1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Is irrotational flow field a conservative vector field?

  1. May 22, 2016 #1
    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. jcsd
  3. May 22, 2016 #2
    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
  4. May 22, 2016 #3
    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?
     
  5. May 22, 2016 #4

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    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. May 22, 2016 #5
    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?
     
  7. May 22, 2016 #6
    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##
     
  8. May 23, 2016 #7

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  9. May 23, 2016 #8
    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?
     
  10. May 23, 2016 #9

    boneh3ad

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  11. May 24, 2016 #10

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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.
     
  12. May 24, 2016 #11
    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?
     
  13. May 25, 2016 #12

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted