Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have derived geometric optics for gravitational waves and I am trying to interpret one of the results. I have

\begin{equation} k_{\rho}k^{\rho}=0 \end{equation}

for the wavevector. For the case of electromagnetic waves, Misner, Thorne and Wheeler say that this means the wavevector is "null". What does this mean?

At the moment, I can see two things. First and most obviously, it's the 'dot' product of a four-vector with itself, which equals zero. I'm not exactly sure what this means.

Secondly, if we take $p=\hbar k$ then we find that

\begin{equation} p_{\mu}p^{\mu}=0 \end{equation}


\begin{equation} m=0. \end{equation}

With $m=0$ we would presumably say that the particle associated with the wave (photon, or graviton in this case) is massless, and thus they travel along null geodesics. Am I missing something or do we need to talk about particles? After all, our original equation involved $k$, which is a *wave*vector. Is there an interpretation from GR itself?

share|cite|improve this question
up vote 3 down vote accepted

The basic classical interpretation of this expression is that EM waves, or any wave described by a null 'wave-4-vector' like gravitational waves, travel at the speed of light. You can easily see this by considering the classical definition of the wave-4-vector $k = (\frac{\omega}{c},\mathbf{k})$, where $\mathbf{k}$ is the usual wavevector in 3-space. This definition is required so that the function $$f(x) = A e^{i k_{\mu} x^{\mu}}$$ describes a wave propagating with frequency $\frac{\omega}{2\pi}$ and wavelength $\frac{2\pi}{|\mathbf{k}|}$. The null condition translates to $$ k_{\mu}k^{\mu} = 0 \Rightarrow \frac{\omega}{|\mathbf{k}|} = c. $$ In other words the phase (and group) velocity of the waves is equal to the speed of light. This makes sense, since the wavevector points along the direction of propagation in spacetime, and you know that light follows null geodesics.

This is obviously independent of the particle interpretation. However, if you want to describe the waves as associated with particles due to QM considerations, it also follows that those particles must be massless.

share|cite|improve this answer

The interpretation from GR itself is your first thing--the inner product of the vector with itself is zero. Namely, you have $k^{a}k^{b}g_{ab} = 0$. In the case of special relativity, $g_{ab} = \eta_{ab}$, and this equation becomes $-(k^{t})^{2} + (k^{x})^{2} + (k^{y})^{2} + (k^{z})^{2} = 0$.

As for the other interpretations, we can associate the wavevector of a particle with that particle's 4-momentum. And general relativity gives us the special relation that, if a particle has four momentum $p^{a}$, then we have $p^{a}p^{b}g_{ab} = -m^{2}$. So, the wavevector of the photon being zero, and the wavevector being equivalent to the 4-momentum of the photon, we conclude that the mass of the photon is zero.

share|cite|improve this answer
Thanks for the reply, but I'm not satisfied with your answer. You've just re-written my question. – user12345 Mar 27 '13 at 17:03
@user16307: there's not much more to it--moentum is the Fourier transform of position, and that's where the wavevector appears when you're doing the Green's function business to solve the wave equation. – Jerry Schirmer Mar 27 '13 at 17:54

In general we talk about the vectors as being spacelike, timelike, or null--which is also called lightlike. The distinction is useful because no Lorentz transformation can turn a spacelike vector into a timelike or lightlike vector, and the same for the others. All timelike observers will agree whether a given vector is spacelike, timelike, or lightlike (null).

In no way do we need to talk about particles here--this property of vectors is valid whether we're talking about a single particle's trajectory or a vector field that may vary over spacetime. For instance, you could talk about the velocity field of some fluid as being timelike everywhere, even though it's not constant everywhere.

It's perfectly valid to talk about the wavevector of an EM wave through spacetime. Consider a simple, linearly polarized wave. The EM tensor has the form

$$F^{ab} = Ak^{[a} n^{b]} = A (k^a n^b - k^b n^a)$$

for some polarization vector $n$ and $A$ a constant. As it turns out, the source equations of Maxwell's equations become $\partial_a F^{ab} = k_a F^{ab}$, which is

$$k_a F^{ab} = A (k_a k^a n^b - k_a k^b n^a)$$

The first term is clearly zero. The second term can be seen as zero by the following logic: if $k_a n^a \neq 0$, then $n^b = (k^a n_a) k^b + \ldots$, where the dots denote other stuff orthogonal to $k^b$. Then $k_b n^b = (k^a n_a) k_b k^b = 0$ because $k^b$ is null.

This shows that no particle interpretation is required. Otherwise, you've made most of the relevant observations. Single particles that are massless travel on lightlight geodesics.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.