# Helmholtz Decomposition for a four-vector field

A big result from vector calculus is the Helmholtz Decomposition: for any vector-valued function $\mathbf{F} : \mathbb{R}^{3} \to \mathbb{R}^3$ that is well-behaved enough, we can always decompose it as follows: $$\mathbf{F}(\mathbf{r}) = - \boldsymbol{\nabla} \Phi(\mathbf{r}) + \boldsymbol{\nabla} \times \mathbf{C}(\mathbf{r})$$

There always exist functions $\Phi : \mathbb{R}^3 \to \mathbb{R}$ and $\mathbf{C} : \mathbb{R}^3 \to \mathbb{R}^3$ for any choice of such $\mathbf{F}$.

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My question is, does there exist a similar result for $4$-vectors? I'm reading about gauge fields in QFT....is there some way to decompose any gauge field $A_{\mu}(x) = A_{\mu}(x_0,x_1,x_2,x_3)$ into a sum similar to the above?

I'm thinking something along the lines of: $A_{\mu}(x) = \partial_{\mu} \lambda(x) + \mathrm{something}$

I can't think of what the curl term would look like here.

• You have Hodge's theorem, $f=\mathrm df_1+\mathrm d^\star f_2+f_3$, where $f_3$ is harmonic. – AccidentalFourierTransform Nov 19 '17 at 21:05
• @AccidentalFourierTransform: Now, please translate into physicist-speak ;) What would the second term look like? Something like $\epsilon_\mu{}^{\alpha\beta\gamma} \partial_{[\alpha}B_{\beta\gamma]}$? – Christoph Nov 19 '17 at 21:52
• @Christoph For many of us, forms are now physicist speak too :) – JamalS Nov 19 '17 at 23:17
• @Christoph, it is customary to use the metric tensor and leave all 4 indices of the Levi-Civita pseudotensor on the same position, i.e. either all „up” or all ”down”. – DanielC Nov 19 '17 at 23:39
• According to chapter 2, page 21 of Prof Wheeler's classical field theory notes, Hodges theorem becomes: $V_\mu=S_\mu + I_\mu + V^0_{\mu} \;\text{where} \;\ \partial^{\mu}S_{\mu} = 0, \;\partial_{\mu}I_{\nu}-\partial_{\nu}I_{\mu}=0, \;\Box V^0_{\mu}=0$ Perhaps someone more competent than myself could show this as the accepted answer? – Larry Harson Feb 14 '18 at 1:04

The Hodge decomposition states that any $1$-form $\omega$ can be written as $$\omega = d\alpha+\ast\, d\ast\beta +\gamma$$ where $\alpha$ is a function, $\beta$ is a $2$-form and $\gamma$ is a harmonic $1$-form. Here $d$ is the exterior derivative and $\ast$ is the Hodge dual.
In coordinates, this means that we can write a covector $A_\mu$ as $$A_\mu = P_\mu + Q_\mu + R_\mu$$ The corresponding conditions are as follows. $P$ corresponds to $d\alpha$ and in Minkowski spacetime $$P = d\alpha\,\,\, \mathrm{for\,\,some}\, \alpha \iff dP=0$$ In coordinates this means that $\partial_\nu P_\mu -\partial_\mu P_\nu=0$. Similarly $Q$ satisfies $d\ast Q =0$, which in coordinates translates as $\partial^\mu Q_\mu=0$. Finally $R$ must be harmonic, so $\partial^\nu \partial_\nu R_\mu = 0$.
• Sorry, I mean the result I quoted in the question - the Helmholtz decomposition. This says that $F_{\mu} = \partial_{\mu} \Phi + \epsilon_{\mu\nu\sigma} \partial_{\nu} C_{\sigma}$ for 3D functions $F$ for some $\Phi$, $C$. Is there a way to see this specific result from the Hodge decomposition? – QuantumEyedea Sep 5 '18 at 13:48
• @Greg.Paul Roughly speaking yes: take your 3D vector field $F$ and apply the Hodge decomposition. You get a gradient plus a curl plus a harmonic term. Then if $F$ vanishes sufficiently fast at infinity (faster than $1/r$) you know that the harmonic function goes to zero faster than $1/r$, so it must be identically zero (this is a standard result in analysis). Hence you're left with gradient + curl. The details are more complicated however; see the question I linked in the answer. This paper may also be of help: arxiv.org/abs/1404.3679 – John Donne Sep 5 '18 at 16:05