- #1
rachmaninoff
"directional derivative" - conventions
Warning: nitpicking notational issues ahead!
So the "directional derivative" in vector analysis - the differential rate of increase of a scalar function φ in a vector direction - is by convention* defined starting from the vector gradient (the 'del' operator): [tex]\nabla \phi \cdot \hat{u}=
\vec{\nabla} \phi(\vec{r}) | _{r_0} \cdot \hat{u}[/tex]
(the directional derivative of φ in the u-direction).
So it's not really fundamental - it's defined in terms of del-φ, which is 'more fundamental' I guess. Big deal.
*"by convention" meaning, in every one of a dozen reference books I looked through...
For example, in cartesian coordinates, a directional derivative (at r) might look like:
[tex]\vec{\nabla} \phi_r \cdot \hat{u} = \left(\hat{i}\frac{\partial{\phi}}{\partial{x}}| _r + \hat{j} \frac{\partial{\phi}}{\partial{y}} | _r + \hat{k}\frac{\partial{\phi}}{\partial{z}} | _r \right) \cdot \left(u_x \hat{i} + u_y \hat{j} + u_z \hat{k} \right)[/tex]
[tex]=\frac{\partial{\phi}}{\partial{x}}u_x +\frac{\partial{\phi}}{\partial{y}}u_y + \frac{\partial{\phi}}{\partial{z}}u_z [/tex]
So, the vector u is decomposed into some coordinate axes, and the derivatives of the scalar function are found with respect to those three axes, and things are scalar-multiplied and added together. Your definition of the directional derivatives goes through a coordinate system, of your preference - through the definitions of the del operator.
But why not define this directly, in the usual manner for defining a derivative:
[tex]\lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}[/tex]
Which is the same thing, without those coordinate decompositions,
[tex]d \phi = \frac{\partial{\phi}}{\partial{x}}dx + \frac{\partial{\phi}}{\partial{y}}dy + \frac{\partial{\phi}}{\partial{z}}dz[/tex]
[tex]\mbox{and, } d \vec{u} = \hat{i} d(u_x) + \hat{j}d(u_y) + \hat{k}d(u_z)??[/tex]
Because [itex]d \phi[/itex] looks well-defined to me. And so does [itex]d \vec{u} = \hat{u} d \lambda[/itex]. So then, why don't we just say, without ambiguity,
[tex]\frac{d \phi}{d \vec{u}} \right| _r \equiv \lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}[/tex]
(with the normalized direction [tex]\hat{u} = \vec{u} / \| \vec{u} \| [/tex] ) ?!
So as to eliminate any reference to the coordinate system?
I mean, when we're taking the usual partial derivatives, with respect to say the "x" axis, that's just a 'directional' derivative in the [itex]\hat{x}[/itex] direction. So, can we not generalize this to any (normalized) vector direction, and so define the derivative of a scalar function with respect to a vector?
That's my question - is this notation rigorous enough?
Warning: nitpicking notational issues ahead!
So the "directional derivative" in vector analysis - the differential rate of increase of a scalar function φ in a vector direction - is by convention* defined starting from the vector gradient (the 'del' operator): [tex]\nabla \phi \cdot \hat{u}=
\vec{\nabla} \phi(\vec{r}) | _{r_0} \cdot \hat{u}[/tex]
(the directional derivative of φ in the u-direction).
So it's not really fundamental - it's defined in terms of del-φ, which is 'more fundamental' I guess. Big deal.
*"by convention" meaning, in every one of a dozen reference books I looked through...
For example, in cartesian coordinates, a directional derivative (at r) might look like:
[tex]\vec{\nabla} \phi_r \cdot \hat{u} = \left(\hat{i}\frac{\partial{\phi}}{\partial{x}}| _r + \hat{j} \frac{\partial{\phi}}{\partial{y}} | _r + \hat{k}\frac{\partial{\phi}}{\partial{z}} | _r \right) \cdot \left(u_x \hat{i} + u_y \hat{j} + u_z \hat{k} \right)[/tex]
[tex]=\frac{\partial{\phi}}{\partial{x}}u_x +\frac{\partial{\phi}}{\partial{y}}u_y + \frac{\partial{\phi}}{\partial{z}}u_z [/tex]
So, the vector u is decomposed into some coordinate axes, and the derivatives of the scalar function are found with respect to those three axes, and things are scalar-multiplied and added together. Your definition of the directional derivatives goes through a coordinate system, of your preference - through the definitions of the del operator.
But why not define this directly, in the usual manner for defining a derivative:
[tex]\lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}[/tex]
Which is the same thing, without those coordinate decompositions,
[tex]d \phi = \frac{\partial{\phi}}{\partial{x}}dx + \frac{\partial{\phi}}{\partial{y}}dy + \frac{\partial{\phi}}{\partial{z}}dz[/tex]
[tex]\mbox{and, } d \vec{u} = \hat{i} d(u_x) + \hat{j}d(u_y) + \hat{k}d(u_z)??[/tex]
Because [itex]d \phi[/itex] looks well-defined to me. And so does [itex]d \vec{u} = \hat{u} d \lambda[/itex]. So then, why don't we just say, without ambiguity,
[tex]\frac{d \phi}{d \vec{u}} \right| _r \equiv \lim_{\lambda \rightarrow 0^+} \frac{\phi (\vec{r} + \lambda \hat{u} ) - \phi (\vec{r})}{\lambda}[/tex]
(with the normalized direction [tex]\hat{u} = \vec{u} / \| \vec{u} \| [/tex] ) ?!
So as to eliminate any reference to the coordinate system?
I mean, when we're taking the usual partial derivatives, with respect to say the "x" axis, that's just a 'directional' derivative in the [itex]\hat{x}[/itex] direction. So, can we not generalize this to any (normalized) vector direction, and so define the derivative of a scalar function with respect to a vector?
That's my question - is this notation rigorous enough?
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