# Physical Meaning Of Divergence

1. Aug 18, 2014

### AbhiFromXtraZ

Currently I'm reading Magnetostatics. While reading Divergence of B, I fell into a confusion that what divergence really means w.r.t. a coordinate system.
Supposed there is a current distribution J at r' w.r.t. some primed coordinate system. And B is defined at position r w.r.t. unprimed coordinate system.
Now if we integrate (Del.B ) over a volume enclosing the whole current distribution w.r.t. the unprimed coordinate system, then it means the total outward flux over the volume.
Upto this line I have no problem. My confusion arises at the following.
WHAT (Del.B ) MEANS (no integration over any volume)??
Here "Del" and B is defined from unprimed coordinate system and inside B, J is from primed coordinate and R (in B ) is equal to (r - r').
If it means flux over an infinitesimal unit volume then what is the position of that volume.

Thanks.

2. Aug 18, 2014

### vanhees71

I don't know, what you mean by the different coordinate systems. It's wise to study a problem in one and only one reference frame. Some textbooks state the Maxwell Laws in different referent frames in the same formula. That's bad, because hinders the understanding of the physics. You should study electromagnetism in a fixed inertial reference frame first. Then you should be aware that Maxwell electromagnetism is a relativistic theory, and that all quantities have a well-defined meaning in terms of Minkowski-space tensors (including scalars and vectors as special cases, of course).

Usually electrodynamics is treated in terms of usual Euclidean 3D vector calculus. That's fine, but only if you are aware that this form is valid for a fixed inertial reference frame.

Now, the divergence of a vector field is defined as a limiting process for a surface integral. Now take a volume around a point $\vec{x}$ in space with boundary $\partial V$. The boundary is, by definition, oriented such that its normal vectors $\mathrm{d}^2 \vec{f}$ all point out of the volume. Then the divergence of the vector field is defined as
$$\vec{\nabla} \cdot \vec{B}(\vec{x})=\lim_{V \rightarrow \{\vec{x} \}} \frac{1}{V} \int_{\partial V} \mathrm{d} \vec{f} \cdot \vec{B}.$$
As you see, this expression is independent of the coordinate system you choose, it's a scalar (in the sense of usual 3D Euclidean vectors; you still have to stick to a fixed inertial reference frame). It's easy to show that for Cartesian coordinates, the divergence is given by
$$\vec{\nabla} \cdot \vec{B}=\frac{\partial B_1}{\partial x_1}+\frac{\partial B_2}{\partial x_2}+\frac{\partial B_3}{\partial x_3}.$$
That's where the notation of the nabla symbol comes from. In Cartesian coordinates it reads
$$\vec{\nabla}=(\partial_1,\partial_2,\partial_3) \quad \text{with} \quad \partial_j=\frac{\partial}{\partial x_j}.$$
Now the physical meaning of the divergence becomes clear: Interpret the vector field as a flow field. Then $\mathrm{d}^2 \vec{f} \cdot \vec{B}$ is the amount of the corresponding flowing quantity that runs through the area element $\mathrm{d} \vec{f}$, with the sign defined by the chosen direction of this area element. Then you integrate over a closed surface. This gives the total amount of the quantity flowing through this closed surface (per unit time). Then you divide by the volume and take the limit. This gives you a local flux of the quantity per unit volume and per unit time.

Now for the magnetic field $\vec{B}(t,\vec{x})$ there's one fundamental law, Gauss's Law for the magnetic field (which in fact is one of the Maxwell equations):
$$\vec{\nabla} \cdot \vec{B}=0.$$
That means that there is no net flow of magnetic flux through any closed surface: From the definition of the divergence, immediately Gauss's integral theorem follows by integrating up all infinitesimal volume elements of the big volume and applying the definition of the divergence for infinitesimal volumes:
$$\int_{V} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{B}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{B}.$$
But according to Gauss's Law for the magnetic field, this is 0, i.e., there are no "sources" of flow for the magnetic field. In other words there are no magnetic charges (monopoles) in nature. So far, indeed nobody has found any!

You should be aware that from Gauss's Law you don't learn anything about the electric current. For this you need the Ampere-Maxwell Law. It's easier to understand for stationary (i.e., time independent) magnetic fields and current densities. Then it simplifies to the simple Ampere Law,
$$\vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j} \quad \text{stationary fields and currents only!}.$$
To understand it, you should look up the definition of the operator $\mathrm{curl}=\vec{\nabla} \times$ in terms of line integrals. Then also this law becomes pretty clear.

3. Aug 18, 2014

### AbhiFromXtraZ

What you have told just ran over my head. Ok let me simplify my confusion for you.
This time I shall introduce only one coordinate system. And for my convenience I'm talking about electric field E instead of B.
Suppose there is a charge distribution at origin O. The field for this distribution at position $\vec{r}$ is $\vec{E}$.
Now what do you mean by (Del.E)?
No volume is involved.

Last edited: Aug 18, 2014
4. Aug 18, 2014

### mal4mac

This thread may be useful, especially the last post:

So the divergence (d) of E is its rate of change in the direction r. (That's exactly what the equation says!) Note that d is a scalar quantity. So if d is high then E is changing quickly at r, in the direction of r.

