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Helmholtz's theorem tells us that a sufficiently smooth vector field $\mathbf{A}$ can be decomposed into curl and gradient free parts as the gradient of a scalar potential plus the curl of a vector potential $$ \mathbf{A} = \nabla \phi + \nabla \times \mathbf{B}$$ Given the constraint that the vector field must be divergence and curl free $$ \nabla \cdot \mathbf{A} = 0$$ $$ \nabla \times \mathbf{A} = 0$$ this leads to $$\nabla^2 \phi = 0 $$ $$\nabla \times (\nabla \times \mathbf{B}) = 0$$ In application, such as potential flow theroy, it is often taken as a given that $\mathbf{B}=\mathbf{0}$ and $\mathbf{A} = \nabla \phi$. I can see no reason that this is absolutely required. Other than simplification, is there a justification for $\mathbf{B}=\mathbf{0}$? For potential flow theory the choice leads to useful solutions, but is it ignoring a subset of possible solutions?

Edit: There are certainly solutions that have non-zero $\mathbf{B}$ that satify the divergence free and curl free constraints. For instance $$\phi = 0$$ $$\mathbf{B}=\alpha y \hat{i}+\beta x\hat{j}$$ which gives $$\mathbf{A} = \beta-\alpha$$ for constants $\alpha$ and $\beta$. Any constant vector field satisfies the divergence and curl free constraint. My intuition is that while $\mathbf{B}=0$ is not necessary, it simply does not add any new unique solution that can't be found by simply satisfying the constraints with $\phi$.

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In electromagentism, the definition of $\mathbf{B}$ through $\mathbf{A}$ is: $\mathbf{B}=\nabla\times \mathbf{A}$, which gives a-priori, the Helmoltz decomposition of $\mathbf{B}$. This is because $\mathbf{A}$ is the vector magnetic potential and $\mathbf{B}$ is the associated magnetic field. So $\mathbf{B}=0$ if $\nabla\times \mathbf{A}=0$.

There is no reason otherwise to take $\mathbf{B}=0$. For example, $\mathbf{B}$ can equal a constant and trivially satisfy all of the conditions above.

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  • $\begingroup$ yes, but can you have $\phi = 0$ and satisfy the curl and divergence free constraint on $\mathbf{A}$ with a $\mathbf{B}$ that leads to a non-constant $\mathbf{A}$. That is basically what I'm asking. $\endgroup$ – SimpleLikeAnEgg Dec 5 '13 at 21:47

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