D'Alembert operator and special relativity

We are currently covering special relativity in the theoretical physics lectures where we defined: $$\mathrm ds^2 := \mathrm dt^2 - \mathrm dx^2 - \mathrm dy^2 - \mathrm dz^2$$

In Road to Reality, this is introduced using a metric tensor $g_{\mu\nu}$ which is $\mathop{\mathrm{diag}}(1, -1, -1, -1)$.

With a scalar product between two (four-row) vectors $x$ and $y$ $$\langle x, y\rangle := g_{\mu\nu} x^\mu y^\nu$$

I would have a norm: $$\|x\| = \sqrt{\langle x, x\rangle}$$

Now I read about the Lorentz gauge in electromagnetism and realized that I could write the d'Alembert operator $\mathop\Box$ like so: $$\mathop\Box = \left\| \left(\partial_t, \nabla \right) \right\|^2$$

So that the $\Box$ operator is basically the Laplace $\triangle$ operator, although not in a 3-dimensional space but in a $(1,3)$-dimensional spacetime?

-
Easy answer here: Yes! You're correct. Though by convention you should have written $\eta_{\mu\nu}$ to indicate you are working in flat spacetime. The generalisation to curved spacetime is somewhat more complicated, but straightforward using covariant derivatives. – Michael Brown Jan 10 '13 at 13:23
You're completely right, but I do advise against using the $\| \cdot \|$ notation for the d'Alembertian. It is a good habit to only apply it to objects that have a definite norm, such that $\| \dotsm \|$ is a real number. – Vibert Jan 10 '13 at 14:38
@Vibert: I think using the inner product like Vladimir Kalitvianski's answers is better, since it avoids the $\| \cdot \|^2$, like you said. – Martin Ueding Jan 10 '13 at 15:52
@MichaelBrown: I think I only saw $g_{\mu\nu}$ in Penrose's book, but he usually goes for the generalized notation right away. I've seen $\eta_{\mu\nu}$ all over Wikipedia and was wondering already why they did not just use $g$. Now I know. Thanks! – Martin Ueding Jan 10 '13 at 15:54
Another comment on your notation: You define $\langle x, y\rangle := g_{\mu\nu} x^\mu y^\nu$. Notice that in standard notation $y_\mu=g_{\mu\nu}y^\nu$ while $(\partial_t,\nabla)\equiv\frac{\partial}{\partial x^\mu}$ is taken to be $\partial_\mu$ with indices down already. In Minkowski coordinates $g_{\mu\nu}$ are the same components as $g^{\mu\nu}$, but in general this might make a difference. Moreover, people are sloppy in distinguishing curved spacetime from curvilinear coordinates. Be cautious about the context of your definitions ($\langle x, y\rangle$ is covariant, why spoil that). – NikolajK Jan 10 '13 at 15:59

Using $c=1$, Cartesian coordinates, and your metric signature, the Laplace operator is

$$\triangle := -g^{ij} \partial_i \partial_j = -g^{ij} \frac{\partial}{\partial x^i} \frac{\partial}{\partial x^j}$$

where Latin indices run over $1,2,3$ and $g^{ij} = \mathrm{diag}(-1,-1,-1)$. Its proper spacetime generalization is the D'Alembert operator defined by

$$\Box := -g^{\mu\nu} \partial_\mu \partial_\nu = -g^{\mu\nu} \frac{\partial}{\partial x^\mu} \frac{\partial}{\partial x^\nu} = \triangle - \frac{\partial^2}{\partial t^2}$$

where Greek indices run over $0,1,2,3$ and $g^{\mu\nu} = \mathrm{diag}(1,-1,-1,-1)$. However, you are defining it in the alternative form

$$g^{\mu\nu} \partial_\mu \partial_\nu = g^{\mu\nu} \frac{\partial}{\partial x^\mu} \frac{\partial}{\partial x^\nu} = \frac{\partial^2}{\partial t^2} - \triangle$$

and this one cannot be considered a mere generalization to 4D because of the minus sign. The key is in your definition of the norm, which lacks a minus sign because you are using a metric signature with trace -2.

-

In your notations $$\mathop\Box = \langle \frac{\partial}{\partial x}, \frac{\partial}{\partial x}\rangle$$

-