## The Goldstone Theorem for Real Dummies November 10, 2007

Posted by dorigo in mathematics, personal, physics, science.

I have been spending the last few days preparing part of a course in particle physics for 5th year students in Physics (the second and last year of what is called “Laurea Specialistica”, like a masters degree in the US). I must say I had forgotten how much I like to study. The last serious time I spent in the company of physics books was over two years ago, but that was a very stressful occasion with an impending exam, burdened by the high stakes of getting tenured. Besides, the prospect of explaining the standard model to students who have at least some familiarity with quantum field theory is really stimulating. I am not a theorist, so in principle I am not qualified to present the theory of particle physics in an impeccable way, but the course I will teach takes a quite phenomenological-experimental point of view, so I think I will not be able to do too much damage to those innocent souls (and the course is taught for three quarters by a more experienced colleague – I only do the last part, on Higgs and collider physics).

Of course this blog has been suffering recently from my involvement in preparing the course… So I decided I would try and kill two birds with one stone, and make an attempt at making available one tiny bit of my course today, aiming at real laypersons, here. I think of it as a challenge to myself to test the inverse Feynman’s grandmother’s conjecture: whether, that is, one can explain things to grandma if one has understood them (the original conjecture states that you haven’t really understood something unless you can explain it to grandma). And since I lost both my grandmothers, you are my guinea pig for today.

Enough chat. Now, what is Goldstone’s theorem and why should you bother ? The theorem is a crucial preliminary to understand the need for a Higgs boson in the Standard Model theory of particle physics, and that would suffice to keep you awake: but it is also a very nice illustration of how the physics of a system can be extracted from a quite abstract concoction – the lagrangian density. If you do not know what a lagrangian density is, worry not: you will not really need to understand what it is in and out, because I intend to present things in a very handwaving way. That will not prevent me from calling things with their real names!

So let us consider a lagrangian density for a real scalar field. What is a scalar field ? Take air temperature, for instance. It is a real number defined in any point of space, a number depending on space coordinates. In quantum physics, however, a scalar field represents a particle capable of moving and interacting with its peers: doing the things that particles do, that is.

Ok but, what is then a lagrangian density ? The lagrangian density is some mathematical scribbling that enshrines the physics of our scalar field. It is defined as L=T-V, where T is the kinetic energy of the field (the particle), and V its potential energy. Think of a ball thrown in the air: Once you’ve kicked it up, it has speed -and that is a form of energy, called kinetic energy- and height -and that, too, is a form of energy: potential energy. You know what potential energy is: it is the reason why you avoid walking under a baby grand piano being lifted to the third floor. So L is just the kinetic energy of our ball subtracted of its potential energy. Despite the simple definition above, the lagrangian may take complicated forms. It is an expression which, handled the right way, can sometimes be squeezed to extract the dynamics of our particle. I will not tell you how today, but deal L with the respect it deserves, since unlike you and me, L knows everything about the scalar particle: its past, present, and future motion. Here is our lagrangian for the real scalar field $\phi$:

Quite a far cry from “T-V”. ain’t it ? But worry not. In the expression for L above $\phi$ is our scalar field, and the curly symbol before it represents a derivative – by putting it there we express the fact that we consider the variation of the field with respect to its position in space. By multiplying together the spacetime derivatives of the field, we are in effect writing the kinetic energy it possesses. The second term is instead the potential in which our particle sits: it depends on two real parameters, $\mu^2$ and $\lambda$. Regardless of their value, we observe that L exhibits a manifest symmetry with respect to the substitution of the field with its opposite value, $\phi(x) \to -\phi(x)$. A symmetry of the lagrangian is something worth noting: it usually reflects the presence in the theory of a conserved quantity, something that the operation of symmetry does not change, that is.

Now let us investigate more our potential V: as a first example we give both parameters $\mu^2$, $\lambda$ a positive value. We then know the form of the potential from calculus you should have had some time in the past at school: a quartic curve, with a minimum where the field is equal to zero. It is shown in the plot below, on the left diagram.

