Problem on divergence and curl

  • Thread starter Thread starter heman
  • Start date Start date
  • Tags Tags
    Curl Divergence
Click For Summary
Divergence and curl are fundamental concepts in vector calculus that describe the behavior of vector fields. The divergence measures the rate at which "density" exits a point in a field, indicating sources or sinks, while curl measures the rotation or twisting of the field around a point. These concepts are crucial in applications like electromagnetism and fluid dynamics, simplifying complex equations by reducing the number of variables involved. For instance, without divergence and curl, Maxwell's equations would become significantly more complex, requiring many more variables and equations. Understanding their significance helps in grasping physical phenomena and solving related problems effectively.
  • #31
pmb_phy said:
I can't make sense out of that. It sounds like you quoted part of a book and without the context (there is a lot that isn't said that there which is probably made clear in diagrams and previous paragraphs and shown equations etc) and as such it gave me a headache reading it. :-p

The divergence of a vector field is simply this - Suppose there is a vector field E. Choose a point r in that field and construct a closed surface, e.g. a sphere, around that point. Take the surface integral of the normal component of E over the closed surface and divide the result by the volume enclosed by that surface. Now let the radius of the spherical surface go to zero. That result is called the "divergence of E."

The curl is kind of like that - Take a point in space and pass a plane through it whose surface normal is parallel to the z-axis. In that plane construct a cirlce whose center is the point of interest. Take the line integral of the vector field around that circle. Let the sense of the line integral be consistent with the right hand rule with the +direction of the z-axis (if you grab the z-axis and your thumb is pointing in the +direction then the direction of the integration around the cirlce will be in the direction your fingers are curling). Now divide the result of that integral by the surface area of the circle. Take the limit and let the radius of the circle go to zero. That gives a number which is called the "z-component of the curl of the vector field E at the point." Do that with two other planes whose surface normals are in the +x and +y direction and you have the components of the curl vector.

Pete

Very nicely defined. I can visualize it precisely with your definition... just have a few questions that I would like to have clarified.

I am just wondering for the divergence, what is the purpose of dividing the closed surface by the volume... is it to find an average outward projection within each point of the enclosed surface?

And simularly for the curl... just the average projection of the xy vector components in the xy plane for the z-component of the curl?

Thanks

-CronoSpark
 
Physics news on Phys.org
  • #32
pmb_phy said:
I can't make sense out of that. It sounds like you quoted part of a book and without the context (there is a lot that isn't said that there which is probably made clear in diagrams and previous paragraphs and shown equations etc) and as such it gave me a headache reading it. :-p

The divergence of a vector field is simply this - Suppose there is a vector field E. Choose a point r in that field and construct a closed surface, e.g. a sphere, around that point. Take the surface integral of the normal component of E over the closed surface and divide the result by the volume enclosed by that surface. Now let the radius of the spherical surface go to zero. That result is called the "divergence of E."

The curl is kind of like that - Take a point in space and pass a plane through it whose surface normal is parallel to the z-axis. In that plane construct a cirlce whose center is the point of interest. Take the line integral of the vector field around that circle. Let the sense of the line integral be consistent with the right hand rule with the +direction of the z-axis (if you grab the z-axis and your thumb is pointing in the +direction then the direction of the integration around the cirlce will be in the direction your fingers are curling). Now divide the result of that integral by the surface area of the circle. Take the limit and let the radius of the circle go to zero. That gives a number which is called the "z-component of the curl of the vector field E at the point." Do that with two other planes whose surface normals are in the +x and +y direction and you have the components of the curl vector.

Pete

..but still 1 thing in case of divergence how to decide the normal compnent becoz we have considered an apherical surface and this was from urs integral point of view..can u explain tthat from differntial point of view too...the previous one was well explained .thx very much..
 
  • #33
It's coming from here
\oint\oint_{\Sigma} \vec{E}\cdot \vec{n} \ dS

You see,that "vec{n}" is the NORMAL UNIT VECTOR EXTERIOR TO THE CLOSED (ARBITRARY,YET ORIENTABLE) SURFACE SIGMA.Taking the scalar product of this unit vector with the vector field,the normal component of the vector field is the only that stands,because the tangential one (which forms a right angle with the (exterior) normal to the surface)) does not give any contribution.

That scalar product is essential...

