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The divergence of an electric field due to a point charge according to Coulomb's law is zero. To resolve this, Dirac applied the concept of a deltafunction and defined it in an unrealistic way the function value is zero everywhere except at the origin where the value is infinity. However the concept was accepted and we became able to show that. Conclusion : The source of the electric field exists although its divergence is zero everywhere except at the source point.

In the case of the magnetic field we are yet to observe its source or sink. However, the zero divergence of this field implies that no magnetic charge exists and since we don't have any real magnetic monopole at hand, there is no question of finding the field at the source point. Isn't this a double standard?

## Del in cylindrical and spherical coordinates

Do we really need to find a non-zero divergence of a field for its source to exist? This become a lot clearer if you consider the integral forms of Maxwell's equations. Further more this behaviour where the value of an integral is given by the value of the integrand at a point is the definition of the Dirac delta. If you find this unsatisfying you can push back to the question of whether point charges actually exist, but this is an empirical, rather than theoretical question.

I Right, the differential form of Gauss's law. It is instead a generalized function. It is possible to give a mathematically consistent treatment of the Dirac delta distribution. To get a flavor of the various intricacies that can arise with distributions, the reader might find this Phys. SE post interesting. II To avoid the notion of distributionsit is more safe and probably more intuitive to work with the equivalent integral form of Gauss's law.

The corresponding Gauss's law for magnetism. Also we should mention the well-known fact that integration theory can be appropriately generalized from non-negative functions to complex-valued functions. Yet, no matter how you feel about the Dirac delta, where there is charge, there is non-zero divergence of the electric field.

And, conversely, where there is non-zero divergence, there is charge. Now, it is not the Dirac delta that is "unrealistic" it is a perfectly well defined distributionit is the concept of a "point charge".

Every charged thing we know has this charge distributed over a - however small - area of space, and the Dirac delta is a way to model that this area is so small that we don't care that its not point-like. And if there truly was a point-like charge, the Dirac delta would exactly describe its charge density - because the volume of a point is clearly zero, and whatever charge the thing has divided by zero is infinite.

Do not take this as a rigorous statement, this is as handwavy as it gets. There is nothing wrong with the Dirac delta as a charge or other density.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up. First, some preliminaries. To do this I'll use the chain rule. I will use that notation myself now. Now that we know how to take partial derivatives of a real valued function whose argument is in spherical coords.

But that part is just linear algebra! Now that everything is set up, we can calculate the divergence. But what is the divergence? What I mean is, how do we write it as an abstract object that acts on functions? Try to convince yourself why the above formula is so.

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Now just substitue all of the expressions we just derived for the basis vectors, and the differential operators. Now, construct the corresponding surface integral from the divergence theorem. Your surface will have six smooth pieces like a cube, but the surfaces are curved, as they follow the coordinate lines.

This is a computation for two of the six faces of this not-exactly-cube-shaped surface. You can also use a symmetric integration volume instead, if you're comfortable keeping around factors of 2. Indeed, this process is sometimes used as a definition of the divergence and other differential operations. Doing so makes the fundamental theorem of calculus divergence theorem, etc. Convert it to spherical coordinates, then take the exterior derivative. Sign up to join this community.Divergence of the vector field is an important operation in the study of Electromagnetics and we are well aware with its formulas in all the coordinate systems.

Generally, we are familiar with the derivation of the Divergence formula in Cartesian coordinate system and remember its Cylindrical and Spherical versions intuitively. This article explains the step by step procedure for deriving the Deriving Divergence in Cylindrical and Spherical coordinate systems. If more and more field lines are sourcing out, coming out of the point then we say that there is a positive divergence.

Ancient hebrew bible pdfWhile if the field lines are sourcing in or contracting at a point then there is a negative divergence. The uniform vector field posses a zero divergence. Go through the following article for complete discussion of the Divergence of the vector field. What is the Divergence of vector field? The divergence formula in cartesian coordinate system can be derived from the basic definition of the divergence. Go through the following article for intuitive derivation.

