3
$\begingroup$

I asked the following question on math.stackexchange but no one seemed to have an authorative answer so I'm posting here hoping that experts will see it.

The gradient is a tensor $\nabla f:\mathbf{V} \to \mathbf{R}$ where the partial derivatives are evaluated at some point $(x_0, y_0, z_0)$. And evaluation of this linear form at some vector $v=(v_1,v_2,v_3)$ gives

$$ (\nabla f)(\mathbf{v}) = \partial_x f v_1 + \partial_y f v_2 + \partial_z f v_3 $$ Furthermore in going to a new coordinate system these partial derivatives transform in the expected way.

But what about a function $g:\mathbf{V} \to \mathbf{R}$ which is defined only using the partial derivative of $f$ in the $x$ direction. $$ g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3 $$ As I understand this is not considered a tensor because in moving to a new coordinate system it does not transform correctly.

This has confused me endlessly. The strict definition considers a map such as $f:\mathbf{V}\times\mathbf{V} \to \mathbf{R}$ a tensor if linearity holds in each parameter. The function $g$ above certainly satisfies that. It seems to me that this definition is not used and that the definition of a tensor that is actually used consists of two parts.

  • linearity in each parameter (i.e. multilinear form),
  • and the algebraic structure of the coefficients is maintained in coordinate transformation

Because once we have calculated $\partial_x f$ it is just a scalar and we just hit $(\partial_x f,\partial_x f,\partial_x f)$ with the usual transformation for a covariant vector to get the new coefficients for $g$ in the new coordinate system. That these new coefficients don't have the right algebraic structure doesn't make multilinearity of $g$ go away. Is this at all correct?

Improve this question
$\endgroup$
9
  • 3
    $\begingroup$ One reason for a lack of authoritative answer could be want of a definition. Any of your $g$s, which is on $V^3$, not $V$, is (canonically identified with an element of $(V^*)^{\otimes3}$, so a mathematician might say (as you do) that it is a tensor. What $g$ you get, and hence with which element it is identified, depends on the coördinates used to take the partial derivative of $f$, so a physicist might care to argue that it is not a tensor. To answer your question of who is right, we need a definition. What definition do you want to use? $\endgroup$
    – LSpice
    Commented Aug 28, 2020 at 2:21
  • $\begingroup$ Indeed, a physicist would balk at more than one of the premises. First, that "$\partial_{x} f$ is just a scalar". It's not, it's the first component of a vector, which is not the same thing. But, let's indulge and interpret this as "in some fixed coordinate system, I have calculated $h=\partial_{x} f $, and I declare this $h$ that I have obtained a scalar." But then, secondly, $(h,h,h)$ is not a vector - after all, it doesn't change upon transformation. You don't get to "hit it with the usual transformation", you must show that it behaves according to that transformation - which it doesn't. $\endgroup$
    – Michael Engelhardt
    Commented Aug 28, 2020 at 3:13
  • $\begingroup$ Your $g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3$ formula is not a well defined function $g:\mathbf{V} \to \mathbf{R}$. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Aug 28, 2020 at 3:36
  • $\begingroup$ The behaviour of tensors (1-forms in your case) under coordinate changes is a necessary condition for your g functions to be well defined (that is: independent of the coordinate system adopted to describe the vector space V=R^3). $\endgroup$
    – Konstantinos Kanakoglou
    Commented Aug 28, 2020 at 3:55
  • $\begingroup$ Is $\mathbf V$ an abstract vector space or is it $\mathbb{R}^n$? $\endgroup$
    – Deane Yang
    Commented Aug 28, 2020 at 3:59

0

Reset to default

Browse other questions tagged .