Imagine a particle sitting at the point $\phi(x)=0$. It is at the minimum value of the potential. If you move it about the minimum, it produces small oscillations – perturbations of its physical state – and we can compute the physics of the field using something called “perturbation series”: the potential difference introduced by the perturbation is small, so its effect is a small modification to the motion of the particle. By grouping together modifications of the same order of magnitude and summing them we can determine the dynamics. Crucially, the particle has a positive mass, corresponding to the resistence it opposes to any attempts at displacing it from the point at $\phi=0$: you may well call it inertia.

Much more interesting is the case arising if we instead take $\mu^2<0$. We then get the form of potential shown on the right diagram in the figure above. The potential term with a negative value of $\mu^2$ is at odds with what you would have learned by browsing the first few chapters of a quantum field theory book: it appears to represent a particle with imaginary mass. It is easy to see why it is so: it gives a “negative resistance” to any attempts of moving it from the origin, where the field is zero. The potential decreases in both directions, so it is energetically favorable for our field to roll down to one of the two saddle points. These lie at the value $\phi = \pm v = \pm \sqrt{-\mu^2/\lambda}$, as is easy to realize by inspecting the form of V – or, if you know better, by just setting to zero the derivative of V with respect to the field.

We like to call the minimum of the potential our “vacuum”: you cannot have less energy than that. In the case of our potential with negative $\mu^2$, the vacuum does not correspond to zero value of the field! Rather, the field takes the value $v$. There is something wrong here: imaginary mass, vacuum containing non-zero fields… Surely, we can mend the situation by redefining our scalar field: we shift it to the minimum at +v by calling $\phi(x)=v+\eta(x)$. The physics cannot depend on the shift of the scalar field by a constant value v, and now the lagrangian takes a different form:

In terms of the shifted field $\eta(x)$, there is nothing wrong with the lagrangian any more: the vacuum has zero value for the field -it is a real vacuum!-, and the field has a mass term of the right sign: $-1/2 m^2 \eta^2 = -\lambda v^2 \eta^2$ so the mass of the scalar is now $m = \sqrt{2\lambda v^2} = \sqrt{-2\mu^2}$, a positive value (forget the terms with cubic and quartic in the field: they describe self-interactions, not the mass).

Even better, we can now do perturbation expansions around the new minimum, and our expansions will converge. We will be thus able to compute the dynamics for the new field $\eta$. All is good, a true success. But there is something we had to give up in exchange: L’ is not symmetric for the operation $\eta \to -\eta$ any more!

It is important to note that the physics described by L cannot have changed as a result of a simple constant shift of the field: so we are brought to conclude that the symmetry is still there, but it is “hidden” by our choice of the vacuum at a value +v for the original field. The symmetry of L generated a degeneracy in the vacuum: two values share the minimum for V. By choosing one of the two possible vacua we have hidden the symmetry from view.

A bit harder: a complex scalar field and the Goldstone theorem

Ok, now we need to make things just a bit more complicated. We want to write a lagrangian which is symmetric under a continuous transformation law of the field, not just the simple mirroring as before. That will allow us to state Goldstone’s theorem. The simplest lagrangian we can write is the one below.

This time we have defined a field $\phi = \frac {1} {\sqrt {2}} ( \phi_1 + i \phi_2)$: this is a complex scalar field, which is actually equivalent to two real scalar fields. Please do not worry about the imaginary symbol $i$ or the complex conjugation operator *: you will not need to even touch those with a stick. Instead, look at L. If you look at it for long enough, you realize it is an expression which remains invariant if we modify the field by a phase transformation $\phi \to \phi'=exp(i\alpha)\phi$ (with $\alpha$ the constant phase shift). That happens because every time you multiply the field by its complex conjugate the two phases annihilate, and L above only depends on such well-behaved products.

