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I How does divergence calculate all of flow through a surface?

  1. May 24, 2017 #1
    I'd like to use 2-d for simplification.

    Divergence is the rate of change of a component of a field F as you travel along that component's direction.

    So Fx represents the part of F flowing the X direction and same with Fy along the Y direction, and so divergence is calculated by dFx/dx + dFy/dy.

    But isn't it possible that flow of Fx, or flow of F in the X direction, changes not as a function of X, but as a function of Y?

    So as you travel along the X axis at a constant Y, the change of Fx is 0. And likewise with Fy: it's possible that the flow of Fy changes as a function of the X direction and so the change of Fy along the Y axis is 0.

    Given this, we get that total divergence is dFx/dx + dFy/dy = 0 + 0 = 0, meaning there is no flow out of the surface.

    But it seems to me that there must be some flow out of the surface, it's just that the rate changes as you move about a different axis.

    So shouldn't divergence actually be adding up how Fx changes as you move along both the X and Y axis, and how Fy changes as you move along X and Y axis?
     
  2. jcsd
  3. May 24, 2017 #2

    BvU

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    Hi yo,
    Is it ? Wasn't it defined differently -- as you use later on: as a sum of 'rates of change' ?
    Sure, why not ? Simple example: ##\vec F = (y,x)##.
    Yep.
    Yep
    What surface was that ?
     
  4. May 24, 2017 #3

    Charles Link

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    To the OP: From what I can tell, you are mixing two quantities, the current density ## \vec{J} ## and the divergence of ## \vec{J} ##, which is ## \nabla \cdot \vec{J} ##. ## \\ ## If ## \nabla \cdot \vec{J} =0 ##, it doesn't mean that there isn't any flow. Instead it means there is no buildup of whatever is flowing. By the continuity equation ## \nabla \cdot \vec{J}+\frac{\partial{ \rho}}{\partial{t}}=0 ##, so that if ## \nabla \cdot \vec{J}=0 ## that means that the density ## \rho ## is constant (doesn't change with time) at a given position, so that ## \frac{\partial{\rho}}{\partial{t}}=0 ##. There can still be a considerable flow as given by the function ## \vec{J} ##. ## \\ ## Editing: You can apply Gauss' law over a volume to get ## \iiint\limits_{V} \nabla \cdot \vec{J} d \tau=\iint\limits_{V} \vec{J} \cdot dS=- \iiint\limits_{V} \frac{\partial{\rho}}{\partial{t}} \, d \tau=-\frac{\partial{Q}}{\partial{t}} ## where ## Q ## is the total amount of material in the volume ## V ## that you integrate over and compute the flux. ## \\ ## Alternatively, ## \iint\limits_{A} \vec{J} \cdot \, dS ## over an area ## A ## gives how much material flows through that area. In this second case, a computation of ## \nabla \cdot \vec{J} ## is unnecessary.
     
    Last edited: May 24, 2017
  5. May 24, 2017 #4
    Yeah, sorry, I meant to say divergence is the sum of rates of change. And let's say the surface is a box. Also as Charles said, I should be thinking that divergence is actually gain/loss.

    But if given what I said is true... how can the gain/loss be zero in the box? I understand that if we take the divergence formula as a measure of buildup, we will record 0. But it doesn't seem to me that it accurately represents total buildup in the box. I feel that the actual gain/loss is actually nonzero since there is Fx exiting, but it's only visible if you examine change in Y direction.

    So I claim 'real divergence' is dFx/dx + dFx/dy + dFy/dx + dFy/dy
     
    Last edited: May 24, 2017
  6. May 24, 2017 #5

    scottdave

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    It's been awhile since I've done vector Calculus. This is an interesting question. Something to consider would be something like fluid flow in a pipe. At the edge of the pipe, essentially there is no flow. Then as you go away from the edge, the flow increases. But all flow is parallel to the pipe wall, so if you move in the same direction as the pipe there is no change in flow rate, and the flow rate in the direction perpendicular to the pipe wall is zero.

    How could this be modeled to show how divergence is calculated to represent total flow rate through a section of pipe?
    I just read that some people have given some answers. One thing to note in my pipe example. If you look at a specific volume of pipe, you can agree that there won't be any accumulation in that section if the density is constant (same amount of fluid exits the section as is entering).

    So you can see that these mathematical tools can have real-world applications.
     
  7. May 25, 2017 #6

    BvU

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    For the example I gave a circle is more sensible (at least it's easier). For a box you need to do some work: a useful exercise !

    Physically, divergence zero means source-free and from that follows conservation. Let Charles' post #3 sink in a bit.
     
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