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I was thinking about the Planck Constant and made some simple calculations and found that a photon with an energy of 1 h would have a wavelenght of c * s and a frequency of 1 Hz. Does that means that 1 Hz is the minimun frequency possible for a photon? if so, why?

Nicolas Caous
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3 Answers3

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Planck's constant doesn't have units of energy, so your calculation doesn't make a lot of sense.

There is such a thing as a Planck energy, though:

$$E_P = \sqrt{\frac{\hbar c^5}{G}} \approx 10^{19}\, \mathrm{GeV} \approx 2\, \mathrm{GJ}$$

and a corresponding Planck frequency:

$$f_P = \frac{E_P}{2\pi\hbar} \approx 3 \times 10^{42}\, \mathrm{Hz}$$

This isn't, as far as we know, some sort of possible maximum energy. Indeed, by macroscopic standards it's pretty big but not that big. What it means is that fundamental processes involving energies of the order of $E_P$ require a theory of quantum gravity to be explained.

Javier
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An EM wave with frequency of $1$ Hz will have a wavelength of $3\times10^8$ m, which is about double the radius of Jupiter. I don't see why it shouldn't be possible to create this sort of radiation. Given that our hearts beat at that frequency, we might be generating these ourselves.

The Planck constant has units of Joules seconds, so it is not an energy scale.

If you want to find a sort of minimal energy, you could consider some very large wavelengths, since frequency is inversely proportional to wavelength, take the size of the universe, about $10^{27}$ m in diameter, then you would get a sort of "minimum imaginable frequecny" of $3.4×10^{-19}$ Hz, which is much less than what you thought.

Andrea
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Nature does not depend on the SI system or whatever system of units we chose. The second is just quite arbitrarily 1/24 times 1/60 times 1/60 the period of Earth's current rotational period - why would it have anything to do with limits on photons?