# Proving properties of curl

1. Jan 25, 2012

### SithsNGiggles

1. The problem statement, all variables and given/known data
The curl satisfies

(A) curl(f+g) = curl(f) + curl(g)

(B) if h is real values, then curl(hf) = hcurl(f) + h'·f

(C) if f is C2, then curl(gradf) = 0

Show that (B) holds.

2. The attempt at a solution
I'm not quite sure how to interpret the "h is real valued" part. Does it mean that, as is, h is a scalar quantity and its derivative is a vector?

Any help is appreciated.

- - - -
So far I've been reviewing the definition of curl.
I let f = f(x, y, z) = (f1(x, y, z), f2(x, y, z), f3(x, y, z))

hf = (hf1(x, y, z), hf2(x, y, z), hf3(x, y, z))

curl(hf) = (D2(hf3) - D3(hf2), D3(hf1) - D1(hf3), D1(hf2) - D2(hf1)),

but I don't see where this route is taking me.

Last edited: Jan 25, 2012
2. Jan 25, 2012

### DrewD

Probably means h is a real valued function. Something like
$h(x_1,x_2...)\vec{f}(x_1,x_2...)$
so the values of h are scalars, but it is not necessarily constant.

3. Jan 25, 2012

### lanedance

yeah I would probably take it as real scalar function of position h = h(x,y,z), multiplying anything else by a vector does not make sense unless it is a vector product

4. Jan 25, 2012

### lanedance

also the fact that h' is in a dot product means it is probably the gradient of h

5. Jan 25, 2012

### SithsNGiggles

I looked at the problems ahead, and there's a similar one involving divergence, where I have to show that

so I'm not sure if h' is the gradient in the curl question.

6. Jan 25, 2012

### Dick

In my reference the h'.f term is the cross product of grad(h) with f.

7. Jan 25, 2012

### SithsNGiggles

I ended up continuing with process and was able to simplify the expression to

curl(hf) = h(D2f3 - D3f2, D3f1 - D1f3, D1f2 - D2f1) + ((D2h)f3 - D3h)f2, (D3h)f1 - D1h)f3, (D1h)f2 - D2h)f1)

curl(hf) = h curl(f) + ((D2h)f3 - (D3h)f2, (D3h)f1 - (D1h)f3, (D1h)f2 - (D2h)f1)

Is there any way to simplify the rest of this expression?

8. Jan 25, 2012

### Dick

The rest of it looks like grad(h) x f. Where 'x' is the vector cross product.

9. Jan 25, 2012

### SithsNGiggles

I suppose there's no way of expressing "grad(h) x f" as "h' f", is there?

10. Jan 25, 2012

### Dick

I have no idea what "h' f" is supposed to mean. It's no standard notation I've ever seen. I wouldn't torture yourself over it. You've got the vector formula right.

11. Jan 25, 2012

### SithsNGiggles

Good enough, I suppose. Thanks everyone for the help.