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Kindly,

Shawn

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In summary, the conversation discusses the relationship between conservative vector fields, incompressible vector fields, and irrotational vector fields. It also delves into the definitions of these terms and how they relate to one another. The question being asked is whether every conservative vector field is also incompressible, given the fact that they are irrotational and have a zero divergence. The expert summarizer explains that while this statement may hold true in some fields of study, it is not necessarily true in general. They provide a specific example to demonstrate this and clarify any confusion on the topic.

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Kindly,

Shawn

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How does one compress a vector field?

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I'm not sure if that comment was sarcasm or a way to get me to think in a more logical sense but i'll explain what I can in hopes it clears up any confusion on what I am asking. I'm coming here for help and motivation. I might be completely confused and not understand properly but I feel I do and this what I understand:

Well I know I can show F is irrotational in any vector field when I have F(x,y,z)=<f(x), g(y), h(z)> while vectors of the form F(x,y,z)=<f(y,z)b g(x,z), h(x,y)> are incompressible by calculating curlF and observing that F does not depend on y or z, g does not depend on x or z, and h does not depend on x or y. As a result the curl is zero; curlF=0 (irrotational). And if f is a conservative vector field then, Py=Qz, Rx=Pz, & Qx=Py.

Also it would be incompressible when divF=0. So, divF= ∂/∂x(f(y,z))+ ∂/∂y(g(x,z)) + ∂/∂z(h(x,y))=0; since the derivatives are being taken with respect to varibles on which the respectivr functions do not depend then we can see for sure divF=0 where F is incompressible.

So given their relationship, div(curlF)=0, and knowning curlF=0 means F is a vector field and also means F is irrotational, does that also mean that every conservative vector field is incompressible?

Well I know I can show F is irrotational in any vector field when I have F(x,y,z)=<f(x), g(y), h(z)> while vectors of the form F(x,y,z)=<f(y,z)b g(x,z), h(x,y)> are incompressible by calculating curlF and observing that F does not depend on y or z, g does not depend on x or z, and h does not depend on x or y. As a result the curl is zero; curlF=0 (irrotational). And if f is a conservative vector field then, Py=Qz, Rx=Pz, & Qx=Py.

Also it would be incompressible when divF=0. So, divF= ∂/∂x(f(y,z))+ ∂/∂y(g(x,z)) + ∂/∂z(h(x,y))=0; since the derivatives are being taken with respect to varibles on which the respectivr functions do not depend then we can see for sure divF=0 where F is incompressible.

So given their relationship, div(curlF)=0, and knowning curlF=0 means F is a vector field and also means F is irrotational, does that also mean that every conservative vector field is incompressible?

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1) Conservative field is defined; The integral ##\int_A^B F \cdot ds## is independent of path?

2) Incompressible field is defined; ##\nabla \cdot F = 0## ?

3) Irrotational field is defined; ##\nabla \times F = 0##?

I'm familiar with the third, the first is borrowed from potential theory and the last I heard somewhere. The expression ##\nabla\cdot(\nabla \times F) = 0## is an identity true for any 3-vector field twice differentiable. This expression is still true if ##\nabla\times F =0## to start with. Is ##F## incompressible given this fact. Answer, No Proof Take ##F = (x,y,z)## clearly ##\nabla\times F = 0## so ##F## is irrotational. However, ##\nabla\cdot F = 3 \ne 0## so it is compressible.

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Paul Colby said:

1) Conservative field is defined; The integral ##\int_A^B F \cdot ds## is independent of path?

2) Incompressible field is defined; ##\nabla \cdot F = 0## ?

3) Irrotational field is defined; ##\nabla \times F = 0##?

I'm familiar with the third, the first is borrowed from potential theory and the last I heard somewhere. The expression ##\nabla\cdot(\nabla \times F) = 0## is an identity true for any 3-vector field twice differentiable. This expression is still true if ##\nabla\times F =0## to start with. Is ##F## incompressible given this fact. Answer, No Proof Take ##F = (x,y,z)## clearly ##\nabla\times F = 0## so ##F## is irrotational. However, ##\nabla\cdot F = 3 \ne 0## so it is compressible.

1) A conservative vector field is defined; ∫cF⋅dr=0 for every closed path of C in D. Not sure if what you said means the same.

2) Yes an Incompressible F is defined ∇⋅F=0 which is also divF=0.

3) Correct again, an irrotational F is defined ∇xF=0 which is also curlF=0.