If the math I invoked above is above your head, do not worry. Suffices to accept that we have managed to put together a lagrangian which is invariant for phase transformations of the complex scalar field . The physics described by L does not depend on the value of $\alpha$, that is. But of course it depends on the two parameters in the potential energy terms, $\lambda$ and $\mu^2$. What happens now if we take as before the former positive and the latter negative ? Again, we get a field with an imaginary mass term, which might not really look like a good idea. Worse, we now have not just two, but a full circle of minima for the potential, lying at the values of the field satisfying $\phi_1^2+\phi_2^2=-\mu^2/\lambda$. An infinity of choices for the vacuum! The situation is pictured in the drawing above.

Having previously worked out the simpler example of one single real scalar field, we are not impressed by the compication, since we know how to get things straight: we choose one of the vacua for a translated field by writing $\phi(x) =1/\sqrt{2} (v+\xi(x) + i\eta(x)$.Don’t worry about yet new greek letters: we just renamed the two components of the original scalar field, after translating them to the point (v,0). In terms of the shifted fields L becomes

If we examine the latter form of L, we recognize kinetic terms (the ones with two derivatives of the field) for the scalar fields $\xi$ and $\eta$. But while we also have a mass term (the one quadratic in the field) for $\eta$, $\xi$ gets no mass term: that means the field is massless. The two fields correspond to orthogonal oscillations about the vacuum we have chosen: and the massless field corresponds to oscillations along the direction where the potential remains at a minimum – along the circle of minima, that is. Because of that, it encounters no resistance – no inertia, no mass.

The spontaneous breaking of the symmetry of the original lagrangian for $\phi$ has generated a mass for one of the two scalars, and a further massless scalar has appeared in the theory. This is the Goldstone theorem in a nutshell: The spontaneous breaking of a continuous symmetry of the lagrangian generates massless scalars. They correspond to fluctuations around the chosen vacuum in the direction described by the neighboring vacua.

Massless scalar particles do not belong to any reasonable theory of nature. Our world would be a quite different place if there were massless scalars around! We do not observe such particles. Indeed, there is a mathematical trick, called the Higgs mechanism, which gets rid of the massless goldstone bosons. The degrees of freedom of the theory associated to the Goldstone bosons reappear as mass terms for the weak vector bosons… But this is stuff for another lesson.

Still here ? I would love to know if among the twentyfive readers of this post there is at least one who has made it to this last paragraph. If you are him or her, and you had no prior knowledge of quantum field theory or lagrangian formalisms, drop me a line. I’d like to know what made you think you could learn these difficult things by reading a blog post… 😉

1. Charles - November 10, 2007

Re Goldstone Theorem, if you think a layman can even begin to follow this, you are smoking pot.

2. Alejandro Rivero - November 10, 2007

Or he hopes _the layman_ is smoking pot 😀
As for the “explain to grandma” I am not sure if it is a good filter neither. For instance I can not imagine Hélène Langevin-Joliot using it effectively as a remark to her classroom, and not specially when speaking about T_c, the Curie critical temperature.

3. Guess Who - November 10, 2007

Hi TD. While I’m not in your target audience, may I offer some constructive criticism? The statement

In quantum physics, however, a scalar field represents a particle capable of moving and interacting with its peers: doing the things that particles do, that is.

is setting up the reader for confusion down the road. I suggest explaining that in quantum physics, a field is a collection of oscillators, and that an excited oscillator is what we call a “particle”. So a field does not represent a particle, but any number of them.

This motivates the shift to the selected ground state: small perturbations about the ground state are low-lying excitations of the oscillators associated with the shifted field coordinates, i.e. a small number of particles. That’s a manageable description, and the one most relevant to particle physicists living in a given ground state and only exploring small deviations from it.

