Proving the Generalized Curl Theorem

The proof for this relies on applying the well-known Regular Curl Theorem (Stokes' Theorem) to a cleverly chosen vector field, effectively demonstrating how core principles of vector calculus can be used to derive new, elegant relationships. The visualization reinforces this by showing how gradient vectors, which point only in the direction of steepest ascent, have no rotational component, thus confirming that their curl is indeed zero.

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✍️Mathematical Proof

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Method of Proof

The proof relies on a common technique in vector calculus: using a known theorem (the regular Curl Theorem, or Stokes' Theorem) as a foundation and applying it to a cleverly chosen, specific vector field. This is a powerful and elegant way to prove new identities.

Choice of Vector Field

The key to the proof is setting the vector field $A$ equal to the product of a scalar field $f$ and an arbitrary constant vector $c$ (i.e., $A=f c$ ). This choice allows you to simplify the equations and factor out the constant vector, leading to the desired result.

Reliance on Product Rule and Scalar Triple Product

The proof utilizes two key vector identities: the product rule for the curl, $\nabla \times(f c)=(\nabla f) \times c$, and the scalar triple product, $a \times b \cdot c=c \cdot a \times b$. Both are essential for manipulating the right-hand side of the equation and simplifying the integral.

Generalizability

The use of an arbitrary constant vector $c$ is crucial. Because the final equality holds true for any constant vector, it proves that the two vector expressions themselves must be equal, thus establishing the generalized theorem.

🎬Demonstration

This demo is that the generalized curl theorem, beautifully illustrates a fundamental principle: the path integral of a scalar field around a closed loop is always zero. Since the curl of a gradient is non-existent, the surface integral of a scalar field’s gradient over any surface will always be zero, which means its boundary integral must also be zero. The demo visually reinforces this by showing that the gradient vectors radiate outward without any swirling or rotational component.

the line integral of a scalar field around a closed path is equal to the surface integral of the curl of its gradient