Is a Vector Field Equal to Zero if Its Contour Integral is Zero?

[kent davidge]

I was thinking about this while solving an electrostatics problem. If we have a vector $\vec V$ such that $\oint \vec V \cdot d\vec A = 0$ for any enclosed area, does it imply $\vec V = \vec 0$?

 

[MathematicalPhysicist]

Yes because you can take your enclosed area to be small enough such that $\vec{V}$ is constant in it, so you can take it outside the integral and $\vec{V}\cdot \vec{\epsilon}=0$ where $\vec{\epsilon} = \int d\vec{A_\epsilon}$; you can take $\vec{\epsilon}=\vec{V}/N$ where you take $N\gg 1$ and you get: $\vec{V}^2=0$ which means that the vector is zero.

It's a bit heuristic and not rigorous but it should do since I don't remember now how to formalize it.

 

[kent davidge]

That was also my line of reasoning. What I did different from you, was that I failed to realize that if the enclosed area is very small, the vector can be taken constant throughout it. Thanks.

 

[WWGD]

This is true only if V is assumed continuous, otherwise its values can vary pretty wildly even in small regions.

 

[MathematicalPhysicist]

This is true only if V is assumed continuous, otherwise its values can vary pretty wildly even in small regions.

Can you give us a wild such case?
I am not saying you're not correct, I am just not that good in finding examples.

 

[WWGD]

Can you give us a wild such case?
I am not saying you're not correct, I am just not that good in finding examples.

Sure, let me think a bit. If you want just a scalar field, consider any subset S of the interior and use its characteristic function, making sure S has measure 0, e.g., for scalar fields, the Char function of ( a finite collection of) Rationals. The integral will be 0 ( and will exist for a finite collection) but the function is not identically zero.
But let me see if I can come up with something better.

 

[WWGD]

For 1D, a non-zero simple example would be that of a function that is 1 at one point and 0 elsewhere. I think we can use this to define a vector field with similar properties.

 

[fresh_42]

I was thinking about this while solving an electrostatics problem. If we have a vector $\vec V$ such that $\oint \vec V \cdot d\vec A = 0$ for any enclosed area, does it imply $\vec V = \vec 0$?

No, not at all.

The condition is called a conservative vector field: $\operatorname{rot}\vec{V}=0$. This is a first hint, because we wouldn't give $\vec{0}$ an extra name.

Example: $\vec{V}(x,y,z)=(0,\sin y, 0)$.

Yes because you can take your enclosed area to be small enough such that $\vec{V}$ is constant in it.

No. This is the other way around. A constant vector field is conservative. You must not conclude from free of rotations to vanishing!

 

[WWGD]

Dont we need simple-connectedness and a simple-closed curve? Maybe this is assumed?

 

[fresh_42]

Dont we need simple-connectedness and a simple-closed curve? Maybe this is assumed?

Even if! My example is simply connected and the area closed if you choose a disc around zero; or I misunderstood the condition. Stokes does not mean that the integrand is zero! If you run around in an electrical potential, you will not gain or lose energy. That doesn't mean there is no field.

 

[Infrared]

@fresh_42 For conservative vectors fields, the integral along a closed loop is zero, not the integral over an area. It is indeed correct that if $V$ is continuous and integrates to zero over every measurable subset of $\mathbb{R}^n$, then $V=0$ everywhere. For a discontinuous counterexample, suppose that $V=0$ everywhere, except at a single point. Then the integral over any area is still zero.

Also, $\text{rot}V=0$ does not imply that $V$ is conservative if the domain is not simply connected.

 

[fresh_42]

i read the circle in the integral sign as contour integral, my fault.