Then you can wet your audience’s appetite for more by dropping dark hints of non-perturbative aspects and maybe by mentioning that the hot early universe was anything but perturbatively close to a ground state…

4. Paul - November 10, 2007

I found it well-written and enlightening. But then, I’m a statistical
outlier (math prof). I assume the same is true of string-theoretic
moduli (they are degenerate directions of some Einstein-Hilbert
action, hence should give rise to massless fields)?

5. Interested reader - November 10, 2007

I’m a relatively educated layman (graduated MIT 30 years ago, did a lot of math and physics when I was there, read all the physics for layman books, can still do calculus and basic differential equations, etc.) and this was mostly over my head….sorry.

6. Doug - November 10, 2007

Hi Tommaso,

A comment about the bowl-like figure with Re, Im, V axes indented in the paragraph beginning “If the math I invoked above is above your head, do not worry.”

This may be the bottom of a ‘horn torus’.
A ‘spindle torus’ seems less likely.
http://mathworld.wolfram.com/Torus.html

7. Domenic - November 10, 2007

Well, I made it to the last paragraph, and felt like it was all completely understandable. Although, the actual physical implications were pretty light on detail—“They correspond to fluctuations around the chosen vacuum in the direction described by the neighboring vacua” doesn’t make much sense to me without some more context.

For reference, I’m a second-year Caltech student who has failed to place out of any classes (and thus am taking wave mechanics, differential equations, and real analysis), but has compensated by reading lots of books (actual textbooks, that is). So while I’ve read the chapters in Goldstein explaining Lagrangian mechanics, and have played around a bit with some quantum field theory stuff during my summer research (enough, for instance, to instantly recognize the first term in the Lagrangian as corresponding to potential energy), I haven’t taken any classes in them…

I think that might be the best you can hope for?

8. Domenic - November 10, 2007

Dammit, I wasn’t supposed to be logged in to that account. Please disregard the Dungeons & Dragons blog that you can find by clicking my name in the previous post :-P.

9. dorigo - November 10, 2007

Hi Charles,

hmmm I haven’t smoked pot in quite a while actually. But of course I was addressing this post to “educated laymen”, that is, people with a curriculum outside physics who are still interested in it. Then, of course if the only math one had stopped before differential calculus, there is little to do in a 2000-word piece.

Cheers,
T.

10. dorigo - November 10, 2007

Well Alejandro, there are grandmas who kick ass, of course. We are speaking of the generalized XXth century grandma here…

Cheers,
T.

11. dorigo - November 10, 2007

Hi GW,

your suggestion is most welcome, although I explicitly avoided talking of many particles. I was trying to sacrifice some accuracy for clarity, but maybe you are right, I did it in the wrong place.

As I was writing this, one of the things I most wanted to introduce was the concept of the negative energy of the vacuum, and the implications for the expansion of the universe. But then I realized it was just too much for a single post… Already, discussing perturbation expansions was probably over the top.

Cheers,
T.

12. dorigo - November 10, 2007

Hi Paul,

now _that_ is over my head 🙂 I confess I have never touched a book on string theory!

Cheers,
T.

13. dorigo - November 10, 2007

Hi Doug,

yes, the bottom of a horn torus. But really, the potential increases forever for arbitrary large fields…

Hi Domenic,

thank you for your valuable input. Yes, the issue of the fluctuations of the field along the circumference of vacua was not explained with the necessary amount of detail… There is always some tension between trying to be as descriptive as possible and avoiding unnecessary lengthening of the explanation.

Cheers all,
T.

14. lius - November 10, 2007

In physics, a field is a perturbation of space-time, a functions which strictly depends on coordinates. We can think it is a collection of quantum oscillators, but it is only our view of nature and it is an approximation.
But how can you explain to your grandma what is a quantum oscillator?
The main problem of the physics today is that we haven’t physicists who want to teach their interests in the easiest way. People doesn t need to understand the molecular composition of milk or Bernoulli’s law to know why milk is important… my grandma will have to be happy to drink milk!

