Talk:D'Alembert operator

Super 2 or not ?
Is it written $$\Box$$ or $$\Box^2$$?

It is defined $$\Box=\partial_t^2-\nabla_x^2$$, written without extra square and with this choice of signs, in most of today's math/physics literature.

Is it equal to $$ \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} $$ or $$ \frac{1}{c^2} \frac{\partial^2}{\partial t^2} - \nabla^2 $$ ? Sources differ again.


 * It should be $$\Delta := (\mathrm{d} + \mathrm{d}^*)^2 =|_{\mathbf{R}^3} -\nabla^2$$ and $$\Box$$ is simply the Laplacian in Minkowski space. Of course signs are subject to conventions, but the square I have never seen before.MarSch 16:18, 18 Mar 2005 (UTC)


 * So far as big physics texts go, Griffiths uses $$\Box^2$$, while Jackson, the physics standard on E&M, uses $$\Box$$. They are always recognized as the same, since the box isn't used for anything else and you never need to square the D'Alembertian.  The sign convention is less clear, but most of the four-vector articles are using the convention where the minus sign goes on the time parts, so then it would be
 * $$\nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} $$
 * I seem to recall this is also Griffiths's definition, although not Jackson's. The answer is that there is no answer, but we can at least try to be consistent within wikipedia.  --Laura Scudder 20:39, 6 Apr 2005 (UTC)

My notes (3rd year undergrad course on Relativistic Electrodynamics, Cambridge University) make the remark in a footnote that strictly '4-Gradient' is $$\Box= \frac{1}{c} \frac{\partial}{\partial t}-\nabla$$ and the d'Alembertian is $$\Box^2 = \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \nabla^2$$. It goes on to remark that some texts refer to the d'Alembertian as $$\Box$$ and some as $$\Box^2$$ MikeMorley 10:08, 12 April 2006 (UTC)


 * I think it should definitely be mentioned that the operator is often written as $$\Box^2$$ since sources like Griffiths and MathWorld use that notation. Currently the notation section does not mention this, so I will add it in --DFRussia (talk) 04:49, 13 December 2009 (UTC)

Klein-Gordon
D^2 + m or D^2 - m is a matter of convention, because the sign of D^2 is. --MarSch 08:56, 23 October 2005 (UTC)

conflicting notations
I've rewritten the article to only mention all alternate notation, but only use the Delta-symbol for the Laplacian which the d'Alembertian is. Comments welcome. --MarSch 09:33, 23 October 2005 (UTC)

Not so obvious?
This may be obvious, but I've had a devil of a time confirming it... The D'Alembert Operator is named for Jean le Rond d'Alembert, correct? Seems like that should be mentioned somewhere if it is (as I assume it to) true.--Falcorian (talk) 19:50, 15 May 2007 (UTC)
 * This is correct. I have added a link to the eponymous fellow. Eutactic (talk) 05:49, 29 July 2008 (UTC)

Symbol looks like an undefined character
When I first visited this page I thought my browser was not capable of displaying the symbol. I eventually figured out that the symbol really is a featureless empty square. I wonder if anyone can suggest a way to make this more obvious. I know it says "the four sides of the box representing..." in the text but that's pretty easy to miss.

I note that some pages have a tag warning readers of this issue:

and this warning is still often necessary (see examples at Unicode_and_HTML; in my browsers the Ge'ez and Runic symbols are not rendered) so it is reasonable to expect that a lot of users are still encountering featureless empty squares in aticles that use special symbols, and will be suspicious of $$\Box$$. Robert Munafo (talk) 21:31, 6 June 2010 (UTC)
 * This bothered me as well. I was hoping that - by analogy with the greek capital delta - there existed some D'Alembertian character with one or two bolded edges, however I could not find one in use or in any of the common LaTeX sets. Whilst a greek-ish D'Alembertian character has been proposed at least once (here), there sadly doesn't really seem to be any precedent beyond speculation that would empower anyone to use a less ambiguous symbol. I will however add a comment in the lead that will point out that the symbol is indeed a square. Eutactic (talk) 06:58, 7 June 2010 (UTC)

I agree, I saw this operator on another page and followed the link to see what it "should" look like. Maybe add some bold text at the start of the article to let those of us who are "too familiar" with computers know that the square box isn't a browser bug! 86.140.176.71 (talk) 16:18, 4 July 2012 (UTC)

I also agree. When I first saw it I was uncertain if it was a symbol or error. Put the text caveat right under the equation or in a similarly obvious spot. Each section needs a caveat because many people don't read the whole article - they skip right to the section they are interested in. — Preceding unsigned comment added by 24.143.251.106 (talk) 23:32, 26 April 2019 (UTC)

=Covariant derivative?== Something is amiss. I found a website claiming that although $$ \partial_{\mu} $$ is allowed, $$ \partial^{\mu} $$ isn't and in principle, one is actually using a covariant derivative in Minkowski space. Some precision or clarification on that issue would be helpful.TonyMath (talk) 08:24, 14 July 2015 (UTC)

= Should c appear=

Here we can read $$ \Box = \frac{1}{c^{2}} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2} $$, but if we have fixed the time metric as $$ g=\mathrm{diag}(1,-1,-1,-1) $$ (ie, with $$ c=1 $$, I think the value of c shouldn't appear. In my notes it doesn't appear. --JBecerra (talk) 12:55, 22 June 2016 (UTC)


 * What do you mean? It is not obvious that $x^{0}≡ ct$ ? Later on, if you keep reading, c is set equal to one. Cuzkatzimhut (talk) 14:05, 22 June 2016 (UTC)

Dead reference
The article has a single reference and the link is dead 169.235.95.203 (talk) 19:27, 17 October 2022 (UTC)