I am reading the paper http://www.goshen.edu/physix/mathphys/gco/TensorGuideAJP.pdf in order to gain a basic understanding about tensors. I had some difficulties about understanding some definitions.

1) At the end of the page 499 and at the start of 500, the definition of a field is made as a function of space coordinates $\textbf{x}=(x_1,x_2,x_3)$. If the function is a vector it is named a vector field, if it is a scalar it is named as a scalar field.

So is this definition equivalent to this: For example a vector field on $\textbf{x}=(x_1,x_2,x_3)$,of 3 dimensions is the set of vectors$$\begin{pmatrix}A_1(x_1,x_2,x_3)\\A_2(x_1,x_2,x_3)\\A_3(x_1,x_2,x_3)\end{pmatrix}$$

And a scalar vector field is just the real function $A_1(x_1,x_2,x_3)$?

2)They define the differential operator as ${\partial_a}=\frac{\partial}{\partial x_a}$ then say the divergence of the vector field $\textbf{A}=\textbf{A}(\textbf{x})=\sum_{a=1}^3 A_a(\textbf{x})\hat{x_a}$ is equal to $$\nabla .\textbf{A} = \sum_{a=1}^3 \partial_aA_a $$

Since they defined the partial derivative in a weird way, I got confused in this statement. This looks like the dot product of differential vector $(\frac{\partial A_1(x_1,x_2,x_3)}{\partial x_1},\frac{\partial A_2(x_1,x_2,x_3)}{\partial x_2},\frac{\partial A_3(x_1,x_2,x_3)}{\partial x_3})$ and vector field evaluated at $(x_1,x_2,x_3)$, namely: $(A_1(x_1,x_2,x_3),A_2(x_1,x_2,x_3),A_3(x_1,x_2,x_3))$. Is this really what it is meant in that definiton?

Thanks in advance

  • $\begingroup$ scalar vector field is just the real function. A scalar field is just a 1-dimensional function, like Temperature. (Maybe in some instances it could also be a complex function). A "scalar vector field" does not make sense, most likely just a typo. $\endgroup$ – NicoDean Jun 24 '14 at 18:01

Right, they make it look like a vector, called the Nabla operator, $\nabla$. You can do a "dot product" with a vector field, yielding a scalar (the divergence), but also a "cross product" to yield a vector (the rotational), or a "scalar product" (with a scalar field) giving a vector (the gradient).

Nabla can be applied to the gradient of a scalar field, giving the Laplacian: symbolically, $\Delta=\nabla\nabla=\nabla^2$.

This is a notational convenience, but it makes sense.

  • $\begingroup$ Sorry for the late answer. As far as I understand, they treat the partial derivative operator ${\partial_a}=\frac{\partial}{\partial x_a}$ as an actual number and use the vector consisting of these operators as an actual numerical vector. So, for example if ${\partial_a}=\frac{\partial}{\partial x_a}$ is multiplied with $A_a$ it actually defines the differentiation of $A_a$ with respect to $x_a$. Is that right? $\endgroup$ – Ufuk Can Bicici Jul 2 '14 at 6:48
  • $\begingroup$ Quite right. This "trick" has some connection with operational calculus, but that's another story. $\endgroup$ – Yves Daoust Jul 2 '14 at 7:16
  • $\begingroup$ Can you elaborate on this calculus trick a little bit? Just out of curiosity. $\endgroup$ – Ufuk Can Bicici Jul 2 '14 at 7:45
  • $\begingroup$ Google is your best friend. $\endgroup$ – Yves Daoust Jul 2 '14 at 7:46

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