Gradient,divergence and curl.

  • Thread starter Urmi Roy
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  • #1
Urmi Roy
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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 across 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.
 

Answers and Replies

  • #2
truth is life
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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.
 
  • #3
Urmi Roy
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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?
 
  • #4
theDRG5
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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.
 
  • #5
Urmi Roy
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I share your problem theDRG5,so could anyone help us out? Please please please!
 
  • #6
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 [tex]\vec{v}=u(x,y,z)\vec{i}+v(x,y,z)\vec{j}+w(x,y,z)\vec{k}[/tex]
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:
[tex]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[/tex], 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:
[tex] (u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2})dydz[/tex]
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:
[tex]\frac{\partial{u}}{\partial{x}}dxdydz[/tex]

Doing the same thing for the y-sides and z-sides yields the total amount of outflow per unit of time to be:
[tex]\nabla\cdot\vec{v}dxdydz=\nabla\cdot\vec{v}dV[/tex]
Since this equals the total amount of volume outflow per time unit, we have:
[tex]\frac{d(dV)}{dt}=\nabla\cdot\vec{v}dV\to\nabla\cdot\vec{v}=\frac{1}{dV}\frac{d(dV)}{dt}[/tex]
 
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  • #7
elect_eng
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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.

[tex]\nabla \cdot \bar D={\rm lim}_{\Delta V\to 0} {{\oint _{ S} \bar D \cdot d\bar S}\over{\Delta V}} [/tex]

[tex]\nabla \times \bar H={\rm lim}_{\Delta S\to 0} {{\oint _{ l} \bar H \cdot d\bar l}\over{\Delta S}} [/tex]
 
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  • #8
arildno
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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.
 
  • #9
Urmi Roy
753
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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 [tex]\vec{v}=u(x,y,z)\vec{i}+v(x,y,z)\vec{j}+w(x,y,z)\vec{k}[/tex]

Do u,v and w give is the velocity components in the x,y,z directions respectively?(Usually I have come across 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:
[tex]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[/tex], 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:
[tex] (u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2})dydz[/tex]

(how?)
 
  • #10
arildno
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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 across 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.
 
  • #11
arildno
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As for why the two later terms cancel, we have, for example:
[tex]\frac{\partial{u}}{\partial{y}}\frac{y^{2}}{2}\mid_{y=-\frac{dy}{2}}^{y=\frac{dy}{2}}=0[/tex]

Remember that the partial derivative is a CONSTANT, (i.e, hvaing been evaluated at the origin in the Taylor expansion)
 
  • #12
Urmi Roy
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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:
[tex] (u(0,0,0)+\frac{\partial{u}}{\partial{x}}\frac{dx}{2})dydz[/tex]

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?
 
  • #13
arildno
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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.
 
  • #14
arildno
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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!
 
  • #15
Urmi Roy
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Thanks a lot for your help. I think I've understood divergence well enough!
 
  • #16
Urmi Roy
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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 realize 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?
 
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  • #17
Lord Crc
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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.
 
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  • #18
Urmi Roy
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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?
 
  • #19
arildno
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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:
[tex]\vec{S}(x,y)=x\vec{i}+y\vec{j}+f(x,y)\vec{k}[/tex]

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:
[tex]\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}[/tex]


I.e, we calculate DERIVATIVES, and we may choose the following two independent tangent vectors:
[tex]\vec{T}_{x}=\vec{i}+\frac{\partial{f}}{\partial{x}}\vec{k}[/tex]
[tex]\vec{T}_{y}=\vec{j}+\frac{\partial{f}}{\partial{y}}\vec{k}[/tex]
A vector normal is readily found by the CROSS product of these two vectors, namely:
[tex]\vec{N}=-\frac{\partial{f}}{\partial{x}}\vec{i}-\frac{\partial{f}}{\partial{y}}\vec{j}+\vec{k}[/tex]
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:
[tex]\nabla{f}=\frac{\partial{f}}{\partial{x}}\vec{i}+\frac{\partial{f}}{\partial{y}}\vec{j}[/tex]

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 [tex]\vec{S}[/tex]
 
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  • #20
Urmi Roy
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Is [tex]\vec{T}_{x}=\vec{i}+\frac{\partial{f}}{\partial{x} }\vec{k}[/tex]
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) '?
 
  • #21
arildno
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Is [tex]\vec{T}_{x}=\vec{i}+\frac{\partial{f}}{\partial{x} }\vec{k}[/tex]
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.
 
  • #22
Lord Crc
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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...
 
  • #23
Urmi Roy
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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.
 