15. fliptomato - November 10, 2007

Hi Tommaso,

I like the post! (Though I’m probably not the intendend audience.) I’m not sure what I think of the following statement, however:

“Massless scalar particles do not belong to any reasonable theory of nature.”

I understand what you mean, but that’s still a bit harsh on scalar fields, no?

One could argue for the possibility of massless scalars that don’t couple directly to the Standard Model (but instead to some undiscovered hidden sector). Or one could point to the nearly-massless axion which is allowed by the Standard Model. If one assumes inflation, there are models with the inflaton as a massless scalar. The last two, for example, could play a very real role in nature addressing the CP and cosmic coincidence problems. And the first option could be related to theories of SUSY breaking (there’s been at least one paper on massless scalar susy breaking).

16. Quasar9 - November 10, 2007

“…treat L with the respect it deserves, since unlike you and me, L knows everything about the scalar particle: its past, present, and future motion.”

Rather than throwing or ‘kicking’ a ball in the air
I was thinking a man or woman (boy or girl) bouncing on a trampolin. The person (particle?) – body of the person; the field – the position the body occupies at any given moment or point in time; The lagrangian density – some mathematical scribbling that enshrines the physics of our scalar field, defined as L=T-V, where T is the kinetic energy, and V its potential energy.

But we know how predictably ‘expected’ the trajectory is inside an atom, and how dynamic (predictably unpredictable) bouncing on a trampoling can be, even for a ‘gold medal’ olympic gymnast.

17. Pioneer1 - November 10, 2007

Tommaso,

Many thanks for this post. Great explanation. I think this is the kind of article internet was created for.

So far I only read it carefully half way through and I have several questions, if you don’t mind.

Maybe you could expand a bit more on the concepts of field and particle? These are crucial concepts but I still don’t understand how you use them in this context. Do you define the field as particle? I understand temperature field but not when you write “field (particle).”

Lagrange density is defined as L = T – V

Ok, but do you use “Lagrange density” and “Lagrange” as synonyms? I think you do. Is this correct?

“Kinetic energy equals potential energy” is a physics term of art which means r^2/t^2 = 1/r. This is kepler’s rule. Therefore, the fundamental concept here is not the Lagrangian, and it is not Newton’s law (Kepler’s rule written as r/t^2 = 1/r^2) but it is Kepler’s rule.

The importance of calling Kepler’s rule Kepler’s rule is that 1/r = potential has no meaning unless it is reunited with r^2/t^2. Kepler’s rule describes a density continuum. Physicists artificially break this density continuum and look at it as if it were two discontinuities: potential and force. Neither potential nor force exist on their own. They are always eliminated to recover the density continuum (Kepler’s rule.) I would appreciate your comments on this.

So when you write

Lagrangian density L knows everything about the scalar particle: its past, present and future motion

you are saying nothing more than the particle moves according to Kepler’s rule. And this makes sense. Planetary orbits move according to Kepler’s rule. Standard model models planetary motions in a different scale. This is also proved by the fact that Standard Model makes use of Fourier transforms, the analytical epicycles, or trigonometric expansions.

A symmetry of the Lagrangian is something worth noting: it usually reflects the presence in the theory of a conserved quantity…

This is true. Physicists call a proportionality symmetry. So here we have a proportionality: Kepler’s rule. It remains the same as the radius and the period of the orbit vary. This is what a proportionality means.

When we accept the fundamental nature of Kepler’s rule as the density continuum everything is simplified and all these high level concepts invented by physicists, such as symmetry and energy conservation, to save Newton’s authority and the authority of physics disappear.

Thanks again. I find that this article is important for me to understand in order to clearly understand and recover Kepler’s rule buried deep in standard model, so I created a couple of pages to discuss your article furhter. I numbered the sentences and put comment functionality as well, and I welcome everyone to comment there if they feel like it.

Thanks again.