Daniel.
 
  • #34
but how do we really decide the surface...on which we do surface integration..and that normal is perpendicular to it..
 
  • #35
I believe i inserted the word ARBITRARY...Orientable (as to be able define the vector field of unit vectors normal to the surface in each point of the surface) and closed.Nothing more.

The fact that this surface (denoted by me with \Sigma) is ARBITRARY is very useful...

Daniel.
 
  • #36
ahh is it...we take an spherical surface..take any arbitrary planar surface and take its normal and then take the dot product with the field and compute ...thx clear now..
 
  • #37
not only sphere, you can take a cube, a cylinder or anythinig you what...
 
  • #38
ok...and how will be explain divergence and curl if we take from differential point of view//
 
  • #39
vincentchan said:
not only sphere, you can take a cube, a cylinder or anythinig you what...

Does the word SMOOTH mean anything to you (mathematically speaking,of course)??How would you define the exterior normal in the 8 corners of the cube and on its 12 sides??

Daniel.
 
  • #40
heman said:
ok...and how will be explain divergence and curl if we take from differential point of view//

In this case,"explanation"----------->"definition"...You define the divergence as the contracted tensor product between the derivative operator nabla and the vector field.And the curl as the cross product between the same diff.operator and the vector field...

Daniel.
 
  • #41
dextercioby said:
Does the word SMOOTH mean anything to you (mathematically speaking,of course)??How would you define the exterior normal in the 8 corners of the cube and on its 12 sides??

what is the total surface area for the 8 corners and 12 sides?
do I have to show you the proof of \nabla \cdot \vec{F} = \frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z} when I apply \nabla \cdot \vec{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec {S}} {\Delta V} to a cube?

EDIT...heman...differential form? you mean you want me to show \nabla \cdot \vec{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec {S}} {\Delta V} =\frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z}?
 
Last edited:
  • #42
i was asking for the geometrical interpretation...
 
  • #43
dextercioby said:
Does the word SMOOTH mean anything to you (mathematically speaking,of course)??How would you define the exterior normal in the 8 corners of the cube and on its 12 sides??

Daniel.
You don't. To find the flux you evaluate over the flat surfaces of the sides of the cube.

Go ahead. Give it a try. Do exactly this. You'll find that the answer you get is

\nabla \bullet E = \frac{\partial E_x}{\partial x} i + \frac{\partial E_y}{\partial y} j + \frac{\partial E_z}{\partial z} k

In fact this is how this expression is usually derived, i.e. by using a cube. I'm surprised that you've never seen this derivation ... or have you? If not then see div grad curl and all that - 3rd ed, by H.M. Shey, pages 36-40. Most Calculus texts do this same derivation as I recall.

Pete
 
  • #44
vincentchan said:
what is the total surface area for the 8 corners and 12 sides?
do I have to show you the proof of \nabla \cdot \vec{F} = \frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z} when I apply \nabla \cdot \vec{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec {S}} {\Delta V} to a cube?

EDIT...heman...differential form? you mean you want me to show \nabla \cdot \vec{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec {S}} {\Delta V} =\frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z}?

yeah ..and also pls explain geometrically..
 
  • #45
heman said:
yeah ..and also pls explain geometrically..
That is a geometrical explanation.

Pete
 
  • #46
pmb_phy said:
That is a geometrical explanation.

Pete

i mean to say ..doesn't we explain in any specific way like we did by view of integral calculus and take the surface do integral and apply limit...and it will be nice to see how does that come..
 
  • #47
pmb_phy said:
You don't. To find the flux you evaluate over the flat surfaces of the sides of the cube.

Go ahead. Give it a try. Do exactly this. You'll find that the answer you get is

\nabla \bullet E = \frac{\partial E_x}{\partial x} i + \frac{\partial E_y}{\partial y} j + \frac{\partial E_z}{\partial z} k

In fact this is how this expression is usually derived, i.e. by using a cube. I'm surprised that you've never seen this derivation ... or have you? If not then see div grad curl and all that - 3rd ed, by H.M. Shey, pages 36-40. Most Calculus texts do this same derivation as I recall.

