# Why *should* the mass of elementary particles theoretically be of the magnitude of the Planck mass?

Why should the mass of elementary particles be theoretically of the magnitude of the Planck mass?

I've read that already a few times but I don't understand why it should be that way.

For example: Zwiebach - A first course in string theory, p.55

If the fundamental theory of nature is based on the basic constants $G$, $c$, $\hbar$ it is a great mystery why the masses of the elementary particles are so much smaller than the "obvious" mass $m_p$ that can be built from the basic constants.

$m_p$= Planck mass

Zwiebach sounds to me as if it would be very logical that the mass of the elementary particles should theoretically be around $m_p$. Could you explain me this connection?

Edit: It is not about the problem that there is a gap between the masses.

• They shouldn't and aren't. See en.wikipedia.org/wiki/Planck_mass – Harry Johnston Nov 17 '11 at 1:21
• @Harry: consider posting that as an answer (and if you can expand on it, better yet) – David Z Nov 17 '11 at 1:34
• Ok my question is maybe formulated a little bit misleading (i will adjust it)! – ungerade Nov 17 '11 at 1:36
• @HarryJohnston: According to Zwiebach it sounds to me that they some kind of "should" (i know that they don't and that seems to be a problem. – ungerade Nov 17 '11 at 2:02
• The usual issue is with fundamental scalars. In this case of the Higgs, which is the only fundamental scalar in the standard model, this is the "Hierarchy Problem" and the wikipedia page explains it adequately. One does not need such fine tuning for fermions for technical reasons. – BebopButUnsteady Nov 17 '11 at 2:17

## 1 Answer

The mass is generally a relevant parameter in the renormalization group sense, so that tuning the mass to zero is going to a special point. When you see a parameter tuned to a special point, you have to ask why is it so?

The Planck scale is very closely analogous to the atomic scale of space and time. For atomic systems, you can ask nearly the same question about the correlation length. If I give you a generic material and you make a density perturbation of small size, you remove a few atoms in a certain microscopic region, you can ask, how far before the density perturbation dies away and the material looks unperturbed?

For a generic solid, the answer is that the perturbations decay exponentially within a few atomic radii. This is generically true, it holds because the atoms set the distance scale in the solid, so the decay rate is just one atomic radius by dimensional analysis, times a coefficient which is generically order 1, unless there is a reason for it to be zero.

But if you tune your material to a critical point, say you make high pressure water at just the right temperature, then the correlations decay as a power law, the mass parameter of the effective theory is tuned to zero, and the fluid has density perturbations on all scales, and it looks milky white because there are wavelength-of-visible-light size perturbations. This is not an idle analogy: the critical limit of a material is the massless limit of the statistical theory describing its fluctuations, and statistical theories are related to quantum field theories by an analytic continuation in time in the path integral. So whatever sets the atomic scale of space-time, this is the quantity that you expect to set the correlation length, the inverse mass, of the particles.

But our universe is filled with particles of inverse mass much larger than the graininess scale, much large than the Planck length. When you find a solid tuned to a critical point, you have to ask why it is tuned to be so. It would be strange to find a material whose density is critical without fine-tuning.

### Typical scalars

To see more clearly how this works, consider the Ising model as a model of a typical scalar field. The average spin in a region is the value of the field, and if you look at the correlations between these average spins, they are described by the following statistical probability

$$P(h(x)) = e^{-\int h^2}$$

This probability distribution is a sum in the exponent (an integral is a sum, really), so it is a product of independent factors at each point x, and this means that h(x) is independent at each point.

A field like this, whose value at any point is completely independent from the values at any other point is called "ultralocal". Ultralocal means that the field correlations decay instantly. In our universe, space and time are not really well defined at the Planck length, so we expect that the analog of the lattice scale will be set by the Planck length.

Now if you tune the Ising model to the critical point, the higher derivative terms in the probability distribution suddenly become visible:

$$P(h) = e^{-\int Z|\nabla h|^2 + t h^2 + \lambda \lambda h^4 }$$

In the limit that t is tuned to the critical value (which is some large negative number, it isn't at zero), the effective mass of the field theory becomes tiny, the correlation scale goes to infinity, and you have a massless limit.

