I have some conceptual problems with divergence...

1.divergence is supposed to be the flux per unit volume at a particular point....again,I saw on wikipedia,that they define divergence as "the derivative of net flow of of the vector field accross surface of a small region relative to the volume of the region...I can't see how these two defineitions are equivalent.

2. The word 'flow' is used in the above definition....all vectors however can't have 'flow',like the displacement vector for example...right?

3.I found on a website, an example illustrating divergence....it said, that for a particular vector field,the change of the vector along x axis,y axis and z axis were 1,-2,3.....subsequently,the author calculated the divergence as 1-2+3=2....how can the divergence be obtained just like that? I don't understand.

4.As I said,the divergence is the volume density of flux...however,in the formulation of the divergence in terms of the del operator does not include 'volume' anywhere...please explain.
5. Again,the divergence is said to be a measure of the expansion of a vector field at a point,but the notion of divergence as "volume density of flux" does not seem to indicate any 'expansion'.

6. The point itself at which the divergence is being measured may not be a source or sink of flux....a vector might just be passing the point...right?

Sorry for asking so many questions! I need to get my head around this.

1: Flux basically is the net flow of something across a surface, so there's no problem there. Your adherence to the volume definition is obviously causing you a lot of problems, and I would advise you to abandon it in favor of something else. I stick with the del dot definition--div(x) = del dot x. It's worked fine for me. Most of your problems will disappear then, since all the other stuff is relatively useless semantics.

3: Well, it may be helpful to think of the del representation of the thing. div(x) = del dot x, and since del is d/dx i + d/dy j + d/dz k, simply summing up the partial derivatives in each direction will indeed give you the divergence.

Thanks truth is life,but I needed to get the intuitive and in depth understanding of these quantities (gradient,curl and divergence)....right now I find myself memorising them,rather than understanding them.

Please could anyone help me by explaining my questions in an intuitive way?

My understanding for gradient is that basically a vector that points in the direction of greatest change. Curl on the the other hand being the uniformity(or lack there of) in the vector field. I have a hard time getting an intuitive understanding of divergence though.

arildno
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Okay, let us take a tiny box, centered at (0,0,0), with side lengths (dx,dy,dz).

