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Largest wavelength of photon


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Is there some lower (or upper) limit to the wavelength of a photon. Like macrowaves or something?

 

 

Twice the width of the universe - so you have a half a wavelenght filling the universe and nodes each edge, the first standing wave you can have in a cavity

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But a maximum wavelength (7kHz, the official lower limit to radiofrequencies) that can propagate between Earth's soil and ionosphere.

The Earth's diameter defines one lower resonating frequency, a bit like an optical fiber diameter but in 3D, where measurements have been made.

 

Transmitting and receiving equipment is already unreasonable at 7kHz (old radionavigation like Loran used similar ones, comms with submarines use them), so much lower frequencies may exceed our technical capabilities.

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Is there some lower (or upper) limit to the wavelength of a photon. Like macrowaves or something?

No.

 

 

Your answer is with direct conflict with Planck work.

 

Quote from

http://en.wikipedia.org/wiki/Planck_constant

 

"Classical statistical mechanics requires the existence of h (but does not define its value).[2] Planck discovered that physical action could not take on any indiscriminate value. Instead, the action must be some multiple of a very small quantity (later to be named the "quantum of action" and now called Planck's constant). This inherent granularity is counterintuitive in the everyday world, where it is possible to "make things a little bit hotter" or "move things a little bit faster". This is because the quanta of action are very, very small in comparison to everyday macroscopic human experience. Hence, the granularity of nature appears smooth to us."

 

"Thus, on the macroscopic scale, quantum mechanics and classical physics converge at the classical limit. Nevertheless, it is impossible, as Planck discovered, to explain some phenomena without accepting the fact that action is quantized. In many cases, such as for monochromatic light or for atoms, this quantum of action also implies that only certain energy levels are allowed, and values in-between are forbidden"

 

It means that photon can have energies:

E=h*1=6.62607*10^-34 J

E=h*2=1.325*10^-33 J etc etc

E=h*xxx

where xxx is any integer number, and floating point numbers of frequency are forbidden.

 

wavelength = c/frequency

so the largest wavelength = c/1 = c

 

for 2 Hz:

wavelength = c/2 = 149896229 m

 

for 3 Hz

wavelength = c/3 = 99930819.3 m

etc.

 

We can speculate and imagine that h is not quantum of energy but instead reduced h is (or other fraction of h), and have energy levels like this:

E=hbar*1

E=hbar*2

E=hbar*3

 

Quantization of energy levels require quantization of frequency (because f=E/h) and quantization of angular frequency w=E/hbar

Without quantization we literally would not be able to calculate h constant, if f is any positive real number.

Without quantization we would return to XIX and early XX century "catastrophe in ultraviolet".

 

 

Transmitting and receiving equipment is already unreasonable at 7kHz (old radionavigation like Loran used similar ones, comms with submarines use them), so much lower frequencies may exceed our technical capabilities.

 

On the other end are photons with the largest energy detected, 3.5 TeV, which is frequency = 3.5e+12/4.135667e-15 = ~8.5 *10^26 Hz

Edited by Sensei
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" In many cases, such as for monochromatic light or for atoms, this quantum of action also implies that only certain energy levels are allowed, and values in-between are forbidden""

"Many" is not the same as "all".

There may be an upper bound to photon energy, (and thus a lower bound to wavelength)

But there's no upper bound to the wavelength.

Also, there are excited states for the hydrogen atom with arbitrarily small separations between them.

Photons absorbed and emitted by transitions between those states have arbitrarily long wavelengths.

Give me a wavelength, and I can calculate the two states required to give a longer wavelength.(and an infinite number of possible longer wavelengths too)

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[The largest possible wavelength of a photon is] Twice the width of the universe - so you have a half a wavelenght filling the universe and nodes each edge, the first standing wave you can have in a cavity

Or equal to the width of the universe of the universe is cyclic rather than a cavity with hard walls. Or lower if it's not like a torus (in which case plane wave momentum eigenstates may not even be the proper basis for the lowest-energy modes). Or any other value if the "edge" of the universe happens to be well described by neither a box with infinite potential outside nor by periodic boundary conditions.

