Energy-momentum vector is orthogonal to itself

Question

Let the energy-momentum covector $k$ be $k_idx^i$ in Einstein summation notation where $x^0=t$. Let ${}^3k=k_1dx^1+k_2dx^2+k_3dx^3$ be the space part of $k$.

Let $Ee^{ik_{\mu}x^{\mu}}$ be the electric field of a plane wave, where $E=E_jdx^j$.

The Maxwell equation implies $${}^3k\wedge E=-ik_0 \star_S E.$$

The subscript $S$ means that the Hodge star is only applied in the space part of Minkowski space-time.

Now, the exercice in my book asks to show the following:

Prove that $k_{\mu}k^{\mu}=0$.

This just means that $\langle k,k\rangle =0$. And the Hodge star is also defined in terms of this inner product: $\omega \wedge \star \mu = \langle \omega, \mu \rangle vol$.

So, I tried to take the wedge of the above equation with an appropriate 1-form, but I did not find one that worked. Another possible approach would be to apply $d$ and then take the wedge with something, but I have no intuition in which direction to go.

Can someone give me a hint?

Of course, a full solution is also appreciated, but I am well capable of doing calculations, so I am mainly interested in seeing what to do and why to do it.

Answer 1

I'm not to familiar with how the hodge star works...but I'm pretty sure you want to try something like this. Let's write this puppy out:

$$ k_\mu k^\mu = k_0^2 + k_1^2 +k_2+ k_3^2 $$

We want to show this guy is zero. So let's proceed with trying to show

$$-k_0^2 = k_1^2 +k_2^2+ k_3^2 $$

We call upon the help of Maxwell's,

\begin{align*} ^3 k \wedge E = & (k_1E_2 -E_1k_2)dx^1 \wedge dx^2 +(k_1E_3 -E_1k_3)dx^1 \wedge dx^3+(k_2E_3 -E_2k_3)dx^2 \wedge dx^3 \\ =&-ik_0 \star_S E \end{align*}

If we take the norm of this "vector" (I wrote out the 2 form above to show the vector form), it gives us the following relation:

$$ (k_1^2 + k_2^2 +k_3^2)|E|^2 =i^2 k_0^2 |E|^2 \implies k_1^2 + k_2^2 +k_3^2 = - k_0^2 $$

This is what we wanted, so we're done :)

Answer 2

I'll notate things in terms of clifford algebra and geometric calculus instead; it's a little cleaner.

Let $k = \omega e_0 + \kappa v$, where $v$ is a unit spatial vector.

In an EM plane wave, we take the magnitudes of the electric and magnetic fields to be the same, and we say that they're orthogonal to each other. In terms of the EM bivector $F$, this imposes a certain structure: Let $p$ be the direction of the electric field. Then we can write the EM field as

$$F = \alpha (e_0 p + vp) \exp(i k \cdot x)$$

where $e_0, p, v$ are all orthogonal, so these geometric products could be replaced with wedge products instead, if you prefer. The $e_0 p$ part corresponds to the electric field; the $pv$ part corresponds to the magnetic field.

Check: the magnitudes of the electric and magnetic fields should be equal. We can check this by taking $F^2$ and looking at the scalar part. Write $F = e_0 E + \epsilon_3 B$ and square it, looking at the scalar part:

$$\langle F^2 \rangle_0 = (e_0 E + \epsilon_3 B)^2 = e_0 E e_0 E + \epsilon_3 B \epsilon_3 B = E^2 - B^2$$

This is a well-known invariant of the EM tensor.

Looking at our posited form of the EM bivector, we get

$$\langle F^2 \rangle_0 = \alpha^2 \cos^2(k \cdot x) (e_0 + v) p (e_0 + v)p = -\alpha^2 p^2 (e_0 + v)^2 \cos^2(k \cdot x)$$

But $e_0$ and $v$ are unit vectors, so $e_0 + v$ is null, and $\langle F^2 \rangle_0 = 0$. This implies $E^2 = B^2$, which is exactly what we should get.

Now let's do some calculus. Maxwell's equations in vacuum reduce to $\nabla F = 0$, but $\nabla F = -ikF$ using our established form of the Faraday bivector. Let's write out those terms:

$$-i kF = -i \alpha [-\omega p + \kappa v e_0 p + \omega e_0 vp + \kappa p] \exp(i k \cdot x)$$

More clearly, let's separate out the vector terms:

$$\langle -ikF \rangle_1 = -i \alpha [-\omega p + \kappa p] \exp(i k \cdot x)$$

For this to be zero, $\omega = \kappa$ always. This means $k = \omega(e_0 + v)$, and we already established that vectors of the form $e_0 + v$ are null, so $k$ is null. The equation for the trivector terms gives the same result.

Incidentally, you might notice that $F = \ldots \alpha (e_0 + v)p \ldots$ also. That is, $F \propto kp \exp(i k \cdot x)$. This actually gives us a stronger result: That $F^2 = 0$, for $F^2 = kpkp = -pkkp = 0$. That also tells us about the pseudoscalar invariant of the Faraday tensor, which is proportional to $E \cdot B$, which we also constructed to be zero.