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So I have seen that the directional derivative can be written as

$$ \frac{df}{d\lambda} = \frac{dx^i}{d\lambda}\frac{df}{dx^i} $$

And we can identify $ \frac{d}{dx^i} $ as basis vectors and $ \frac{dx^i}{d\lambda} $ as components. What I don't understand is why is $\frac{df}{d\lambda} $ considered a vector? It's a derivative of a function w.r.t. a parameter and surely that's not a vector?

I.e. In vector notation the directional derivative is given by a dot product

$$ \frac{df}{d\lambda} = \hat{n} \cdot \nabla f $$ which is a scalar but in tensor notation that seems to not be the case?

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  • $\begingroup$ Are you familiar with Einstein summation notation? $\endgroup$ – probably_someone Jan 2 '17 at 3:45
  • $\begingroup$ It is important to note that $f'(\lambda)$ is not a vector, but $d/d\lambda$ is. In particular, it's the vector which corresponds with the "arrow" pointing in the $\lambda$ direction. The reason it's a vector appeals to the fact that partial derivatives of coordinates are in one-to-one correspondence with column vectors. $\endgroup$ – Ultima Jan 2 '17 at 3:51
  • $\begingroup$ Also, where is $df/d\lambda$ considered a vector? Do you have a source for this? $\endgroup$ – probably_someone Jan 2 '17 at 3:52
  • $\begingroup$ physicspages.com/2013/02/10/… $\endgroup$ – Matt0410 Jan 2 '17 at 3:59
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I think the physicspages author is just confused. $df/d\lambda$ is a scalar, not a vector. It's the scalar product of the covector $\nabla f$ with the vector $d\mathbf{x}/d\lambda$. They say, "Regarding the partial derivatives as basis vectors, ..." and go on as if $\partial f/\partial x$ and $\partial f/\partial y$ were the basis vectors. This is wrong. In the notational convention they have in mind, it's the operators $\partial/\partial x$ and $\partial/\partial y$ that are used as basis vectors.

In this notational convention, the partial derivative operators are never actually applied to anything. They never have anything written to the right of them. The convention is a notational trick that exploits an isomorphism between vectors and derivative operators, but it doesn't involve actually taking the derivative of anything.

The thought that they're probably trying to express is that in their example, their paraboloid is embedded in a higher-dimensional space (which would not normally be the case in general relativity). They're being sloppy/confused with their notation, because they're using the symbol $f$ to mean a scalar field defined on the $(x,y)$ plane, but they're also treating $f$ as if it were a position vector in $(x,y,z)$ space.

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    $\begingroup$ Does the idea of basis vectors as partials also stem from the idea that we can write basis vectors in flat Cartesian coordinates as $ \hat{e}_\alpha = \frac{\partial \vec{R}}{\partial x^\alpha} $ where $ \vec{R} $ is the position vector? However in curved space there is no concept of a position vector so we omit it? $\endgroup$ – Matt0410 Jan 2 '17 at 11:52
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The directional derivative $\frac{df}{d\lambda}$ is not actually a vector in the space spanned by the $x^i$. What the source was trying to say was that in the abstract vector space spanned by the partial derivative operators, $\frac{d}{d\lambda}$ can be thought of as a vector.

*http://www.physicspages.com/2013/02/10/tangent-space-partial-derivatives-as-basis-vectors/

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  • $\begingroup$ This is not quite right. $\partial f/\partial\lambda$ is not an operator, $\partial/\partial\lambda$ is. $\endgroup$ – Ben Crowell Jan 2 '17 at 4:27
  • $\begingroup$ Fixed, thanks. You get a vector operator from this space which you can apply to $f$. $\endgroup$ – probably_someone Jan 2 '17 at 4:29
  • $\begingroup$ I don't think the new version of the answer is right, either. The physicspages author doesn't say that $d/d\lambda$ is a vector (which possibly could make some sense), they imply that $df/d\lambda$ is a vector (they say it has components), which is simply a mistake. $\endgroup$ – Ben Crowell Jan 2 '17 at 5:04
  • $\begingroup$ I suppose so. Hence the answer claiming that this was what the authors were "trying to say" rather than what they actually said. $\endgroup$ – probably_someone Jan 2 '17 at 5:05

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