 

@Sensei: I advice having a look at the units of your presumed calculation. Keeping track of the units is not foolproof (in particular, having the units correct does not guarantee a statement being correct). But checking the units still is the simplest and most helpful tool for sanity-checking. Particularly for people not familiar with physics. In case you do not understand what I say: The Planck Constant does not have units of energy, but of energy times time (which is usually called "action").

Edited by timo
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Your answer is with direct conflict with Planck work. [...]

 

It means that photon can have energies:

E=h*1=6.62607*10^-34 J

E=h*2=1.325*10^-33 J etc etc

E=h*xxx

where xxx is any integer number, and floating point numbers of frequency are forbidden.

Mamma mia!

 

Since when is h an energy?

And: do you imagine a quantization independently of an object?

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Your answer is with direct conflict with Planck work. [...]

 

It means that photon can have energies:

E=h*1=6.62607*10^-34 J

E=h*2=1.325*10^-33 J etc etc

E=h*xxx

where xxx is any integer number, and floating point numbers of frequency are forbidden.

Mamma mia!

 

Since when is h an energy?

 

I wrote E=h*f

where

h = 6.62607*10^-34 J*s

and replaced f by 1 Hz = 1 s^-1

so after multiplication it will be

E=6.62607*10^-34 J*s * 1 s^-1 = 6.62607*10^-34 J

That's quite obvious, isn't?

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I wrote E=h*f

where

h = 6.62607*10^-34 J*s

and replaced f by 1 Hz = 1 s^-1

so after multiplication it will be

E=6.62607*10^-34 J*s * 1 s^-1 = 6.62607*10^-34 J

That's quite obvious, isn't?

How obvious is it that you can also do that calculation with 0.5 Hz?

What about 0.0000000001 Hz?

 

The error is where you say "where xxx is any integer number, and floating point numbers of frequency are forbidden."

There's no call for it to be an integer, fractions (and irrationals) are allowed.

 

Consider the energy levels of a hydrogen atom ('cos the maths is easy)

 

 

67ca16c2a44e8d6c6532121efe92775b.png

 

 

I can make lambda as big as I want by choosing two very large values for n1 and n2

Say I choose them to be 1000000 and 1000001

 

R is about 10^7

That gives me a wavelength of about 50 km .

Larger values of n1 and n2 will give me correspondingly longer wavelengths.

Why do you think there is a limit?

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Rydberg formula is from 1888 year, pre-quantum physics era.
Rydberg had no idea about E=h*f equation, nor any quantization.
http://en.wikipedia.org/wiki/Rydberg_formula

 

I wouldn't mind if it would be just 50 km, it's still within radiowave frequencies with ~6 kHz.

 

But your example n1=1,000,000 and n2=1,000,001 gives me 45.5 million km wavelength, not 50 km. And frequency = 0.00658 Hz.

How do you want to detect such photon with frequency that can't be even called long radiowave?

ELF = Extremely Low Frequency

http://en.wikipedia.org/wiki/Extremely_low_frequency

They're starting at 3 Hz...

Above link contain issues in detection and emission of such photons.

 

324px-Light_spectrum.svg.png

Edited by Sensei
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"Rydberg formula is from 1888 year, pre-quantum physics era.

Rydberg had no idea about E=h*f equation, nor any quantization."

Yes, I know, and it clearly doesn't matter.

His formula still predicts the wavelength of emitted photons.

Why did you bother to mention that his work is part of the foundations of QM?

​Did you think it made some difference to the conclusion?

 

"But your example n1=1,000,000 and n2=1,000,001 gives me 45.5 million km wavelength, not 50 km. And frequency = 0.00658 Hz."

Oops, I must have pushed the wrong buttons on my calculator, but it doesn't affect the conclusion does it?

.

"How do you want to detect such photon with frequency that can't be even called long radiowave?"

For a start, it can be called long wave radio.

("They're starting at 3 Hz.." is entirely arbitrary: the name doesn't make any difference to the outcome)

For a finish I don't care how you detect it. The point is that it exists.

 

However, in the interests of science, I will tell you how, in principle, you can detect it.

You shine it onto a hydrogen atom in an excited state that's , for example, the n=1000002 state.

The energy of the photon will be enough to ionise that atom and if it's in an electric field, you get a current flow.

OK the detector would be difficult to build (and it would need to be big), but that's just an engineering problem.

The photons still exist.

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