# Does it mean “ scalar field changes in a particular direction” by that the curl of the gradient of a scalar field is zero in practice? [closed]

What does mean that the curl of the gradient of a scalar field is zero in practice?

My effort: the gradient means the direction at which the magnitude of vector change maximally and the curl says how it changes its direction. So that means a scalar field changes in the same direction at all the points of level curve. How is that wrong?

• The divergence of a vector field is a scalar field. You either mean that the curl of the gradient of a scalar field is zero, or that the divergence of the curl of a vector field is zero. – J. Murray Jun 18 '18 at 15:58
• @J.Murray is the curl of gradient of a scalar field zero? Just for confirmation – TIME RUB Jun 18 '18 at 16:10
• Yes. Strictly speaking, it is the zero vector field, whereas the divergence of the curl of a vector field is the zero scalar field. – J. Murray Jun 18 '18 at 16:13
• @J.Murray thanks.... I've now edited the question to give it bit of relevance with your help.... Thanks – TIME RUB Jun 18 '18 at 16:15
• You are of course free to revert my edits, but just so you know, "maximumly" is not an English word. The correct word in this context is "maximally." You also removed several articles (the curl and the vector) which are grammatically necessary. – J. Murray Jun 18 '18 at 16:22

It is $0$ in practice as a result of being mathematically $0$ as well, in this case :).
Calculate: If $\varphi$ is a physical scalar field (Temperature, say. When represented as a real number). Mathematically, we have a function $\varphi :R^3 \to R$ giving a real number for each point in $R^3$. It is not a vector field, yet. It is a scalar field.
If we assume the 2nd derivatives exists and are continous (In classical theory of fields we enjoy this mathematical property), then we have mathematically: $$\frac{\partial^2 \varphi}{\partial x_i \partial x_j} = \frac{\partial^2 \varphi}{\partial x_j \partial x_i}$$ for any $i,j \in \{1,2,3\}$ being cartesian indices as a nice choice. It is called second derivative symmetric.
Consequently, $\nabla \times (\nabla \varphi)$ is of the form $$\frac{\partial^2 \varphi}{\partial x_i \partial x_j} - \frac{\partial^2 \varphi}{\partial x_j \partial x_i}=0$$ in each of $x,y,z$ directions. Write it down as a 'determinant' to see it more clearly.