Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Continuity/Gauss' divergence

  1. Mar 14, 2009 #1
    I have followed a derivation of the continuity equation, and it uses the divergence theorem at some point, but looking at the actual meaning of the equation, it almost seems like it is saying the same thing as the theorem. That is, local conservation. So my question is...

    Can the continuity equation be considered a formulation of Gauss' divergence theorem?
  2. jcsd
  3. Mar 14, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    There are 5 distinct steps in the derivation of the continuity equation:

    1. Conservation of mass for EVERY particle-based (i.e), material system.
    A material system consists of the SAME particles throughout the observation period.

    Thus, for an arbitrary material system with density p, and occupying spatial region [tex]V_{M}(t)[/tex], with volume V(t), we have:
    [tex]0=\frac{d}{dt}\int_{V_{M}}\rho{dV} (1)[/tex], where we sum up the masses for the whole spatial region.
    The mass conservation principle for material systems is AXIOMATIC in classical physics (it is, as it happens, untrue; a material system under the influence of relativistic effects, for example, may well experience changes in its mass!)
    2. Transition from a material system perspective, to a coincident geometric perspective:
    A material system, which labels its constituent particles and tracks them meticulously over time is a rather cumbersome way of dealing with the time evolution of a continuum.
    Generally, we prefer to use a GEOMETRIC perspective, in that we keep under control and designation a particular geometrical region, IRRESPECTIVE of whether individual particles choose to remain within that region, or choose to to leave (or enter) it.

    (We call the material approach "Lagrangian", the geometric approach "Eulerian").

    Thus, we are no longer concerned with individual particle quantities per se, like particle density, particle velocity, and so on; rather, we are interested in the evolution of FIELD quantities, i.e, the evolution of the density field and the velocity field.

    The field density function, [tex]\rho(x,y,z,t)[/tex], is then related to to the material density distribution as being equal at a particular time t to the density to whichever material particle that happens to be at (x,y,z) at the time t. Time evolution of the density function, then, must be understood as the difference in density of the DIFFERENT particles occupying (x,y,z) at different times. ((x,y,z) itself, of course, must be understood as the centre of a tiny region over which we have averaged quantities).

    At any time t, the region occupied by some material system is clearly coincident with some geometrically defined region [tex]V_{G}(t)[/tex], whose surface velocity is defined as [tex]\vec{v}_{s}(x,y,z,t)[/tex]. A fixed geometric region has surface velocity 0 everywhere, and we'll use that in the following.

    Thus, (1) may be rewritten as:
    [tex]0=\frac{d}{dt}\int_{V_{G}}\rho{dV}(2)[/tex] where with a slight abuse of notation, I have retained dV as designating the (geometric) volume element, and the same letter for material and geometric density functions.

    3. DIfferentiation of the integral.
    This is given by Reynold's transport theorem, i.e we generally have:
    [tex]\frac{d}{dt}\int_{V_{G}}\rho{dV}=\int_{V_{G}}\frac{\partial\rho}{\partial{t}}dV+\int_{S_{G}}\rho\vec{v}\cdot\vec{n}dA(3)[/tex], where S_G is the surface of the region, and dA the area element.

    Thus, inserting (3) in (2), we get:
    [tex]0= \int_{V_{G}}\frac{\partial\rho}{\partial{t}}dV+\int_{S_{G}}\rho\vec{v}\cdot\vec{n}dA(4)[/tex]

    Note that this simply says that the increase of mass within the specified region (first integral) must balance the outward mass flux (second integral).
    That this should be equivalent to the mass conservation principle should be obvious.

    4. Application of Gauss' divergence theorem.
    This is now utilized on the second integral on the left hand side of (4):
    Adding together the volume integrals we now have, we get by manipulations:
    where [tex]\frac{D\rho}{dt}[/tex] is called the material derivative of [tex]\rho[/tex]

    5.Use of arbitrariness of chosen region.
    Now, since (6) is to hold for any, and every region, the integrand has to equal 0 everywhere.

    Therefore, since [tex]\rho[/tex] is not, in general zero everywhere, the continuity equation follows:

    So, as answer to your question:
    No, the continuity equation is not merely Gauss' theorem in disguise, but it certainly depends on it for its own validity!
  4. Mar 14, 2009 #3
    Firstly, wow! You really went above and beyond give me a thoughtful and comprehensive response, so thank you very much. Secondly, I meant to post this in the quantum forum, as that is what I am studying at the moment. Of course having understood the majority of your post, I also see that this will hold for the quantum case where it is continuity of probability current that is being being described by the equations.

    So, if we were to make a measurement that violated the continuity equation, could we also say that gauss' divergence theorem has failed? Or could the discrepency arise from somewhere else? I guess this is one of those 'If unicorns existed, what colour would they be?' kind of question, so I'm not sure if it is really a valid question, but I thought I'd put it out there.
  5. Mar 14, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    No problem. :smile:
    Absolutely not. Empiricism has no place in judging the validity of mathematical propositions!

    Rather, it would testify either of two things:
    a) The measurement was false
    b) The measurement is correct, but the conditions required for the appliciability of Gauss' theorem are not present. (I.e, you have a case of misapplication)

    Which of these would be the right one in your particular case, I don't know.
  6. Mar 14, 2009 #5
    yep. Makes sense.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook