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Why is it not possible to have waves with high frequency (high energy) and great wavelength and converse?

What physical quantity ties frequency to wavelength in an inversely proportional way?

Qmechanic
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kurito
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  • You can have a high frequency wave with a great/large wavelength. For example a classical electric field like $E=E_0 sin(\omega t)$, where $E_0$ is constant and $\omega$ is "large". The wavelength is infinite (is that great enough?) and the frequency is as high as you would like. – hft Oct 20 '21 at 07:40
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    What physical quantity ties frequency to wavelength - speed (in this case, speed of light) – slebetman Oct 20 '21 at 09:32

6 Answers6

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The wavelength is how long each individual little wave is. The frequency is how many of these little waves can pass through in one second.

They are automatically inversely related if the speed at which they move is the same for waves everywhere.

Suppose a two-lane street. The left lane is for 1-meter short motorcycles. The right lane is for 5-meter long limousines.

Suppose we see that in 1 second, 5 motorcycles can pass through the left lane.

This means, on the right lane, only 1 limousine can pass by in one second, right? Because the limousine is longer.

In technical terms, this means the wavelength of the motorcycle is 1 meter. The frequency the motorcycles pass through the street is 5 vehicles per second.

The wavelength of the limousine is 5 meters. The frequency they pass through the street is 1 vehicle per second, assuming the speed in both lanes is always the same.

So the longer the wavelength, automatically the less number of those waves can pass through in a second, and therefore the lower the frequency counted becomes.

Hope it helps!

  • Ahem. They are automatically inverse iff the phase velocity is constant. With your car analogy, your math assumes constant velocity. For example, in your analogy, a 1 meter long cyclist would not have the same frequency as 1 meter long motorcyclists (even if both cases, they draft as aggressively). – Aron Oct 21 '21 at 03:39
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The equation for the speed of an EM wave is speed = wavelength x frequency. So for a given speed, larger frequency means smaller wavelength.

To motivate this equation, frequency = 1/period, where the period is the time for one cycle. So if you watch one wavelength of a wave pass you by, it does it in a time equal to the period. So the speed, which is distance divided by time, is wavelength/period which equals wavelength x frequency.

Not_Einstein
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  • Why should the speed be a "given"? Note that you're describing phase velocity, which is not limited by the speed of light (the speed of light is a limit on the group velocity). – Glen O Oct 21 '21 at 00:58
  • But for e-m waves in a vacuum, and to a very good approximation in air at atmospheric pressure, Group velocity = Phase velocity. – Philip Wood Mar 13 '22 at 17:47
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Wave length and period vs. wave vector and frequency
It is a little bit as comparing apples and oranges: it is more meaningful to compare either

  • wave length and its period, which characterized the periodicity of the wave in space and time

or

  • wave vector and frequency, which characterize the rapidity of this change, and which are respectively inverse of the wave length and wave period: $$ \omega=2\pi f=\frac{2\pi}{T}, k=\frac{2\pi}{\lambda}.$$

These pairs of quantities are related via dispersion relation, which in case of simplest wave equation (describing EM wave in vacuum, but also many other different types of waves of various nature) has simple form: $$\omega=ck\Leftrightarrow \lambda = cT.$$

Dispersion relations
For waves propagating in media the dispersion relation can be more complex , and is usually expressed by a function relatingf requency and wave vector, $$\omega(\mathbf{k})$$(see here for an amusing example). In some cases this function may actually have a negative derivative, as, e.g., in case of optical phonons, which are ubiquitous in solids, see here.

Remark
More generally the dispersion relation is written as $$f(\omega, \mathbf{k})=0,$$ since, in principle, there are may be multiple spatial frequencies corresponding to the same temporal frequency and vice versa. This is what we see, e.g., in the case of optical and acoustic phonons cited earlier.

Note that in more than one dimension, $\omega(\mathbf{k})$ may also correspond to different modes of the same wave length, propagating in different directions.

Roger V.
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  • is there a difference in frequency and periodicity in more complex wave theories? –  Oct 20 '21 at 07:47
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    @James not sure I understand your question: could you expand it a bit? – Roger V. Oct 20 '21 at 07:57
  • i am wondering if frequency = periodicity in all wave theories, or there are cases where this is not so? –  Oct 20 '21 at 08:02
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    @James It depends on what one calls waves. One usually means something periodic in space and time, but occasionally it is used for non-periodic waves, like shock waves or solitons, which are also solutions of wave equation. See also a bit more expanded version of this comment: https://physics.stackexchange.com/a/546346/247642 – Roger V. Oct 20 '21 at 08:07
  • @ Roger thank you! –  Oct 20 '21 at 09:14
  • Why can each spatial frequency only be supported at one temporal frequency? – Jagerber48 Oct 21 '21 at 02:36
  • @Jagerber48 this is not correct - we can have many waves of the same spatial and temporal frequencies propagating in different directions. For more complex dispersion relations, like the cited example of phonons, there is more than one spatial mode for each frequency even in one dimension. Eventuy, it depends on whether the dispersion equation has multiple solutions. – Roger V. Oct 21 '21 at 04:44
  • Yes, in the case of phonons there may be multiple values of spatial frequency $|\boldsymbol{k}|$ for any temporal frequency $\omega$, but I believe it is only a few, like maybe a half dozen at most. Why not more? Or why can't a continuous range of $|\boldsymbol{k}|$ values be allowed for a given value of $\omega$?

