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Is the current a kind of circulation?

  1. Jun 7, 2014 #1
    I know what is circulation by a mathematical point of view, but in the reality (in the physical world), what is circulation? I'm thinking that the electrical current through of a circuit, that the water current through of an pipe or of an river, that the water vortex formed over a drain or that the wind around of a twister are example of circulation. I'm right? Detail: the term "circulation" presuppose that the trajectory of the flux/current is a closed loop, but I don't think so, if the water current through of a pipe or of a river follows an open loop, so the circulation can be computed simply how the flux/current through of path and not necessarily of a circle (a closed loop), correct!?
  2. jcsd
  3. Jun 7, 2014 #2
    By circulation do you mean the rotational of the flow?
  4. Jun 7, 2014 #3


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    The water in a river is not an open loop system
    google water cycle

    current flows through a closed circuit ... eg out of the negative terminal of a battery through the closed switch through the wires and circuit components and back to the positive terminal

    if you open the switch, the current stops flowing and the circuit stops operating

  5. Jun 8, 2014 #4
    By circulation I understand the flux/current of some quantity through of a path (closed or open), for example:



    By line integral I understand:

    And by circulation I understand:

    The flux Φ is the surface integral of a vector field f over an oriented surface S, open or closed. In the same way, the circulation Γ is the line integral of a vector field f over an oriented curve s, open or closed. Just because the name is "circulation" I think that is not necessary that the curve s must be closed. The theorem of the rectangle triangle has the name of "pythagorean theorem", but the name "Pythagoras" has nothing to do with the computation...

    What do you want mean with it?

    Attached Files:

  6. Jun 8, 2014 #5
    A bit off topic, but where did you get that animation from? Its awesome!

  7. Jun 9, 2014 #6


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    To understand the various operations in vector calculus, it's good to have the intuitive example of fluid dynamics (i.e., the theory of the flow of gases or liquids) at hand. It's the most simple example for a classical field theory which makes sense in both the non-relativistic and the relativistic realm. Let's concentrate on the non-relativistic case for now.

    To characterize the fluid, we introduce some of the most important physical quantities related with it. We usually describe the fluid in terms of Euler coordinates. Here we have in mind an observer, who looks at the momentary state of the fluid at a given position [itex]\vec{x}[/itex] as a function of time. It is in important to keep in mind that in this picture at any instant of time a different portion of fluid molecules is located at this point.

    To describe the motion of the fluid, we introduce the velocity field, i.e., the momentary velocity of the fluid cell at any position, [itex]\vec{v}(t,\vec{x})[/itex]. Another important quantity is the density of the fluid at any position, [itex]\rho(t,\vec{x})[/itex]. This gives the mass per volume of a little fluid element at this position as a function of time.

    The first law we want to describe is the conservation of mass (which is a necessary conclusion from the space-time properties of Newtonian mechanics, i.e., a very fundamental law). It tells us that a given amount of fluid never changes its total mass. To see, how to formulate the corresponding law we put a fixed time-independent volume [itex]V[/itex] into the fluid. Then the mass-conservation law tells us that the mass of the matter contained in it can only change by the flow of fluid particles into or out of this volume through its boundary [itex]\partial V[/itex]. For this boundary we also have to specify the orientation relative to the volume enclosed by it. The standard choice is to direct each area-normal vector [itex]\mathrm{d}^2 \vec{A}[/itex] out of the volume [itex]V[/itex].

    Then through such a surface element in an infinitesimal time [itex]\mathrm{d} t[/itex] there flows the amount of mass that is contained in the volume element [itex]\mathrm{d} t \mathrm{d} \vec{F} \cdot \vec{v}(t,\vec{x})[/itex]. Fluid flowing out has to be counted negative, so we have
    [tex]\mathrm{d} m_V=\mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} \int_V \mathrm{d}^3 \vec{x} \rho(t,\vec{x})=-\mathrm{d} t \int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \rho(t,\vec{x}) \vec{v}(t,\vec{x}).[/tex]

    Using Gauss's integral theorem we can write
    [tex]\int_V \mathrm{d}^3 \vec{x} \partial_t \rho(t,\vec{x})=-\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \rho(t,\vec{x}) \vec{v}(t,\vec{x})=-\int_V \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot [\rho(t,\vec{x}) \vec{v}(t,\vec{x})].[/tex]

    To also understand the curl operation we investigate, how the relative vectors of two close fluid cells within the fluid change with time. In the infinitesimal time increment [itex]\mathrm{d} t[/itex] each point changes its position by
    [tex]\vec{\Delta}(t,\vec{x})=\vec{v}(t,\vec{x}) \mathrm{d} t.[/tex]
    The relative deformation for two infinitesimally close points [itex]\vec{x}[/itex] and [itex]\vec{x}+\mathrm{d} x[/itex] thus is
    [tex]\mathrm{d} \vec{x} \cdot \vec{\nabla} \vec{\Delta}=\vec{e}_k \mathrm{d} t \mathrm{d} x_j \partial_j v_k.[/tex]
    Now we can write
    [tex]\partial_j v_k=\frac{1}{2} (\partial_j v_k-\partial_k v_j)+\frac{1}{2} (\partial_j v_k+\partial_k v_j)=\epsilon_{jkl} \omega_{j} + \epsilon_{jk}.[/tex]
    Now let's investigate the change of the distance of the two adjacent points:
    [tex]\mathrm{d} l(t)=|\mathrm{d} \vec{x}|, \quad \mathrm{d} l(t+\mathrm{d} t)=\sqrt{(\mathrm{d} \vec{x}+\Delta)^2}=\mathrm{d} l(t)+\mathrm{d} \vec{x} \cdot \Delta + \mathcal{O}(\mathrm{d}t^2)=\mathrm{d} l(t) + \mathrm{d} x_k \mathrm{d} x_j \partial_j v_k \mathrm{d} t=\mathrm{d} l(t)+\epsilon_{jk} \mathrm{d}x_j \mathrm{d} x_k \mathrm{d} t.[/tex]
    The part of the motion due to the antisymmetric part thus corresponds to an infinitesimal rotation, because it doesn't give rise to a length change.
    [tex]\vec{\Delta}_{\text{rot}}=\mathrm{d} t \vec{\omega} \times \mathrm{d} \vec{x},[/tex]
    of the fluid cell as a whole and thus is not part of an intrinsic deformation.

