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Adel Makram
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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?
Adel Makram said:with a constant velocity
this you will see from definition of the potential functionAdel Makram said:what will be the potential of that field?
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?wrobel said:this you will see from definition of the potential function
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?
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: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]
this is wrong. The standard counterexample is as follows. Consider a domainboneh3ad said:saying a flow field is irrotational is equivalent to saying it is conservative.
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##
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?
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?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.
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 bywrobel 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##
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?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.
An irrotational flow field is a type of vector field in which the curl, or rotational component, of the vector field is equal to zero at every point. This means that the flow of the field is smooth and does not have any swirling or rotating motion.
A conservative vector field is a type of vector field in which the line integral of the vector field along any closed path is equal to zero. This means that the work done by the field on any closed path is zero, and the path taken does not affect the final result.
An irrotational flow field is always a conservative vector field. This is because the curl of an irrotational flow field is equal to zero, and the line integral of a conservative vector field is also equal to zero. Therefore, all irrotational flow fields are conservative vector fields, but not all conservative vector fields are irrotational flow fields.
Some examples of irrotational flow fields include the velocity field of a non-spinning object, the flow of an ideal fluid, and the flow of heat in a stationary medium.
The concept of irrotational flow fields is important in many areas of science and engineering, such as fluid mechanics, electromagnetism, and thermodynamics. It helps us understand and analyze the behavior of vector fields in these fields, and allows us to make accurate predictions and calculations.