What ties high frequency electromagnetic waves to short wavelength and converse?

Question

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?

Answer 1

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!

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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:

(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.

Answer 6

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.