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Became a little confused over last hour trying to understand the units of the Heaviside vector. So far I have come up with this from reading online:

The Heaviside vector allows us to write the energy density flux as

$\mathbf{U} = \frac{1}{4 \pi G}(\Gamma^2 \times c^2\Omega^2)$

where this time $\frac{1}{4 \pi G}$ is the permittivity. Being a flux, I expect this to mean it has units of energy density per unit of time.

So I just take a look at the dimensions, it appears gamma is defined as an acceleration - we have $\Gamma = c\ \Omega$ with units of acceleration, but I am not understanding something because this divided by Newtons constant has units of mass over length squared and so it seems I missing some crucial understanding.

To check units further, we take a look at a Heaviside relationship for momentum density as

$\mathbf{p} = \frac{1}{c^2}\mathbf{H} = \frac{1}{4 \pi G} [\Gamma \times \Omega]$

and so the energy (related to the stress energy of the system is)

$\mathbf{U} = \mathbf{p}c = \frac{1}{c}\mathbf{H} = \frac{1}{4 \pi G} [\Gamma \times c\ \Omega]$

This relationship nearly made sense to me, the units are $a/G = m/r^2$and then with the additional omega (torsion) if there had been an inverse speed of light, then it would be another space derivative gathering a density (at least). But the speed of light coefficient is not inverse, it coupled to torsion like an acceleration term. Suppose I write that inverse speed of light in

$ = \frac{1}{4 \pi G} [\Gamma \times \frac{1}{c}\ \Omega]$

Then ok, we have the acceleration, divided by Newtons constant and an extra space derivative, we have a density... but if this was the right way to write it, it would be inconsistent with

$\mathbf{U} = \mathbf{p}c = \frac{1}{c}\mathbf{H}$

Because we started with a momentum density, not a mass density. Say it had been a mass density, the energy density I get would be

$\rho c^2 = \frac{c^2}{4 \pi G} [\Gamma \times c^{-1}\ \Omega]$

Have I successfully muddled myself up? To be fair, I am sure this is close to saying the same thing as

$\mathbf{U} = \mathbf{p}c = \frac{1}{c}\mathbf{H} = \frac{1}{4 \pi G} [\Gamma \times c\ \Omega]$

But then I don't get this

$\mathbf{U} = \frac{c^4}{4 \pi G}(\Gamma^2 \times \Omega^2)$

as the flux.

Can anyone shed some light on my confusion concerning the units? Thanks up front.

reference:

https://en.wikiversity.org/wiki/Heaviside_vector

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  • $\begingroup$ Can anyone answer this at all? $\endgroup$ – Gareth Meredith Mar 22 at 7:46
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Since it has approached a month, I have decided to provide my own answer. After some consideration, I think the wiki article has the wrong representation of the units.

This was based on Heavisides equation listed in a wiki article, but I am sure the dimensions are wrong. Let's take a dimensional look at the equations. The gravitational analogue of the electromagnetic field density taken from wiki as:

$\mathbf{H} = - \frac{c^2}{4 \pi G}(\Gamma \times \Omega)$

where

$-\frac{c^2}{4 \pi G}$

is the gravitational permeability $\frac{1}{\mu_G}$. The Heaviside vector allows us to write the energy density

$\mathbf{U} = \rho c^2 = \frac{c^2}{4 \pi G} [\Gamma \times c^{-1}\ \Omega]$

where this time $\frac{1}{4 \pi G}$ is the permittivity. The equation we would obtain is

$\gamma_0 \mathbf{U}_{\mu \nu} = \frac{c^2}{\mu_G}\ [(\mathbf{\nabla}_{\mu} \cdot \Gamma_{\nu} + \nabla_{\nu} \cdot \Gamma_{\mu})\gamma_0 + 2i \vec{\sigma} \cdot (\Gamma_{\mu} \times c^{-1}\ \Omega_{\nu})]^k \gamma_k$

where $\mathbf{U}$ is the energy density of the gravitational field. To make sense of the dimenions, the gravitational field has units of acceleration, when weighted by G, it gives a mass over radius squared, the remaining derivative gives a density.

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