content_image/351931.jpg content_image/351930.jpg content_image/351929.jpg content_image/351928.jpg content_image/351927.jpg
Finally, we need to consider the scalar bidoublet and the scalar left doublet. In this model, ( φ 0 $_{1}$, φ − 1 ) will be heavy, and the two doublets ( φ + $_{2}$, φ 0 $_{2}$), ( φ + $_{L}$, φ 0 $_{L}$) are similar to the usual two Higgs doublets considered in the SM. The linear combination ( v 3 Φ 2 − v 2 Φ$_{L}$) / √ v 2 2 + v 2 3 = [ H + $_{L}$, ( H 0 $_{L}$+ iA 0 $_{L}$) $_{/}$√$_{2] is physical and light.}$ 

## IV. GAUGE SECTOR 

Since e has L = 1 and n has L = 0, the W + R of this model must have L = S − T 3 R = 0 − 1 = − 1. This also means that unlike the conventional LRM, W ± R does not mix with the W ± L of the SM at all. This important property allows the SU (2)$_{R}$ breaking scale to be much lower than it would be otherwise, as explained already 22 years ago [2, 3]. Assuming that g$_{L}$ = g$_{R}$ and let x ≡ sin $^{2}$θ$_{W}$ , then the neutral gauge bosons of the DLRM (as well as the ALRM) are given by 

<img src='content_image/351931.jpg'>

Whereas Z couples to the current J 3 L − xJ$_{em}$ with coupling e/ √ x (1 − x ) as in the SM, Z ′ couples to the current 

<img src='content_image/351930.jpg'>

with the coupling e/ √ x (1 − x )(1 − 2 x ). The masses of the gauge bosons are given by 

<img src='content_image/351929.jpg'>

<img src='content_image/351928.jpg'>

where zero Z − Z ′ mixing has been assumed, using the condition [3] v $_{2}$/ ( v $_{2}$+ v $_{3}$) = x/ (1 − x ). Note that in the ALRM, Δ$_{R}$ is absent, hence v 6 = 0 in the above. Also, the assignment of ( ν, e )$_{L}$ there is different, hence the Z ′ of the DLRM is not identical to that of the ALRM. At the LHC, if a new Z ′ exists which couples to both quarks and leptons, it will be discovered with relative ease. Once M$_{Z}$ ′ is determined, then the DLRM predicts the existence of W ± R with a mass in the range 

<img src='content_image/351927.jpg'>

