Is every conservative vector field incompressible?

  • #1
So I have found that everyone conservative vector field is irrotational in a previous problem. Based on the relationship irrotational vector fields and incompressible vector fields have, div(curl*F)=0, does that also imply every conservative vector field is incompressible?

Kindly,

Shawn
 

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  • #3
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 im 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?
 
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  • #4
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I wasn't being sarcastic I was hoping for a definition of terms.

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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  • #5
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Okay, I'm still trying to parse your question completely. I have to add that the ##F## above may be written ##F = \nabla (x^2+y^2+z^2)/2## so it is conservative, irrotational and compressible. Hence the answer to your question is ##F## is not necessarily incompressible.
 
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  • #6
I wasn't being sarcastic I was hoping for a definition of terms.

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
 
  • #7
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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.
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.
 
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  • #8
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
 
  • #9
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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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  • #10
lavinia
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If a gradient ##∇f## is incompressible then ##f## is said to be a harmonic function..
 

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