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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?

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    $\begingroup$ Hi goo, please don't edit the answer into the question as you did. I've removed it for you. $\endgroup$
    – David Z
    Commented Aug 11, 2016 at 21:38

2 Answers 2

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

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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?

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    $\begingroup$ Unit vectors are dimensionless. $\endgroup$
    – David Z
    Commented Aug 11, 2016 at 21:37

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