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Divergence of radial unit vector field

  1. Jul 15, 2015 #1
    Sorry if this was addressed in another thread, but I couldn't find a discussion of it in a preliminary search. If it is discussed elsewhere, I'll appreciate being directed to it.

    Okay, well here's my question. If I take the divergence of the unit radial vector field, I get the result:

    [itex]\vec \nabla \cdot \hat r = \vec \nabla \cdot \frac{\vec r} {|\vec r|} = \frac {2} {|\vec r|}[/itex]​

    Now since the direction of the vector field at any point is dependent on position, the non-zero result is not surprising. What is surprising however is why the constant magnitude vector field has a divergence that is inversely proportional to distance from the origin.

    The fact that this vector field is unphysical might be why I'm having such a hard time trying to make sense of it. But here's the thinking that's getting me in trouble. If I try to interpret this vector field as say a flow of air moving away from the origin, then regardless of where I measure, I should find the exact same magnitude of flow of the air. Let's say I choose to measure at a point 1m away from the origin, and then measure at a point 2m away from the origin, then I should get the exact same magnitude for the flow at both points. Further, let's choose the two points such that the line connecting the two points also passes through the origin. So now there is absolutely no difference in the flow from one point to the next (in both magnitude and direction). Why then does the divergence of the two points differ by a factor of two? It seems to me that both points in the vector field are indistinguishable, and there isn't really any way to identify one point from the other. In other words, if I move the origin anywhere along the line connecting the two points, then the value and direction of the vector field at both points remain the same and is independent of the location of the new origin (provided the new origin does not lie between the two points!) Yet the divergence would change!! What the heck is going on?

    One of the ways Wikipedia defines the divergence is like this, "In vector calculus, divergence is a vector operator that measures the magnitude of a vector field's source or sink at a given point, in terms of a signed scalar. More technically, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point." So in the above example, if we take two separate but equal infinitesimal volumes around each point, why should we expect half (or twice) the volume density of the outward flux at one point versus the other point?

    Okay well I've laid out my thought process as best I could. I appreciate any insight into what I'm missing regarding what exactly the divergence represents.
    Last edited: Jul 15, 2015
  2. jcsd
  3. Jul 15, 2015 #2


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    Ah, but there is a difference. Assume the point where you are measuring divergence is directly to the right of the origin ('on the x axis') and consider a very small disc ##D1## of radius ##\epsilon## around that point, the radial vectors spread out as they move through that disc from left to right. They 'diverge'. A radial 'flow line' that enters ##D1## on the left above (below) the x axis at a point with y coordinate ##h_{e}\equiv\frac{\epsilon}{4}## will leave ##D1## at a point with y coordinate ##h_{12}## whose absolute value is greater than ##\frac{\epsilon}{4}##, ie further from the x axis than where it entered. Then ##\theta_1\equiv\frac{h_{12}}{h_{e}}## is an indicator of the 'divergence' in that disc.

    Now consider a new disc ##D2## that is twice as far from the origin. Again consider a radial flow line that enters and leaves ##D2## at points with y coordinates ##h_{e}## and ##h_{22}## and let ##\theta_2\equiv\frac{h_{22}}{h_{e}}##. Then it is easy to see that the radial flow lines diverge less across ##D2## than they did across ##D1## because they are closer to parallel, and that ##\theta_2<\theta_1##. In fact, it may well be the case that ##\frac{\theta_1}{\theta_2}=2##, although I have not attempted to prove that.

    So if you think of the divergence as in some sense representing how much the flow lines diverge as they cross a small fixed-size disc from one side to the other, it makes sense that the divergence decreases as we get further from the origin because the flow lines become closer to parallel.
  4. Jul 15, 2015 #3
    Awesome!! Thanks andrewkirk. That does make sense. That's the fundamental piece I was overlooking. Further away from the origin, the less the field lines "diverge" from one another!!
  5. Jul 21, 2015 #4

    So I've done some more thinking about the divergence and there are two things I'd like a little clarification with. Firstly, if instead of the divergence of the radial unit vector field, the divergence of the actual radial vector field is found, that gives a constant.