    In addition usually the fluid cell can be moving as a whole in a parallel motion. This "rigid translation" is cancelled completely in considering the relative vector

    The intrinsic deformation of the fluid element is totally due to the symmetric part [itex]\epsilon_{jk}[/itex] of the derivatives of the velocity field. This symmetric strain tensor can be decomposed further in a homogeneous part and a traceless part,
    [tex]\epsilon_{jk}=\frac{1}{3} \epsilon_{ll} \delta_{jk} + \eta_{jk}.[/tex]
    To analyze the physical meaning of this decomposition, we evaluate the change of the volume of the infinitesimal fluid element.

    To that end we remember that we can always choose a Cartesian coordinate system such that the symmetric deformation tensor [itex]\epsilon_{jk}[/itex] becomes diagonal. An infinitesimal box of fluid parallel to the corresponding principal coordinate axes changes its volume according to
    [tex]\mathrm{d} V(t+\mathrm{d} t)=(\mathrm{d} x_1+\Delta_1)(\mathrm{d} x_2+\Delta_2) (\mathrm{d} x_3+\Delta_3)=\mathrm{d} V(t) [1+(\partial_1 v_1 + \partial_2 v_2 + \partial_3 v_3) \mathrm{d} t]+\mathcal{O}(\mathrm{d} t^2)=\mathrm{d} V(t) [1+\mathrm{d} t \vec{\nabla} \cdot \vec{v}]+\mathcal{O}(\mathrm{d} t^2)=\mathrm{d} V(t) [1+\mathrm{d} t \mathrm{tr} \epsilon]+\mathcal{O}(\mathrm{d} t^2).[/tex]
    Now the trace is an invariant under coordinate transformations and thus can be evaluated in any coordinate system. Thus the trace of the strain tensor describes a change of the volume.

    The remaining deformation of the fluid element doesn't change the volume and is thus to be interpreted as a shear.

    This completely visualizes the various covariant first-order derivatives of the velocity field.
  8. Jun 9, 2014 #7
    ^ you're traveling...



    Anyway... someone can answer me if the examples that I give in the 1st post are examples of circulation, please!?!?
  9. Jun 12, 2014 #8


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    I'm not so sure whether I understand correctly what you mean. That's why I tried to answer your question in a general form.

    Mathematically "circulation" has different meaning in different context. Usually it's a functional of vector fields depending on closed lines. It is defined as
    [tex]\Gamma[\vec{V},C]=\int_{C} \mathrm{d} \vec{x} \cdot \vec{V}.[/tex]
    Often you are interested in the local form. Then you investigate the curl of a vector field.

    There is a somewhat subtle difference in the meaning of the global (integral) point of view and the local (differential, curl) point of view about circulation or vortex flows. A paradigmatic example is the famous potential vortex which is a flow field that is curl free everywhere, where it is well defined but has a non-vanishing circulation for a class of closed lines.

    The example goes as follows. It's most simply stated in cylinder coordinates [itex]\rho,\varphi,z[/itex]. We start from the innocent looking potential
    [tex]U(\vec{x})=-A \varphi.[/tex]
    The corresponding flow field is
    [tex]\vec{v}=-\nabla U=\frac{A}{\rho} \hat{\varphi}.[/tex]
    Superficially seen this is a curl free flow, because obviously
    [tex]\vec{\nabla} \times \vec{v}=-\vec{\nabla} \times (\vec{\nabla} U)=0.[/tex]
    On the other hand, it's clearly a kind of "vortex motion", i.e., it's "curling" around the [itex]z[/itex] axis, and the circulation along a curve encircling the [itex]z[/itex] axis once, gives the same non-zero circulation, namely [itex]\Gamma=2 \pi A[/itex], as one immediately evaluates by taking an arbitrary circle in any plane perpendicular to the [itex]z[/itex] axis with its center on the [itex]z[/itex] axis. That all other closed curves around the [itex]z[/itex] axis yield the same result directly follows by using Stokes's Law together with [itex]\vec{\nabla} \times \vec{v}=0[/itex].

    The point of this example is that the field is singular along the entire [itex]z[/itex] axis, and that's not due to the coordinate singularity of cylinder coordinates, because rewritten in Cartesian coordinates, which are free of coordinates singularities it reads
    [tex]\vec{v}=\frac{A}{x^2+y^2} \begin{pmatrix}
    -y \\ x
    It thus clearly has a true singularity along the [itex]z[/itex] axis.

    This makes the domain [itex]\mathbb{R}^3 \setminus \{z-\text{axis} \}[/itex] multiply connected, because you cannot shrink a closed curve that encircles the [itex]z[/itex] axis to a single point by only doing continuous deformations and staying in the domain.

    This is why it is possible to have a field that is curl free everywhere in its domain of definition but still can have a finite circulation for some closed paths. Poincare's Law, stating the independence of line integrals from the vanishing of the curl of a vector is thus only generally valid in simply connected parts of the domain (i.e., locally)!
  10. Jun 12, 2014 #9
    I just was waiting for one "yes" or "not"...
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