The formulas of the Divergence with intuitive explanation! Later by analogy you can work for the spherical coordinate system. Where do they come from? What is the logic behind them. Because thinking intuitively, one might expect the formula similar to cartesian one. So one can think of getting partial derivatives w. But it is not like that. The answer for this can be found in the steps for deriving the divergence in cylindrical system.

Actually we can reach up to the result by two approaches. In this approach, you start with the divergence formula in Cartesian then convert each of its element into the cylindrical using proper conversion formulas.

Possible unhandled promise rejection webviewBut, for deriving Divergence in Cylindrical and Spherical, I am going to explain with another approach discussed below. If I take the del operator in cylindrical and dotted with A written in cylindrical then I would get the divergence formula in cylindrical coordinate system.

Crj 200Read here: — How to convert the Del operator from Cartesian to Cylindrical? Be careful. This is not the simple dot product consisting of the multiplication of the respective components. You have to take an account of the derivative. For example let us consider the first term. Because cylindrical and spherical unit vectors are not universally constant.

Though their magnitude is always 1, they can have different directions at different points of consideration. So unlike the cartesian these unit vectors are not global constants.

I am going to cover the derivatives of the unit vectors in the independent article. Which is our required divergence operator in cylindrical. Similar steps can be followed for deriving the Divergence in Spherical. Tag: Electromagnetism.By using our site, you acknowledge that you have read and understand our Cookie PolicyPrivacy Policyand our Terms of Service.

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It only takes a minute to sign up.

First, some preliminaries. To do this I'll use the chain rule. I will use that notation myself now. Now that we know how to take partial derivatives of a real valued function whose argument is in spherical coords.

But that part is just linear algebra! Now that everything is set up, we can calculate the divergence. But what is the divergence? What I mean is, how do we write it as an abstract object that acts on functions?

Try to convince yourself why the above formula is so. Now just substitue all of the expressions we just derived for the basis vectors, and the differential operators.

Now, construct the corresponding surface integral from the divergence theorem. Your surface will have six smooth pieces like a cube, but the surfaces are curved, as they follow the coordinate lines. This is a computation for two of the six faces of this not-exactly-cube-shaped surface. You can also use a symmetric integration volume instead, if you're comfortable keeping around factors of 2. Indeed, this process is sometimes used as a definition of the divergence and other differential operations.

Side view of hip bone diagram diagram base website boneDoing so makes the fundamental theorem of calculus divergence theorem, etc. Convert it to spherical coordinates, then take the exterior derivative. Sign up to join this community. The best answers are voted up and rise to the top.

Home Questions Tags Users Unanswered. Derivation of divergence in spherical coordinates from the divergence theorem Ask Question.

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### The Divergence

Active Oldest Votes.Hot Threads. Featured Threads. Log in Register. Search titles only. Search Advanced search…. Log in. Forums Physics Classical Physics. JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding. Divergence in spherical polar coordinates. Thread starter Idoubt Start date Aug 18, Related Classical Physics News on Phys.

I like Serena Homework Helper. Hi Idoubt! The reason that confused me was Electric fields can have a non zero divergence. Idoubt said:. Last edited: Aug 19, Isn't the field due to just one charged assumed to be a field created by a point charge? I mean it makes no difference if we make that assumption right? But, how can an electric field have a non zero divergence then?

It would only make a difference when you get very close to where the point charge is supposed to be.

Take for instance 2 point charges and you'll have non-zero divergence. Right on both counts. Only that a real electric field has no singularity. Insights Author. Gold Member. It's not too tricky to prove with help of Green's first integral theorem applied to the whole space with a little sphere around the origin taken out and then taking its radius to 0.

This by definition is a singularity isn't it? Galron said:. E X being the Energy of the system. A vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied "scaled" by numbers, called scalars in this context.

Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex numbers, rational numbers, or even more general fields instead. The operations of vector addition and scalar multiplication have to satisfy certain requirements, called axioms, listed below. An example of a vector space is that of Euclidean vectors which are often used to represent physical quantities such as forces: any two forces of the same type can be added to yield a third, and the multiplication of a force vector by a real factor is another force vector.Remember Me?

