Meaning of directional derivative of a vector field

Question

Suppose I have a vector field $ \vec{B} (x,y,z)$ then do $ \frac{ \partial B}{ \partial n}$ where n is the direction vector of a line denote the directional derivative of the vector in the direction of $n$?

The reason I ask is that I recently encountered this in a physics textbook but all the gradients and directional derivatives that I've seen till now were defined for scalar fields.

Edit:

The real quantity that I started with was the one from this mse post:

$$ (\nabla B_i) n_i $$

I thought this would be the directional derivative since it looked like one but then I later realized this is actually a vector field.

A picture from the book:

Page-158, I.e irodov basic laws of electromagnetism

Answer 1

Suppose I have a vector field $\vec{B}(x,y,z)$ then do $\frac{\partial\vec B}{\partial n}$ where $n$ is the direction vector of a line denote the directional derivative of the vector in the direction of $n$?

Yes. So is also stated for Force by Dipole in electric field in the same book. Page number 20, I.E. Irodov. Although the author explicitly states that the operation is rather complex and will not be discussed in the book.

The formal definitions can be noted from here and $\downarrow$.

Derivatives of vector-valued functions of vectors:

Let ${\displaystyle \mathbf {f} (\mathbf {v} )}$ be a vector-valued function of the vector ${\displaystyle \mathbf {v} }$ . Then the derivative of ${\displaystyle \mathbf {f} (\mathbf {v} )}$ with respect to ${\displaystyle \mathbf {v} }$ (or at ${\displaystyle \mathbf {v} }$) in the direction ${\displaystyle \mathbf {u} }$ (which maybe $\hat i,\hat j,\hat k$) is the second-order tensor defined as:

$${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =D\mathbf {f} (\mathbf {v} )[\mathbf {u} ]=\left[{\frac {d}{d\alpha }}~\mathbf {f} (\mathbf {v} +\alpha ~\mathbf {u} )\right]_{\alpha =0}}$$ for all vectors ${\displaystyle \mathbf {u} }$ .

Properties:

1. If ${\displaystyle \mathbf {f} (\mathbf {v} )=\mathbf {f} _{1}(\mathbf {v} )+\mathbf {f} _{2}(\mathbf {v} )}$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}+{\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\right)\cdot \mathbf {u} .}$
2. If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{v})\times\mathbf{f}_2(\mathbf{v})$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} =\left({\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right)\times \mathbf {f} _{2}(\mathbf {v} )+\mathbf {f} _{1}(\mathbf {v} )\times \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right).}$
3. If $\mathbf{f}(\mathbf{v}) = \mathbf{f}_1(\mathbf{f}_2(\mathbf{v}))$ then ${\displaystyle {\frac {\partial \mathbf {f} }{\partial \mathbf {v} }}\cdot \mathbf {u} ={\frac {\partial \mathbf {f} _{1}}{\partial \mathbf {f} _{2}}}\cdot \left({\frac {\partial \mathbf {f} _{2}}{\partial \mathbf {v} }}\cdot \mathbf {u} \right).}$

Answer 2

Keeping it intuitive in $\mathbb{R}^2$ and you can easily extrapolate to $\mathbb{R}^n$.

In cartesian coordinates you can write the direction vector $\vec{e_r}$ from the origin to a point in a circunference of radius $1$. Direction vectors are unit vectors.

$$\vec{e_r} = \cos(\theta)\vec{e_x} + \sin(\theta)\vec{e_y}$$

Recall that $\vec{e_x}$ and $\vec{e_y}$ are fixed in our coordinates, so they will not be affected by the derivative, but $\theta$ do. Generally, in real life problems, $\theta$ changes over implicit time $t$ such that $\theta(t)$. With this in mind we can now take the derivative of the vector $\vec{e_r}$:

$$\frac{d}{d\theta}\vec{e_r} = \frac{d}{d\theta}\cos(\theta)\vec{e_x} + \frac{d}{d\theta}\sin(\theta)\vec{e_y}$$ $$\frac{d}{d\theta}\vec{e_r} = (-\sin(\theta)\vec{e_x} + cos(\theta)\vec{e_y})\frac{d}{d\theta}\theta$$

Can you see that they are perpendicular to each other? In fact we can prove it with the dot product:

$$\vec{e_r} \cdot \frac{d}{d\theta}\vec{e_r} = -\cos(\theta)\sin(\theta) + \sin(\theta)\cos(\theta) = 0$$

Geometrically, why are they perpendicular? Consider the vector $\vec{e_r}$ for some $\theta$. It changes with infinitesimal $d\theta$ to a new position of $\theta$, say, $\vec{e_r}(\theta + d\theta)$. Note that in the limit $\vec{e_r}(\theta + d\theta) -\vec{e_r}(\theta)$ is a vector perpendicular to $\vec{e_r}(\theta)$.

Answer 3

This answer is the general meaning of gradient of vector field, here we take the directioncal derivative of this gradient.

However, I'll give a standalone interpretation:

When we do the derivative of $B$ field in the normal direction, we take consider a point in space, and move in the direction of the normal of the loop. Here it looks like along the z-axis, we see take the difference between the vector we get after moving along the z-axis a teeny bit and the original vector. This gives us a vector difference, dividing this difference vector by the amount of distance we moved in space gives the required value.

So, since we talk about vector difference $\frac{dB}{dn}$ is infact a vector again.