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So, for a scalar field $T(x,y,z)$, the derivative along $d\vec l$ is given by $$\frac {dT}{|d\vec l|} = |\vec \nabla T| \cos\theta$$where $\theta$ is the angle between $\vec \nabla T$ and $d\vec l$

For a vector field $\vec V (x,y,z)$, I understand that $\vec \nabla . \vec V$ and $\vec \nabla \times \vec V$ give the Divergence and the Curl respectively.

But, is there a way in which $\vec \nabla$ can act on $\vec V$ to give an expression for $\frac {d \vec V}{|d\vec l|}$, the directional derivative of $\vec V$ along $d\vec l$?

PS: I've only just started to learn vector calculus, so pardon me if this question comes out as silly.

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    $\begingroup$ What static means here? $\endgroup$ Commented Aug 23, 2020 at 10:17
  • $\begingroup$ When I say static, I mean that the field remains constant with time. @enzotib $\endgroup$ Commented Aug 23, 2020 at 10:21
  • $\begingroup$ So, given that time does not appear anywhere, it is not useful, and misleading, to talk at all of time variable, and to use the word static. $\endgroup$ Commented Aug 23, 2020 at 10:23
  • $\begingroup$ Thanks for the tip! @enzotib $\endgroup$ Commented Aug 23, 2020 at 10:26
  • $\begingroup$ It is not a silly question. I will summarize. To take directional derivatives of vector fields, you need the notion of a connection. Given a Riemannian metric $g$ on your smooth manifold, you then have a notion of directional derivative using the associated Levi-Civita connection. $\endgroup$ Commented Aug 23, 2020 at 10:55

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The nabla operator is not the correct tool for vector fields. In this case, it is better to recognize that the gradient $\nabla T$ is just a special case of the total differential $\mathrm DT$, which is just the Jacobian of $T$. The directional derivative in $x_0$ in direction $v$ is then

$$\mathrm DT(x_0)\cdot e_v,$$

where $e_v$ is the unit vector in $v$-direction. This will be a vector, since the Jacobian $\mathrm DT$ is a matrix, but this is expected, since $T$ is vector valued. Going back to the special case of a scalar field, where $\mathrm DT=\nabla T$, this becomes

$$\mathrm DT\cdot e_v=\nabla T\cdot e_v=\vert\nabla T\vert\vert e_v\vert\cos\theta=\vert\nabla T\vert\cos\theta,$$

which is exactly what you started with.

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    $\begingroup$ The question was about the directional derivative of a vector field $\endgroup$ Commented Aug 23, 2020 at 10:43
  • $\begingroup$ @enzotib: Thanks for the heads up. Should be better now. $\endgroup$ Commented Aug 23, 2020 at 10:52
  • $\begingroup$ I arrived at $\frac {dT}{|d\vec l|} = |\vec \nabla T| \cos\theta$ from the expression: $dT=\frac {\partial T}{\partial x} dx +\frac {\partial T}{\partial y} dy+\frac {\partial T}{\partial z} dz$, which I believe is the total differential. I wrote the same equation as $dT=\biggr{(}\frac {\partial T}{\partial x} e_{x} +\frac {\partial T}{\partial y} e_{y}+\frac {\partial T}{\partial z} e_{z}\biggr{)}.(dx\space e_x+dy\space e_y+dz\space e_z)$ where $e_x,e_y,e_z$ are the unit vectors in the x,y,z directions respectively. This is the same as $dT=\vec \nabla T.d\vec l$ $\endgroup$ Commented Aug 23, 2020 at 12:36
  • $\begingroup$ How do I do the same for a vector field? $\endgroup$ Commented Aug 23, 2020 at 12:36
  • $\begingroup$ If you really have to use coordinates, you should at least use index notation, because otherwise it will be atrocious to read. Then you can write $\mathrm dT_j=\mathrm dx_i\partial_i T_j=\mathrm JT\cdot\mathrm d\vec l$, where $\mathrm JT=(\partial_i T_j)_{ij}$ is the Jacobian. So $\frac{\mathrm d\vec T}{\vert\mathrm d\vec l\vert}=\mathrm JT\cdot \vec e_l$ with $\vec e_l$ the unit vector in direction $\vec l$. $\endgroup$ Commented Aug 23, 2020 at 13:19

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