Can the Cross Product of a Vector Field and Its Conjugate Be a Gradient?

[Mentia]

Is it possible to nontrivially represent the cross product of a vector field $\vec{f}(x,y,z)$ with its conjugate as the gradient of some scalar field $\phi(x,y,z)$?

In other words, can the PDE

$\vec{\nabla}\phi(x,y,z) = \vec{f}(x,y,z)\times\vec{f}^\ast(x,y,z)$

be nontrivially (no constant field $\vec{f}$) solved?

If not, why? If so, can you give an example of such a scalar field? This problem has popped up in my research and I'm afraid my PDE skills are lacking.

 

[meldraft]

If I remember correctly, you cannot take the cross product of two vector fields, you need the external product. It is difficult to otherwise define which vectors you are operating on.

Does your question stem from Laplace's equation?

 

[Mentia]

Let $\vec{f}(x,y,z)=f_x(x,y,z)\hat{x}+f_y(x,y,z)\hat{y}+f_z(x,y,z)\hat{z}$. Then $\vec{f}\times\vec{f}^\ast =(f_yf_z^\ast -f_y^\ast f_z)\hat{x}+(f_zf_x^\ast -f_z^\ast f_x)\hat{y}+(f_xf_y^\ast -f_x^\ast f_y)\hat{z}=2i\left [ \text{Im}(f_yf_z^\ast)\hat{x} + \text{Im}(f_zf_x^\ast)\hat{y} + \text{Im}(f_xf_y^\ast)\hat{z} \right ]$.

Does there exist a non-constant and necessarily complex $\vec{f}$ for which there exists a $\phi$ that satisfies $\vec{\nabla}\phi=\vec{f}\times\vec{f}^\ast$?

 

[meldraft]

Hmmm, complex vector spaces, nice

Here's some tips that may help you: In order for a potential function to exist, it must satisfy Laplace's equation. Try calculating the divergence of the formula you derived and see whether it suggests something meaningful.

As for it being necessarily complex, the condition is that:

$$f\neq f^*$$

Try starting with the condition $f=f^*$ in the divergence of $\nabla φ$. It should lead you to a non-true statement.

 

[blue_raver22]

I am not an expert in PDEs, but I can provide some insights on this topic. The cross product of a vector field \vec{f}(x,y,z) with its conjugate is a mathematical operation that produces a vector perpendicular to both \vec{f} and its conjugate. This is commonly used in physics, engineering, and mathematics to calculate torque, angular momentum, and other quantities.

However, it is not possible to represent the cross product as the gradient of a scalar field in a nontrivial manner. This is because the cross product produces a vector, while the gradient of a scalar field produces a vector field. In order for the cross product to be represented as a gradient, the resulting vector field would need to be equivalent to the original vector field \vec{f}. This would require the scalar field to have a gradient that is perpendicular to \vec{f}, which is not possible in a nontrivial manner.

To further illustrate this, let's take the example of a 2D vector field \vec{f}(x,y) = (x,y). The cross product with its conjugate is \vec{f}(x,y) \times \vec{f}^\ast(x,y) = (x,y) \times (-x,-y) = (0,0). This is a constant vector field, which means that the gradient of any scalar field representing it would also be a constant vector field. This is not possible in a nontrivial manner, as the gradient of a scalar field should vary with position.

In conclusion, the PDE \vec{\nabla}\phi(x,y,z) = \vec{f}(x,y,z)\times\vec{f}^\ast(x,y,z) cannot be nontrivially solved in a manner that represents the cross product as the gradient of a scalar field. This is due to the fundamental differences between the two mathematical operations and the resulting vector and vector field. I hope this helps in your research and understanding of PDEs.