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hanson

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Can anyone tell me how to interpret the divergence of a vector field? Please kindly help.

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- Thread starter hanson
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hanson

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Can anyone tell me how to interpret the divergence of a vector field? Please kindly help.

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I find no clue why the divergence of an electric field measures the charge denity?

That's essentially Gauss' Law [relating the field to the source]. It's a law of physics.

(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.

Have you seen the definition as a limit of flux per unit volume?Can anyone tell me how to interpret the divergence of a vector field? Please kindly help.

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- #3

mdelisio

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Can anyone tell me how to interpret the divergence of a vector field? Please kindly help.

I would say that the divergence of E at a point tells you how much the field is "spreading out" or "emerging" (diverging?) from the neighborhood of that point. If the divergence is large, you must have some sources (charges) there.

- #4

billiards

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If the divergence of a velocity field is zero, it means the fluid is incompressible.

- #5

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This (and the subsections) might be useful:

http://www.math.gatech.edu/~carlen/2507/notes/vectorCalc/divcurlMean1.html [Broken]

http://www.math.gatech.edu/~carlen/2507/notes/vectorCalc/divcurlMean1.html [Broken]

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- #6

Claude Bile

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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 non-zero, 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.

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 non-zero, 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

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.

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hanson

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Thank you all.

Let me read through the replies carefully.

Let me read through the replies carefully.

- #8

rbj

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... 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.

as other's pointed out, this has to do with Gauss's Law. it is a differential form of Gauss's Law.

Gauss's Law is applicable in 3-dimensional space for any inverse-square field. and, in 3-space, inverse-square fields are quite natural for any conserved physical quantity.

imagine a 100 watt ([itex]P[/itex]) light bulb radiating energy equally in all directions (omni-directional). 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 inverse-square 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 inverse-square because we imagine such E-field as being proportional to some imagined

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

- #9

hanson

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Thank you all.

I think a get a better understanding now.

Thanks!

I think a get a better understanding now.

Thanks!

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