# Is any constant vector field conservative?

#### Caio Graco

Is a constant vector field like F = kj conservative? Since the work of F for any closed path is null it seems that F is conservative but for a force to be conservative two conditions must be satisfied:

a) The force must be a function of the position.
b) The circulation of force is zero.

My question is about the first condition. For the given vector field I believe that there is no dependence on the position, because the field does not depend on any of the x, y, or z coordinates. And if so, the given field is non-conservative. The relation of this doubt with the Physics is given as far as the uniform gravitational field or electric.

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#### andrewkirk

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For the given vector field I believe that there is no dependence on the position, because the field does not depend on any of the x, y, or z coordinates. And if so, the given field is non-conservative.
I think that rule has been mis-stated above. The correct statement is that force can only depend on position. That is, it cannot depend on anything else, such as velocity.

A spatially-constant field in flat space-time depends only on position. The dependence is the function that maps every point in space to the same force vector. It gets a bit more complex when spacetime is curved, because then it no longer makes sense to talk about the force being the same at two different points.

• Caio Graco and Orodruin

#### Orodruin

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Just to add to #2, it is relatively easy to show that being conservative is equivalent to being curl free. Clearly a constant field has zero curl.

• Caio Graco

#### PeroK

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Is a constant vector field like F = kj conservative? Since the work of F for any closed path is null it seems that F is conservative but for a force to be conservative two conditions must be satisfied:

a) The force must be a function of the position.
b) The circulation of force is zero.

My question is about the first condition. For the given vector field I believe that there is no dependence on the position, because the field does not depend on any of the x, y, or z coordinates. And if so, the given field is non-conservative. The relation of this doubt with the Physics is given as far as the uniform gravitational field or electric.
A constant function is still a function. What condition a) means is that it does not change over time. If a vector field is, say, diminishing over time, then it not conservative.

• Caio Graco

#### Ray Vickson

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Is a constant vector field like F = kj conservative? Since the work of F for any closed path is null it seems that F is conservative but for a force to be conservative two conditions must be satisfied:

a) The force must be a function of the position.
b) The circulation of force is zero.

My question is about the first condition. For the given vector field I believe that there is no dependence on the position, because the field does not depend on any of the x, y, or z coordinates. And if so, the given field is non-conservative. The relation of this doubt with the Physics is given as far as the uniform gravitational field or electric.
If ${\bf F} = \langle a,b,c \rangle$is a constant field (with $x,y,z$-components $a,b,c$ then ${\bf F} = \nabla V,$ where $V = ax+by+cz.$ Any field of the form $\nabla V$ is conservative.

• Caio Graco and PeroK

#### Orodruin

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Ok, it has been given in bits and pieces already, but not explicitly so let me summarise.

The following three statements about a vector field $\vec v$ are equivalent (as long as time independence holds):
1. $\vec v$ has a scalar potential $\phi$ such that $\vec v = - \nabla\phi$.
2. $\vec v$ is conservative, i.e., $$\oint_\Gamma \vec v \cdot d\vec x = 0$$ for all closed curves $\Gamma$.
3. $\vec v$ is curl free, i.e., $\nabla \times \vec v = 0$.
It is rather easy to show that these are equivalent. 2 and 3 are directly related to the curl theorem, 1 directly implies 3 since $\nabla \times \nabla\phi = 0$ regardless of $\phi$, and using 2 you can easily construct an explicit potential, thereby implying 1.

• andrewkirk, Caio Graco and PeroK

#### vanhees71

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3. is only equivalent of the domain of the vector field is simply connected.

The paradigmatic counter example is the "potential vortex"

$$\vec{V}=A \begin{pmatrix} -y \\ x \\ 0 \end{pmatrix} \frac{1}{x^2+y^2}.$$
It's easy to see that everywhere, where $\vec{V}$ is well defined (i.e., everywhere except along the $z$ axis) is curl free but that any loop around the $z$ axis gives $2 \pi A \neq 0$.

You can find a potential in any simply connected part of the domain, i.e., in $\mathbb{R}^3$ minus one arbitrary half-plane with the $z$ axis as boundary.

• Caio Graco

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