- #1

- 614

- 13

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Adel Makram
- Start date

- #1

- 614

- 13

- #2

wrobel

Science Advisor

- 886

- 618

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?

Last edited:

- #3

- 614

- 13

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

- #4

- 3,299

- 977

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]

- #5

- 614

- 13

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]

- #6

wrobel

Science Advisor

- 886

- 618

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##

- #7

- 3,299

- 977

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.

- #8

- 614

- 13

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.

- #9

- 3,299

- 977

- #10

- 19,082

- 9,887

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.

- #11

- 614

- 13

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?

- #12

- 19,082

- 9,887

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

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.

Share:

- Replies
- 18

- Views
- 2K