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