## January 25, 2007

### Gravitational Leptogenesis?

I was very excited when I saw a recent paper by Alexander, Peskin and Sheikh-Jabbari (which, I guess, is based on this older paper). They claimed to produce an acceptable lepton-asymmetry during inflation if the inflaton is a (CP-odd) pseudoscalar field, $\phi$, with a coupling

(1)$F(\phi) R\tilde{R}$

to the curvature, where $F(\phi)$ is an odd function of $\phi$.

Fluctuations generated during inflation drive an expectation-value for $R\tilde{R}$, which generates a lepton-asymmetry via the anomaly,

(2)$\partial_\mu j^\mu_L = \frac{N}{16\pi^2} R\tilde{R}$

For reasons, that will become apparent shortly, however, they proceed to say something very unconventional about the anomaly coefficient, $N$:

In general, when heavy right-handed neutrinos are also added to the Standard Model, as is done in the seesaw mechanism for explaining the smallness of the neutrino mass, (2) will be correct in an effective theory valid below a scale $\mu$, of order of the right-handed neutrino mass. More concretely, $N$ can in general be a function of energy. At low energies, below the right-handed neutrino mass scale $N=3$. At higher energies, $N$ could be anywhere between zero to three, depending on the details of the particle physics invoked.

Umh … no. That’s not how anomalies work. The anomaly depends only on the massless spectrum and the anomaly coefficient does not run with energy.

Now, it’s true that, above the scale $\mu$, one can treat the right-handed neutrino, $\Psi$, as massless. In that theory, there’s a new current which is non-anomalous. $\partial_\mu (j^\mu_L + j^\mu_\Psi) = 0$ However, once one includes the Majorana mass for $\Psi$, $j^\mu_\Psi$ is not conserved, even classically. The correct, classically-conserved, current in the theory which includes the mass term, $\mu$ is $j^\mu_L$. And the anomaly in that current is the same, whether one calculates it in the high- or low-energy theory.

The motivation for the seemingly bizarre statement about the anomaly (2) becomes apparent a little later on. One would naïvely expect the gravitational contribution to the lepton asymmetry, during inflation, to be tiny, suppressed by at least a factor of ${\left(\tfrac{H}{M_{\text{pl}}}\right)}^2$. They, however, claim that there’s an enhancement over the naïve answer by a factor of ${\left(\tfrac{\mu}{H}\right)}^6$. This enhancement comes about because, in their calculation, $\langle R\tilde{R}\rangle$ is UV-divergent, dominated by short-distance effects near the cutoff, $\mu$.

I don’t really understand their calculation. But, if they are correct that, in the effective theory, $\langle R\tilde{R}\rangle \sim \mu^6$, then what this means is that, in the full theory, where one includes loops of $\Psi$, the contribution of those extra diagrams cancels the erstwhile UV divergence, yielding a finite (but $\mu$-dependent) answer. I don’t see why there should be such a cancellation, here.

Posted by distler at January 25, 2007 9:15 AM

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### Re: Gravitational Leptogenesis?

Slightly unrelated but …
I am unable to get windows binaries for the latest version (1.1.8) of itex2mml. Could you help me with this?

Posted by: Harry H on January 27, 2007 3:33 AM | Permalink | Reply to this

### itex2MML

This is one of the reasons I placed itex2MML in BZR. I don’t have access to a Windows machine on which I could compile itex2MML. But, if someone else does, who was willing to place a compiled binary in their own BZR repository, I’d be happy to “pull” it from them, and package it up for all to use.

I know that people have, in the past, been successful at compiling itex2MML under CygWin and under VisualStudio.

I don’t know whether anyone has compiled the Ruby bindings (that’s what’s new in 1.1.8) under Windows. But compiling the commandline tool should be the same as before.

Posted by: Jacques Distler on January 27, 2007 9:09 AM | Permalink | PGP Sig | Reply to this

### Re: itex2MML

I’ve tried to do that. But I’m unable to get itex2mml to generate equation numbers - even after properly installing (I think) ruby and maruku.

Posted by: harrh h on January 27, 2007 4:15 PM | Permalink | Reply to this

### Maruku

Ah!

That’s a different story.

Maruku offers a choice of MathML engines (the default is 'none'). You need to edit lib/maruku/defaults.rb to set itex2mml as the engine to use.

You also need to style the equation labels in your CSS stylesheet. I use

.maruku-eq-number {float:right}

All of this (except for the installation of itex2mml) is set up in my Instiki distribution.

Posted by: Jacques Distler on January 27, 2007 4:54 PM | Permalink | PGP Sig | Reply to this

### Re: Maruku

Thanks a lot.

Posted by: harry h on January 27, 2007 11:36 PM | Permalink | Reply to this

### Re: Gravitational Leptogenesis?

I read the paragraph on the anomaly as you did. If this is really what they meant, it is clearly wrong. But I somehow shy away from that conclusion, because Peskin is one of the authors. Do you think he misunderstands basic facts about anomalies in QFT?

Posted by: Dan on January 27, 2007 3:20 PM | Permalink | Reply to this

### Anomalies

Do you think he misunderstands basic facts about anomalies in QFT?

