Given a and b as standard i j k vectors and asked to solve at a given point.
Surely just solve divergence inside both brackets, which I dont think can depend on the ordering? then get the outcome, times every part of b by the result and plug in the numbers.
1) What do you mean by the gradient of a vector?
2) What do you mean by the gradient dotted with a vector?
My vector calculus is a bit rusty, but I'm guessing what you mean by grad . a is the divergence of the vector? If that's true it's written del . a or just div a.
As for (a . grad)b I don't think this makes sense. I would read it as a dotted with the gradient of b, but b is a vector and you only take the gradient of a scalar function. The del operator doesn't commute, so you can't say a . grad = grad . a, and I don't think (a . grad)b, a and b vectors, even makes sense. I would go back and check your definitions.
thylacine,
Is (a . del), a is a vector, defined? I've never seen it written like that.
And if it is, I would think it would be defined such that (a . del) = (del . a), so the two operations in OP would be equal, right?
I don't believe (a . del) is defined, but your example threw me off. I don't quite see how they are analogous since we have two operations (div and grad) in OP, and only one (differentiation) in your example.
grad is the name we gave the upside down triangle denoting the gradient function, hence TRIANGLE THING dot producted with a vector field gives you its divergence.
I thought grad was standard, apologies;.
SO afaik:
grad . a if a = Xi + Yj + Zk = d(x)/dx + d(y)/dy + d(z)/dz
I think you need to clarify what you mean by (a . grad)b then. LLY's definition makes sense to me, though, and if he's right the difference b/w your two statements would be:
(grad . a)b = [d/dx (ax) + d/dy (ay) + d/dz (az)]b where a = ax i + ay j + az k
So it's your divergence of a (a scalar) times your vector b.
(a . grad)b = (ax * d/dx + ay * d/dy + az * d/dz)b
In this would you would distribute (a . grad) into each component of b. (sorry I think it would be more clear if I could use better notation) But what it boils down to is in the former you are differentiating the components of a, adding, then multiplying by b, but in the latter you are going to end up differentiating the components of b because the ordering is different (which is what thylacine was getting at).
Matt R is correct. Calling the nabla operator "grad" is going to be deliberately confusing; it's non-standard. "Grad a" is typically shorthand for "the gradient of a", but that does not imply the nabla operator means "grad". Call the nabla operator del (as is standard) and you won't have a problem. For example, "Del dot b" is the divergence of vector b.
Call the nabla operator del (as is standard) and you won't have a problem.
let's not generalize too much, no one says 'del' in math (i personally say grad/div or 'D', the latter being the notation Evans uses for the gradient in his graduate pde text). that being said, the notation used by the OP is bad.
let's not generalize too much, no one says 'del' in math (i personally say grad/div or 'D', the latter being the notation Evans uses for the gradient in his graduate pde text). that being said, the notation used by the OP is bad.
I say 'del' on the very rare occasions I talk about this.
I have looked high and low through every textbook I have and the internet, and nothing turns up.
Now remembering my amateur-ness of describing my last problem, I state this exactly.
Find equations of the field line C of: a=yi+xj+xyk through (1,1,1)
Im sure this is really simple, but I dont recall ever using the phrase "field line" of a vector and I cannot find a definition of it ANYWHERE.
(I have used italics just to highlight the vectors, as I have already used bold!)
Find equations of the field line C of: a=yi+xj+xyk through (1,1,1)
Im sure this is really simple, but I dont recall ever using the phrase "field line" of a vector and I cannot find a definition of it ANYWHERE.
Quote:
Originally Posted by Wikipedia
A field line is a locus that is defined by a vector field and a starting location within the field. A vector field defines a direction at all points in space; a field line may be constructed by tracing a path in the direction of the vector field. More precisely, the tangent line to the path at each point is required to be parallel to the vector field at that point.
