# Correct expression for D'Alembert operator in $c=1$ units

In QFT texts with $c=1$ units (most of them), D'Alembert operator is written as:

$$\Box ={\partial^2 \over \partial t^2} - \nabla^2$$

For pedagogical purposes, however, some texts don't set $c=1$, and then

$$\Box =\frac{1}{c^2} {\partial^2 \over \partial t^2} - \nabla^2$$

Is that right? I am a bit confused because Wikipedia uses the second form here despite the fact that, in the very next line it is stated

We have assumed units such that the speed of light $c=1$

Am I missing something very basic, or is that page in Wikipedia simply wrong?

EDIT: In other words, when $c=1$ is assumed, $c$ NEVER appears in the equations. Is that correct in QFT, just as it happens in GR?

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One should always keep in mind that the marvelous Wikipedia is constantly being updated by different users. Most of the time an update constitute an overall improvement, but there is always a danger that previous coherence gets lost when a new user change something in one place but not in the rest of the text.

So concretely, the sentence We have assumed units such that the speed of light $c=1$ should of course have been removed after $c$ was restored in $\Box$. Also note that in the current version, Wikipedia mixes notations $g_{\mu\nu}$ and $\eta_{\mu\nu}$ for the flat metric.

Such minor flaws should always be expected of Wikipedia until a page has matured.

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Yes, Wikipedia is marvelous and I use it a lot (see my first closed question for instance) I had noticed the "wrong" notation for the metric, but it is not critic, as long as you understand that you are in Minkowski space. – Eduardo Guerras Valera Feb 1 '13 at 15:09
Wikipedia is simply wrong!  They must have made a simple typo.

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That is, $c$ NEVER appears in the equations when using $c=1$ units. Is that right? (Please confirm that only if you are sure and you are used to work with different books of QFT). – Eduardo Guerras Valera Feb 1 '13 at 14:51
Yes, this is correct. At least in all QFT books I partially read, and these are a lot. – Noldig Feb 1 '13 at 14:53
Of course it doesn't appear. It's actually ONE if you choose these units. – Bzazz Feb 1 '13 at 15:01
@twistor59, thanks for your quick answer. I have upvoted it. I had to give the "answered" mark to someone, and I choosed the other question because it had additional information (the other, not critical, typo with the notation for the metric) but thanks very much. – Eduardo Guerras Valera Feb 1 '13 at 15:13
@EduardoGuerrasValera no problem. I think you can upvote Qmechanic's answer in addition to accepting it. – twistor59 Feb 1 '13 at 15:18

$c$ is never required to appear in the questions when using a system of units where $c = 1$. Strictly speaking, it can appear as a multiplicative factor, just as you could write in a multiplicative factor of 1, but that would be very unusual, and also confusing because the presence of $c$ in an equation is conventionally taken as a sign that the equation is valid for systems of units where $c$ is not necessarily equal to 1.

As other answers said, this is one of the pitfalls of Wikipedia while an article is developing, which is why you have to use a bit of sense when evaluating the statements there.

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