18. Paolo - November 11, 2007

Hi Tommaso. Just noticed your article, I’m sure I will like it… One comment at the outset about the title: I understand your intention of citing the “for dummies” books as a joke, but I don’t buy it. I mean, I don’t find the title funny and I would never buy those books!

19. Paolo - November 11, 2007

… indeed, I liked the article. Instead of writing here my almost-trivial personal opinions, I would add a reference, maybe appropriate for those students: Ch. 8 of Ho-Kim, Pham, “Elementary Particles and Their Interactions”, Springer

20. Chris Oakley - November 11, 2007

On my web site I say that the Higgs Mechanism is “contrived”.

Although – as far as I know – no-one is listening, I am going to expand on that here. At least a sceptical view will then be on record.

This playing around with V(φ) potentials means nothing at all in the context of quantum field theory. Such manipulations are only meaningful for classical fields. If there is mathematical justification for spontaneous symmetry breaking (= the Higgs Mechanism), then it is in considering it as pre-processing stage, i.e. before field quantization takes place.

The problem being solved is that massive vector bosons are observed in nature (specifically the W’s and the Z – the carriers of the Weak force), whereas Feynman-Dyson perturbation theory explicitly forbids them on the grounds of non-renormalizability. ‘t Hooft and Veltman demonstrated that gauge theories – which have only massless vector particles – are renormalizable. Not very interesting if we want massive particles, except that masses can be given to them by breaking the symmetry via the Higgs mechanism, and if this is done, the theory remains renormalizable. So the Higgs Mechanism gives us an escape route. The problem is that “renormalizability” is a swamp. Yes – you can get the right answers by subtracting infinity from infinity. But you can also get the wrong answers. You can get any answers you want. So why is anything that involves renormalization a scientific theory? And even if one accepts it (I obviously do not) – the ways in which the Higgs mechanism can be applied are legion: hence the search for “underlying” principles such as GUTs or supersymmetry or both to guide model building. None of these principles seem to work. Well – maybe that is because the Higgs Mechanism is an edifice built on quicksand.

But, hey, at least I get paid more in an Investment Bank.

21. chiara bortignon - November 11, 2007

i think i got the point of it..that’s not bad anyway, huh?!
oh well, just one thing, i’m pretty sure it was Einstein the one who said “use grandma’s language”!
have a nice sunday night

cheers :D,
chiara

22. Marius - November 12, 2007

Hi, I am a mathematician who knows very little about the SM. However the post, which I have been able to follow, surprized
me for the following reason.

There is a mathematically sound and physically predictive theory
of appearance of microstructures in materials ranging from trivial
steel alloys to nematic elastomers.

This theory is based on lagrangians of basically same nature and the appearance of microstructure is a consequence of that.

The mathematical explanation seems different though.

For an introduction see this, for example: http://www.mis.mpg.de/preprints/ln/lecturenote-0298-abstr.html

What do you think, (seriously)?

Marius

23. Guess Who - November 12, 2007

Hi Marius. When you add an electromagnetic field to the Goldstone model described in this post, you get the Abelian Higgs model, which is essentially the relativistic equivalent of the phenomenological Ginzburg-Landau model of superconductivity:

http://en.wikipedia.org/wiki/Ginzburg-Landau_theory

Both support tubes of quantized magnetic flux, a simple example of topologically stable defects. In one of those, the complex scalar field winds around the potential valley one or more full loops as you walk along a spatial loop about the axis of the tube; the topological stability argument is simply that you can not continuously deform this configuration to a constant one without lifting the field out of the potential valley, over the top of the “Mexican hat”, so at low energy, the configuration is classically stable (this works without electromagnetic field too, but then the total energy of the configuration diverges). Cosmologists know them as cosmic strings.

With other potentials you can also get topologically stable monopoles (point defects), walls (Bloch walls; the two-well potential will give you that) and textures. Non-trivial, dynamically stable field configurations can also occur absent topological stability, but then you have to work much harder to prove their existence.

