Talk:D'Alembert operator

Is is 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)

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)