Difference between $\nabla T$ and $\nabla \cdot E$

Question

Why is $\nabla T = (\frac{\delta T}{\delta x},\frac{\delta T}{\delta y},\frac{\delta T}{\delta z})$, but $\nabla \cdot E \neq (\frac{\delta E}{\delta x},\frac{\delta E}{\delta y},\frac{\delta E}{\delta z}) $ So what difference does the dot between the nabla and the letter make here? How about just $\nabla E $? Is $\nabla E \neq (\frac{\delta E}{\delta x},\frac{\delta E}{\delta y},\frac{\delta E}{\delta z})$?

Answer 1

The dot indicates a dot product. For instance, $(1,2) \cdot (3,4) = (1)(3) + (2)(4) = 3 + 8 = 11$.

So your two quantities live in different dimensional domains.

Answer 2

T is a scaler quantity while E is a vector quantity. $$\eqalign{ & T = T(x,y,z) \cr & \nabla = \left( {{\partial _x},{\partial _y},{\partial _z}} \right) \cr & E = \vec E = \left( {{E_x},{E_y},{E_z}} \right) \cr & \nabla T = \left( {{\partial _x}T,{\partial _y}T,{\partial _z}T} \right) \cr & \nabla \bullet E = {\partial _x}{E_x} + {\partial _y}{E_y} + {\partial _z}{E_z} \cr} $$

Answer 3

for a scalar field $T, \nabla T$ represents the direction in which most of the change in $T$ will occur. for a vector field $E, \nabla \cdot E = div\ E$ represent the rate of change of flux of $E$ over a infinitesimal close surface to the volume. that is $$ dT = \nabla T \cdot dx,\, div \, E = \nabla \cdot E = \frac{\int_S E.dS}{dV} $$

Answer 4

The notation $\vec{\triangledown} \cdot \vec{E}$ is taking a bit of a liberty with the mathematical 'del' operator and the dot product operation.

As you probably know, the dot product of two vectors is defined by:

$\vec{\mathbf{a}} = (a_{1}, a_{2}, ..., a_{n}) $ ; $\vec{\mathbf{b}} = (b_{1}, b_{2}, ..., b_{n}) $

$\vec{a} \cdot \vec{b}$ = $a_{1}b_{1} + a_2b_2 + ... + a_{n}b_{n} $

Furthermore the 'Del' operator is loosely defined as:

$\vec{\triangledown } = (\frac{\delta }{\delta x_{1}}, \frac{\delta }{\delta x_{2}}, ... , \frac{\delta }{\delta x_{n}})$

So, $\vec{\triangledown} \cdot \vec{E}$ can be interpreted to mean:

$ \frac{\delta }{\delta x_{1}}E_{1} + \frac{\delta }{\delta x_{2}}E_{2} + ... +\frac{\delta }{\delta x_{n}}E_{n} $

The key difference being that the result of this form is a scalar, while the typical gradient operator ($\vec{\triangledown} \vec{E}$) returns a vector. Also notice, as is pointed out in the comments earlier, that the gradient operation produces a vector function in t variables when it acts on a scalar function of t variables, but the operation $\vec{\triangledown} \cdot \vec{F}$ always acts on a vector field which can be in any number of variables.