A question on the definition of the curl of a vector

[user1139]

Homework Statement

See below.

Relevant Equations

See below.

The curl is defined using Cartersian coordinates as

\begin{equation}
\nabla\times A =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
A_x & A_y & A_z
\end{vmatrix}.
\end{equation}

However, what are the physical consequences, if any, if I were to define the curl instead as

\begin{equation}
\nabla\times A =
\begin{vmatrix}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
A_x & A_y & A_z \\
\hat{x} & \hat{y} & \hat{z}
\end{vmatrix}\,?
\end{equation}

 

[PeroK]

The definition of a vector operator cannot have physical consequences! That said, depending on how you interpret that determinant, your new definition looks the same as the old one.

 

[user1139]

But the second definition will incur an overall minus sign relative to the first definition after calculating the determinant.

 

[PeroK]

But the second definition will incur an overall minus sign relative to the first definition after calculating the determinant.

I'm not sure about that. They look the same to me. In any case, a minus sign won't make a significant difference mathematically. It's effectively just a right-hand, left-hand convention.

 

[user1139]

Does it then make sense to use a left-hand convention for defining the curl in a right-handed coordinate system?

 

[PeroK]

Does it then make sense to use a left-hand convention for defining the curl in a right-handed coordinate system?

It may not make much sense, but all it will do is introduce a negative sign in any existing identity for curl.

 

[Steve4Physics]

But the second definition will incur an overall minus sign relative to the first definition after calculating the determinant.

Swapping two rows (or columns) of a determinant multiplies the determinant's value by -1.

You need two row-swaps to move between your two definitions. So they are identical.

 

[kuruman]

Look at Faraday's law $$\vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t}.$$ The induced current will flow in such a way as to oppose a proposed change in magnetic flux regardless of how the curl is written down. The induced current doesn't care whether you have swapped the rows of the curl determinant or not. So the answer to your question is that no physical consequences take place. If I say "the sun is shining" is English and you say "el sol está brillando" in Spanish, changing the way you formulate physical reality does not change the physical reality itself.

 

[robphy]

Look at Faraday's law $$\vec \nabla \times \vec E=-\frac{\partial \vec B}{\partial t}.$$

Note that there are two curls here, one on the left and one on the right (\color{blue}{\vec B} as the curl of \vec A)

So, one should be aware of all of the implications of a change in definition and/or a change in convention.
One should also look at the real definition of curl as circulation per unit area.

 

[vanhees71]

Despite the fact that both "definitions" give the same result, namely
$$\vec{\nabla} \times \vec{A}=\begin{pmatrix} \partial_2 A_3 -\partial_3 A_2 \\ \partial_3 A_1-\partial_1 A_3 \\ \partial_1 A_2 -\partial_2 A_1\end{pmatrix},$$
it's not a good way to get an intuition, what the curl of a vector field indeed means.

That you get with the coordinate independent definition of the curl via a line integral. To that end let $F$ be a surface with boundary $\partial F$. Define unit-surface-normal vectors $\vec{n}(\vec{x})$ along the surface and then orient the boundary curve according to the right-hand rule. Then let the curve shrink to a point $\vec{x}_0$ keeping $\vec{n}(\vec{x}_0)=\vec{n}_0=\text{const}$. Then the corresponding component of the curl is defined by
$$\vec{n}_0 \cdot [\vec{\nabla} \times \vec{A}(\vec{x}_0)]=\lim_{F \rightarrow \{\vec{x}_0 \}} \int_{\partial F} \mathrm{d} \vec{x} \cdot \vec{A}(\vec{x}).$$
It's good to draw a picture of this to see that indeed the curl measures how much the vector field "curls" around an axis in direction of $\vec{n}_0$ at the point $\vec{x}_0$, i.e., it describes the "vortex density" at this point.

Another intuitive physical case is in fluid flow. Here
$$\vec{\omega}(t,\vec{x})=\frac{1}{2} \vec{\nabla} \times \vec{v}(t,\vec{x})$$
is the momentaneous angular velocity of a fluid-volume element around $\vec{x}$ as a whole, describing a rigid rotation of the fluid element, while the symmetric tensor,
$$\epsilon_{jk} = \frac{1}{2} (\partial_j v_k + \partial_k v_j),$$
is the strain rate, which describes the change in the shape (deformation) of the fluid-volume element per unit of time. It can be further decomposed in the trace-less part which describes volume-preserving shearing deformations and the bulk or stretching deformation $\vec{\nabla} \cdot \vec{v}$ related to the change of volume of the fluid element per unit of time.

https://en.wikipedia.org/wiki/Strain-rate_tensor

It's always good to keep in mind that physical significance is in invariant geometrical properties, i.e., in scalars, vectors, and tensors, which are independent on any choice of a coordinate system or arbitrary curvilinear coordinates, while our calculations are most easily done with the corresponding vector and tensor components, but these refer to the specific choice of coordinates and basis vectors describing the invariant objects.