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I Do time-dependent and conservative vector fields exist?

  1. Oct 27, 2016 #1
    Hello Forum,

    A conservative vector field G(x,y,z) is one that can be expressed as the gradient of a scalar field P(x,y,z).

    Could a time-varying vector field like D(x,y,z,t) be a conservative vector field? If not, why not? Can it be conservative (or not) at different time instants?

    Thanks!
     
  2. jcsd
  3. Oct 27, 2016 #2

    dextercioby

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    Science Advisor
    Homework Helper

    The spatial gradient doesn't act on the time variable. Can you reread the definition of "conservative vector field" from your book?.
     
  4. Oct 27, 2016 #3
    Thanks.

    I know that the gradient only operates on the spatial variables. But I also think there are some particular conditions that forbid a time-changing vector field to be conservative....Don't know what they are though...
     
  5. Oct 27, 2016 #4

    fresh_42

    Staff: Mentor

    What does it mean for a vector field to be conservative? This directly poses strict conditions on a time dependent one.
     
  6. Oct 27, 2016 #5
    Based on what I know, a conservative field is one whose line integral does not depend on the path but only on the starting and ending points. Hence, a closed path integral will always be zero regardless of the shape and length. I also know that the curl of a conservative field is zero.
     
  7. Oct 27, 2016 #6

    fresh_42

    Staff: Mentor

    Yes, and therefore any time dependent path, over a changing field has to add up to zero energy. Especially if you choose the same path and move along at any different, possibly varying speeds. This can only be done if constant in time. The only way out is to let time be part of space, then the question is meaningless.
     
  8. Oct 27, 2016 #7
    Thanks fresh_42.
    hope you have patience. I would like to understand your explanation.
    Just to make sure. The paths, in space, are fixed. What changes over time is the vector field. The vector field can be vary in space (i.e. be non homogeneous) but change in time at every spatial point....

    I am still not getting it :)
     
  9. Oct 27, 2016 #8

    fresh_42

    Staff: Mentor

    If you consider a special path, it is a parameterized function on the manifold, i.e. ##f: [0,1] \rightarrow M##. But it is not fixed, that you have to move by ##t \mapsto t##. You could as well move by ##t \mapsto t^2##. Now to always get the same path integral, ##0##, it cannot depend on where you are when. So there cannot be a change in time.
     
  10. Oct 27, 2016 #9
    Thanks. But I don't know about manifolds yet. Sorry. I can handle vector calculus for now.
     
  11. Oct 27, 2016 #10

    fresh_42

    Staff: Mentor

    That doesn't matter here. The vector field has to be somewhere. This somewhere is a manifold, e.g. a circle.
    If it changes in time, the paths depend on when and where. A varying speed along a path will therefore effect the integral, which is forbidden.
    Conservative means: nothing gained and nothing lost on a closed path. If you could gain something at a certain time, and gain nothing the next time you cross the point, then the field cannot be path independent.
     
  12. Nov 18, 2016 #11
    I think to understand exactly what you have meant - have been in situation of late to prove that the natural orbital systems, starting from the Solar (Sun-Earth) - to atomic and galactical ones - are essentially non-conservative, while having been erroneously and totally incorrectly/misleading modeled as conservative (both regarding the alleged time-independent energy and angular momentum, as the zero-valued work over a closed path encompassing center - for introduction, please see
    andor send me an e-mail at nedic.slbdn@gmail.com). I am quite positive that the line integral over unit-circle expressed in time-parameterized form would reveal non-zero line integral in the field of "central force" proportional to 1/r^2, whereby r is expressed as time-parameterized Keplerian ellipse, or in other words that the macro-circulation macro-curl/rotor) of such an time-dependent field would turned out to be non-zero ... Keen to cooperate on this?!
     
  13. Nov 18, 2016 #12
    In relation to the just preceding post, I decided to upload the the file with the related article ...
     
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