOT "... contains all different frequencies of light..." WE do not have a detection scheme to detect an infinite range of frequencies of light, so how in the world can we claim that such a range exists, and exists for sunlight?

What you were PROBABLY told was that VISIBLE light contains a range of frequencies. This is distinctly different than claiming that sunlight contains an infinite range of frequencies of light.

If you have a concrete source that told you this from your school, then you should cite the source. Otherwise, we can only guess that you misunderstood or misinterpret what you were told.

Zz.
 
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  • #8
PeroK
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Yes but the issue of 'continuity' is not the same as the conclusion that the power should be infinite. A few lumps and bumps in the spectrum don't affect the basic misconception.
I assume the OP came to PF to find where the misconception was, rather than be flamed for not already knowing the answers.

I assume he's a school kid, not a PhD student.
 
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  • #10
sophiecentaur
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I assume the OP came to PF to find where the misconception was,
And the misconception was dealt with at a simple level in the first few replies. What was in post #2 that "flamed" the OP? The issue of holes and peaks in the spectrum of sunlight was not particularly relevant for the OP. Post #5 was possibly a bit harsh but it was criticising the teaching and not the OP. The same question as in the OP would also apply to the light from a halogen bulb.
 
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  • #11
PeroK
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So, if I understand this correctly the finite energy is explained by the fact the sun's spectrum is not originally continuous and gets continuous in the way??
If you have a very large number of things, you can often model this as a continuous distribution.

One example is that we can model an object as either a very large number of point particles, or as a continuous mass distribution.

The Sun's EM spectrum has so many different frequencies that it looks like a continuous spectrum.

To do this mathematically you need the integral calculus.
 
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  • #12
another_dude
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If you have a very large number of things, you can often model this as a continuous distribution.

The Sun's EM spectrum has so many different frequencies that it looks like a continuous spectrum.
Aha, so it is modelled as a continuous spectum, but it isn't really. Thanks for the replies!
 
  • #13
sophiecentaur
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So, if I understand this correctly the finite energy is explained by the fact the sun's spectrum is not originally continuous and gets continuous in the way??
Nothing happens to the Sun's spectrum "on the way" to affect the finite amount of Power. Perhaps you are misunderstanding the meaning of the word 'Continuous' in this context. It doesn't mean uniform / flat over an infinite range. It means that a spectrometer with a high resolution would measure nearly the same power in a given wavelength interval, as in the next and the next. There is, of course, a hump in the spectrum of visible sunlight arriving at the atmosphere and, even there, it is not a pure black body spectrum (see this link) and note the visible range. The atmosphere introduces a number of deep nulls (but not in the visible range). But the presence or absence of the atmosphere does nothing to make the power infinite - that is an entirely separate (and more basic) issue.
The Sun produces a finite amount of EM Power so bear that in mind when you are looking for an answer to your original question. Infinite doesn't come into it. The spectrum curve from MF Radio waves to Gamma Rays has a finite area (i.e total power)
 
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  • #14
sophiecentaur
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The Sun's EM spectrum has so many different frequencies that it looks like a continuous spectrum.
Further to this: If you were to look at the power arriving from the Sun (or a lightbulb) in a very narrow range, that power would be constantly varying in a noise like way. In a larger bandwidth / range, the level would measure as constant - averaged out.
 
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  • #15
BvU
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Dear dude, :welcome:

I watched your thread because I thought it was a good question and I didn't have a simple, clear answer (even with a PhD, that can happen :smile:).

We are indeed being bombarded by photons that arrive in huge numbers per second on a given area. When we measure a spectrum, we count photons that fall in a small range of wavelengths (the resolution of our spectrometer). The better the resolution, the smaller the number of photons that are detected per unit of time. If (theoretically) we would "increase" the resolution to zero, the counting rate would also go to zero. And that's what offsets the infinity you deducted -- thats's my attempt to explain it in a non-calculus way (don't try this in homework)
 
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  • #16
sophiecentaur
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The better the resolution, the smaller the number of photons that are detected per unit of time.
I don't think the photons actually help in this sort of discussion. The real live 'granularity' of many quantities is not bound to the calculus methods that describe macroscopic relationships. In Calculus, the limit we take is δx→0 and not δx→hf, which is far more specific and arbitrary.
 
