Existence of Scalar Potential for Irrotational Fields

[center o bass]

Hi I know it's easy to prove that if a vectorfield is the gradien of a potential, $$\vec F = \nabla V$$, then $$\nabla \times F = 0.$$ But how about the converse relation? Can I prove that if $$\nabla \times F = 0,$$ then there exist a salar potential such that $$\vec F = \nabla V?$$

I get as far as proving the existence of a potential function

$$V(\vec r) = \int_{\vec r_0}^{\vec r} \vec F \cdot d\vec r,$$
but how do I now establish that

$$F = \nabla V?$$

What is the gradient of a line integral?

Homework Equations

The Attempt at a Solution

 

[LCKurtz]

Hi I know it's easy to prove that if a vectorfield is the gradien of a potential, $$\vec F = \nabla V$$, then $$\nabla \times F = 0.$$ But how about the converse relation? Can I prove that if $$\nabla \times F = 0,$$ then there exist a salar potential such that $$\vec F = \nabla V?$$

I get as far as proving the existence of a potential function

$$V(\vec r) = \int_{\vec r_0}^{\vec r} \vec F \cdot d\vec r,$$
but how do I now establish that

$$F = \nabla V?$$

What is the gradient of a line integral?

To answer your last question first, that line integral evaluates to a scalar function of x,y, and z (it's the integral of a dot product). Now, if you have

$$V(\vec r) = \int_{\vec r_0}^{\vec r} \vec F \cdot d\vec r,$$

you automatically have independence of path. Let me rewrite that equation:

$$V(x,y,z) = \int_{P_0(x_0,y_0,z_0)}^{P(x,y,z)}\langle S(x,y,z),T(x,y,z),U(x,y,z)\rangle \cdot\langle dx,dy,dz\rangle$$

You want to show, for example, that V_x(x,y,z) = S(x,y,z)

Look at:
$$\frac {V(x+h,y,z)-V(x,y,z)}{h}=\frac 1 h\left(\int_{(x_0,y_0,z_0)}^{(x+h,y,z)} \vec F\cdot d\vec r - \int_{(x_0,y_0,z_0)}^{(x,y,z)} \vec F\cdot d\vec r\right)$$

Now that first integral can be taken along any path. So let's use the path from
(x₀,y₀,z₀) to (x,y,z) to (x+h,y,z) for the first integral. The first section of the path will cancel with the second integral leaving

$$\frac {V(x+h,y,z)-V(x,y,z)}{h}=\frac 1 h\left(\int_{(x,y,z)}^{(x+h,y,z)} S(x,y,z)\,dx\right)$$

since dy and dz are 0 on this path segment. Now apply the mean value theorem for integrals on the x variable:

$$\frac 1 h\left(\int_{(x,y,z)}^{(x+h,y,z)} S(x,y,z)\,dx\right) = \frac{hS(c,y,z)}{h}$$
for some c between x and x+h. Now if you let h→0 you get V_x=S(x,y,z).

You do the other two partials similarly.

 

[center o bass]

Thank you! That was a very clear explanation.

So one may then conclude that for a irrotational field

$$\vec F = \nabla \int_{\vec r_0}^{\vec r} \vec{F} \cdot d \vec r ?$$

 

[LCKurtz]

Thank you! That was a very clear explanation.

So one may then conclude that for a irrotational field

$$\vec F = \nabla \int_{\vec r_0}^{\vec r} \vec{F} \cdot d \vec r ?$$

Yes, as long as all relevant hypotheses are satisfied. I don't have a reference handy, but I think appropriate continuity of partials on an open connected set was mentioned if I remember correctly.