They have a relationship div(curlF)=0 or also ∇⋅(∇f)=0 or ∇^2f=0 (which is essentially Helmholtz's Theorem or also a Laplace operator).

And knowing every conservative vector field is irrotational i wasn't sure if, because of that relationship.

But I just came across some information that the condition of zero divergence is satisfied whenever a vector field F has only a vector potential component, because the definition of the vector potential.

And also you're proofs seem to help clarify to me that there is in fact a possibility that not all conservative vector fields are incompressible.

I do want to thank you greatly for taking the time to assist me with this. the wording has been causing me confusion and not really knowing how to prove it was the bigger challenge.

Kindly,

Shawn

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Your question is in Mathematics > Calculus. Different fields of study make different assumptions and have different givens. I think that these are getting all run together in these posts in a way that even I find confusing. For example, the terms "incompressible" and "irrotational" are borrowed from fluid mechanics while the term "conservative" is from EM / mechanics. You're asking general vector calculus questions and it's easy for truths from these other fields to bleed into the discussion. I have provided a specific example where ##F = \nabla (x^2+y^2+z^2)/2## has only a "potential component" yet is not incompressible. In potential theory or in electrostatics people often are discussing solutions of ##\nabla^2 \phi = \rho## which for ##F=\nabla \phi## the quoted statement is true. However, in general not every potential component is incompressible in vector calculus.Shawn Huetter said:But I just came across some information that the condition of zero divergence is satisfied whenever a vector field F has only a vector potential component, because the definition of the vector potential.

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Paul Colby said:Your question is in Mathematics > Calculus. Different fields of study make different assumptions and have different givens. I think that these are getting all run together in these posts in a way that even I find confusing. For example, the terms "incompressible" and "irrotational" are borrowed from fluid mechanics while the term "conservative" is from EM / mechanics. You're asking general vector calculus questions and it's easy for truths from these other fields to bleed into the discussion. I have provided a specific example where ##F = \nabla (x^2+y^2+z^2)/2## has only a "potential component" yet is not incompressible. In potential theory or in electrostatics people often are discussing solutions of ##\nabla^2 \phi = \rho## which for ##F=\nabla \phi## the quoted statement is true. However, in general not every potential component is incompressible in vector calculus.

Yes I agree, and my Multi-variable Calculus book states that it is pulled from fluid mechanics. And again, I greatly appreciate your help. This has cleared up a lot of confusion. I got lost in the confusion of all the other sources online that were talking more specifically about mechanics rather than vector calculus. Do you happen to know when this will studied more in depth after multi-variable calculus? Possibly Differential Equations? I'm certain it will be covered in Dynamic Vector Mechanics and Engineering Physics II when we go over Gauss's Law of Magnetism. I just want to be fully prepared for this sections.

Kindly,

Shawn

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Shawn Huetter said:Do you happen to know when this will studied more in depth after multi-variable calculus? Possibly Differential Equations? I'm certain it will be covered in Dynamic Vector Mechanics and Engineering Physics II when we go over Gauss's Law of Magnetism. I just want to be fully prepared for this sections.

The good thing about working through a text say on vector calculus is it should be self contained. If they draw on other fields it should be clear that this is what they are doing. Being new to a subject it might well be confusing to do too much web searching. There not too much quality control out there. I have found the Wikis on math an physics very helpful and usually well written. They often summarize a narrow subject well and allow you to get the lay of the land before you get too much in detail. It's been quite some time since I've been in school. In physics most of ones vector calculus was learned from Jackson's book on EM at least in the schools I attended. I still use Jackson as a reference. Best of luck in your studies.

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If a gradient ##∇f## is incompressible then ##f## is said to be a harmonic function..

A conservative vector field is a type of vector field in which the work done by the force is independent of the path taken. This means that the force is always pointing in the direction of decreasing potential energy.

A vector field is conservative if its curl is equal to zero. Mathematically, this can be represented as ∇ × F = 0. Additionally, if the vector field can be expressed as the gradient of a scalar function, then it is conservative.

An incompressible vector field is one in which the divergence is equal to zero. This means that the vector field has no sources or sinks, and the flow of the vector field is uniform throughout.

No, not every conservative vector field is incompressible. While all incompressible vector fields are conservative, there are some conservative vector fields that are compressible (have non-zero divergence).

Conservative vector fields are commonly used in physics and engineering to model and analyze fluid flow, such as in aerodynamics and hydrodynamics. They are also used in electromagnetism to describe electric and magnetic fields.

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