atdotde: February 2008

Thursday, February 28, 2008

Zeit Sudoku Bookmarklet

As you know, I am a Sudoku addict. For a nice ten minute break I often download the daily sudoku from Die Zeit. Until recently, from this bookmarked page, I had to change the level from "leicht" to "schwer" and then press the button for the PDF version to print out.

But now, they introduced even another page before that (where you have to choose between the Flash version and the puzzle from the printed (only weekly) newspaper or the oldstyle version that allows the download. You could no longer bookmark the next page as there the URL already contained some sort of session ID.

Of course, they want me to go through all these pages to make sure I do not miss any of the ads they want to present me. But I think that this new version requires a few mouse clicks too much and so I decided to have a look at the page's source code.

It turns out that the URL for the PDF no longer contains a session ID but instead contains today's date, slightly more than you can do with a static bookmark. But that got me thinking that one might be able to solve this problem with a bookmarklet, a bookmark that makes use of the fact you can have JavaScript in a URL and in a bookmark.

Last time I looked into JavaScript was roughly ten years ago and at that time it seemed like a very stupid idea to have some crippled language where you have to transmit all of the source code to the browser which then slowly interprets it if you can do much much more powerful things on the server side with CGI scripts.

Since then, a lot of time has passed and I have heard many interesting things (let me mention only AJAX) suggesting I should maybe reconsider my old dismissal of JavaScript. I had a look though a number of reference sheets and here it is: My first own JavaScript sniplet: A sudoku download bookmarklet. Clicking on this link (or bookmarking it and retrieving the bookmark) brings you directly to the PDF version on the latest "schweres" sudoku! Here is the source:


javascript:var d=new Date();
  var m=d.getMonth()+1;
  if(m<10){m="0"+m};
  var t=d.getDate();
  if(t<10){t="0"+t};
 open('http://sudoku.zeit.de/sudoku/kunden/die_zeit/pdf/sudoku_' +  
      d.getFullYear()+'-'+m+'-'+t+'_schwer.pdf')

Thursday, February 07, 2008

Geometric Hamilton Jacobi

Today, over lunch, togther with Christian Römmelsberger, we tried to understand Hamilton-Jacobi theory from a more geometric point of view.

The way this is usually presented (in the very end of a course on classical mechanics) is in terms of generating functions for canonical transformations such that in the new coordinates the Hamiltonian vanishes. Here I will rewrite this in the laguage of symplectic geometry.

As always, let us start with a 2N dimensional symplectic space M with symplectic form $\omega$ . In addition, pick N functions $q^i$ such that the submanifolds $L(x^i)=\{m\in M|q^i(m)=x^i\}$ are Lagrangian, that is $T_mL(x^i)$ is a Lagrangian subspace of $T_mM$ (meaning that the symplectic form of any two tangent vectors of $L(x^i)$ vanishes). If this holds, the $q^i$ can be regarded as position coordinates.

Starting from these Lagrangian submanifolds, we can locally find a family of 1-forms in the normal bundle $\theta\in N^*L(x^i)$ such that $\omega=d\theta$ . You should think that $\theta=p_idq^i$ for appropriate momentum coordinates $p_i$ on the Lagrangian leaves of constant $q^i$ . But here, these are just coefficient funtions to make $\theta$ a potential for $\omega$ .

Now we repeat this for another set of position coordinates $Q^i$ which we assume to be "sufficiently independent" of the $q^i$ meaning that $TM=TL(q^i)\oplus TL(Q^i)$ . This implies that locally $(q^i, Q^j)$ are coordinates on M. With the $Q^i$ comes another 1-form $\Theta$ and since $d\Theta=\omega=d\theta$ are locally related by a "gauge transformation". We have $\theta = \Theta + dF$ for a function F.

