Divergence in spherical polar coordinates

[Idoubt]

I took the divergence of the function 1/r²\widehat{r} in spherical coordinate system and immediately got the answer as zero, but when I do it in cartesian coordiantes I get the answer as 5/r³.

for \widehat{r} I used (xi+yj+zk)/(x²+y²+z²)^1/2

what am i missing?

 

[IttyBittyBit]

What you're missing is the handling of the singularity at r=0.

 

[I like Serena]

Hi Idoubt!

If I take the cartesian derivative with respect to x I get:
{\partial \over \partial x}{x \over (x^2+y^2+z^2)^{3/2}} = {-2 x^2+y^2+z^2 \over (x^2+y^2+z^2)^{5/2}}
Of course the derivatives wrt to y and z have to be added yet...

How did you get 5/r³?

 

[Idoubt]

sorry I made a silly error in my calculation



\frac{\partial}{\partial x} x(x²+y²+z²)^-\frac{3}{2} =

(x²+y²+z²)^-\frac{3}{2} - 3x²(x²+y²+z²)^-\frac{5}{2}

When i add y & z parts it cancels and becomes zero, so the div of \frac{1}{r^2}\widehat{r} is zero.

The reason that confused me was Electric fields can have a non zero divergence. Is that because of the singularity at r=0?

 

[I like Serena]

The reason that confused me was Electric fields can have a non zero divergence. Is that because of the singularity at r=0?

Not every electric field is generated by exactly 1 symmetrical charge.
And in practice there is no singularity, since the charge would have to be a point charge for that.

 

[Idoubt]

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?

 

[I like Serena]

Yes. 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.

 

[Idoubt]

But when we take two charges we no longer have the \frac{1}{r^2} function ( unless we approximate at large distances)

It is my understanding that even the electric field of a point charge has non zero divergence is it not so?

 

[I like Serena]

Right on both counts. :)

My point about point charge is unrelated to divergence.
Only that a real electric field has no singularity.

 

[Idoubt]

my question is , how can a field which is scalar multiple of the function \frac{1}{r^2}\hat{r} have a divergence when i just proved that the divergence of this function is zero?

 

[vanhees71]

The divergence is not 0, it's 4 \pi \delta^{(3)}(\vec{x}). 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.

 

[Idoubt]

The divergence is not 0, it's 4 \pi \delta^{(3)}(\vec{x}). 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]

This by definition is a singularity isn't it?

In physics it's often modeled thusly AAMOI:

I know not relevant but...

 

[Idoubt]

In physics it's often modeled thusly AAMOI:

I know not relevant but...

greek to me :)

 

[Galron]

greek to me :)

meh a and b are just constants and in this case F(X) is obviously the Function of X.

The Greek symbol that looks like an O with a hat means Tensor.

E(X) I would assume means the Energy of the system in question in terms of the vector.

 

[Idoubt]

meh a and b are just constants and F(X) is obviously the Function of X.

E(X) being the Energy of the system.

greek with a few constants and a function of x now =)

 

[Galron]

greek with a few constants and a function of x now =)

If you've done matrices you'll know a tensor is expressed as a matrix of a matrix.

A vector in this case is simply a matrix with a scalar.

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. In the same vein, but in more geometric parlance, vectors representing displacements in the plane or in three-dimensional space also form vector spaces.

Vector spaces are the subject of linear algebra and are well understood from this point of view, since vector spaces are characterized by their dimension, which, roughly speaking, specifies the number of independent directions in the space. The theory is further enhanced by introducing on a vector space some additional structure, such as a norm or inner product. Such spaces arise naturally in mathematical analysis, mainly in the guise of infinite-dimensional function spaces whose vectors are functions. Analytical problems call for the ability to decide if a sequence of vectors converges to a given vector. This is accomplished by considering vector spaces with additional data, mostly spaces endowed with a suitable topology, thus allowing the consideration of proximity and continuity issues. These topological vector spaces, in particular Banach spaces and Hilbert spaces, have a richer theory.

Historically, the first ideas leading to vector spaces can be traced back as far as 17th century's analytic geometry, matrices, systems of linear equations, and Euclidean vectors. The modern, more abstract treatment, first formulated by Giuseppe Peano in the late 19th century, encompasses more general objects than Euclidean space, but much of the theory can be seen as an extension of classical geometric ideas like lines, planes and their higher-dimensional analogs.

Today, vector spaces are applied throughout mathematics, science and engineering. They are the appropriate linear-algebraic notion to deal with systems of linear equations; offer a framework for Fourier expansion, which is employed in image compression routines; or provide an environment that can be used for solution techniques for partial differential equations. Furthermore, vector spaces furnish an abstract, coordinate-free way of dealing with geometrical and physical objects such as tensors. This in turn allows the examination of local properties of manifolds by linearization techniques. Vector spaces may be generalized in several directions, leading to more advanced notions in geometry and abstract algebra.

http://en.wikipedia.org/wiki/Vector_space

So to steal more wiki imagery a tensor is:


[URL]http://upload.wikimedia.org/wikipedia/commons/4/45/Components_stress_tensor.svg[/URL]

 

[Idoubt]

hmm ok ok so that's a little advanced for me, but my question was about the divergence of the \frac{1}{r^2}\hat{r} function. In simpler terms can you tell me the reason for non-zero divergence in the electric field of a point charge?

 

[vanhees71]

This by definition is a singularity isn't it?

No, it's a distribution, the socalled Dirac \delta distribution. It's a very useful concepts. It's even so useful that the mathematicians built a whole new subfield of mathematics to define this concept rigorously, functional analysis!

 

[I like Serena]

I suspect you're referring to Maxwell's law that says:
\textrm{div }\vec E = \frac {\rho} {\epsilon_0}

The divergence of a spherically symmetric charge is zero everywhere except at the place where the charge actually is.
A point charge is a special case, since the charge density would be infinite (actually a Dirac delta-function).
In practice the charge would take in a certain volume.
The divergence of the electric field is not zero at the place where the charge density is not zero.

 

[Idoubt]

So can I go one more step and state that, at the points in space where the charge 'is' the electric field can no longer be defined by the function \frac{1}{r^2}\hat{r} ?

 

[I like Serena]

Correct.
In the case of a solid sphere with constant charge density within, the electric field is proportional to r \hat {\boldsymbol r} (inside the sphere).

 

[Idoubt]

thank you, this helped a lot