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  • #17
OmCheeto
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I don't think the photons actually help in this sort of discussion.
It did for me.
The real live 'granularity' of many quantities is not bound to the calculus methods that describe macroscopic relationships. In Calculus, the limit we take is δx→0 and not δx→hf, which is far more specific and arbitrary.
I have no idea what this means.

Anyways, @another_dude 's question reminded me of an old unsolved problem of mine.

Homework Statement


Given that black body radiation yields values at all wavelengths, guesstimate how many 700 nm photons per time period are emitted from a human body.

constants:
surface area: 1 m^2 (spherical, of course)
temp: 310 K
k_B 1.3806485E-23 J/K Boltzmann constant
h 6.62607E-34 J⋅s Planck constant
c 299792458 m/s speed of light
e 2.71828182845905
λ 700 nm (the limit of human vision on the red end)​

Homework Equations

[/B]

Planck's law broken down into bits, as I was sure I would transcribe something wrong
a = 2*h*c^2/λ^5
b = h*c/(λ*k_B * T)
B_λ(λ,T) = a * 1/(e^b - 1) [power output per (area steradian) @ frequency]
Photon energy = hc/λ

The Attempt at a Solution


[/B]
Find B_λ(λ,T) at 700 and 701 nm, ie power output at each point

700 nm --> 1.14e-23 W m-2 sr-1 nm-1
701 nm --> 1.24e-23 W m-2 sr-1 nm-1

Average those two values.

1.19e-23 W m-2 sr-1 nm-1

multiply by the difference in wavelength, 1 nm, yielding bandwidth power output

1.19e-23 W m-2 sr-1

since the subject is 1 m2, power output will be
1.19e-23 W sr-1

Determine the energy of photons in that range, ie joules/photon @ 700.5 nm

Photon energy = hc/λ =
2.84E-19 joules/photon

I know a watt second = 1 joule ---> watt = joule/sec
so
1.19e-23 W sr-1 =
1.19e-23 joules sec-1 sr-1

therefore, since we want photons/second, we rearrange:
1/2.84e-19 photons/joule * 1.19e-23 joules sec-1 sr-1 = 4.19E-05 photons sec-1 sr-1
which is a very tiny number, so I took the reciprocal, yielding
23866 seconds per photon
which is still too large to recognize, so converted that, and came up with:

A 1 m2 human @ SOT* will emit a 700 nm photon every 6.6 hours.

So, in answer to the OP's question: although the line is continuous from 700 to 701 nm, the energy of the individual photons in that range limits the output.
Which I am sure is also true if one were to do these types of calculations from 700 nm to 56700 nm.
Which, of course, I did.

Obligatory graph:

2018.03.12.black.body.output.of.a.human.body.png


Output in this range, per my calculations: 163 watts per steradian (for a 1 m2 human)
Total output: 525 watts

*SOT = Standard Operating Temperature (98.6°F)
 

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  • #18
OmCheeto
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multiply by the difference in wavelength, 1 nm, yielding bandwidth power output

1.19e-23 W m-2 sr-1

This is somewhat off, per an online calculator, which said the value was supposed to be: 1.23e-23

But after looking at the maths involved:oldsurprised:, I decided to ignore it.

As we plebeians say; "Close enough for government work!"
 
  • #19
sophiecentaur
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I have no idea what this means.
It means that, if you don't like the idea of a continuous function then you can either decide to use photons, which correspond to different amounts of energy for each frequency [hf] ( and the step size - granularity - varies over the spectral range) or you can learn about Calculus which does its thing with intervals (validly) that are infinitesimally small so the actual size of interval is not relevant. As in many cases, using photons gives people the illusion of 'understanding a sort of mechanical model.
When someone decides that the Maths is too hard for a problem then, very often, it's the actual; problem that's too hard but they think they have got it sussed. That's fine but they should then avoid trying to 'explain' the situation to a third party in over simplified terms.
It's a bit like thinking you 'understand' football when you remember a football score but you don't know anything about the offside rule. (I don't know the offside rule so I avoid trying to explain that stuff and I cannot remember a single football score I have ever been told.)
 
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Ever hear of an integral?

It's part of analysis that resolves all these kind of queries about apparent infinities such as the famous Zeno's paradox.. The integral of the frequencies being emitted by the sun is finite - not infinite.