Although the massless limit is defined by a free theory in which the parameter correpsonding to t is zero, the presence of the self-interaction $\lambda$ shifts the critical value away from t=0 to some other value. This makes it doubly mysterious, because to get the mass to be zero, the more fundamental t parameter has to be tuned to some absurdly small distance away from a random looking number.

### Typical uncharged Fermions

For Fermionic field theories, there is a natural mass scale. If you have one Fermionic representation of the Lorentz group, the Lagrangian is:

$$\int \bar{\psi}^\dot\alpha \sigma^\mu_{\dot\alpha\beta} \partial_\mu \psi^{\beta} + M \epsilon_{\alpha\beta}\psi^\alpha\psi^\beta + M \epsilon_{\dot{\alpha}\dot{\beta}}\bar\psi^\dot{\alpha}\bar\psi^\dot{\beta}$$

in two-component notation. This is massive with mass M. In order to have a massless Fermion, you need to tune the parameter M. I wrote this down (schematically--- you should choose a good two-index convention and check it) because people often wrongly claim that Weyl fermions can't get a mass. This mass is called a "Majorana mass", because you can rewrite two component complex spinors in 4d as real 4 component spinors, and then the mass term looks like a Dirac equation mass.

### Naturally massless scalars

In a solid, you are surprised when the statistical fluctuations of the density are massless. But you are not surprised that the sound-modes (the transverse movements of the atoms) have power law correlations. The sound waves are naturally massless, because the solid breaks the translation symmetry.

The naturally massless scalar fields are the Goldstone bosons. These are the analog of sound waves. These are described in the Wikipedia article on Goldstone bosons.

### Naturally massless Fermions

Although Fermions can have masses, the mass term mixes the particle with the antiparticle. If you give the field $\psi$ a phase, $\psi\psi$ is not invariant, only $\bar\psi\psi$ is. This is why the Majorana mass term is often ignored.

In order to give a charged Fermion a mass, you must have a partner Fermion of opposite helicity and the same charge to couple it to. The standard model is entirely built from fermions which do not have a partner with the same charges, so these Fermions are naturally massless.

### Natural small-mass scales from gauge fields

The last ingredient you are allowed to naturally use is gauge fields, and these can have couplings which are small, but not small masses. But a confining gauge field in 4d has a confinement length which is exponentially large in the inverse coupling. This means that you can make extremely small masses by starting even with not-so-small couplings at the Planck scale.

The running of the QCD coupling in our universe sets the mass scale for the proton, and for atoms. It is independent of the Higgs mechanism, which is believed to set the mass for the quarks and leptons.

• Thanks for your answer. I thought the answer would be much easier because Zwiebach just mentions it without further explanation (and he explains a lot of easy stuff quite detailed). I currently try to understand the first five paragraphs and would like to see it more explicit. Could you suggest me a good starting point to learn this? – ungerade Nov 18 '11 at 23:03
• I went into detail, but it is possible to just say "the inverse mass is a length scale, and what is the natural length scale in the universe?" and leave it at that. The argument is second nature to most physicists today. The theory of renormalization is covered well in an encyclopedic way in the reference collection Domb & Green, but I suppose you are interested only in the original exposition, which is Wilson, and there is a 1974 Review of Modern Physics article which is a classic. You can also read a mongraph of Parisi called "Statistical field theory", which avoids epsilon expansion. – Ron Maimon Nov 19 '11 at 3:49
• Ron, presumably G "should be" subject to renormalization like any other coupling constant; would it be the bare value of G or the observed value that should be used in calculating what the mass of elementary particles "should be"? – Harry Johnston Nov 23 '11 at 18:32
• @Harry: The running of G is different from the running of other coupling constants, because it is dimensionful, so it runs as a power of scale, not as a log. The power-running is so much stronger than any residual logarithmic running that you might as well ignore the log running corrections, they can't change the Planck scale in any significant way. – Ron Maimon Nov 23 '11 at 20:47