How to interpret physically the divergence of vector field?by hanson Tags: divergence, field, interpret, physically, vector 

#1
Feb807, 10:39 AM

P: 320

Hi all. I have difficulty in visualizing the concept of divergence of a vector field. While I have some clue in undertanding, in fluid mechanics, that the divergence of velocity represent the net flux of a point, but I find no clue why the divergence of an electric field measures the charge denity?
Can anyone tell me how to interpret the divergence of a vector field? Please kindly help. 



#2
Feb807, 10:46 AM

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P: 4,107

(It's somewhat akin to asking why the acceleration of an object is proportional to the net force on the object.) Arguably, one could regard Gauss' Law as but one of the laws of electromagnetism...and find a deeper formulation of the laws where Gauss' Law arises as a particular aspect. 



#3
Feb807, 12:17 PM

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#4
Feb807, 12:29 PM

P: 745

How to interpret physically the divergence of vector field?
If the divergence of a velocity field is zero, it means the fluid is incompressible.




#5
Feb807, 12:38 PM

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This (and the subsections) might be useful:
http://www.math.gatech.edu/~carlen/2...curlMean1.html 



#6
Feb807, 05:46 PM

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P: 1,465

Consider what I like to call the "bath tub" analogy.
In this 2D analogy, regions of positive divergence are "taps" (sources), regions of negative divergence are sinks, water represents electric flux. At any point where the divergence is zero (no sources or sinks), the water flowing in must equal the water flowing out correct? Hence if the divergence is zero, we can conclude that there is no net flux and thus there are no sources or sinks present at that point. Conversely, if the net flux is nonzero, that indicates that water is being added by a source (positive divergence) at that point, or being taken away (negative divergence) by a sink. Of course, electromagnetically, the sources are positive charges and the sinks are negative charges. Gauss' Law however states that the charge density determines the divergence at a given point. How do we incorporate charge density into the "bath tub" analogy? By considering that a source or sink is not a point source (or sink), but rather distributed over space. Keeping with the "bath tub" analogy, the divergence at a point is therefore dependant on the amount of water being added (or removed) at that point, which in turn is equal to the total amount of water added by the entire (distributed) source, divided by the surface area of which the source is distributed. This is essentially a crude expression of Gauss' Law in differential form. Using our electrostatic analogue, this is equivalent to saying the divergence is equal to the total charge of a given distibuted charge, divided by the volume of the charge (i.e. the charge density), which is what Gauss' law says in differential form. I hope that long winded spiel helped rather than hindered Claude. 



#7
Feb807, 07:09 PM

P: 320

Thank you all.
Let me read through the replies carefully. 



#8
Feb807, 10:09 PM

P: 2,265

Gauss's Law is applicable in 3dimensional space for any inversesquare field. and, in 3space, inversesquare fields are quite natural for any conserved physical quantity. imagine a 100 watt ([itex]P[/itex]) light bulb radiating energy equally in all directions (omnidirectional). we consider intensity of radiation to be the amount of power that falls on a unit area held perpendicular to the imaginary line connecting the source to that unit area. so intensity is watts per square meter. imagine that this light bulb is surrounded by a series of concentric spheres, all centered with that light bulb in the middle. since energy is conserved, all of this power radiating outward has to be the same whether it is the power escaping from one of the smaller spheres (where the intensity is higher, but the surface area of the sphere is less) or from one of the larger spheres. for each sphere, the sum of the power crossing each little segment of surface area must add up to the total power. and, for a sphere, each little segment of surface area is perpedicular to the origin of the radiant power moving out and equidistant from the origin. the power that crosses a little segment of surface area is equal to the intensity, [itex]I[/itex], times the area of that little segment. being a sphere, all little areas add up to the total area and the intensity is the same for all of these little areas because of symmetry (they are all equidistant from the source at the origin). so the total power, [itex]P[/itex], must be [tex] P = (4 \pi r^2) \cdot I [/tex] that total power moving from inside the sphere to outside is the same whether the sphere is a small one or a large one. then the intensity must be: [tex] I = \frac{P}{4 \pi r^2} [/tex] so the intensity must be an inversesquare field since energy and power are conserved and the area of a sphere is [itex]4 \pi r^2 [/itex]. now, our understanding is that (static) electric field from a point charge (or a collection of them) is naturally inversesquare because we imagine such Efield as being proportional to some imagined "flux" density, called [itex]D[/itex]. imagine these lines of flux as proportional to the amount of charge emanating out from a point charge. since these lines of flux are a conserved quantity (no new lines of flux are created out of nothing or destroyed into nothing), then the density of these lines of flux (flux per unit area) must also be an inverse square field. integrating the flux density (or the Efield) over the whole surface area of the sphere is proportional to the charge at the origin. what about flux coming from other sources? well, when those flux lines cross from outside the sphere to the inside, that counts as negative and gets canceled by the same flux lines when they cross from inside back to outside. Gauss's Law also shows that even for sources not at the origin, but inside the sphere, when you count only the component of the flux density vector that is perpendicular to the surface when it crosses from inside to out, that all of this flux adds up to the charged contained inside. now, divergence is this same thing except that now the containing volume is getting smaller and smaller to, in the limit, an infinitesimally small volume. so now, if you consider that, when you get this small, that charge is distributed evenly in the space and the amount of charge contained inside the sphere is proportional to the volume (charge density times volume). but even in this small volume, the net amount of flux moving from inside the volume to outside is still proportional to that charge (which is proportional to the volume). the divergence is the net amount of flux per unit volume moving from inside a differential amount of volume to outside of it and it is proportional to the charge per unit volume which is the charge density. 



#9
Feb907, 12:52 AM

P: 320

Thank you all.
I think a get a better understanding now. Thanks!! 


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