Question

For a scalar field \(\psi(\vec{r})\) and a vector field \(\vec{A}(\vec{r})\), \(\nabla\times(\vec{A}\,\psi)\) is equivalent to the expression

• A. \(\psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi)\)
• B. \(\psi(\nabla\times\vec{A}) + \vec{A}\times(\nabla\psi)\)
• C. Null vector
• D. \(\vec{A}\times(\nabla\psi) - \psi(\nabla\times\vec{A})\)

Hint:

Use the product rule \(\nabla\times(\psi\vec{A}) = \psi(\nabla\times\vec{A}) + (\nabla\psi)\times\vec{A}\), then flip the order of the second cross product.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The curl of a scalar times a vector is not a basic operation you can look up on its own; it comes from applying the product rule for the curl operator, in the same spirit as the ordinary product rule for derivatives, but with an extra cross product because curl involves \(\nabla\times\).

Step 2: Key Formula or Approach:
The standard vector identity for a scalar field \(\psi\) and a vector field \(\vec{A}\) is:
\[ \nabla\times(\psi\vec{A}) = \psi(\nabla\times\vec{A}) + (\nabla\psi)\times\vec{A} \]
This second term needs to be rewritten so it matches the form used in the answer options.

Step 3: Detailed Explanation:
Write out the identity component by component using index notation, with \(\epsilon_{ijk}\) the Levi-Civita symbol and a sum over repeated indices understood:
\[ [\nabla\times(\psi\vec{A})]_i = \epsilon_{ijk}\partial_j(\psi A_k) = \epsilon_{ijk}\left[(\partial_j\psi)A_k + \psi(\partial_j A_k)\right] \]
The second piece is just \(\psi[\nabla\times\vec{A}]_i\). The first piece, \(\epsilon_{ijk}(\partial_j\psi)A_k\), is exactly the \(i\)-th component of \((\nabla\psi)\times\vec{A}\). So:
\[ \nabla\times(\psi\vec{A}) = \psi(\nabla\times\vec{A}) + (\nabla\psi)\times\vec{A} \]
Now use the anti-commuting property of the cross product, \(\vec{u}\times\vec{v} = -\vec{v}\times\vec{u}\), to flip the order in the second term:
\[ (\nabla\psi)\times\vec{A} = -\vec{A}\times(\nabla\psi) \]
Substituting back:
\[ \nabla\times(\vec{A}\psi) = \psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi) \]

Step 4: Why the other options are wrong.
Option (B) has the wrong sign on the second term, the mistake of forgetting to flip the cross product order after moving the gradient term out front. Option (C) would need \(\psi(\nabla\times\vec{A})\) and \(\vec{A}\times(\nabla\psi)\) to cancel exactly for every possible \(\psi\) and \(\vec{A}\), which only happens in special cases, not as a general rule. Option (D) is the correct expression with an overall sign flip, which does not match the derivation above.

Final Answer:
The identity gives \(\psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi)\), matching option (A). \[ \boxed{\nabla\times(\vec{A}\psi) = \psi(\nabla\times\vec{A}) - \vec{A}\times(\nabla\psi)} \]