# What is the difference between path independence and a conservative ve

1. Nov 7, 2013

### ainster31

What is the difference between path independence and a conservative vector field?

It seems like both definitions given in the textbook mean the same. What is the difference between the two? According to wikipedia, path independence is a consequence of a conservative vector field, but according to my textbook, path independence and a conservative vector field are only equivalent in an open region.

2. Nov 8, 2013

### UltrafastPED

They are the same; there are three equivalent ideas each of which implies the other:

Curl of the vector field is zero
Path independence
Conservative force field

Whenever you impose "closed" boundary conditions things happen: a marble falls off of the table, a ball takes a bounce off of the wall.

3. Nov 8, 2013

### HallsofIvy

Staff Emeritus
The concepts are the same. I would not say that "a conservative field" and "an integral independent of the path" were "the same" gramatically. A "conservative field" is a vector field satisfying certain conditions. The integral from one point to another in such a field is independent of the path between those points.

4. Nov 8, 2013

### lavinia

As has already been said a conservative field will always be path independent.

In an open domain this is the same as saying that the field is irrotational. The intuition I have is that the work that a test particle does against a field along a closed path can only be non-zero if the field rotates along the curve. If it doesn't rotate then it is path independent.

On surface such as a cylinder, a field can be irrotational yet not be path independent. Take for instance the tangent field to the circles around the cylinder. Still in any simply connected domain on the surface, the field is path independent but along these circles it is not. Perhaps this is what your text was talking a bout.

5. Nov 9, 2013

### vanhees71

Well, there is one caveat. These are equivalent for simply connected regions in space.

The reason is simple. To prove [a]$\rightarrow$, you need Stokes's integral theorem. Suppose your vector field $\vec{V}$ is curl free, i.e., $\vec{\nabla} \times \vec{V}=0$ in some region of space. Then to show that the line integral is path independent it's equivalent to show that the integral along any closed curve is 0. Now let $F$ be an arbitrary surface lying in the region, where $\vec{\nabla} \times \vec{V}$ is defined and integrable with an arbitrary boundary curve $\partial F$. Then you can use Stokes's theorem:
$$\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{V}=\int_F \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{V})=0.$$
Again, this only holds, if for any closed line you can find a surface with this curve as boundary which lies entirely in the region of space, where $\vec{\nabla} \times \vec{V}=0$.

The most simple counter example is the potential vortex. It's given by
$$\vec{V}=\frac{1}{x^2+y^2} \begin{pmatrix} -y \\ x \\ 0 \end{pmatrix}.$$
You can easily see by direct calculation that
$$\vec{\nabla} \times \vec{V}=0$$
everywhere except on the entire $z$ axis, where the vector field is singular and thus its curl undefined.

Then, using a circle $C$ parallel to the $xy$ plane with center on the $z$ axis, you get
$$\int_C \mathrm{d}r \cdot \vec{V}=2 \pi.$$
The region is not simply connected, because you cannot contract this circle (or any other curve surrounding the $z$ axis) continuously to a point staying all the time in the region of the vector field, where it is defined.

There's also not a potential in the entire region of definition of the field, i.e., the $\mathbb{R}^3$ with the $z$ axis taken out. There is, however a potential everywhere with the exception of a semi-plane with the $z$ axis, which you can choose arbitrarily.

Take, e.g., the semi-plane $x<0$, $y=0$ out of $\mathbb{R}^3$. In this region any closed curve can be contracted to a point, because you are no longer allowed to draw a closed curve around the $z$ axis, because it would necessarily hit this semi-plane! In this region also $\vec{\nabla} \times \vec{V}=0$, and it's simple to see that the corresponding potential is
$$\Phi=-\mathrm{sign} y \, \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right ).$$
This is even easier seen by using cylinder coordinates $(\rho,\varphi,z)$ with $\rho>0$, $\varphi \in (-\pi,\pi)$, $z \in \mathbb{R}$, which parametrizes the entire space with the semi-plane taken out. Then the vector field is
$$\vec{V}=\frac{1}{\rho} \vec{e}_{\varphi}$$
and the potential
$$\Phi=-\varphi.$$
Indeed in cylindrical coordinates we have
$$\vec{\nabla} \Phi=\begin{pmatrix} \partial_{\rho} \Phi \\ 1/\rho \partial_{\varphi} \Phi \\ \partial_z \Phi \end{pmatrix}=-\frac{1}{\rho} \vec{e}_{\varphi}=-\vec{V}.$$
Along the semi-plane, we have chosen as the semi-plane where the potential has a singularity, indeed this potential has a jump by $2 \pi$ (it's approaching $\pi$ if you go to the negative $x$ axis from $y>0$ and $-\pi$ if you come from $y<0$).

You can also chose any other semi-plane or even an arbitrary surface with the $z$ axis as a boundary to define a potential with is well-defined everywhere except on this surface. The potential can then be always constructed by choosing an arbitrary path from a fixed point in the remaining part of the space to any other point $\vec{r}$ in this region, but never intersecting the surface. The choice of such path is again unimportant, because the result of the line integral depends again only on the initial and final point but not on the shape of the path, as long as you stay with pathes not intersecting the arbitrarily chosen surface. The so defined scalar function is a potential for $\vec{V}$ that, however is singular at any point of the surface, where it makes a jump by $2 \pi$ when approaching the point on the surface from the one or the other side.

With this you can also show that all paths encircling the $z$ axis once give the same line integral, $2 \pi$, as the circle around the $z$ axis.

So you can say that if [a] holds in a multiply connected region of space then there is a uniquely defined potential only in every simply connected subregion of space but not in the entire domain, where the original vector field is defined. Thus [c] then holds only locally in open regions of the domain where $\vec{V}$ is defined that are simply connected.

This topological subtleties have very funny implications in quantum theory. E.g., the Aharonov-Bohm effect can be understood in a topological way. It's closely related to gauge invariance of electromagnetism. You can read about this in many textbooks of quantum mechanics, e.g., in

J. J. Sakurai, Modern Quantum Mechanics, Addison-Wesley

Last edited: Nov 9, 2013