There is an interesting posting on Tommaso Dorigo's blog A Quantum Diaries Survivor entitled The Goldstone Theorem for Real Dummies. Quoting from his posting the Goldstone theorem is: "The spontaneous breaking of a continuous symmetry of the lagrangian generates massless scalars". Despite its claim to be for real dummies his derivation of the Goldstone theorem assumes familiarity with a lot of theory, which ensures that the "dumb" reader will not develop an intuitive feel for the masslessness of the particles (the Goldstone bosons, that is).

I am personally at ease with the level of theory that he uses, but I am in a very small minority in this respect. Wouldn't it be nice to write an intuitive description that does not assume that the reader has prior knowledge of much theory, but where the writer ensures that the intuitive description faithfully conforms to the underlying theory that is known to him. The Goldstone theorem can be understood intuitively without having to follow the detailed algebraic steps of its proof.

Here I will attempt to "prove" the Goldstone theorem using only words (i.e. no pictures, and no maths). Pictures would greatly improve the presentation, and maths would very succinctly summarise and generalise what is going on, but it is interesting to see what you can do with words alone.

Mass is *defined* as the propensity of the vector-valued field value *alone* (i.e. *not* including any field derivatives, or equivalently the field values at neighbouring points) to contribute to a restoring force that pushes the field towards an equilibrium value of the field. If within a local neighbourhood of field space there are direction(s) that have *zero* restoring force then each of these directions corresponds to a *zero* mass excitation.

It suffices to visualise a potential in field space whose (negative) gradient defines the field restoring force, and to ask ourselves what sort of shapes this potential could have in field space. We need to find places in field space where this potential has zero gradient, and then enquire whether the gradient of the potential is actually zero over the whole of a local neighbourhood rather than just at a single point, so that displacement of the field value within such a neighbourhood would encounter zero restoring force, and would thus correspond to a zero mass excitation.

The visualisation problem reduces to the classification of the stationarities of a scalar potential function in field space, which is something that is very easy to visualise (e.g. minima, maxima, saddle points, etc). The particular case that is relevant to the Goldstone theorem is when the potential function is constrained to be symmetric, e.g. under the continuous group of rotations about the origin of the vector-valued field. What sort of stationarities are allowed under such symmetry constraints? For a rotation symmetry the potential must be constant on each origin-centred spherical shell in field space, but different spherical shells have independent values of the potential. The potential is therefore fully described by its variation as a function of distance from the origin in field space (i.e. a 1-dimensional potential).

To find an equilibrium point you need to find a stationary point of the 1-dimensional version of the potential, then constancy of the full version of the potential over each spherical shell gives us zero gradient of the potential over the entire spherical shell that intersects this stationary point, which thus gives as many zero-mass degrees of freedom of the vector-valued field as there are directions within the spherical shell (i.e. one less than the dimensionality of the field). This is the Goldstone theorem.

Rereading what I have written above I am now uncertain whether it is of much use to anyone other than me. It is just the internal chatter (minus the pictures) that goes on inside my head when I think "Goldstone theorem", so it isn't guaranteed to be as intuitive for other people as it is for me.

## 3 comments:

It's been a while, post-Goldstone, post-Martin and post-Forster, so I'll mull it over. At first sight, though, it looks good.

Steve, well done - when you wrote it you probably realized how hard it is to explain things in a really, really simple way! I sometimes have not the energy to do it...

Cheers,

T.

I think we are so comfortable with explaining things algebraically to ourselves and to each other that we usually don't try to find other ways of explaining things to people who are less algebraically sophisticated (i.e. almost everybody else!).

I strongly feel that I haven't understood a piece of maths if I can't find a way of visualising it, and I blogged about this point here. A useful side-effect of seeing a piece of maths visually

myselfis that I then have a visual way of explaining the maths tootherpeople.My attempt to use only words (and

notpictures) in this posting was sheer laziness on my part. I gave a verbal description of the picture inmyhead, thus forcing the reader to rebuild the same picture intheirhead.Post a Comment