No, I don’t.

Which is why I wrote this post.

The subsequent analysis (as I explained) relies on this paragraph. Since the “obvious” interpretation of this paragraph is plainly wrong, I’d like to know what the “correct” interpretation is.

Posted by: Jacques Distler on January 27, 2007 3:47 PM | Permalink | PGP Sig | Reply to this

### Re: Gravitational Leptogenesis?

I confess I’m a bit confused by your comment:

The anomaly depends only on the massless spectrum

In the Standard Model, the fermions are not massless, getting masses by the couplings to the Higgs. However, we still check for anomaly cancelation.

Possibly there are subtleties when the fermions get their masses by the Higgs mechanism…

Anyway, what Alexander et al might want to say is that some of the chiral fermions have very large masses but they get them by the Higgs mechanism so they should still count as massless when computing the anomaly coefficient.

Posted by: Sidious Lord on January 28, 2007 2:17 AM | Permalink | Reply to this

### The Dark Lord saith

Lord Sidious [a very silly nom-de-plume (nom-de-clavier?)] said:

Possibly there are subtleties when the fermions get their masses by the Higgs mechanism…

I’m afraid I don’t know what you are driving at, so I guess you will have to elaborate.

Lepton number is (almost) an exact classical symmetry of the low energy effective Lagrangian. It is violated only at the level of the dimension-5 operator

(1)$- \frac{c^2}{M} h h \psi_L \psi_L + \text{h.c.}$

(here and in the following, I use 2-component Weyl spinor notation).

Since $\langle h\rangle \ll M$, it makes sense to neglect this term and treat lepton number as a nearly exact classical symmetry.

In the high energy theory, the situation is a little (but only a little) more subtle. The Lagrangian contains terms $\mathcal{L} = \dots + M \Psi \Psi + 2c\, h \psi_L \Psi + \text{h.c.}$ The second term violates lepton number.

Again, since $\langle h\rangle \ll M$, this term induces a small off-diagonal entry in the $\Psi$-$\nu$ mass matrix. [Going to a mass eigenstate basis induces the neutrino mass term, (1).] In the same spirit as before, it makes sense to neglect the small off-diagonal mass term, and treat lepton number as a nearly exact classical symmetry of the high energy lagrangian as well.

In either case, the anomaly coefficient, $N=3$, irrespective of whether one calculates it in the low- or high-energy theory.

Posted by: Jacques Distler on January 28, 2007 9:29 PM | Permalink | PGP Sig | Reply to this

### Re: The Dark Lord responds

After picking on my nom de clavier, Jacques Distler said:

I’m afraid I don’t know what you are driving at, so I guess you will have to elaborate.

My comment was triggered by the following facts:

• You said that the anomalies depend only on the massless spectrum.
• The fermions in the Standard Model are not massless.
• The computation that the anomalies cancel in the Standard Model is very well known.
• When discussing anomalies either by Fujikawa’s method or by computing the triangle diagrams, the case of spontaneous symmetry breaking does not seem to be included (I’m reffering explicitely to the couplings that give masses to the fermions). Then, the result for the massless fermions is used for the computation of anomalies in the Standard Model. Before I read your post I never questioned whether this is legitimate or not…
Posted by: Lord Sidious on January 29, 2007 9:08 AM | Permalink | Reply to this

### Re: The Dark Lord responds

Massive fermions cancel out of any anomaly computation. Indeed, any fermions — which can be given a mass, compatible with the (gauge or global) symmetries of the model — cancel out of the anomaly computation.

It really doesn’t matter whether the mass, in question, arises from spontaneous symmetry-breaking.

There is, I will grant you, a subtlety associated to spontaneous breaking of the (gauge or global) symmetry. Namely, integrating out the fermions can induce a Wess-Zumino term for the Goldstone bosons, which then reproduces the anomaly.

Consider the following example.

1. Let $\psi$ be a chiral fermion in the fundamental $\mathbf{N}$ of $SU(N)$ (which we will, ultimately, try to gauge but which, for the moment, think of as just a global symmetry).
2. Let $\chi_i$ be $N$ chiral fermions which are singlets of $SU(N)$.
3. Let $\phi_i$ be $N$ scalars, each of which is in the anti-fundamental $\overline{\mathbf{N}}$ of $SU(N)$.

With a generic symmetry-breaking potential, we can arrange that all of the $\phi_i$ have expectation-values, which break $SU(N)$ completely. Moreover, the Yukawa coupling, $\phi_i \psi \chi_i +\text{h.c.}$ gives masses to all the fermions.

Whither the anomaly in the $SU(N)$?

It turns out that the anomaly is reproduced by a Wess-Zumino term for the Goldstone bosons (the massless fluctuations of the $\phi_i$ tangent to the space of minima of the scalar potential).

This effect is well-known, and was first (to my knowledge) worked out by d’Hoker and Farhi for the case of integrating out the top quark in the Standard Model.

Posted by: Jacques Distler on January 30, 2007 1:56 PM | Permalink | PGP Sig | Reply to this

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