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- Thread starter Adel Makram
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wrobel

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I suspect that "irrotational flow field" means "##\mathrm{rot} =0"## but not necessarily

Assume that a vector field v is defined in simply connected domain and its rotor equals zero. Then v has a potential function.

with a constant velocity

Assume that a vector field v is defined in simply connected domain and its rotor equals zero. Then v has a potential function.

this you will see from definition of the potential functionwhat will be the potential of that field?

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What sort of potential function that its gradient yields a constant velocity field? If I integrate a constant velocity ##v(x)=c## with respect to ##x##, this gives ##cx##. So what physical potential has this form?this you will see from definition of the potential function

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What sort of potential function that its gradient yields a constant velocity field? If I integrate a constant velocity ##v(x)=c## with respect to ##x##, this gives ##cx##. So what physical potential has this form?

Didn't you just answer your own question? And potential of the form ##\phi = Ax + By## will give you a constant velocity field.

Regarding your irrotationality question: saying a flow field is irrotational is equivalent to saying it is conservative. Consider that a conservative field can always be described as a gradient of a potential, and rotational is measured by the curl. Therefore,

[tex]\nabla \times \nabla\phi \equiv 0[/tex]

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So if fluid is pumped through a pipe and flows at a constant velocity, what is the name of physical potential which is measured at any point along the length of the pipe? In general, does the potential gradient of a conservative field have to be a force field or it may be just a constant velocity field?Didn't you just answer your own question? And potential of the form ##\phi = Ax + By## will give you a constant velocity field.

Regarding your irrotationality question: saying a flow field is irrotational is equivalent to saying it is conservative. Consider that a conservative field can always be described as a gradient of a potential, and rotational is measured by the curl. Therefore,

[tex]\nabla \times \nabla\phi \equiv 0[/tex]

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wrobel

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this is wrong. The standard counterexample is as follows. Consider a domainsaying a flow field is irrotational is equivalent to saying it is conservative.

$$D=\{(x,y,z)\in\mathbb{R}^3\mid 1<x^2+y^2<2\}$$ and the following field ##v## in it

$$v=\Big(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2},0\Big).$$ It is easy to see that ##\mathrm{rot}\,v=0##. However there is no function ##f## in ##D## such that ##\mathrm{grad}\,f=v##

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this is wrong. The standard counterexample is as follows. Consider a domain

$$D=\{(x,y,z)\in\mathbb{R}^3\mid 1<x^2+y^2<2\}$$ and the following field ##v## in it

$$v=\Big(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2},0\Big).$$ It is easy to see that ##\mathrm{rot}\,v=0##. However there is no function ##f## in ##D## such that ##\mathrm{grad}\,f=v##

That's fair. I took it a step too far with the "iff" relationship. I'll amend my original post to say that any flow field that can be described by a potential function is also, by definition, irrotational. Irrotationality is necessary but not sufficient for a field to be expressible as a potential.

So if fluid is pumped through a pipe and flows at a constant velocity, what is the name of physical potential which is measured at any point along the length of the pipe? In general, does the potential gradient of a conservative field have to be a force field or it may be just a constant velocity field?

I am not sure that I fully understand what you are asking here. What do you mean by the "name of the physical potential"? Further, you can't really measure potential in a fluid flow. A potential gradient is, by definition (at least in this case) a velocity field (##\vec{v} = \nabla \phi##). The velocity in field does not have to be constant.

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The convention in physics is that the potential is just a short name of potential energy. But the unite of the potential in this example is velocity times distance which is not a unite of energy? So my question, what does this potential represent?I am not sure that I fully understand what you are asking here. What do you mean by the "name of the physical potential"? Further, you can't really measure potential in a fluid flow. A potential gradient is, by definition (at least in this case) a velocity field (##\vec{v} = \nabla \phi##). The velocity in field does not have to be constant.

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This is the famous example of the potential vortex. It has a potential in any domain with some half plane along the z-axis taken out. Such a potential is given in cylinder coordinates bythis is wrong. The standard counterexample is as follows. Consider a domain

$$D=\{(x,y,z)\in\mathbb{R}^3\mid 1<x^2+y^2<2\}$$ and the following field ##v## in it

$$v=\Big(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2},0\Big).$$ It is easy to see that ##\mathrm{rot}\,v=0##. However there is no function ##f## in ##D## such that ##\mathrm{grad}\,f=v##

$$V=-\varphi$$

since then

$$-\vec{\nabla}V=1/r \vec{e}_{\varphi}=(-y,x,0)/(x^2+y^2).$$

Depending on which open interval of length ##2\pi## you have taken out a corresponding half-plane, which restricts the domain to a single connected part.

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I don`t understand this famous example because I am confusing about definitions. First what is the meaning of "some half plane along z taken out"? If the vector field can be represented as a gradient of a potential as required by the definition of conservative field, why isn`t it conservative?This is the famous example of the potential vortex. It has a potential in any domain with some half plane along the z-axis taken out. Such a potential is given in cylinder coordinates by

$$V=-\varphi$$

since then

$$-\vec{\nabla}V=1/r \vec{e}_{\varphi}=(-y,x,0)/(x^2+y^2).$$

Depending on which open interval of length ##2\pi## you have taken out a corresponding half-plane, which restricts the domain to a single connected part.

Also, wrobel said that the rot of that field =0 but in wikipedia is equal to ##2\pi##.

Finally, by Stokes theorem the microcirculation in the form of curl should equal to macrocirculation in the form of rot, but here it is not the case?

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$$\vec{\nabla} \times \vec{A}=2 \pi \delta(x) \delta(y) \vec{e}_z.$$

For a detailed discussion see

https://www.physicsforums.com/threads/struggling-with-ab-effect.872156/#post-5477281

Note that there I discuss the negative of the field discussed here.

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