We have a velocity field $$\vec{v}=u(x,y,z)\vec{i}+v(x,y,z)\vec{j}+w(x,y,z)\vec{k}$$
Now, at the wall (dx/2,y,z), $-\frac{dy}{2}\leq{y}\leq\frac{dy}{2},-\frac{dz}{2}\leq{z}\leq\frac{dz}{2}[/tex], we have that the local OUTWARD velocity can be written as: $$u(\frac{dx}{2},y,z)=u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2}+\frac{\partial{u}}{\partial{y}}y+\frac{\partial{u}}{\partial{z}}z$$, derivatives being measured at the origin, non-linear expansions ignored. Note that over the whole side, integrating over the y and z, the net contribution to the outflow rate from the latter two terms will cancel, and we sit back with the contribution to the outflow rate from this side to be: $$(u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2})dydz$$ Now, at the side x=-dx/2, the net OUTFLOW rate will be -u(-dx/2,y,z) (think about it!) A similar expansion, along with addition yields the total contribution to the outflow rate from sides with fixed x, namely: $$\frac{\partial{u}}{\partial{x}}dxdydz$$ Doing the same thing for the y-sides and z-sides yields the total amount of outflow per unit of time to be: $$\nabla\cdot\vec{v}dxdydz=\nabla\cdot\vec{v}dV$$ Since this equals the total amount of volume outflow per time unit, we have: $$\frac{d(dV)}{dt}=\nabla\cdot\vec{v}dV\to\nabla\cdot\vec{v}=\frac{1}{dV}\frac{d(dV)}{dt}$$ Last edited: I share your problem theDRG5,so could anyone help us out? Please please please!! When I first studied electromagnetic fields as an undergraduate, our professor gave us the definitions of divergence and curl in their limit-form, and then he derived the explicit differential forms in cartesian, cylindrical, spherical and general curvilinear coordinate systems. I took it upon myself to understand and reproduce these derivations in detail. I can still do it today, 25 years later. It is very simple in cartesian coordinates, but can be a little tricky in general curviliner coordinates. Anyway, at the time this made the meanings of divergence and curl very clear to me. I recommend this approach. Also, I find the definitions themselves to be intuitively satisfying. It's like understanding that derivative is slope when you first study calculus. If you understand what a limit means, then the concepts of divergence and curl, at a point, are crystal clear. $$\nabla \cdot \bar D={\rm lim}_{\Delta V\to 0} {{\oint _{ S} \bar D \cdot d\bar S}\over{\Delta V}}$$ $$\nabla \times \bar H={\rm lim}_{\Delta S\to 0} {{\oint _{ l} \bar H \cdot d\bar l}\over{\Delta S}}$$ Last edited: arildno Science Advisor Homework Helper Gold Member Dearly Missed Note: elect eng is presenting the real story here! For Cartesian coordinates, my approach is easy to see holds, but in order to prove that the formulae for the divergence and curl holds irrespective of the shapes of the sequence of volumes, you'll need to go with e.e's definition. Thanks,arildno,for your reply....I understood the basic flow of the derivation,if you know what I mean,but there are some places in between that I don't understand... We have a velocity field $$\vec{v}=u(x,y,z)\vec{i}+v(x,y,z)\vec{j}+w(x,y,z)\vec{k}$$ Do u,v and w give is the velocity components in the x,y,z directions respectively?(Usually I have come accross the notation V vector = Uxi + Uyj + Uzk) Also,is this is so,why is the velocity u,along x axis also a function of y and z? Now, at the wall (dx/2,y,z), [itex]-\frac{dy}{2}\leq{y}\leq\frac{dy}{2},-\frac{dz}{2}\leq{z}\leq\frac{dz}{2}[/tex], we have that the local OUTWARD velocity can be written as: $$u(\frac{dx}{2},y,z)=u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2}+\frac{\partial{u}}{\partial{y}}y+\frac{\partial{u}}{\partial{z}}z$$, derivatives being measured at the origin, non-linear expansions ignored. Why did you consider dx/2? (sorry if this is a silly question). Note that over the whole side, integrating over the y and z, the net contribution to the outflow rate from the latter two terms will cancel, and we sit back with the contribution to the outflow rate from this side to be: $$(u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2})dydz$$ (how?) arildno Science Advisor Homework Helper Gold Member Dearly Missed Thanks,arildno,for your reply....I understood the basic flow of the derivation,if you know what I mean,but there are some places in between that I don't understand... Do u,v and w give is the velocity components in the x,y,z directions respectively? Yes. (Usually I have come accross the notation V vector = Uxi + Uyj + Uzk) That's also fairly normal. Also,is this is so,why is the velocity u,along x axis also a function of y and z? The horizontal velocity component might well change with height and breadth. Think of water flowing in a channel, in one direction. If you are close to the side-walls, the horizontal velocity is less there than in mid-channel, similarly if you are close to the bottom of the channel. Why did you consider dx/2? (sorry if this is a silly question). The box has length dx, with x=0 being at the centre. Thus, we have that x varies between -dx/2 and dx/2, and similarly for the two other variables. arildno Science Advisor Homework Helper Gold Member Dearly Missed As for why the two later terms cancel, we have, for example: $$\frac{\partial{u}}{\partial{y}}\frac{y^{2}}{2}\mid_{y=-\frac{dy}{2}}^{y=\frac{dy}{2}}=0$$ Remember that the partial derivative is a CONSTANT, (i.e, hvaing been evaluated at the origin in the Taylor expansion) Note that over the whole side, integrating over the y and z, the net contribution to the outflow rate from the latter two terms will cancel, and we sit back with the contribution to the outflow rate from this side to be: $$(u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2})dydz$$ The above expression gives the flow along the x-axis,coming out of the face with area yz...right? Now,I think I've understood what arildno said....so let me try and answer one of my own questions based on it... 1.3.I found on a website, an example illustrating divergence....it said, that for a particular vector field,the change of the vector along x axis,y axis and z axis were 1,-2,3.....subsequently,the author calculated the divergence as 1-2+3=2....how can the divergence be obtained just like that? The values 1,-2 and three are the flow out of the surface yz,xz and xy respectively...along the x,y and z axes respectively....right? However,it would be nice if you paid special attention to the following two questions,which I still haven't got clear. 