I haven’t looked closely at the notes you linked to, but my first impression is that they seem to be about the kind of hard work which you have to do absent a topological stability argument. Is that right?

24. DB - November 12, 2007

Re fliptomato (#15) and your statement, “Massless scalar particles do not belong to any reasonable theory of nature”:

Well, pions are light (relative to protons) because they are the pseudo-Goldstone bosons of chiral symmetry. Not massless, because this symmetry isn’t exact, but still it’s not a bad real-life example.

25. dorigo - November 12, 2007

Hi all,

Lius, that is right, it is not possible to explain everything without math. I think it is useful to explain whatever can be explained and understood, however, because filling the gap between science and the general public must be done whenever possible.

Hi Fliptomato,

you are right – I oversimplified things. Massless scalars do not belong to the SM as we know it, but they might still play a role somehow.

Quasar9, maybe the gymnast is a good way to describe dynamics of motion in a gravitational field – but to me it is rather complex. For instance, it is not even easy to understand how angular motion can arise out of the blue, when somebody jumps and _then_ starts spinning.

Pioneer1,

a quantum field, as GW correctly discusses above, is a collection of quantum oscillators. You really need to get trained on quantum mechanics to understand this however. I know no simple way of explaining it… As for “lagrangian density”, it is a function which, integrated over the space-time coordinates, gives the lagrangian proper. In that sense it is a density. Also, do not forget that the lagrangian formalism works with any potential, not just a central one.

Hi Paolo, thanks for visiting… The “for dummies” books are indeed a bit dumb, I agree. But the idea is fine – they provide introduction to unarmed people.

Chris, what do you mean when you say “no one is listening” ? For god’s sake, this place is not unattended 🙂
Anyway, thank you for your comment. I am nobody to provide constructive criticism to it. But I just want to say that I think the Higgs mechanism is not only a way to keep the theory renormalizable while endowing bosons with mass: it also provides the extra bonus of fermion masses, free of charge. I think I understand why you say the SM is built on quicksand, but you could say the same of basically every other physical theory man has put together since aristotelian motion.

Ciao Chiara, very good! And you are probably right about EInstein, I must have gotten confused.

Marius, thanks for visiting, I think GW answered you better than I could ever have (thanks GW!).

Cheers all,
T.

26. Chris Oakley - November 12, 2007

Hi Tommaso,

OK – no-one agrees (at least – no-one with a PhD), not no-one is listening.

A trilinear fermion-fermion-Higgs coupling becoming the same plus a fermion mass when the Higgs acquires a VEV is not IMHO very deep, especially as fermion masses are not the problem here.

If you can show me where ∞ – ∞ is assigned to the number you want to get in elementary quantum mechanics, or in calculating planetary motions, then I will happily take your point about other theories in physics. But as far as I know, it is not – QFT is uniquely flawed in this respect.

27. dorigo - November 12, 2007

Hi Chris,

I understand your point, but I think the finite answers arising from infinities subtracting from each other means that we should not really think at individual diagrams as carrying any real physical significance without the others renormalizing them.

I prefer to think at QFT methods as useful computational tools, which work because of something we still do not fully grasp. They work nonetheless, as does newtonian mechanics despite its negligence of a finite speed of light.

Cheers,
T.

28. Coin - November 12, 2007

Wow, this is the clearest explanation of spontaneous symmetry breaking I have ever seen! I actually have seen this exact set of phenomena explained before at least twice, but for some reason this is the only time it has ever “clicked” for me, I think because you did not try to hide the math from us*. I do have a few questions after reading this though.

1. You summarize the Goldstone theorem as “The spontaneous breaking of a continuous symmetry of the lagrangian generates massless scalars.” Am I reading correctly that the massless scalars you say are produced in this sentence are the “Goldstone bosons” I hear about?