Pete

I've seen it...In my 10-th grade and i still remember it.However,it defines the flux through the LATERAL SURFACES OF THE CUBE/RECTANGULAR PARALLELIPIPED.And yet my problem is unswered.In the corners and on the sides u cannot define normals to the surface... :rolleyes: The flux through these 0 and 1-D manifolds is still zero,but the normal cannot be defined...

Daniel.
 
  • #48
even the normal @ the corner is undefine... you can still take the divergence of a cube... so why are you bring it up here?



i mean to say ..doesn't we explain in any specific way like we did by view of integral calculus and take the surface do integral and apply limit...and it will be nice to see how does that come..

the maths is straight forward and easy, and most vector calculus textbook has it... no one will waste his time to type it here, why don't you google a little bit and see if you have luck
 
Last edited:
  • #49
okay...its also not good to ask every minute thing..thx very much
 
  • #50
dextercioby said:
I've seen it...In my 10-th grade and i still remember it.However,it defines the flux through the LATERAL SURFACES OF THE CUBE/RECTANGULAR PARALLELIPIPED.And yet my problem is unswered.In the corners and on the sides u cannot define normals to the surface... :rolleyes: The flux through these 0 and 1-D manifolds is still zero,but the normal cannot be defined...

Daniel.
Perhaps you can find your answer in a question. Ask yourself how you can calculate the flux through the entire surface of the cube and yet you can't define it. If you can answer that question to your own satisfation then you have the answer to your question.

Pete
 
  • #51
I can define the flux but not with a scalar product.Or not with the unit vector...

So the question is:must the manifold be smooth or not?

Daniel.
 
  • #52
dextercioby said:
I can define the flux but not with a scalar product.Or not with the unit vector...

So the question is:must the manifold be smooth or not?

Daniel.
Choose a criteria to which you wish to define what you're looking for. If you can find the flux then is that adequate for the purpose which you are interested in?

Pete
 
  • #53
No,Pete,u missunderstood me...I have hothing against the idea & the definition of a flux of a vector field...It's just that in the case of a nonsmooth manifold (like the cube) it doesn't make any sense,unless u consider the cube without its corners and sides...

Daniel.
 
  • #54
dextercioby said:
No,Pete,u missunderstood me...I have hothing against the idea & the definition of a flux of a vector field...It's just that in the case of a nonsmooth manifold (like the cube) it doesn't make any sense,unless u consider the cube without its corners and sides...

Daniel.
No, Daniel, I did understand you. My response was intended to say that when you used the term "must" that one has to have a definition in hand before answering your question. So you must first choose a definition for the divergence and then you can address your question.

Let me give you an example; suppose we choose the definition for divergence above (flux per unit volume) and also demand that the surface normal exist at all points on the surface for the diverence to be defined. Then the answer to your question is yes, it must. However, suppose that we define the divergence as the sum that I gave above (i.e. as expressed in Cartesian form - div E = parial E/parial x + etc... oops! I made a mistake in that definition - I gave the gradient :redface: ). Then the answer too your question is no. And I've seen a ton of places/texts which define the divergence in this fashion. E.g. Kaplan's "Advanced Calculus" text.

Pete
 
  • #55
Okay,i didn't look at your formula... :-p

The divergence has indeed a differential involving definition...The part with the flux comes just as an application to the Gauss-Ostrogradski formula and,by considering the scalar product in the surface integral,it restrains the applicability.


Daniel.
 
  • #56
When it's important to worry about edges, or points with surface integrals, then the best idea is to take the limit of a problem you can solve -- like a sphere -> cube. You can convince yourself that the contributions to flux from lines or points is zero -- unless there are some very singular field characteristics. The area of a line is rather small, so it takes a big field to drive some flux through a line. Not to worry, cubes are just fine.

regards,
Reilly Atkinson
 
  • #57
if I'm looking at a graph of a vector field, how do I recognize positive divergence or negative curl, for example?
 
  • #58
Hi,
I was revising my Electrodynamic notes and i came up with some queries!

Why can't an Electric Field rotate or act like a whirlpool!
I know Mathematically it is Zero but what will be the physical explanation of this??

and one more thing while deriving the differential form of Gauss Law,we shrink the body to differential element ,obviously its volume decreases but what happens to the charge!
Does all the charge concenterate in that small differential element or we cut the body and keep on removing its contents till we reach differential element!
 

Similar threads

Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
9K
Replies
1
Views
1K
Replies
17
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
35K