1 Again,the divergence is said to be a measure of the expansion of a vector field at a point,but the notion of divergence as "volume density of flux" does not seem to indicate any 'expansion'. 2 The point itself at which the divergence is being measured may not be a source or sink of flux....a vector might just be passing the point...right? arildno Science Advisor Homework Helper Gold Member Dearly Missed The above expression gives the flow along the x-axis,coming out of the face with area yz...right? It gives the infinitesemal flow RATE out of the side x=dx/2, yes. The total flow rate in the x-direction must also include the flow rate through the side x=-dx/2 Now,I think I've understood what arildno said....so let me try and answer one of my own questions based on it... 1.3.I found on a website, an example illustrating divergence....it said, that for a particular vector field,the change of the vector along x axis,y axis and z axis were 1,-2,3.....subsequently,the author calculated the divergence as 1-2+3=2....how can the divergence be obtained just like that? . The values 1,-2 and three are the flow out of the surface yz,xz and xy respectively...along the x,y and z axes respectively....right? With the caveat given above, YES! I'll deal with the others later on. arildno Science Advisor Homework Helper Gold Member Dearly Missed 1 Again,the divergence is said to be a measure of the expansion of a vector field at a point,but the notion of divergence as "volume density of flux" does not seem to indicate any 'expansion'. It sure does! "Flux" is how much "flows out" of the box per unit time. This must equal the volume expansion per time. Dividing this quantity by the volume, yields the volume density flux, and that is equal to the divergence. 2 The point itself at which the divergence is being measured may not be a source or sink of flux....a vector might just be passing the point...right? Sure enough. However, think of the divergence AT a point as the relative volume expansion of a tiny box centered at that point. That is easier to visualize! Thanks alot for your help. I think I've understood divergence well enough! Okay,as you'll have noticed,I started this post to finally get myself clear about the three operations-gradient,divergence and curl. Since I think I'm done with divergence,I'd like to move onto gradient. 1. How can we realise the fact the gradient id the direction of maximum increase of a function? (Is this because gradient is the vector sum of the partial derivatives of the function along the x,y,z directions?) 2.How is the gradient perpendicular to the surface in concern? 3. What does the scalar potential of a factor mean? Why and how is it related to the gradient of the function? 4. divergence.divergence= gradient squared= laplace operator......what does this mean? what is the laplace operator anyway? 5. What is a level surface....are these planes parallel to the xy plane.....what exactly is the relation of gradient to level surfaces? 6. The gradient vector gives the direction of maximum change of the function at a point( as per definition)...but the vector arrow representing gradient only gives the direction of change,it does not give us the distance we have to travel to get the maximum value of the function.....what's the point of having a vector like that that doesn't tell us practically anything? Last edited: 2.How is the gradient perpendicular to the surface in concern? If you have a surface z = f(x, y), then the gradient of f is NOT perpendicular (as it's per definition tangential). However the gradient of F(x, y, z) = z - f(x, y) IS perpendicular. 5. What is a level surface.... A level surface (in 3D) or more generally a level set, is the set of points for which a function f(x_1, x_2, ..., x_n) = c, for some arbitrary constant c. You've probably seen them in maps as contour lines (level curves), ie the line tracing out which points on the ground are N meters above the sea. Last edited: If you have a surface z = f(x, y), then the gradient of f is NOT perpendicular (as it's per definition tangential). However the gradient of F(x, y, z) = z - f(x, y) IS perpendicular. Okay,how does this happen? A level surface (in 3D) or more generally a level set, is the set of points for which a function f(x_1, x_2, ..., x_n) = c, for some arbitrary constant c. You've probably seen them in maps as contour lines (level curves), ie the line tracing out which points on the ground are N meters above the sea. This means that the curve is a constant valued function 'parallel to the ground',i,e parallel to the xy(or yz or xz) plane,right? arildno Science Advisor Homework Helper Gold Member Dearly Missed Now, let me try to sort out a few issues first: Suppose we have a surface in 3-D, so that the z-coordinate for a point upon that surface can be written as z=f(x,y). Then, the VECTORIAL REPRESENTATION of that surface can be given as: $$\vec{S}(x,y)=x\vec{i}+y\vec{j}+f(x,y)\vec{k}$$ In order to find tangent VECTORS to this surface at some point, this is really simple: We pick two points ON the surface, calculate a secant between them, and let the distance between our two points shrink. For example, we may start with a secant written as follows: $$\frac{\vec{S}(x+\bigtriangleup{x},y)-\vec{S}(x,y)}{\bigtriangleup{x}}=\vec{i}+\frac{f(x+\bigtriangleup{x},y)-f(x,y)}{\bigtriangleup{x}}\vec{k}$$ I.e, we calculate DERIVATIVES, and we may choose