    [itex]\vec \nabla \cdot \vec r = 3 [/itex]
    This result means that the divergence is independent of position. Now this baffles me. I admit this is sort of like the opposite of my last post. At that time, I was expecting a constant divergence and was shocked at a divergence that decreased with distance from the origin. Now, this vector field has a constant divergence but trying to use your example of a disc on the x-axis, it seems the lines again should diverge less as we move away from the origin. The difference between this vector field and the previous one was that the magnitudes are changing here while they were constant for the previous vector field. But as far as "diverging" from each other, the two vector fields point in the same direction everywhere!! So again back to the drawing board in terms of making sense of what this divergence quantity is actually representing!!

    Secondly, kind of along the same lines, the divergence of the x-vector field is one. But there is absolutely no divergence of the field lines. While the divergence of the x-unit vector field is zero. So unfortunately there doesn't seem to be any real analogy between the radial unit vector and any of the Cartesian unit vectors. It's possible I might have taken the divergence wrong, and if so I'd be happy to be corrected. But if not, then I really really would love some insight into what's going on again.

    [itex] \vec \nabla \cdot (x \hat i) [/itex] = 1
    [itex] \vec \nabla \cdot \hat i =\vec \nabla \cdot(\frac { \vec x}{x} )[/itex]=0​
  6. Jul 21, 2015 #5


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    I can answer the first one. Then I have to have breakfast. I'll try to look at the other one later.
    You've sort of answered the question yourself by pointing out that in this case the magnitudes change. In fact they change in the opposite direction of how the div of the unit field changes, so that the increase of magnitude exactly offsets the reduction in the unit div.

    To get an intuitive feel, try this: The divergence is in a sense a measure of the irregularity of a vector field near a point. A whirlpool is an example of an irregularity. The extent to which a whirlpool 'disrupts' an object is a function of the accelerations it imparts upon it, which are a function of (1) the curvature of the whirlpool flow lines where the object is and (2) the speed of flow at that place. You experience far more Gs executing a tight turn fast than executing it slowly.

    Now the divergence of the unit vector field focuses only on the curvature of the flow lines, and that curvature decreases with distance. But the div of the non-unit field also takes into account the 'speed' of flow. THe whirlpools get less curvy but are traversed at a faster speed as you get further from the origin. These offset so that the amount of 'disturbance' remains constant, and that's what the div reflects.
  7. Jul 21, 2015 #6


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    The second question: The div of the x vec field is not a sideways (in y direction) spreading of the field like the prev example. Rather it is a spreading in the x direction - an acceleration of the field in that direction. For any small disc, the flow is entering the disc at the left more slowly than it is exiting at the right. If the flow were of a gas, it would mean that the gas gets progressively less dense as it moves to the right. Think of how a falling torrent of water thins out as it gets lower (gains velocity), or how cars in a traffic stream spread out as the traffic speeds up. So the molecules of gas are diverging.

    Another way of thinking about divergence that I find useful involves conservation laws, and is crucial in relativity theory. The conservation laws of both energy and momentum can be summarised in a single tensor equation saying that the 4D tensor-divergence of the stress-energy tensor T is zero. Given a coordinate system, this can be expressed in a less jargony fashion as that, for each of the four 4D vectors x-direction momentum, y-direction momentum, z-direction momentum and mass-energy (which is 't-direction momentum'), the divergence is zero. The components of each vector are (1) density (of energy or momentum) (2) flux in x direction (3) flux in y direction (4) flux in z direction. Then, writing ##ME## for mass-energy, the statement that the ME divergence is zero is that:
    $$\frac{\partial (\textit{ME density})}{\partial t}
    +\frac{\partial (\textit{ME x-flux})}{\partial x}
    +\frac{\partial (\textit{ME y-flux})}{\partial y}
    +\frac{\partial (\textit{ME z-flux})}{\partial z}=0$$

    (We can write a similar statement replacing ME by each of x-mom, y-mom, z-mom)

    This says that the time rate of increase in ME density in an infinitesimal volume must equal the sum of the rates of net fluxes of ME into that volume from the three co-ord directions. That is the same as saying that ME can be neither created nor destroyed.
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