Let's look at $\theta$ a little bit closer. A general normal 1-form would look like $f_i(q^j,Q^k)dq^q$ . But since we started from Lagrangian leaves, there is no $dq^i\wedge dq^j$ in $\omega$ and thus $\theta=p_i(Q^j)dq^i$ . But expressing this in coordinates yields

$p_i(Q^j)dq^i=\theta=\Theta + dF= P_idQ^i + \frac{\partial F}{\partial q^i}dq^i + \frac{\partial F}{\partial Q^i}dQ^i$

Comparing coefficients we find $p_i = \frac{\partial F}{\partial q^i}$ and $P_i = -\frac{\partial F}{\partial Q^i}$ . You will recognize the expressions for momenta in terms of a "generating function".

What we have done was to take two Lagrangian foliations given in terms of $q^i$ and $Q^i$ and compute a function $F$ from them. The trick is now to turn this procedure around: Given only the $q^i$ and a function $F(q^i,X^j)$ of these $q^i$ and some N other variables $X^j$ , one can compute the $Q^i$ as functions on M: Take a point $m=(q,p)\in M$ and define $Q^i(m)$ by inverting $p_i = \frac{\partial F(q,X=Q)}{\partial q^i}$ . For this remember that $p_i$ was defined implicitly above: It is the coefficient of $dq^i$ in $\theta$ .

Up to here, we have only played symplectic games independent of any dynamics. Now specify this in addition in terms of a Hamilton function h. Then the Hamilton-Jacobi equations are nothing but the requirement to find a generating function $F$ such that the $Q^i$ are constants of motion.

Even better, by making everything (that is h and Q and F) explicitly time dependent, by the requirement that the action 1-form is invariant: $\theta- hdt = \Theta - Hdt$ giving $H = h +\frac Ft$ we get a transforming Hamilonian and we can require this to vanish:

$0= H = h +\frac Ft$

If we think of the Hamiltonian h given in terms of the coordinates $(q^i,p_i)$ this is now a PDE for F which has to hold for all $m\in m$ . That is, writing F as a function of $q^i$ and $X^i$ it has to hold for all (fixed) $X^i$ as a PDE in the $q^i$ and t.

Tuesday, February 05, 2008

Breaking it softly with this song

Today, I would like to discuss a topic that came up in our lunch seminar. The issue however is not specific to this paper but applies to many model building constructions including KKLT. It is about the nature of supersymmetry breaking in many string theory constructions. At least for me the issue was not completely resolved during the seminar. So I dug a bit into the literature. This made it a bit clearer to me but still I would not say I completely understand it.

Maybe somebody more knowledgeable out there reads this and can give me some hints. Others might find interesting what I found out so far. This twofold motivation for writing this up should explain the nature of this post: I will start out with some well known background on susy breaking which everyone who for example has read Wess and Bagger should know. But then it will get more and more technical and end up right in the middle of my confusion.

As you know supersymmetry is a beautiful idea if you want to solve the hierarchy problem of why the electroweak scale of 100GeV given by the Higgs mass is relatively stable to quantum corrections. In plain vanilla QFT in 4d, the mass term of a scalar is quadratically divergent. If you happen to regulate your divergent integrals with a cut off you would expect your renormalised scalar masses generically to be given by the cut-off scale (which in case of the standard model would probably as high as the Planck scale).

In a supersymmetric theory, however, the leading divergeces of bosons and fermions running in the loop cancel. The running of the scalar mass is then only logarithmic, something you can live with.

The problem with supersymmetry is only that it is not a symmetry of nature. There is no mass degeneracy between bosons and fermions. In fact, the difference in mass is so big that we have so far seen only half of the multiplets. Thus at best, supersymmetry is broken in the real world.

This breaking of susy can be realised to different degrees: The simplest form is that it is only spontaneously broken: The theory (given in terms of a Lagrangian) is supersymmetric, applying a susy transformation to a solution again gives a solution. It just happens that the ground state is not invariant under such a transformation and thus susy is not a symmetry of this state.

For simplicity of exposition, let us restrict ourselves to rigid supersymmetry. Then the action is given in terms if two functions, the Kähler potential K and the superpotential W. Let us focus on the superpotential (the argument could be extended to a non-trivial Kähler potential as well). It is a function of the chiral superfields.

