Is any constant vector field conservative?

  • Thread starter Caio Graco
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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.
 

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.
 

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.
 

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.
 

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.
 

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.
 

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.
 

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