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

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer

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

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.