  • #24
elibj123
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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)
 
  • #25
arildno
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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:
[tex]\vec{S}(u,v)=x(u,v)\vec{i}+y(u,v)\vec{j}+z(u,v)\vec{k}[/tex], so that we have, for ALL (u,v):
[tex]F(\vec{S}(u,v))=0(*)[/tex]

3. Because [itex]\vec{S}[/itex] is identified with all points in R, it follows that the tangent vectors of [itex]\vec{S}[/itex] 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:
[tex]\nabla{F}\cdot\vec{S}_{u}=0[/tex], i.e, the vector [itex]\nabla{F}[/itex] is ORTHOGONAL to the tangent vector [itex]\vec{S}_{u}[/itex].
Similarly, we find that the gradient of F is orthogonal to the other tangent vector, and THUS, [itex]\nabla{F}[/itex] is proven to be along the vector normal to the surface [itex]\vec{S}[/itex]



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:
[tex]\vec{S}(u,v)=u\vec{i}+v\vec{j}+(-\frac{au+bv}{c})\vec{k}[/tex]

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

We now have:
[tex]\vec{S}_{u}=\vec{i}-\frac{a}{c}\vec{k},\vec{S}_{v}=\vec{j}-\frac{b}{c}\vec{k}[/tex]
Note that the vector normal can be written as:
[tex]\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}[/tex]
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:
[tex]\nabla{f}=-\frac{a}{c}\vec{i}-\frac{b}{c}\vec{j}[/tex]
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. [tex]\vec{S}[/tex])
 
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  • #26
arildno
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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 realize 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?)


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?

Let [tex]\vec{n}[/tex] be a unit normal in some direction, and [tex](x_{0},y_{0},z_{0})[/tex] some point in the domain of a function g(x,y,z).

Consider the ONE-variable function:
[tex]L(t)=g(x_{0}+tn_{x},y_{0}+tn_{y},z_{0}+tn_{z}), \vec{n}=n_{x}\vec{i}+n_{y}\vec{j}+n_{z}\vec{k}[/tex]

Thus, L(t) measures the change in the value of g as we walk along the chosen direction!

Now, let us compute the rate of change of L AT t=0, i.e, the rate of change of g AT [tex](x_{0},y_{0},z_{0})[/tex] in the direction of [tex]\vec{n}[/tex]

This is simply, by the chain rule:
[tex]\frac{dL}{dt}=\nabla{g}\cdot\vec{n}[/tex]
where dl/dt is evaluated at t=0, [tex]\nabla{g}[/tex] at [tex](x_{0},y_{0},z_{0})[/tex]

Since n is of unit length, this expression is, of course, maximized when n is parallell to [itex]\nabla{g}[/tex]

Thus, the gradient of g gives the direction of fastest growth of g!


Now, why is this useful, how for example can we now calculate where the maximum/minimum of g is?

AT a maximum, g is neither growing or shrinking; i.e, it is STATIONARY (no direction for fastest growth exists). Similarly for minima!

Thus, the equation we need to solv in order to determine stationary points is:
[tex]\nabla{g}=\vec{0}[/tex]

This is the generalization of the one-variable case, where within the set of zeroes of a function's derivative are where the extrema of the function can be found.
 
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  • #27
Urmi Roy
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Thanks arildno,for your posts,which must have taken a lot of effort and time...but I still have this lingering doubt...
I understand that-
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

I realize that since we have an expression in 3 variables,and it is equated to zero,eventually,it has to be represented as a surface in 3 dimensions(z can be explicitly defined in terms of x and y)...however,you also showed that--

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. [tex]\vec{S}[/tex])

Since F(X,Y,Z)=0 and z=f(x,y) represent the same surface in 3D,why should the gradient vector have different relation with these curves?

Also,in regard to elibj123's post on level surfaces,he says that they can be described in terms of a mountain,and if we consider a flat region at a certain level of the mountain,we have a level surface.That's perfectly clear...at that level,we have a constant value of height...but what about the hyperbole thing? Surely a hyperbola cannot have a constant valur of a certain function all along it?
 
  • #28
arildno
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The gradient of f tells you along which direction in the xy-plane f (that is z) will increase most strongly along at a given point.

Thus, the gradient of f at the surface F(x,y,z)=0, i.e, z=f(x,y) tells you the direction of the STEEPEST SLOPE of the surface.

That is something totally different than the direction of the normal vector at the same point on the surface!
 
  • #29
Urmi Roy
753
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It's going to take me a while to get used to this...I realize that there can't be a better explanation than the ones you have already given,so its upto me now to finally realize this.
In the mean time,I'll try to end up this part on gradient by asking a final question...
(Referring to wikipedia,I found:
The physical meaning of the scalar potential depends on the type of the field. For a velocity field of a fluid or gas flow, the definition of the scalar potential implies that the direction of the flow at any point coincides with the direction of the steepest decrease of the potential at that point, and for a force field the same is true of the acceleration at a point. The scalar potential of a force field is closely related to the field's potential energy."

Now,
Is the definition that "scalar potential gives the direction of steepest increase" merely by convention?

