Problem on divergence and curl

[heman]

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..

 

[vincentchan]

not only sphere, you can take a cube, a cylinder or anythinig you what...

 

[heman]

ok...and how will be explain divergence and curl if we take from differential point of view//

 

[dextercioby]

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.

 

[dextercioby]

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.

 

[vincentchan]

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 $$\nabla \cdot \vec{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec {S}} {\Delta V} =\frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z}$$?

 

[heman]

i was asking for the geometrical interpretation...

 

[pmb_phy]

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

 

[heman]

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..

 

[pmb_phy]

yeah ..and also pls explain geometrically..

That is a geometrical explanation.

Pete

 

[heman]

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..

 

[dextercioby]

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... The flux through these 0 and 1-D manifolds is still zero,but the normal cannot be defined...

Daniel.

 

[vincentchan]

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

 

[heman]

okay...its also not good to ask every minute thing..thx very much

 

[pmb_phy]

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... 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

 

[dextercioby]

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.

 

[pmb_phy]

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

 

[dextercioby]

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.

 

[pmb_phy]

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 ). 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

 

[dextercioby]

Okay,i didn't look at your formula...

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.

 

[reilly]

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

 

[heman]

if I'm looking at a graph of a vector field, how do I recognize positive divergence or negative curl, for example?

 

[heman]

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!