What's New? Results 1 to 18 of Thread: HW help: divergence of electric field. Thread Tools Show Printable Version. Join Date Nov Posts HW help: divergence of electric field "the electrostatic field of a point charge q iswhere e r is the radial unit vector. Find the divergence of E away from the origin, which is the point source of the field.

Invitation letter for exhibitionJoin Date Oct Posts 5, Originally Posted by Roobydo. They can't both be true. Originally Posted by macaw. You need to consider two things: and.

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Yes, but I would prefer not to change everything to cartesian coordinates; see below. I would also prefer to work with the unitized radial vector. I see that now that I have three components. But intuition tells me that I shouldn't need three: I know the field is spherically symmetrical about the origin, so the divergence should be a function of radius alone. No, it isn't. If you insist on working in polar coordinates, then you need to remember that: where.

If you're going to give me equivalences to remember, I ask that you tell me how you arrived at them. If you work the problem in cartesian coordinates you should get the same answer. Join Date Oct Posts 27, Indeed, Gauss' law is the statement that charges are what create a local divergence of the field, so surround the charges with a "Gaussian pillbox" that conforms to the symmetry of the charge distribution to get the field that threads the surface of the pillbox.

But that's how you get the field, which is what you usually want to know-- this problem asks you to demonstrate that you get that field by integrating a divergence that here is nonzero only at the origin. You can also see that by using two spheres of different radius, both centered at the origin. That means there is zero divergence of the field between the spheres. Since anywhere but the origin could be "between the spheres", everywhere but at the origin the divergence is zero.

Yes, the divergence is infinite at the origin in this idealized charge distribution. Where does come from? From herefor example. If you study physics, you should have had some advanced calculus.Home When you describe vectors in spherical or cylindric coordinates, that is, write vectors as sums of multiples of unit vectors in the directions defined by these coordinates, you encounter a problem in computing derivatives.

The unit vectors themselves change as you change coordinates, so that the change in your vector consists of terms arising from the changes of the multiples and also those from changes in the unit vectors. We can find neat expressions for the divergence in these coordinate systems by finding vectors pointing in the directions of these unit vectors that have 0 divergence.

Then we write our vector field as a linear combination of these instead of as linear combinations of unit vectors. By the product rule, the expression for the divergence we seek will be a sum over the three directions of the dot product of one of these vectors with the gradient of its coefficient.

The second terms in the product rule will all be 0.

You may very well encounter a need to express divergence in these coordinates in your future life, so we will carry this approach out with spherical coordinates. First please notice: whenever you differentiate functions in polar coordinates you must treat the origin in them separately and carefully.

The coordinates themselves are singular there! Often it will NOT hold at the origin of your coordinates. Such caviats are omitted below but you should assume that they are present whenever differentiation by a polar parameter is involved. The only non-trivial step in doing this is finding vectors in the various required directions that have 0 divergence. This can be done by finding the divergence of any vectors in these directions and figuring out what multiple you need apply in each case to cancel its divergence out, again using the product theorem for divergence.

The vector x, y, z points in the radial direction in spherical coordinates, which we call the direction. Its divergence is 3. It can also be written as or as. A multiplier which will convert its divergence to 0 must therefore have, by the product theorem, a gradient that is multiplied by itself.

The function does this very thing, so the 0-divergence function in the direction is. Exercise What function of r should you multiply it by to get a vector with divergence 0? The vector -y, x points in the direction and has 0 divergence already. It can be written as. The direction is normal to both of these and you can get a vector in it by taking the cross product of -y, x, 0 and x, y, zwith result xz, yz, -r 2.

This vector has divergence 2z, and the form rz u r - r 2 u z. It is the first of these two terms, rz u r which is xz, yz that has the non-vanishing divergence and it is the x and y which lead to it, not the z factor. We can invoke the result of the last exercise to introduce a multiplier that will get rid of that divergence.

The resulting vector has the form whose length is It can therefore be written as. To summarize, the vectors have 0 divergence. If we define these combinations to be respectively, a vector of the form is also writable as. In the last line here we used the form of the gradient in spherical coordinates : recall that is a polar variable with radius r and is a polar variable with radius. Also recall that r is sin. This last expression is not very pretty but it is quite important in physical applications.

It appears in particular in the combination which is called the Laplacian of f.

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