How can I find the scalar field from its gradient?

[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!

 

[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 $$\phi$$

 

[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.

 

[HallsofIvy]

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!

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:
[tex]\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.<br /> <br /> Notice that is exactly what we do to find integrals of "exact differentials" and to solve first order "exact" differential equations.<br /> <br /> <b>Warning</b>, A function of two or three variables is not necessarily equal to a gradient.<br /> 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 <br /> $$\frac{\partial\phi}{\partial x}= 2y$$ <br /> so $\phi(x,y)= 2xy+ g(y)$. Differentiating that with respect to y, <br /> $$\frac{\partial\phi}{\partial y}= 2y+ g'(y)= x$$<br /> which is impossible!<br /> <br /> 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.[/tex]

 

[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!