Fields like velocity,displacement must also have scalar potentials,but it doesn't really make sense,does it?
Besides...what is this scalar potential of a field really,beyond the conventional definition?
 
  • #30
arildno
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It's going to take me a while to get used to this...I realize that there can't be a better explanation than the ones you have already given,so its upto me now to finally realize this.
There is another way:
That we methodically go through, point after point what I have written, and that YOU, instead of writing multiple questions limit your post to ONE question at time, and that we work on that until you get it.

I'm willing to do that, are you? :smile:
 
  • #31
Urmi Roy
753
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That's fine with me,but you see,I have been rushing a little since I have exams every now and then...after this I have some points to get cleared about line integrals,surface integral and the theorems of vector calculus that I have just read through from Kreyzig(I don't have sound understanding of these..and believe me my teacher's can't help me!)...so I do have a little problem with time.
Will you guide me through what's remaining of vector calculus?...then I can take it easy and slowly,and rely on the fact that at the end of a couple of days, I'll be thorough in my understanding of vector calculus(I'll try not to bother you too much after that.)
 
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  • #32
Urmi Roy
753
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Everyone forgotten me? :(
 
  • #33
Urmi Roy
753
1
Okay,since I don't have much time left before my exams,I need to get my last set of questions cleared on curl...please please please help!

1. Is it possible to have Curl of velocity? Intuitively,what would that be?

2. I have noticed that the del operator used to define divergence and gradient fit in perfectly with the definition of these quantities...how does the determinant formulation of curl fit in with the true meaning of curl?...why do we use this determinant formulation?

3. In wikipedia,they interpret curl with the help of a ball with rough edges,and how the fluid moving past it makes it rotate...the magnitude of the rotational velocity is the value of curl...but this cannot be applied to any other vector(all vectors wouldn't make a ball rotate,e.g. displacement vector).

4.Talking of displacement vector,what is the curl of this vector...how would we interpret it intuitively?

5."The curl of a vector field F at a point is defined in terms of its projection onto various lines through the point. If is any unit vector, the projection of the curl of F onto is defined to be the limiting value of a closed line integral in a plane orthogonal to as the path used in the integral becomes infinitesimally close to the point, divided by the area enclosed."- this is from wikipedia...what does it mean?

My last and perhaps most important question:
The curl is defined as the 'microscopic circulation per unit area' (ref: http://www.math.umn.edu/~nykamp/m2374/readings/circperarea/) ...how does this definition tally with the usual way of defining curl that I mentioned earlier in the post?
 
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  • #34
elibj123
240
2
1. It will tell you about the angular velocity/accleration at that point, but that's not very accurate.

2. If we jump ahead to 5 (which I'll explain later) and look at its definition, we can now define the components of the curl field along the xyz axes. Since the shape of the infinitesemal loop is not important (assuming everything is smooth enough), I would suggest taking a rectangular with sides parallel to the different axes (while computing you will have three of this loops). Calculating the circulation density:

[tex]curl(F) _{x}=lim \frac{\oint_{C_{yz}}\vec{F}\vec{dr}}{A_{yz}}[/tex]
(C_{yz} is a rectangular loop parallel to the yz plain)

With this specific shape, will give you exactly the definitions of the derivatives. Summing all the components, you will recognize the structure of a cross product, or as you said, the determinant formulation.

3 & 4. This is just an explanation with analogy. However I don't see how a displacement vector can play a role of a vector field. A displacement vector is usually a function of t. With curl we are talking about functions of (x,y,z).

5. The projection of the curl against a specific direction, will tell you how much the field circulates around that direction at some point.
 
  • #35
Urmi Roy
753
1
1. It will tell you about the angular velocity/accleration at that point, but that's not very accurate.

So it doesn't really have any physical existence,right?

2. If we jump ahead to 5 (which I'll explain later) and look at its definition, we can now define the components of the curl field along the xyz axes. Since the shape of the infinitesemal loop is not important (assuming everything is smooth enough), I would suggest taking a rectangular with sides parallel to the different axes (while computing you will have three of this loops). Calculating the circulation density:

[tex]curl(F) _{x}=lim \frac{\oint_{C_{yz}}\vec{F}\vec{dr}}{A_{yz}}[/tex]
(C_{yz} is a rectangular loop parallel to the yz plain)

With this specific shape, will give you exactly the definitions of the derivatives. Summing all the components, you will recognize the structure of a cross product, or as you said, the determinant formulation.

Thanks,I've understood this.

A displacement vector is usually a function of t. With curl we are talking about functions of (x,y,z).

I should have taken this into account earlier!

5. The projection of the curl against a specific direction, will tell you how much the field circulates around that direction at some point.

Lastly,from what I get (especially from your answer to 3 and 4,)curl doesn't really have to make something (like a ball or a paddle) rotate...it's just a measure of how much the vector arrows are twisting,right?
 
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