Vector Calculus

September 5, 2006 7:10 PM

Is The dot product (vector operation) commutative?

This would seem to be the case with normal numerical vectors, but say i switched around a divergence operation, and did A dot del, is this still divergence? What about (A dot del)B? would the be the divergence (scalar) times b?

I ask because i came across this while doing homework and am wondering if i should regard it as a misprint of divergence or not.

(things in bold should be read as vectors)

posted by I_am_jesus to Science & Nature (12 answers total)

As for all math questions I've ever had, the answer can be found here.

And is, yes.
posted by metaculpa at 7:13 PM on September 5, 2006

Aaaand ... I didn't really read the question. Dunno.
Damn, now I'm n-1/n on mathworld.
posted by metaculpa at 7:15 PM on September 5, 2006

Surely someone who knows better will chime in, but I'm pretty sure that del dot A doesn't really have anything to do with a dot product. Just like writing something like (d/dy)*f(x) doesn't really have anything to do with multiplication. It's just a notation.
posted by Xalf at 7:22 PM on September 5, 2006

Yes, the dot product is still commutative even for divergence. Well, taking a look at this mathworld page, I'm not sure. See especially equation 6 and below.
posted by muddgirl at 7:25 PM on September 5, 2006

We're thinking typo here.
posted by Eringatang at 7:27 PM on September 5, 2006

Best answer: A dot del -- this becomes a scalar OPERATOR, so that when A has 3 components A_x A_y A_z (assuming Cartesian coordinates) and same for B, you get

(A dot del) B = (A_x d/dx + A_y d/dy + A_z d/dz) B
= A_x dB/dx + A_y dB/dy + A_z dB/dz
posted by bread-eater at 7:46 PM on September 5, 2006

(And in case you are not used to this notation, A_x is "A with subscript x" i.e. the x component of the A vector.)
posted by bread-eater at 7:50 PM on September 5, 2006

Oh god, no.

The divergence is not a dot product. Please re-read that: The Divergence Operator is Not a Dot Product.

It looks like one, and it is convenient notation, but that's all. In actuality, the Nabla symbol is not a vector. It's often written as one for simplicity, but please realize that this is simply a case of inventing notation to make things simpler.

But if you don't believe me, try calculate it out yourself, pretending that the Nabla thingy is in fact a vector. That is, calculate on one side

(Nabla) dot (a(x, y, z), b(x, y, z) c(x, y, z))

and on the other

(a(x, y, z), b(x, y, z) c(x, y, z)) dot (Nabla)

Are these even remotely similar objects?
posted by vernondalhart at 7:51 PM on September 5, 2006

Also, what bread-eater said above.
posted by vernondalhart at 7:55 PM on September 5, 2006

Response by poster: ok, bread-eater That makes sense, and makes more sense in the context of the problem.
posted by I_am_jesus at 8:07 PM on September 5, 2006

The real issue, as other people have pointed out, is that operators are not, in general, commutative in the way that numbers tend to be. AB and BA in general differ when operators like the nabla are concerned. You don't need to consider dot products at all to see this, consider just multiplication. For instance, d/dx x = 1 and x d/dx is still an operator. The first one is an operator, too, but all it does is multiply things by one. And, by the way, understanding what happens when you commute two operators is very important in things like quantum mechanics.

As an aside, I see no problem with nabla being a vector operator and having the dot product defined in its usual way. Perhaps it's a bit abstract for just taking a divergance, but the notation can be handled in a perfectly consistent way.
posted by Schismatic at 8:24 PM on September 5, 2006

Aha, vector operator, sure. Vector, no. At least not in the usual sense that one would see at a second (or perhaps third) year undergraduate level.

And that being the case, the divergence is not the dot-product of the Nabla operator and a vector-valued function; the dot product is an inner product, which doesn't even make sense in this case. It's purely notational, and while it's a good shorthand, it can often lead to misconceptions such as this one here.
posted by vernondalhart at 11:01 PM on September 5, 2006