the following two independent tangent vectors: $$\vec{T}_{x}=\vec{i}+\frac{\partial{f}}{\partial{x}}\vec{k}$$ $$\vec{T}_{y}=\vec{j}+\frac{\partial{f}}{\partial{y}}\vec{k}$$ A vector normal is readily found by the CROSS product of these two vectors, namely: $$\vec{N}=-\frac{\partial{f}}{\partial{x}}\vec{i}-\frac{\partial{f}}{\partial{y}}\vec{j}+\vec{k}$$ Is this OK so far? In particular: 1. Note that neither of the TANGENTS have anything to do to with the gradient of f, i.e the vector: $$\nabla{f}=\frac{\partial{f}}{\partial{x}}\vec{i}+\frac{\partial{f}}{\partial{y}}\vec{j}$$ 2. Note that the gradient of the SCALAR function, F(x,y,z)=z-f(x,y) IS, indeed, equal to the vector normal to the surface $$\vec{S}$$ Last edited: Is $$\vec{T}_{x}=\vec{i}+\frac{\partial{f}}{\partial{x} }\vec{k}$$ along a plane parallel to the xz plane (since it is the derivative w.r.t x only,id musn't have any component along the y axis,so I just wanted to confirm my idea) Also,though the normal vector seems to be similar in its form as the normal vector,I can't imagine why it has to be. As Lord Crc pointed out, "for a surface z = f(x, y), the gradient of f is NOT perpendicular (as it's per definition tangential). However the gradient of F(x, y, z) = z - f(x, y) IS perpendicular." What difference is there between the two situations 'z = f(x, y)' and 'F(x, y, z) = z - f(x, y) '? arildno Science Advisor Homework Helper Gold Member Dearly Missed Is $$\vec{T}_{x}=\vec{i}+\frac{\partial{f}}{\partial{x} }\vec{k}$$ along a plane parallel to the xz plane (since it is the derivative w.r.t x only,id musn't have any component along the y axis,so I just wanted to confirm my idea) Correct. Also,though the normal vector seems to be similar in its form as the normal vector,I can't imagine why it has to be. As Lord Crc pointed out, "for a surface z = f(x, y), the gradient of f is NOT perpendicular (as it's per definition tangential). However the gradient of F(x, y, z) = z - f(x, y) IS perpendicular." What difference is there between the two situations 'z = f(x, y)' and 'F(x, y, z) = z - f(x, y) '? A world of difference! f(x,y) is a function of two variables, F(x,y,z) a function of three variables. I was a bit quick when I replied, sorry about that. If you have a surface z = f(x, y), you can rewrite it as the level set F(x, y, z) = z - f(x, y) = 0. These are just two different ways of expressing the same thing. You can always express a surface y = f(x1, x2, ..., xn) as the level set F(x1, x2, ..., xn, y) = y - f(x1, x2, ..., xn) = 0. The directional derivative in the direction v is given by "grad(F) . v" where . is the scalar or dot product. Obviously the rate of change of F "along the surface" (directional derivative) is zero since F is constant there. The direction t "along the surface" is (of course) tangent at the point in question. Thus grad(F) . t = 0, and thus the gradient of F must be perpendicular to the surface. Not really well explained but... I understand that this notion of the gradient vector being perpendicular is applicable to only F(x, y, z) = z - f(x, y) = 0,since as Lord Crc justified,the F is a level surface (my concept about level surfaces is shaky...I presume its a flat surface parallel to xy/xz/yz plane??) and the rate of change of the function has to be zero along it....but the function F is in 4 dimensional space!!! What am I wrong about. I understand that this notion of the gradient vector being perpendicular is applicable to only F(x, y, z) = z - f(x, y) = 0,since as Lord Crc justified,the F is a level surface (my concept about level surfaces is shaky...I presume its a flat surface parallel to xy/xz/yz plane??) and the rate of change of the function has to be zero along it....but the function F is in 4 dimensional space!!! What am I wrong about. A level surface is a region in space (the domain of the function) where the function is constant, that is, not changing. For mountains for example, you can find a level curve, on which you will have the same height all along. This is genralized to surfaces for function of 3 vars. and even volumes for functions of 4 vars. Another name for this regions, is "isothermic curves/surfaces" which comes from heat conduction study. This region doesn't have to be parallel to any plane, it may be a curved surface. For example, the level curves of F(x,y)=xy are hyperboles, certainly not straight lines. Since you know the gradient will point to the greatest change, you will want no component to point at a direction where the function is not changing (the direction of the level surface) since that's just a waste of a component ( :) ). Therefore you conclude that the gradient vector will be perpendicular to the level surface/curve at any point. (This is an intuitive proof for this. An actuall proof will not require the assumption of the first sentence) arildno Science Advisor Homework Helper Gold Member Dearly Missed I understand that this notion of the gradient vector being perpendicular is applicable to only F(x, y, z) = z - f(x, y) = 0,since as Lord Crc justified,the F is a level surface (my concept about level surfaces is shaky...I presume its a flat surface parallel to xy/xz/yz plane??) and the rate of change of the function has to be zero along it....but the function F is in 4 dimensional space!!! What am I wrong about. 1. The GRAPH of F as a function of (x,y,z) would, indeed, be a 4-dimensional beast, with 3 degrees of freedom (i.e, the number of independent variables). 2. The region R in (x,y,z)-space so that for any element (X,Y,Z) in R, we have F(X,Y,Z)=0, IS a 3-dimensional structure, usually writable as some (two-dimensional) SURFACE: $$\vec{S}(u,v)=x(u,v)\vec{i}+y(u,v)\vec{j}+z(u,v)\vec{k}$$, so that we have, for ALL (u,v): $$F(\vec{S}(u,v))=0(*)$$ 3. Because [itex]\vec{S}$ is identified with all points in R, it follows that the tangent vectors of $\vec{S}$ can be found by taking the partial derivatives with respect to u and v.