5. Aug 18, 2014

### AbhiFromXtraZ

I think this is called Gradient.
Divergence is the measure of outward flux of a vector from a volume. i.e. Diverging of a vector.
http://en.m.wikipedia.org/wiki/Divergence

6. Aug 18, 2014

### Staff: Mentor

Take the limit of that as the volume goes to zero. Then you have divergence is the measure of outward flux from a point.

7. Aug 18, 2014

### AbhiFromXtraZ

Which point ?? My purpose was only to ask "Which Point?"
At origin O or at the point P whose position vector is $\vec{r}$?

If the infinitesimal volume be at P then (Del.E) must be zero because this volume is not enclosing any charge.
If you say that volume is at O then why it is necessary to define the field E at position $\vec{r}$ from O?

8. Aug 18, 2014

### Staff: Mentor

You can evaluate $\nabla \cdot E$ at every point. It gives you a scalar field representing the outward flux at each point.

9. Aug 18, 2014

### mal4mac

No Gradient is a vector, note I did not say d had a direction, I stressed it was a scalar! Meir Achuz's definition, from the thread I mentioned, seems to fit the facts, but maybe I am missing something...

10. Aug 18, 2014

### Staff: Mentor

The gradient of a scalar field is a vector field. The divergence of a vector field is a scalar field.

11. Aug 18, 2014

### AbhiFromXtraZ

You are surely missing something.
You said "d" is scalar, but "E" is a vector... so the resultant is a Vector... But resultant of Divergence is scalar.
Divergence is defined as the dot product of Del(vector) and other vector quantity.

12. Aug 18, 2014

### AbhiFromXtraZ

Well after thinking for a while, I came to the conclusion that $\nabla \cdot E$ can be evaluated at every point as you said.
I don't know how you would explain this. My explanation is as follows..
$\nabla \cdot E$ integrated over a volume means the flux for the total charge enclosed by the volume (actually the volume integral involves only the volume of the charge distribution ).
While $\nabla \cdot E$ (no volume ) means the flux for the infinitesimally small charge distribution of unit volume.
In both the case $\nabla \cdot E$ is defined at P whose position vector is r.

13. Aug 18, 2014

### Staff: Mentor

I think that is what I said in post 6.

One thing that I didn't mention in 6 which may be helpful is to think of the divergence as a flux density. Even if the enclosed charge is infinitesimal as the bounding volume goes to zero the flux density can be finite.

14. Aug 18, 2014

### vanhees71

Have a look at my posting again. This is the fundamental definition of the divergence of a vector field, which gives a scalar field at any point. Without understanding the classical vector operations, gradient of a scalar field, divergence and curl of a vector field and the associated integral theorems by Gauss, Stokes, and Green, you'll have no chance to anderstand hydrodynamics or electrodynamics! Together with vector (linear) algebra, vector calculus is one of the most important topics in the introductory math lectures you can get! You find very good summaries in many textbooks, e.g.,

A. Sommerfeld, Lectures on Theoretical Physics, vol. II (Hydrodynamics)
M. Abraham, R. Becker, The Classical Theory of Electricity and Magnetism

A. Zangwill, Modern Electrodynamics

The latter book is a marvelous treatise on classical electromagnetism.

Of course, the divergence also occurs for the electric field:

One of the fundamental laws of electrodynamics is Gauss's Law for the electric field,
$$\vec{\nabla} \cdot \vec{E}(t,\vec{x})=\rho(t,\vec{x}),$$
where $\rho$ is the charge density (i.e., electric charge per unit volume at the point $\vec{x}$ at time $t$). This is, of course, a scalar field.

In integral form, which you get by using Gauss's integral theorem, you find
$$\int_{V} \mathrm{d}^3 \vec{x} \vec{\nabla} \cdot \vec{E}=\int_{\partial V} \mathrm{d}^2 \vec{f} \cdot \vec{E}=\int_V \mathrm{d}^3 \vec{x} \rho = Q_{\text{inside V}}.$$
It tells you that the flux of the electric field through a closed volume $\partial V$ (the boundary surface of the volume $V$) equals the total electric charge contained in the enclosed volume. In other words: The sources of electric fields are electric charge distributions.

15. Aug 18, 2014

### AbhiFromXtraZ

Ok...Got it..
Thanks to all for being with me.

16. Jan 18, 2015

### RajagopalanNair

Your Question 1: What divergence really means w.r.t. a coordinate system?

My Answer: Divergence of vector results in a scalar. The same being defined at any arbitrary point and the divergence of a vector being a scalar, it remains same for any coordinate frame. A scalar always implies a quantity remaining same for all coordinate frames.

Your statement: Supposed there is a current distribution J at r' w.r.t. some primed coordinate system. And B is defined at position r w.r.t. unprimed coordinate system.
Now if we integrate (Del.B ) over a volume enclosing the whole current distribution w.r.t. the unprimed coordinate system, then it means the total outward flux over the volume.

