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:
\frac{\partial\phi}{\partial y}= -x+ g&#039;(y)= -x[/itex] so we must have g&#039;(y)= 0. Since g is a function of y only, g&#039; is an ordinary derivative and g&#039;(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 &quot;exact differentials&quot; and to solve first order &quot;exact&quot; 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&amp;#039;(y)= x<br /> which is impossible!<br /> <br /> The problem is that (2y)_y\ne (x)_x so, in ansrivas&#039; terms, \nabla\cross\nabla f= curl(grad f)\ne 0 which is impossible.

 

[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!