$\begingroup$

I have a very basic question regarding the $\nabla$ (Del) symbol / gradient on a field (which I have called f). I have seen two definitions of this symbol:

1. $\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$
2. $\nabla f = \frac{\partial f}{\partial x} \hat{\imath} + \frac{\partial f}{\partial y} \hat{\jmath} + \frac{\partial f}{\partial z} \hat{k}$ (where $\hat{\imath}$, $\hat{\jmath}$ and $\hat{k}$ are unit vectors)

Are (1) and (2) basically the same thing? If so, how do you get from (2) to (1)?

Many thanks!

$\endgroup$ 7

2 Answers

$\begingroup$

Yes, they're the same. In the first definition, you're writing the gradient in standard 3D coordinate notation $(a,b,c)$. In vector algebra/calculus, you start to think about these three numbers as being components of a vector with respect to some basis (these are linear algebra terms). The "standard basis" of 3D vector calculus consists of the three vectors

$$ \hat{\boldsymbol{\imath}} = \left[\begin{array}{c} 1\\0\\0\end{array}\right],\quad \hat{\boldsymbol{\jmath}} = \left[\begin{array}{c} 0\\1\\0\end{array}\right],\quad \hat{\boldsymbol{k}} = \left[\begin{array}{c} 0\\0\\1\end{array}\right] $$ Then, the coordinates $(a,b,c)$ correspond to a column vector as follows: $$ \left[\begin{array}{c} a\\b\\c\end{array}\right] = a\left[\begin{array}{c} 1\\0\\0\end{array}\right] + b\left[\begin{array}{c} 0\\1\\0\end{array}\right]+c\left[\begin{array}{c} 0\\0\\1\end{array}\right] = a \hat{\boldsymbol{\imath}} +b\hat{\boldsymbol{\jmath}} + c\hat{\boldsymbol{k}} $$ Thinking of the gradient as a column vector, you'll see why they're the same!

$\endgroup$ $\begingroup$

You have already gotten good explanation by @icarays1 on the notation so I'll skip that part.

---

$\nabla$ is an operator. It contains spatial derivatives with respect to all basis vectors. If operating on a scalar valued function it creates a vector.

The divergence of a vector field is the inner product between the nabla operator and a vector valued function:

$$\nabla \cdot {\bf v} = \left(\begin{array}{ccc}\frac{d}{dx}&\frac{d}{dy}&\frac{d}{dz}\end{array}\right){\bf v} = \left(\begin{array}{ccc}\frac{d}{dx}&\frac{d}{dy}&\frac{d}{dz}\end{array}\right) \left(\begin{array}{c}{v_x}\\v_y\\v_z\end{array}\right) = \sum_{k \in \{x,y,z\}}\frac{d(v_k)}{d_k}$$

The gradient of a scalar field is the outer product between the nabla operator and a scalar valued function:

$$\nabla f = \left(\begin{array}{ccc}\frac{d}{dx}\\\frac{d}{dy}\\\frac{d}{dz}\end{array}\right)f = \left(\begin{array}{ccc}\frac{d(f)}{dx}\\\frac{d(f)}{dy}\\\frac{d(f)}{dz}\end{array}\right)$$

$\endgroup$ 2

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy