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Searching for Higgs in Excel

Simple linear expression

Plot of Mh against Mt and Mw

Before the discovery of the Higgs particle was comfirmed, as sumarised (NS, 07-Sep-2013, Instant Expert No.35), a much earlier issue of New Scientist (NS, 30-Mar-2002, p28) gave an excellent top-level summary of the justification for belief that the Higgs particle exists, and a later issue (NS, 03-Mar-2007, p8) gave a summary of the then present state of the hunt. The former explains how the photon, W+, W- and Z particles are all related, but that the Higgs particle interacts asymmetrically with them, so that photons are massless, but the other three are massive. Meanwhile, the latter article includes a graph, showing the contours of a three-dimensional plot of the expected mass of the Higgs boson, mh, against any pairing of masses of top quark, mt, and W boson, mw. Unfortunately, it only explicitly shows two contours (mh=114 GeV and mh=400 GeV).

Here is a somewhat simplistic expression for reading off other values of mw, given any pair of values of mt (in the vicinity of 173.34 GeV) and mh (in the vicinity of 132 GeV, which is the mean of 115 and 150 (NS, 20-Feb-2010, p5)):

mw = Ka*(mt-Kto) + Kb*ln(mh/Kho) + Kwo

In this radiative correction expression for the W boson, Ka (a dimensionless constant of about 0.006) is the gradient of mw against mt for constant mh; Kb (a dimensional constant of about -0.07 GeV) is the gradient of mw against ln(mh) for constant mt.

The calculations on the rest of this page are based on data obtained from:

  • Pour la Science, No.240, Oct-1997, p73
  • A web page, giving more detailed data than the New Scientist, 03-Mar-2007, p8 article
  • A web page, giving further data, dated 23-Feb-2012

Specifying a reference point

The expression needs a reference point (Kto, Kho, Kwo) to fix the position of the plane in its three-dimensional space. Conventionally, two of the coordinates are chosen to be zero, and the remaining one is set as the intersect on the corresponding axis.

Thus, setting Kto and ln(Kho) to zero, and hence setting Kho to one (namely, 1GeV), Kwo, as the intersect on the mw axis, has a value of about 80.83 GeV.

The convenience of choosing one of the points of intersection with the axes as the reference is that two of the constants do not need to appear in the final equation (since subtracting zero, or dividing by unity, is an identity operation).

Although (0,1,80.83) (all in GeV) is the obvious choice of constants, it is not the only possibility. For example, using the reference point (0.0,1e-9,82.28) puts the mh axis at 1eV rather than 1GeV, but begs the question "why eV, rather than J, or Planck-energy units, or some other unit?"

Using the intercept on the ln(mh) axis instead would have been tempting, giving the reference point (0,Kho,0). Unfortunately, Kho then has a value of around 10510, and Excel refuses to work with it directly.

Using the intercept on the mt axis, namely the reference point (Kto,1,0), gives a value of Kto of -13109 GeV (and is little changed for the reference point (Kto,1e-9,0))

There is one further interesting point for this particular function: the one where Kt=Kh=Kw (since, although there are then three constants to handle in the equation, they all have the same value). This leads to the following expression:

mw = Ka*(mt–Kc) + Kb*ln(mh/Kc) + Kc

An inverse function, to find mh from given values of mt and mw, is therefore:

mh = Kc * exp( ( mw–Kc – Ka*(mt–Kc) ) / Kb )

Relevance post Higgs

Now that the discovery of the Higgs particle is confirmed, the role of this expression can be reversed. The observed masses of the three particles can be used as the input (mt,mh,mw) and the closest fit values can be found for the three constants (Ka,Kb,Kc).

Date Ka Kb Kc
Oct-1997 0.0065 -0.0695 79.70
  -7% -3% +0.16%
Mar-2007 0.006052 -0.06760 79.83
  -3% +4% +1.5%
Feb-2012 0.005855 -0.07025 80.99
  +7% -6% -1.4%
Post Higgs 0.006263 -0.06611 79.83

For convenience, here is a short Excel spreadsheet that has been coded up to evaluate the two expressions given earlier.

It should be noted, though, that care should be taken when using the inverse function (mh in terms of mt and mw) because the exponential function makes it very sensitive to small rounding errors in its argument.


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