Gradient,divergence and curl.

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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 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?)


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
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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 realise 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
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It's going to take me a while to get used to this....I realise that there can't be a better explanation than the ones you have already given,so its upto me now to finally realise 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 realise that there can't be a better explanation than the ones you have already given,so its upto me now to finally realise 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
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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
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Everyone forgotten me? :(
 
  • #33
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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
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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
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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 existance,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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  • #36
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Please forgive me for extending even further....just my last two questions..

1. When we calculate curl,we reduce the area over which the path integral is being integrated to a single point--however,due to this, the magnitude of curl should always be very small,as the path integral over an area as small al that would be near to zero!!
Am I saying something wrong?

2.Direction of curl is defined as the direction of maximum rotation...what doe the 'maximum' imply?
 
  • #37
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1. When we calculate curl,we reduce the area over which the path integral is being integrated to a single point--however,due to this, the magnitude of curl should always be very small,as the path integral over an area as small al that would be near to zero!!
But you divide by the (near-zero) area as well. This is similar to how the derivative is defined. The difference f(x+h)-f(x) will also be nearly zero (for continuous functions) as h goes to zero, but you divide by h, and so you get something else for the limit.
 

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