4
$\begingroup$

Suppose I have a function $f(r, \theta) = r\sin\theta$ with $r$ being in meters and $\theta$ being in radians. The gradient of $f$ is $\nabla f = \frac{\partial f}{\partial r}\hat{r} + \frac{\partial f}{\partial \theta} \hat{\theta} = (\sin\theta) \hat{r} + (r\cos\theta)\hat{\theta}$.

Would I be correct to say that, once evaluated, the unit of the first component of $\nabla f$ is $dimensionless$ and the unit of the second component of $\nabla f$ is the $meter$?

If I am correct, the conclusion would contradict this answer. But, if I am wrong, the conclusion would contradict $[\frac{\partial a}{\partial b}] = [\frac{a}{b}]$, taken from this answer.

To be more succinct: What is resultant unit of the function $f$ after the del operator is applied?

$\endgroup$
  • 1
    $\begingroup$ Hi goo, please don't edit the answer into the question as you did. I've removed it for you. $\endgroup$ – David Z Aug 11 '16 at 21:38
7
$\begingroup$

I think, your definition of the gradient is wrong, have a look here. You should have some prefactors which fix the dimensions.

$\endgroup$
2
$\begingroup$

It seems to me that you're excluding the unit vectors from your dimensions. Would the first component of $\nabla f$ not have units of meters, and the second have units of meters $\cdot$ radians?

$\endgroup$
  • 2
    $\begingroup$ Unit vectors are dimensionless. $\endgroup$ – David Z Aug 11 '16 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.