My comment on your statement: Integration of Del.B (= divergence ofB) over a volume in space means the total outward flux through out the surface enclosing the volume. However, divergence of B (B being magnetic flux density) integrated will be always zero. In analogy to divergence of electric field integrated over a volume (enclosed by a surface over which the associated flux integrated remains equal to the mentioned integral of Del.E) equaling the [net electric charge/ (permittivity of vacuum)], Del.B being always zero is indicative of magnetic mono poles being absent. However, Del.B never give the information on the current generating it, though B separately does.It is curl of B that is proportional to current density. However, it is an area enclosed by a closed curve over which curl B is integrated and this will be equal to the line integral of B evaluated along the closed curve enclosing (bordering) the surface that is associated with integral of curl B integrated over the mentioned surface. Again, the value of the mentioned integral is associated with the net current flowing across the area mentioned, in analogy with the previously discussed volume integral of Del.B remaining associated with any magnetic monopole (which is invariably zero) [Also note that the surface(on which the curl B is integrated), as enclosed by a closed curve can be of any arbitrary shape (in our 3 D space with the only restriction regarding the curve being its border) whereas in our normal 3D space, the volume enclosed by a given closed surface remains fixed by volume as well as shape].
Summary: Linking divergence of B with current elements enclosed is entirely wrong - they are not related.

Your Question 2: WHAT (Del.B ) MEANS (no integration over any volume)??

The meaning of (Del.B) at a given point comes from the fact that (Del.B) multiplied by an elementary volume (small volume enough that Del.B remains uniform) becomes the total flux flowing out of the microscopic surface enclosing it (note that the surface integral of (flux of B) still remains expressed as an integral and cannot be expressed as a simple product of the area of the surface and the flux, however microscopic the area becomes (because, flux means at a point on the surface scalar product of B with the unit normal vector to the surface at the point and the unit normal vector varies in its direction from 0 to 2*pi radians along all the planes containing the vector B, which spans the elementary surface enclosing the elementary volume)).
This means, Del.B defined at a point in space is equal to the ratio of (integral of flux of B taken over the area of a microscopic surface enclosing the point) to the (microscopic volume contained by the microscopic surface); the smallness of the area and the volume being such that both B (mind you, not the flux of B over the elementary surface) and Del.B remains uniform (a mathematician will say limit of the volume tending to zero).
Any proposal to the meaning of div B simpler than this can be a mistake.

Your Question 3: Here "Del" and B is defined from unprimed coordinate system and inside B, J is from primed coordinate and R (in B ) is equal to (r - r').
If it means flux over an infinitesimal unit volume then what is the position of that volume.

My interpretation/comment of your question: You meant to say B as a vector field is expressed as a function of position vectors of various points as defined in the primed coordinate system and so also is the current density J (as a vector). Next, it is not clear what do you mean by 'R(=(r-r')) in B.' It is also not understood what it means by 'infinitesimal unit volume' - there is no such thing in general, especially with the types of permanent magnets/electromagnets used in a lab. However, if B as in the case of Earth's magnetic field showing any appreciable deviation only over a few thousand kms, a unit volume can be Infinitesimally small, provided B does not vary much (both as per direction as well as magnitude, since it is a vecto
r
) and within a region of unit volume, bound by surface also not having any of its dimensions too large (be careful, in the case of Earth's magnetic field a cube of 1 cubic meter may be small enough; however, if you think of a cylinder of 10000 km length and radius 0.117 mm as the shape enclosing the volume, B is not constant as a vector all along).
My Answer (if I interpreted the question correctly):Flux flowing out of an infinitesimally small volume (shape of the volume chosen in such a way that none of the dimensions of the shape enclosing the volume is too large too) is independent of its position vector. This means the question on the position vector of the elementary volume considered does not matter to the flux flowing out of the volume. Here, therelative orientationbetween the magnetic field, and unit normal vectors along various points of the surface enclosing the concerned elementary volume are invariant and so also the total flux flowing out of the elementary volume. Therefore, the evaluation of flux over the elementary surface enclosing elementary volume is in an affine vector space (the space in which the vectors are not having their end points fixed). The divergence of B being a scalar (remaining invariant of the coordinate frame chosen) linked to the mentioned flux by Gauss divergence theorem, the same becomes independent of its position vector. It is much the same way as the amount of water flowing out of a tap does not depend on the position vector of the tap as seen by observers keeping their position as the origin.
Note: I am available in face book too.

17. Jan 18, 2015

### RajagopalanNair

Some excellent reference(for beginners)s for learning the meaning of vector operators used in Maxwell's equations;
(1) Berkeley Physics Course - vol. 2 by E M Purcel
(2) Introduction to Electrodynamics by D. J Griffiths
After reading from these two books, if you want to be an expert, read
(3) Classical Electrodynamics by J D Jackson
Recently there is a book
(4) Introduction to Electrodynamics by Capriati and Panat (if you read its Indian edition, be careful about the printing mistakes (authors - not responsible), otherwise this book is much lauded).

18. Feb 3, 2015

Thanks