    It seems that wave equations put tight constraints on allowed dispersion relations for a given medium and I'd like to get better intuition for that. I think what I'm asking here gets at the heart of what is asked in the OP as well.

     – Jagerber48 Oct 21 '21 at 13:45
  • @Jagerber48 It depends on what one calls wave equations and waves. If something is periodic in time, but not in space, shoudl we call it a wave or oscillations or something else? If wave equation is $\partial_t^2u=v^2\nabla^2 u$ then the dispersion is $\omega=ck$, but there are many other equatiosn producing wave-like solutions. – Roger V. Oct 21 '21 at 13:49
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Classical answer - it is a consequence of the invariant speed of light.

Maxwell's equations in vacuum lead wave equations of the form $$ \nabla^2 \vec{E} = \mu_0 \epsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}\ , $$ and something similar for the B-field.

The solutions to this equation are of the form $$\vec{E} = \vec{E_0} f(\vec{k}\cdot \vec{r} - \omega t)\ $$ and it is easy enough to show that these solutions only work if $$ \frac{\omega}{|\vec{k}|} = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\ . $$

If we then recognise that $2\pi/|\vec{k}|$ is the wavelength of the solution and that the frequency of the wave is $\omega/2\pi$, then it is clear that the product of wavelength and frequency must be constant (and that this constant is the speed of light, $(\mu_0 \epsilon_0)^{-1/2}$).

If the waves are not travelling in vacuum and where the speed of light can be less than $c$ and frequency-dependent, then the relationship between frequency and wavelength can be more complicated.

ProfRob
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Strictly speaking, nothing ties "high" frequencies to "short" wavelengths.

The confusion, here, which is also found in some of the other answers, is that there are two distinct velocities involved in electromagnetic waves (and, for that matter, other waves).

The phase velocity describes the speed (and direction) at which a peak of the wave moves. This is given by $v_{phase}=\lambda f$ - that is, wavelength times frequency. For a fixed phase velocity, wavelength will be inversely proportional to frequency.

The group velocity describes the speed (and direction) that the energy moves. In the context of EM waves, this would be the velocity of the photon.

The distinction between phase and group velocity can be visualised like this:

Phase and Group velocities

(source) - the red dot shows the phase velocity, while the green dots show the group velocity.

Now, group velocity is limited by the speed of light - this is the speed at which information can travel, and thus limits movement of energy. However, the phase velocity can be higher than the speed of light. And because of this, there is no inherent bound on the product of wavelength and frequency.

In fact, a related concept in EM waves is also closely related to the phase velocity - the refractive index. The refractive index, $n$, for EM waves is given by $n=c/v_{phase}$, and so, to find a context in which the phase velocity will be greater than the speed of light, you need to find a context in which the refractive index is less than 1.

And we don't actually need to look very far to find an example of this. The ionosphere of the earth has a refractive index of less than 1. Indeed, the formula for the refractive index of the ionosphere is given by the Appleton-Hartree equation, which for the ionosphere is $$ n=1-\frac{40.30N}{f^2} $$ where $N$ is the electron number density and $f$ is the EM frequency. And it's easy to see that this will always be less than 1.

That being said, in most situations, the refractive index will be 1 or higher. And that means that the phase velocity will be capped at the speed of light, which means that the product of wavelength and frequency must also be capped in this way. To this end, high frequencies are tied to short wavelengths in most situations.

Glen O
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The wavelength gets shorter with momentum, hence with kinetic energy and thus, in general, with frequency. So kinetic energy is the answer to your question.

Contrary to the statement in the title and the first sentence, a free electron nearly at rest has a high frequency, $f = mc^2 / h$, yet a long wavelength, $\lambda = h/p$. Its wavelength still decreases with increasing frequency.

In solids, electrons can have negative effective mass in which case the wavelength, though still decreasing with momentum, decreases with decreasing electron energy and thus with decreasing frequency.

Fully analogous effects occur in photonic crystals.

my2cts
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