Calculus/Helmholtz Decomposition Theorem

Template:Calculus/Top Nav The Helmholtz Decomposition Theorem, regarded as the fundamental theorem of vector calculus, dictates that any vector field $𝐅$ can be expressed as the sum of a conservative vector field $𝐆$ and a divergence free vector field $𝐇$: $𝐅=𝐆+𝐇$.

Approach #1

Given a vector field $𝐅$, the vector field $𝐆(𝐪)=\frac{1}{4\pi }\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{(\mathrm{\nabla }\cdot 𝐅){|}_{{𝐪}^{\prime }}(𝐪-{𝐪}^{\prime })}{|𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ has the same divergence as $𝐅$, and is also conservative: $\mathrm{\nabla }\cdot 𝐆=\mathrm{\nabla }\cdot 𝐅$ and $\mathrm{\nabla }\times 𝐆=\mathrm{𝟎}$. The vector field $𝐇=𝐅-𝐆$ is divergence free.

Therefore $𝐅=𝐆+𝐇$ where $𝐆(𝐪)=\frac{1}{4\pi }\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{(\mathrm{\nabla }\cdot 𝐅){|}_{{𝐪}^{\prime }}(𝐪-{𝐪}^{\prime })}{|𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ and $𝐇=𝐅-𝐆$. Vector field $𝐆$ is conservative and $𝐇$ is divergence free.

Approach #2

Given a vector field $𝐅$, the vector field $𝐇(𝐪)=\frac{1}{4\pi }\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{(\mathrm{\nabla }\times 𝐅){|}_{{𝐪}^{\prime }}\times (𝐪-{𝐪}^{\prime })}{|𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ has the same curl as $𝐅$, and is also divergence free: $\mathrm{\nabla }\times 𝐇=\mathrm{\nabla }\times 𝐅$ and $\mathrm{\nabla }\cdot 𝐇=0$. The vector field $𝐆=𝐅-𝐇$ is conservative.

Therefore $𝐅=𝐆+𝐇$ where $𝐇(𝐪)=\frac{1}{4\pi }\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{(\mathrm{\nabla }\times 𝐅){|}_{{𝐪}^{\prime }}\times (𝐪-{𝐪}^{\prime })}{|𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ and $𝐆=𝐅-𝐇$. Vector field $𝐆$ is conservative and $𝐇$ is divergence free.

Approach #3

The Helmholtz decomposition can be derived as follows:

Given an arbitrary point ${𝐪}^{\prime }$, the divergence of the vector field $\frac{𝐪-{𝐪}^{\prime }}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}$ is ${\mathrm{\nabla }}_{𝐪}\cdot \frac{𝐪-{𝐪}^{\prime }}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}=\delta (𝐪;{𝐪}^{\prime })$ where $\delta (𝐪;{𝐪}^{\prime })$ is the Dirac delta function centered on ${𝐪}^{\prime }$ (The subscript ${}_{𝐪}$ clarifies that $𝐪$ as opposed to ${𝐪}^{\prime }$ is the parameter that the differential operator is being applied to). Since ${\mathrm{\nabla }}_{𝐪}(\frac{-1}{|𝐪-{𝐪}^{\prime }|})=\frac{𝐪-{𝐪}^{\prime }}{|𝐪-{𝐪}^{\prime }{|}^3}$, it is the case that $\underset{𝐪}{\overset{2}{\mathrm{\nabla }}}\frac{-1}{4\pi |𝐪-{𝐪}^{\prime }|}={\mathrm{\nabla }}_{𝐪}\cdot \frac{𝐪-{𝐪}^{\prime }}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}=\delta (𝐪;{𝐪}^{\prime })$

Alongside the identities $\mathrm{\nabla }\cdot (f𝐆)=(\mathrm{\nabla }f)\cdot 𝐆+f(\mathrm{\nabla }\cdot 𝐆)$, and $\mathrm{\nabla }\times (f𝐆)=(\mathrm{\nabla }f)\times 𝐆+f(\mathrm{\nabla }\times 𝐆)$, and most importantly $\mathrm{\nabla }\times (\mathrm{\nabla }\times 𝐅)=\mathrm{\nabla }(\mathrm{\nabla }\cdot 𝐅)-{\mathrm{\nabla }}^2𝐅$, the following can be derived:

$𝐅(𝐪)=\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\delta (𝐪;{𝐪}^{\prime })𝐅({𝐪}^{\prime })d{V}^{\prime }$ $=\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }(\underset{𝐪}{\overset{2}{\mathrm{\nabla }}}\frac{-1}{4\pi |𝐪-{𝐪}^{\prime }|})𝐅({𝐪}^{\prime })d{V}^{\prime }$ $=\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }(\underset{𝐪}{\overset{2}{\mathrm{\nabla }}}\frac{-𝐅({𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }|})d{V}^{\prime }$

$=\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }({\mathrm{\nabla }}_{𝐪}({\mathrm{\nabla }}_{𝐪}\cdot \frac{-𝐅({𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }|})-{\mathrm{\nabla }}_{𝐪}\times ({\mathrm{\nabla }}_{𝐪}\times \frac{-𝐅({𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }|}))d{V}^{\prime }$

$=\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }({\mathrm{\nabla }}_{𝐪}(\frac{(𝐪-{𝐪}^{\prime })\cdot 𝐅({𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3})-{\mathrm{\nabla }}_{𝐪}\times (\frac{(𝐪-{𝐪}^{\prime })\times 𝐅({𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}))d{V}^{\prime }$

$={\mathrm{\nabla }}_{𝐪}\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{𝐅({𝐪}^{\prime })\cdot (𝐪-{𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }+{\mathrm{\nabla }}_{𝐪}\times \underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{𝐅({𝐪}^{\prime })\times (𝐪-{𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$

$𝐆(𝐪)={\mathrm{\nabla }}_{𝐪}\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{𝐅({𝐪}^{\prime })\cdot (𝐪-{𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ is the gradient of a scalar field, and so is conservative.

$𝐇(𝐪)={\mathrm{\nabla }}_{𝐪}\times \underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{𝐅({𝐪}^{\prime })\times (𝐪-{𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ is the curl of a vector field, and so is divergence free.

In summary, $𝐅=𝐆+𝐇$ where $𝐆(𝐪)={\mathrm{\nabla }}_{𝐪}\underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{𝐅({𝐪}^{\prime })\cdot (𝐪-{𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ is conservative and $𝐇(𝐪)={\mathrm{\nabla }}_{𝐪}\times \underset{{𝐪}^{\prime }\in {ℝ}^3}{\iiint }\frac{𝐅({𝐪}^{\prime })\times (𝐪-{𝐪}^{\prime })}{4\pi |𝐪-{𝐪}^{\prime }{|}^3}d{V}^{\prime }$ is divergence free.