Reflecting on the groups of units (natural and derived) herein is introduced a function \([\:]\) that is a type of commutator for an operator utilizing elements from the group \(\mathcal{G}\) by an observer at one location and how they “commute” to an observer at another location.
NUVO Commutator
A function to evaluate an operator’s (+,-,*) resultant value of two units when compared between observers. Consider the function of the multiplication operator \((\cdot)\). The NUVO commutator results in either a zero value or a non-zero value.
\begin{equation}
[a,b]=\overbrace{(a\cdot b)}^\text{Observer 1}- \overbrace{(a’\cdot b’)}^\text{Observer 2}
\end{equation}
Where \(a,b\) are an element of \(\mathcal{G}\).
Consider the function of multiplication \((\cdot)\). In the subgroup \(\mathcal{I}\), the product of any element with another element of the subgroup commutes with a value of zero. Let \(a,b\) be an element of \(\mathcal{I}\) then
\begin{equation} \label{eqCommutaton}
[a,b]=\overbrace{(a\cdot b)}^\text{Observer 1}- \overbrace{(a’\cdot b’)}^\text{Observer 2} = 0
\end{equation}
When \(a,b\) is an element of \(\mathcal{N}\)
\begin{equation}
[a,b]=\overbrace{(a\cdot b)}^\text{Observer 1}- \overbrace{(a’\cdot b’)}^\text{Observer 2} \ne 0
\end{equation}
Members of the group \(\mathcal{I}\) with multiplication are invariant as measured locally to all observers. Members of group \(\mathcal{I}\) must be handled carefully when making observations and comparing results with observations from other locations. For example, velocity is a member of the group but length and time are not a member. Thus, when working with velocity (like the speed of light), the value c is equal to:
\begin{equation}
c= \frac{1}{\sqrt{\epsilon_0 \mu_0}} = \nu \lambda
\end{equation}
While c, as a member of the group is invariant, the elements that make up c may or may not be invariant. As the value \(\lambda\) is a length and not a member of \(\mathcal{I}\), thus the value of energy as derived by:
\begin{equation}
E_\gamma = \frac{h c}{\lambda}
\end{equation}
is not invariant. But one may say that energy is a member of \(\mathcal{I}\)! Therefore the definition of what makes up subgroup \(\mathcal{I}\) must be adjusted to be more precise.
Measurement Group \(\mathcal{I}\) Refined
Transformation scalar for \(\phi\) equals 1, and the NUVO Commutation must equal 0.
To be a member of \(\mathcal{I}\) the commutation must equal zero for the measurement. This definition allows some derived measurements, such as energy, to be excluded from the group \(\mathcal{I}\) because their local measurement is not always invariant. Other measurements of energy commute with zero value, for instance:
\begin{equation}
E_0=m c^2
\end{equation}
Commutes to a value of zero and is invariant by all observers. But the setting equal of two equations and solving for a solution does not guarantee the solution to the equation is invariant, thus it doesn’t guarantee the equation is a member of \(\mathcal{I}\). Consider:
Example
This example demonstrates how some derived units in \(\mathcal{I}\) do not pass the NUVO Commute requirement and, therefore, must be handled with care when transforming.
\begin{equation}
\begin{split}
E_{\gamma} = \frac{h c}{\lambda} = E_0 = m c^2 \
\text{therefore } \
\frac{h c}{\lambda}=m c^2 \textbf{ is not invariant}
\end{split}
\end{equation}
One may remark in the example the transformation of energy from mass to photons is invariant with no known exception. If so, then why is there a red shift or a blue shift in the energy of a photon when moving radial in a gravity well, but there is no shift in the energy of a mass moving radial in a gravity well? The energies are not invariant when transferred from observer to observer, thus, they do not fit the definition of invariance for a transformation.
The LHS of \(\frac{h c}{\lambda}=m c^2\) is not a member of \(\mathcal{I}\) while the RHS is a member of \(\mathcal{I}\). Another consequence of units of measure not being a part of \(\mathcal{I}\) is when an observer in GR tries to reduce a local space (using GR terms) down to the level of an infinitesimal patch so that Minkowski space can be applied such that the laws of physics are the same for all observers. The author contends it is impossible to locally measure the acceleration of an observed particle accurately. At the infinitesimal level, the particle observed accelerating is in a different location on the manifold (the value of \(\phi\) has changed and is in a mixed state) due to its acceleration and will have a local effect (discussed later in this chapter). The commutation between the units as measured by the particle accelerating and the units measured by the observer at rest relative to the accelerating particle is not zero. Thus the observation will have an uncertainty proportional to the value of the commutation. The system will be considered in a mixed state. It should be noted (and discussed later in the book) that constants like Planck’s constant \(h\) have units attached to them. It will be shown in order to maintain the invariance of velocity with a NUVO transformation, the units for any constant must also be transformed. This provides a transformation for photon energy to transfer invariant from lab to lab, but the invariance is only for the local ratios. In other words, the photon transferred from one lab to another does not have its energy transfer invariant.