1. Aug 8, 2008

### Omri

Hi,

There is some issue about gradients that disturbs me, so I'd be glad if you could help me figure it out.
Say I have a scalar field $$\phi(\mathbf{r})$$, that is not yet known. What I know is a function that is the gradient of $$\phi$$, so that $$\mathbf{F}(\mathbf{r}) = \nabla\phi(\mathbf{r})$$. I want to find $$\phi$$ from $$\mathbf{F}$$, ignoring the constants of course. What I thought of was:
$$d\phi = \sum_{i=1}^{3}\frac{\partial\phi}{\partial x_i} dx_i = \sum_{i=1}^{3} F_i dx_i$$
And therefore:
$$\phi = \int d\phi = \int\sum_{i=1}^{3} F_i dx_i$$
But if you try that with the two-dimensional example $$\phi = x^2 - xy$$, it doesn't work, and gives and gives $$x^2 - 2xy$$.
Can anyone please explain that to me?

Thanks!

2. Aug 8, 2008

### ansrivas

Look at the definition of a line integral in this case and also note that

curl (grad $$\phi$$) =0

so grad $$\phi$$ defines a vector field which is conservative. Look at how this links to path independence and defines a consistent $$\phi$$

3. Aug 8, 2008

### tiny-tim

Hi Omri!

Suppose your line integral is from (0,0) to (0,y), then from (0,y) to (x,y)

The first part is ∫-x dy = -xy evaluated with x = 0, so it's zero (you thought it was -xy, didn't you? )

And the second part is ∫(2x - y) dx evaluated with y = y, so it's x2 - xy, as expected.

4. Aug 8, 2008

### HallsofIvy

If you are given $\nabla \phi$ and want to find $\phi$ you wouldn't start from a given $\phi$ would you? Well, anyway, if $\phi= x^2- xy$ the $\nabla \phi= (2x- y)\vec{i}- x\vec{j}$. If you were given $(2x- y)\vec{i}- x\vec{j}$ and asked to find $\phi$, its anti-derivative, you would start from the definition
$$\nabla \phi= \frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}$$
Comparing that general formula to the specific example, we must have
$\frac{\partial \phi}{\partial x}= 2x- y$
Rememberting that the partial derivative with respect to x treats y like a constant, we can integrate "with respect to x" and get $\phi(x,y)= x^2- xy+ C$ except that, since we are treating y as a constant, that "C" may in fact be a function of y: call it g(y) so we have $\phi(x,y)= x^2- xy+ g(y)$. Now, differentiate that with respect to y:
$$\frac{\partial\phi}{\partial y}= -x+ g'(y)= -x[/itex] so we must have g'(y)= 0. Since g is a function of y only, g' is an ordinary derivative and g'(y)= 0 means g(y)= C, a constant. That is, $\phi(x,y)= x^2- xy+ C$ exactly what we started with except for the constant of integration, C. Notice that is exactly what we do to find integrals of "exact differentials" and to solve first order "exact" differential equations. Warning, A function of two or three variables is not necessarily equal to a gradient. For example, if I were to claim that $\nabla\phi(x,y)= 2y\vec{i}+ x\vec{j}$ and try to do the same thing, I would start with [tex]\frac{\partial\phi}{\partial x}= 2y$$
so $\phi(x,y)= 2xy+ g(y)$. Differentiating that with respect to y,
$$\frac{\partial\phi}{\partial y}= 2y+ g'(y)= x$$
which is impossible!

The problem is that $(2y)_y\ne (x)_x$ so, in ansrivas' terms, $\nabla\cross\nabla f= curl(grad f)\ne 0$ which is impossible.

5. Aug 9, 2008

### Omri

Thanks to all of you for your help. My mistake was that I tried to solve the integral as a regular integral and not a line integral, so I didn't pay attention to choosing a curve and taking the limits. I think I did it right now, I took your advice and used the simplest curve - first a line from (0,0) to (x,0) and then a line from (x,0) to (x,y) (or I could go to (0,y) first, it doesn't really matter). Now I got it, thanks!