If the ground state of the theory were supersymmetric it would have to have zero energy. But the scalar potential is given by $V=\sum_i|\partial_i W|^2$ (the sum is over all chiral superfields). This can only be zero if there is a point in field space where $W$ is stationary w.r.t. all fields $\partial_iW=0$ . If these equations do not have a simultaneous solution, the energy cannot be zero and susy is spontaneously broken. The loop calculations are not affected by this and thus the mass still runs logarithmically. There are good reasons to believe that this is not how susy breaking works in the real world.

The other extreme is that the theory is supersymmetric right form the start. The Lagrangian contains all sorts of terms and there is no hint of susy at all. The theory is a supersymmetric as generic rock is spherically symmetric. Of course in this situation there are no cancellations between divergences and down the drain is our solution of the hierarchy problem.

There is however a possibility in the middle between the two extremes: The theory is not supersymmetric, there are terms in the Lagrangian that break this symmetry right from the start. But these are not the most general terms but only those which do not destroy the logarithmic running of the mass. This situation dubbed "soft breaking" is what people think nature has chosen and for example it is realised in the MSSM.

Before you start calculating loop diagrams to check if in some non-susy theory the breaking happens to be soft let me tell you about an alternative characterisation: A theory is softly broken if it is possible to replace some coupling constants (the numerical coefficients of the susy breaking terms in the Lagrangian) by dynamical (super)-fields and add some other terms for these fields in such a way that you end up with a supersymmetric theory. You arrange the extra terms in such a way that in this theory the supersymmetry is spontaneously broken in such a way that the derivative in the direction of the new fields (called the "hidden sector") of the superpotential does not vanish in the ground state. Thus the theory is only softly broken if it can be augmented by a hidden sector such that the bigger theory is only spontaneously broken.

For example if $\Phi$ is one of the (super) fields we started with and $H$ is a hidden sector superfield. If the ground state now has $\partial_HW=f\neq0$ (a so called F-term) then a term $\Phi^2H$ in the augmented superpotential contains $f\phi^2$ , a mass term only for the scalar part of $\Phi$ but not for its fermionic part. This example thus shows that a scalar mass term (leading of a splitting of the susy mass degeneracy) is a soft term.

Another nice thing about a theory with soft breaking is that (at least after augmenting the theory) you can still express everything in the supersymmetric language including Kähler- and superpotential which is not the case for a general non-supersymmetric theory.

So much for background. Now to the issue that shoed up in the seminar: In many phenomenological string constructions you start from some compactification that you know and which happens to be supersymmetric. But then you want to make it more realistic (and for example have a positive cosmological constant). To this end you then for example you add some branes to the background that break the supersymmetry completely (for example by adding anti-branes if you already have branes). Now we would like to understand the the physical consequences of these additional branes (which are often put deep in some warped throat to make their effect small). What is now often done is to analyse this set-up still using the language of supersymmetry (for example by computing some super-potential of the Gukov-Vafa-Witten type). But this is at best only satisfied if the susy breaking is soft (and can be made spontaneous by augmenting the theory). If there are however hard (i.e. not soft) breaking terms you will never see them in this treatment even though they are most likely deadly for phenomenology.

Thus the question is: If I add these non-susy branes (or fluxes for that matter) by hand, will susy be broken only softly? I cannot imagine a reason why this should be true in general. But these people who write these papers are much more clever than I and most likely I am missing something. Please tell me what that could be!

What I found was arxiv:hep-th/0209206 where Matthias Klein studies a typical (but simple enough) example: He considers a theory on $\mathbb{R}^4\times [0,\pi R]$ which is supersymmetric in the bulk. This theory is then rewritten in terms of four dimensional fields treating the coordinate on the interval as a continious parameter on the bulk fields. In five dimensions, the minimal number of supercharges is eight, thus from the four dimensional perspective this theory has ${\cal N}=2$ . This bulk theory is then coupled to ${\cal N}=1$ theories on the two boundaries. But this is done in a way such that the two bulk boundary couplings preserve different ${\cal N}=1$ parts and thus the whole set-up of bulk with two boundaries is not supersymmetric. You can think of this as a toy model for a space-time with a brane and an anti-brane. Both these branes by themselves (including the bulk) preserve some susy but both of them at once do not.

