# How does the result $k = \sqrt{\mu \epsilon} \omega$ relate to the symmetries of spacetime?

Plane electromagnetic waves propagating in space obey the following relationship between the frequency and wave vector:

$k = \sqrt{\mu \epsilon} \omega$

where $k$ and $\omega$ can also be understood as eigenvalues of the time and space translation operators. It means that two seemingly independent observables, tied to two completely different dimensions, somehow get related by a straightforward mapping which depends only on the properties of the medium.

I understand that it is probably related to the similarity between time translation and space translation. However, they don't behave similarly, but they are proportional to each other. How does this mapping come about? How can I understand it in terms of symmetries?

• Can you clarify your question? Are you asking why $\omega \propto k$ for light waves, from symmetry? – knzhou Aug 24 '16 at 2:16
• I don't think that I can say that k and omega form a symmetry, rather why the symmetries that correspond to time and space and give observables k and omega have such a simple one-to-one and onto mapping. – MsTais Aug 24 '16 at 2:20
• For example, in simple case total angular momentum and orbital angular momentum of a photon are independent, meaning the observable $\ell$ knows nothing about the observable $m$, even though both are related to spatial rotations and similar symmetries. But for the observable $k$ and the observable $\omega$ we have an elegant relationship. Why is that? – MsTais Aug 24 '16 at 2:22
• All waves have a relationship similar to $\omega \propto k^{\alpha}$, where $\alpha$ can certainly be any positive or negative integer (at least as far as I know) and probably can be any real number, in principle. I guess I am not sure what you are asking. – honeste_vivere Aug 25 '16 at 12:25

The two quantities are linked by Lorentz symmetry. Consider a plane wave, $$A \exp\left(i (\mathbf{k} \cdot \mathbf{x} - \omega t )\right).$$ Since $x^\mu = (\mathbf{x}, t)$ is a 4-vector, the phase factor is only Lorentz-invariant if $k^\mu = (\omega, \mathbf{k})$ is a 4-vector too, so it must be.
Another way to see this is to note, as you've seen, that $\mathbf{k}$ is the eigenvalue of $\partial_{\mathbf{x}}$ and $-\omega$ is the eigenvalue of $\partial_t$. Since we know $\partial_\mu = (\partial_t, \partial_{\mathbf{x}})$ is a covector, this implies $k^\mu$ is a 4-vector, again.
Now consider writing down a general linear wave equation in Fourier space. By linearity and Lorentz invariance, we may only consider combinations of $k^\mu$ and constants. Since E&M is scale-invariant, there is no constant. Then the only nontrivial wave equation we can write down is $$k^\mu k_\mu = 0.$$ When you expand this equation, you find $$\omega^2 = \mathbf{k}^2$$ which gives $\omega = |\mathbf{k}|$, as you found.
• @MsTais Hmm. Yeah, that is a deep question. For example, saying that $\omega \propto k$ is equivalent to saying that the speed of light is constant. And having a constant, maximum speed automatically gives you Lorentz invariance. But maybe there's an even deeper way of looking at it? – knzhou Aug 24 '16 at 2:37