Thanks
Bill
 
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  • #21
doglover9754
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In school we are taught that sunlight contains all different frequencies of light. Also that each frequency has it's own unique wavelength and energy (per photon). So my question is that if there are infinitely many different wavelengths of light (much like infinitely many numbers in an interval) and each wavelength has it's own non-zero energy, how don't all these different and infinitely many energies add up to infinity?
I see where you are going but as I have just learned this spectrum stuff this year, I am curious to know how you think. I’ve never considered this before so I’d like to see some more responses on this topic to clear up what I am now wondering. Good question!
 
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  • #22
another_dude
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I see where you are going but as I have just learned this spectrum stuff this year, I am curious to know how you think. I’ve never considered this before so I’d like to see some more responses on this topic to clear up what I am now wondering. Good question!

Well, my thinking was that if we cut a continuous spectum into infinitely many frequencies - and each frequency has its own energy- the infinitely many energies of all frequencies when added up would result in infinity. But this thinking is flawed. As said above - and thanks for the many responses - this looks more like an integral with a finite value.
 
  • #23
sophiecentaur
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Well, my thinking was that if we cut a continuous spectum into infinitely many frequencies - and each frequency has its own energy- the infinitely many energies of all frequencies when added up would result in infinity. But this thinking is flawed. As said above - and thanks for the many responses - this looks more like an integral with a finite value.
If we're going to explain / invent Integral Calculus then it's better do deal with it on a mathematical basis, rather than tying it to a particular 'mechanical' situation. (Cart before horse etc.) But, of course you are basically right. There are functions with finite integrals over +/- infinity and there are infinite integrals over an infinite range. One of the tests of a good mathematical model of a Physical phenomenon is that it should give you a finite result.
 
  • #24
russ_watters
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I thought it was a good question. There's an analysis here about how continuous the sun's spectrum is:

https://physics.stackexchange.com/q...um-obtained-by-sunlight-said-to-be-continuous
I agree. This is a misconception we've seen before so it shouldn't be surprising or difficult to understand, even if explained in a cumbersome way. I believe it was even a historically real conundrum. Here's perhaps a better way of describing it:

Premise: The light from the sun is an essentially continuous spectrum, and therefore contains an essentially infinite number of frequencies. This part is, I believe, true(but includes a critical hedge).

Why then, does it not have infinite intensity?

There is an additional hidden assumption or two required to get t o the conclusion. It requires each of those infinite frequencies to be continuously represented at a certain energy or amplitude: either an infinite number of photons or continuous waves.

So what is the resolution? The photon flux is finite, not infinite, and the spectrum is only *approximately* continuous because of it. Like the coverage of your lawn with a sprinkler.

Note: this has a Zeno's paradox type feel to it and probably has a similar basis, but *unlike* Zeon's paradox ( @bhobba ) where the infinite number of points can be covered in one step, the spectum *not* fully covered because the number of photons (steps) received is finite. It is not the same paradox.
 
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  • #25
russ_watters
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If you have a continuous spectrum, the total energy would be like an integral, not a sum. It would be like the finite area under a curve. Even though the curve has a height at an infinite number of points, the area under the curve is finite.
While this is true, I believe part of the conundrum is reconciling something continuous and therefore containing an infinite number of points (the spectrum) with something not continuous/infinite (photons). In other words I think it comes from believing the infinite number of points corresponds with an infinite number of photons, each of finite energy.

Another example that comes up that may make the conundrum easier to see: when a telescope takes photos of distant stars/galaxies, pixels on the camera may receive photons at a rate of a few per second or even per minute. Even today I find that shocking. It can be really hard to wrap ones' mind around the idea that photons really are discrete and separated.
 
  • #26
russ_watters
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I don't think the photons actually help in this sort of discussion. The real live 'granularity' of many quantities is not bound to the calculus methods that describe macroscopic relationships. In Calculus, the limit we take is δx→0 and not δx→hf, which is far more specific and arbitrary.
I think it does, particularly since it is true! Otherwise we are left with a continuous stream of waves that are of infinitesimally small amplitude. Essentially, it is trying to find a finite result from infinity*1/infinity. In reality, it is finite*finite.

This is one of the rare cases where the calculus is an approximation and the numerical method is the reality. Usually it is the other way around!
 

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