# B Is irrotational flow field a conservative vector field?

1. May 22, 2016

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?

2. May 22, 2016

### wrobel

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.

this you will see from definition of the potential function

Last edited: May 22, 2016
3. May 22, 2016

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?

4. May 22, 2016

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,
$$\nabla \times \nabla\phi \equiv 0$$

5. May 22, 2016

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?

6. May 22, 2016

### wrobel

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 $\mathrm{grad}\,f=v$

7. May 23, 2016

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.

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. May 23, 2016

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?

9. May 23, 2016

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. May 24, 2016

### vanhees71

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. May 24, 2016

I dont 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 isnt it conservative?
Also, wrobel said that the rot of that field =0 but in wikipedia is equal to $2\pi$.