Given a scalar field $$f : \mathbb{R}^n \rightarrow \mathbb{R}$$, its gradient, denoted $$\nabla f$$ is the vector of partial derivatives $\nabla f = \begin{bmatrix} \partial_1 f\\ \vdots\\ \partial_n f \end{bmatrix}$ Where $$\partial_i f = \frac{\partial f}{\partial x^i}$$ denotes the partial derivative of $$f$$ in the direction of the coordinate $$x^i$$, formally defined as $\frac{\partial f}{\partial x_i } = \lim_{\epsilon \rightarrow 0} \frac{f(x_1, \cdots x_i + \epsilon \cdots x_n) - f(x_1, \cdots, x_n)}{\epsilon}$

To emphasize on the variables held constant, sometimes the notation $$\big( \frac{\partial f}{\partial x_i} \big)_{x_1,\cdots,x_{i-1},x_{i+1},\cdots,x_n}$$ is used.

In this form, the gradient can be rewritten as $$\nabla f = \partial_i f \mathbf{e}_i$$, thus the above definition implicitly chooses the standard Euclidean basis.

The symbol $$\nabla$$ can be thought of as an operator as well, in type notation this would be $$\nabla :: (\mathbb{R}^n \rightarrow \mathbb{R}) \rightarrow \mathbb{R}^n$$ such that $$\nabla = \mathbf{e}_i \partial_i$$.

Note that the gradient returns a vector field. The gradient can be used to rewrite directional derivative of $$f$$ in the direction of the unit vector $$\mathbf{\hat{n}}$$ as $$D_{\mathbf{\hat{n}}}(f) = \nabla f \cdot \mathbf{\hat{n}} = |\nabla f||\mathbf{\hat{n}}| \cos(\theta)$$ - this dot product measures the rate of change of $$f$$ in the direction of $$\mathbf{\hat{n}}$$ thus we can see it is maximized when $$\theta=0$$ i.e. $$\mathbf{\hat{n}}$$ points in the same direction as $$\nabla f$$.

## Thoughts

• Physicists claim the gradient is a real vector, what do they mean by this?

Created: 2022-03-13 Sun 21:45

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