4. Since (*) in 2. holds for ALL values of (u,v), we may differentiate both sides with respect to either u or v, getting for example in the first case:
$$\nabla{F}\cdot\vec{S}_{u}=0$$, i.e, the vector $\nabla{F}$ is ORTHOGONAL to the tangent vector $\vec{S}_{u}$.
Similarly, we find that the gradient of F is orthogonal to the other tangent vector, and THUS, $\nabla{F}$ is proven to be along the vector normal to the surface $\vec{S}$

Let us take an example:

Let F(x,y,z)=ax+by+cz, and let c be different from 0.

Now, consider the level surface described by the equation F=0 (**)

We may now create a vectorial representation of this surface, for example:
$$\vec{S}(u,v)=u\vec{i}+v\vec{j}+(-\frac{au+bv}{c})\vec{k}$$

We may now verify:
$$F(\vec{S}(u,v))=au+bv+c*(-\frac{au+bv}{c})=0$$
irrespective of the values of (u,v)!

We now have:
$$\vec{S}_{u}=\vec{i}-\frac{a}{c}\vec{k},\vec{S}_{v}=\vec{j}-\frac{b}{c}\vec{k}$$
Note that the vector normal can be written as:
$$\vec{N}=\vec{S}_{u}\times\vec{S}_{v}=\frac{a}{c}\vec{i}+\frac{b}{c}\vec{j}+\vec{k}=\frac{1}{c}(a\vec{i}+b\vec{j}+c\vec{k}=\frac{1}{c}\nabla{F}$$
which is, indeed, a normal vector to the plane described by (**).

5. Now, let us look at the relevant f(x,y), as in z=f(x,y), i.e, f(x,y)=-1/c(ax+by)
We have that:
$$\nabla{f}=-\frac{a}{c}\vec{i}-\frac{b}{c}\vec{j}$$
Note that the dot products with either of the surface tangents are NOT 0 in general, i.e, the gradient of f is NOT along the surface normal to the level surface of F (i.e. $$\vec{S}$$)

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