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I am trying to decide whether the covariant derivative and the directional derivative are any different for a scalar field. Perhaps I’m misinterpreting something, but I seem to get a covector field when applying the del operator to a scalar field.

If a scalar field is a (0, 0) tensor, then its covariant derivative will be a (0, 1) tensor. And the del operator is defined $\nabla = \bf{e^i} {\partial \over \partial c^i } $. So then:

$$ \nabla f = \bf{e^i} {\partial f \over \partial c^i } $$

Now this seems to make sense, but I get a covector. On the other hand, the gradient is usually defined as:

$$ \nabla f = g^{ij} {\partial f \over \partial c^j} \bf{e_i} $$

So is the covariant derivative of a scalar field supposed to be a vector field, or a covector field?

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  • $\begingroup$ On a side note, the summation indices on your second equation don't match: one of the lower $i$s should be $j$. I tried to fix it but I can't edit less than 6 characters haha. $\endgroup$ – Jackozee Hakkiuz Apr 30 at 2:24
  • $\begingroup$ whoops, thanks for pointing that out :) $\endgroup$ – user2662833 Apr 30 at 5:07
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Without a metric

you are immediately able to take the covariant derivative $\nabla f$ of a scalar field, which coincides with its exterior derivative $\mathrm{d}f$ $$\nabla f = \mathrm{d}f = \sum_i \partial_i f \,\omega^i$$ where $\omega^i$ are the basis covector fields. Obviously this is a covector field.

Then the derivative of $f$ in the direction of a vector $v$ admits the following notations:

$$vf = \nabla_{v}f = (\nabla f)(v) = (\mathrm{d}f)(v) \tag{1}$$

If you have a metric

say $g$, then it induces the so-called musical isomorphisms $\sharp$ (which maps covector fields to vectors fields) and $\flat$ (which maps in the other direction). So then you can define the gradient vector field of a scalar field as $$\vec{\nabla}f := (\nabla f)^\sharp$$ In this case, the directional derivative $vf$ can be expressed (apart from the notations in $(1)$) by

$$g(\vec{\nabla}f,v)$$.

In short

"Del operator" may be a bit ambiguous. When applied to functions, in my experience people use it to refer to the grandient vector, but judging by what you wrote

the del operator is defined $\nabla = \bf{e^i} {\partial \over \partial c^i } $

you have found a place where "del operator" is used to talk about the covariant derivative.

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  • $\begingroup$ Awesome. So to be clear, it seems that there is no difference between the covariant derivative and the directional derivative for a scalar field. Would that be correct? Also to confirm, the the exterior derivative isn't always the same as the covariant derivative right, that's only true for scalar fields right? $\endgroup$ – user2662833 Apr 30 at 5:13
  • $\begingroup$ See the trouble comes from various places, for example the wikipedia article on the exterior derivative states that $\nabla f = (df)^\#$, so in that case, $\nabla$ is producing a vector field right. So the del operator for the covariant derivative is not the same as the gradient operator (sorry about the confusion here) $\endgroup$ – user2662833 Apr 30 at 5:15
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    $\begingroup$ "Gradient" only makes sense when you have a metric. Thing is, $\nabla$ is used with two different meanings. In this context, perhaps ${\rm grad}(f)$ is a more adequate notation for the gradient vector. As answered above, the covariant derivative $\nabla f$, which is a covariant 1-tensor, coincides with the differential of $f$. So ${\rm grad}(f) = (\nabla f)^\sharp$, if you want. People in basic calculus courses do not pay attention to those subtleties. Use ${\rm d}f$ for the covariant 1-tensor, and save $\nabla f$ for the gradient vector (to avoid writing ${\rm grad}(f)$ every time). $\endgroup$ – Ivo Terek Apr 30 at 5:53

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