Now Klein computes corrections to the mass of scalars on the boundary at $z=0$ say which are massless at tree level. The simplest diagram where we can expect that there are no susy cancellations has to involve fields also on the boundary at $z=\pi R$ and is thus at two loop level. A typical diagram looks like this:

Here, the fields with horizontal lines are on the first brane, the vertical fields live in the bulk and the fields in the circle live on the second brane. Let us denote the momentum in the circle by q and the momentum in the lower loop by k. For this diagram we compute

$\sum_{k_5, k_5'}\int d^4k\int d^4q\,\frac{(-1)^{k_5+k_5'}k_\mu k_\nu k_\lambda(k_\rho+q_\rho) \tr(\gamma_\mu\gamma_\nu\gamma_\lambda\gamma_\rho)} {(k^2+k_5^2/R^2) k^2 (k^2+{k_5'}^2/R^2) (k+q)^2 q^2}$

Note the discrete component of the momentum for the bulk fields. As this component is not conserved on the branes (as the branes break translation in variance) there are two independent components for the two lines. The signs in the nominator are remnants of phases $\exp(i k_5x^5/R)$ evaluated on the branes (depending on the exact boundary conditions of the bulk fields at the branes). The $k_5$ components should have appeared in the nominator as well but give vanishing contributions upon $k_5\mapsto -k_5$ .

Gamma gymnastics brings the nominator to the form $k^2 k\cdot q$ which is then rewritten via $2k\cdot q= (k+q)^2 - k^2 -q^2$ . The first and third term cancel then upon a shift in the q integration. We are left with

$\sum_{k_5, k_5'}\int d^4k\int d^4q\,\frac{(-1)^{k_5+k_5'} k^2} {(k^2+k_5^2/R^2) (k^2+{k_5'}^2/R^2) (k+q)^2 q^2}$

We find that the q integration diverges logarithmically. That's not too bad. Next, we do the sum over the $k_5$ 's. With a little help of mathematica, we find

$\sum_{k_5=-\infty}^\infty \frac{(-1)^{k_5}}{k^2+k_5^2/R^2}=\frac {\pi R}{k\sinh(\pi R k)}$

This is the real surprise: The integrand is exponentially suppressed for large k. Thus, it is actually UV finite! Thus the only UV divergence of this diagram is the logarithmic divergence of the q integration. Of course, this is only one of many diagrams. But the claim is that this is generic: At least for two loops, one loop is running only on one brane including bulk-brane vertices and is thus protected by the bulk-single brane susy to diverge only like a log. The other loop involves a propagator across the bulk which makes this integration finite. Thus all contributions to the scalar (and other) masses on one brane are only log divergent. The claim seems to be that this is also true at higher loops although I do not quite see how to substantiate this. But if this is true, although there is no supersymmetry from the outset, the hierarchy is still protected.

From this point of view, this set-up in which susy is preserved locally but which breaks it only globally is an example of soft breaking. I must say, this is against my intuition as it seems to me that constructing some branes that in total break susy is as bad as it can get and I see no reason why this should be soft in any way.

But if this is true is there a possibility to augment this theory and find that the breaking is spontaneous in that bigger theory?

Even more important: Is the breaking soft enough that we can write down a low energy effective action in the supersymmetric language in terms of a K&aauml;hler and a super potential? Or are there other terms as well?

atdotde

Thursday, February 28, 2008

Zeit Sudoku Bookmarklet

Thursday, February 07, 2008

Geometric Hamilton Jacobi

Tuesday, February 05, 2008

Breaking it softly with this song

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atdotde

Thursday, February 28, 2008

Zeit Sudoku Bookmarklet

Thursday, February 07, 2008

Geometric Hamilton Jacobi

Tuesday, February 05, 2008

Breaking it softly with this song

About Me

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