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Physical interpretation of divergence

  1. Dec 8, 2014 #1
    I'm trying to figure out what the physical meaning of divergence is for a vector field.

    My textbook offered the following example: if v = <u, v, w> represents the velocity field of a fluid flow, then div(v) evaluated at P = (x, y, z) represents the net rate of the change of mass of the fluid flowing from the point P per unit volume.

    How is this possible? Velocity has units of [m/s]. Based on the definition of divergence, div(v) would have units of [m/s^2]. Where in the world do we get a unit of mass?

    I've been scouring the internet trying to find some clarification, but every source has been using either words like amount and volume interchangeably or talking about things like source/sink without clarifying what is meant.

    Any help would be much appreciated.
  2. jcsd
  3. Dec 8, 2014 #2
    kg/(m^3)s is the unit for mass flow.

    you cannot get kg from anywhere so probably the book actually mentions about the unit density flow. which has the unit of 1/s.

    when you take divergence, you multiply the entity with m^2 and divide to m^3, thus unit changes by 1/m on overall.

    apply this over m/s, you will get 1/s, which has the same unit with unit density flow.
  4. Dec 8, 2014 #3
    Divergence is defined as

    lim (1/V)(∫∫F⋅dS)=div(F)
    where ∫∫F⋅dS is a flux integral over some surface, and V is the volume contained within that surface.

    This can actually been proven with a little simple algebra if we assume S is a cube of infinitesimal size.

    Once this is proven, The divergence theorem ∫∫∫div(F)⋅dV=∫∫F⋅dS becomes beautifully and intuitively obvious.

    Lol, i just noticed this goes perfect with my name.
  5. Dec 9, 2014 #4
    Ozgen and DivergentSpectrum, thanks for the replies.

    What exactly is meant by "unit density flow?" I cannot find a reference to this anywhere else.
    The resource I was using is this: http://tutorial.math.lamar.edu/Classes/CalcIII/CurlDivergence.aspx
    The discussion on divergence is copied almost verbatim from the Stewart Calculus textbook. I don't see a mention of unit density flow anywhere.

    One problem I was having for the longest time was with the definition of flux. Turns out, there are two very different definitions of flux, and people seem to be using them almost interchangeably.

    When dealing with Transport Phenomena, flux refers to [flow rate of some quantity]/[area] so that the surface integral of flux dot dA gives you a net flow rate. (This is what you referred to as a "flux integral.")
    When dealing with electromagnetism, however, flux refers to [electric field]*[area] so that the surface integral IS the flux.

    This distinction really confused my until I figured out what was going on.
  6. Dec 9, 2014 #5


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    I think the assumption is that the density is constant, so [itex]\nabla \cdot (\rho \mathbf{v}) = \rho\nabla \cdot \mathbf{v}[/itex]. Thus [itex]\nabla \cdot \mathbf{v}[/itex] is proportional to the mass flux, although obviously for dimensional reasons is not actually equal to it.
  7. Dec 9, 2014 #6

    Stephen Tashi

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    Now you've confused me. If div(v) involves a change in velocity with respect to a measure of length, why would divergence have units of [m/s^2} ?
  8. Dec 9, 2014 #7
    Yes. The units of div(v) are 1/s. div(v) represents the net rate of fluid volume leaving a differential spatial volume divided by the differential spatial volume. As Pasmith noted in #5, if you're talking about mass, then the density should be in there too, and div(ρv) is the net rate of fluid mass flow leaving a differential spatial volume divided by the differential spatial volume.

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