Conservative vector field conditions

If the region were, instead, {(x,y) : x^2+ y^2< 1 and x< 0}, then the region is not "simply connected" because a closed curve that includes (-1,0) in its interior cannot be shrunk to a point without passing through (-1,0). So what conditions must be met for a vector field to be conservative? As the site says, the curl should be zero and the domain should be simply connected. Your calculus book probably implies that you work in R^n or some other simply connected space. A conservative field is usually defined as one that is a gradient of some scalar field. Curl of gradient is automatically zero. On
  • #1
understand.
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My calculus book states that a vector field is conservative if and only if the curl of the vector field is the zero vector. And, as far as I can tell a conservative vector field is the same as a path-independent vector field.

The thing is, I came across this: http://www.math.umn.edu/~nykamp/m2374/readings/pathindex/"

The site shows a vector field where the curl is equal to the zero vector, yet the vector field is not conservative.

As far as I can tell, saying "F is conservative iff Curl(F) = 0" contradicts the claims of the site I posted.

What conditions must be met for a vector field to be conservative?
 
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  • #2
Like the site says, the curl should be zero and the domain should be simply connected. Your calculus book probably implies that you work in R^n or some other simply connected space.

A conservative field is usually defined as one that is a gradient of some scalar field. Curl of gradient is automatically zero. On the other hand, if you have a field whose curl is zero and its domain is simply connected, you can smoothly deform any closed path into a point, and then it's possible to prove that the field is path-independent, therefore you can construct the scalar field out of the vector field.
 
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  • #3
understand. said:
My calculus book states that a vector field is conservative if and only if the curl of the vector field is the zero vector. And, as far as I can tell a conservative vector field is the same as a path-independent vector field.
"Conservative" is really a physics term, from conservative force fields. A more mathematical term is "exact differential".

The thing is, I came across this: http://www.math.umn.edu/~nykamp/m2374/readings/pathindex/"

The site shows a vector field where the curl is equal to the zero vector, yet the vector field is not conservative.

As far as I can tell, saying "F is conservative iff Curl(F) = 0" contradicts the claims of the site I posted.

What conditions must be met for a vector field to be conservative?
"F is conservative iff Curl(F) = 0" is not true. What is true is that if Curl(F)= 0 on a simply connected set then F is conservative. In the example given on that site, F is not defined at (0, 0) so is not conservative on any set that includes (0,0). If you were to integrate around any curve that does NOT have (0,0) in its interior, you would get an integral of 0- conservative field. But if the path does include (0,0) in its interior, its interior is not "simply connected" so you cannot expect to get a 0 integral.

"Simply connected" means that any closed curve can be "shrunk" to a single point while staying inside the region. Since (0,0) is not in the region (F is not defined at (0,0) so Curl(F)= 0 is not true there), any closed curve that includes (0,0) in its interior cannot be "shrunk" to a single point without passing through (0,0) which is not in the region.
 
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Related to Conservative vector field conditions

What is a conservative vector field?

A conservative vector field is a type of vector field in which the line integral between any two points is independent of the path taken between them. This means that the work done by the vector field is only dependent on the starting and ending points, and not the actual path taken.

What are the conditions for a vector field to be conservative?

The conditions for a vector field to be conservative are that it must be a continuous and differentiable function, and its partial derivatives must be equal. This is known as the conservative vector field condition, or the curl-free condition.

What is the significance of a conservative vector field?

A conservative vector field has many practical applications, including in physics, engineering, and economics. It can be used to model and analyze the behavior of various systems, such as fluid flow, electrical fields, and cost-minimizing processes.

How can one test if a vector field is conservative?

To test if a vector field is conservative, one can use the gradient test. This involves taking the partial derivatives of the vector field and checking if they are equal. If they are, then the vector field is conservative. Additionally, one can also perform line integrals along different paths to see if they yield the same result.

What is the difference between conservative and non-conservative vector fields?

The main difference between conservative and non-conservative vector fields is that the line integral of a conservative vector field is independent of the path taken, while the line integral of a non-conservative vector field is dependent on the path. Additionally, conservative vector fields satisfy the curl-free condition, while non-conservative vector fields do not.

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