# Why is the divergence like that?

1. Feb 15, 2010

### LucasGB

Why is the concept of divergence defined to be the sum of the partial derivatives of the x, y and z components of a vector field E with respect to x, y and z, instead of being defined as the sum of the partial derivatives of E itself with respect to x, y and z? What would this operation I just defined measure, and why is it not defined nor employed?

2. Feb 15, 2010

### FallenRGH

Divergence is defined as such because were it the sum of the partial derivatives of E itself, that would yield a vector quantity as opposed to a scalar, and the direction portion would not be of use in describing the source or sink of a field at the point, not to mention it would not describe the same quantity. As far as what your operation measures, I do not know.

3. Feb 15, 2010

### Pengwuino

To add to that, I don't even think that would be a well defined operation. Del is a vector operator so it can't act on another vector.

4. Feb 15, 2010

### LucasGB

But what if we defined divergence to be the sum of the partial derivatives of the magnitude of E?

5. Feb 15, 2010

### FallenRGH

Well, take for example a velocity field of a fluid. What divergence measures is the amount of fluid that is exiting the infinitesimal volume vs. the amount that enters. In order to do this, we take the partial derivative of each component in it's respective direction(for example we take the partial of the i component, the unit vector in the x direction, with respect to x). This demonstrates the amount of fluid that flows through the volume in the direction of the field.

The operation you are describing really is a summation of the components of the gradient of the magnitude, which does not, at least at first glance to me, have any meaning.

6. Feb 15, 2010

### LucasGB

I think I get it now. Thanks!