Is irrotational flow field a conservative vector field?

AI Thread Summary
An irrotational flow field is considered conservative if it can be expressed as the gradient of a potential function. While a flow with constant velocity can be derived from a potential function of the form φ = Ax + By, not all irrotational fields are conservative, as demonstrated by counterexamples like the potential vortex. The discussion highlights that a field can have zero curl but still lack a potential function in certain domains. The relationship between potential and velocity fields is clarified, emphasizing that potential does not always equate to potential energy. Overall, the nuances of irrotationality and conservativeness in fluid dynamics are explored, revealing complexities in their definitions and applications.
Adel Makram
Messages
632
Reaction score
15
For a flowing fluid with a constant velocity, will this field be described as conservative vector field? If it is a conservative field, what will be the potential of that field?
 
Physics news on Phys.org
I suspect that "irrotational flow field" means "##\mathrm{rot} =0"## but not necessarily
Adel Makram said:
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.

Adel Makram said:
what will be the potential of that field?
this you will see from definition of the potential function
 
Last edited:
wrobel said:
this you will see from definition of the potential function
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?
 
Adel Makram said:
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,
\nabla \times \nabla\phi \equiv 0
 
boneh3ad said:
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,
\nabla \times \nabla\phi \equiv 0
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?
 
boneh3ad said:
saying a flow field is irrotational is equivalent to saying it is conservative.
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##
 
wrobel said:
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.

Adel Makram said:
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.
 
boneh3ad said:
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.
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?
 
Potential is not, in general, a short name for potential energy. It often works out that way (an started out that way), but that is not a general rule, as scalar potentials have much broader application than just gravity and electrostatics. A scalar potential is a scalar-valued function that can be used to completely describe a conservative vector field.
 
  • #10
wrobel said:
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##
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.
 
  • #11
vanhees71 said:
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.
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?
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?
 
Back
Top