2. Usually when one hears about the scalar fields in symmetry breaking, it is stated the other way around from how you describe the goldstone theorem– that massless scalars are hanging around, and they cause the spontaneous breaking of a continuous symmetry, not that they are generated by it. For example I never hear people describing the massless scalar that is the Higgs field as being “generated” by EWSB, rather they talk about the Higgs causing EWSB somehow. What is happening here? Am I mixing two kinds of massless scalars here?

3. Assuming the answer to (1) is “yes”: whenever anyone talks about Goldstone bosons they inevitably refer to Goldstone bosons being “eaten”. What on earth does this mean? (Or I guess this is the “mathematical trick, called the Higgs mechanism, which gets rid of the massless goldstone bosons… But this is stuff for another lesson.” you refer to?)

Anyway, I don’t know if anyone else got anything out of this, but I at least hope you continue with posts in this series. I want to know who’s been eating all those Goldstone bosons…

* So I guess the target audience for this blog post would in fact be someone like me who DOES have prior knowledge of quantum field theory or lagrangian formalisms, but has so far completely failed to actually understand them? 😛

29. Chris Oakley - November 12, 2007

I prefer to think at QFT methods as useful computational tools

The “computational tools” are not Quantum Field Theory. They are the Renormalization Group Equations. These do not logically follow from Quantum Field Theory on account of inadmissible infinite subtractions. They are the clapped-out Lada that you drive because your Ferrari was delivered without an engine.

30. Jonathan Vos Post - November 13, 2007

Dear Tommaso,

Excellent explanation!

However, having published a couple of co-authored very speculative conference papers on the question of whether imaginary mass is physical or nonphysical, can you summarize why, as you’ve said in this thread: “Again, we get a field with an imaginary mass term, which might not really look like a good idea.”

31. Pioneer1 - November 13, 2007

OK – no-one agrees (at least – no-one with a PhD), not no-one is listening.

Chris Oakley,

I am not a PhD but I agree. I am proposing to call the 4th clause here “Oakley Clause” in your honor. Please feel free to add to the list.

Thanks.

32. dorigo - November 13, 2007

Hi Chris,

hmm, I guess I do not agree. But you should not worry about me :), my understanding of the matter is not much more than basic.

Cheers,
T.

33. dorigo - November 13, 2007

Dear Jonathan,

🙂 my remark had nothing to do with my own beliefs (which are not all that meaningful) and a lot to do with the standard way to interpret terms quadratic in the field in the lagrangian density.

Cheers,
T.

34. Chris Oakley - November 13, 2007

Hi Pioneer.

Well .. let’s have a look at the oath.

1. I will not call myself or any particle God (Newton is the only God)

I’m not sure about the divine status of Sir Isaac, but calling the Higgs the “god particle” is certainly dumb.

2. I will not claim to read God’s mind

Which God? Newton? Actually I don’t object to this … providing that one accepts that one may have got it wrong!

3. I will not call other doctors crackpots

A nice idea, but insults are such an important part of scientific research that I don’t think you could require this.

4. I will not divide infinity by infinity.
5. If I divide infinity by infinity I will not cover up by calling it renormalization.

These should be combined; but I don’t want the clause named after me unless it is reworded thus: “In the case of those eventualities in which my theory seemeth not to work, and when integrals that I expect to be finite convergeth not and there is lamentation amongst the People of the Study of the Elementary Particle on account of the fact that they convergeth not, I swear that I will not introduce spurious and meaningless cutoffs in the offending integrals; I swear that I will not introduce spurious and meaningless extra dimensions and I swear that I will not introduce spurious and meaningless counterterms to get me out of this tight spot, for I am cognizant of the basic mathematical fact that infinity minus infinity may get me the answer that I want, but then again, it may get me any other answer as well”
There. Sorry – it’s a bit of a mouthful, but I think that it is worth the extra effort.

6. I will not cite automatically without reading what I am citing

Agreed.

7. I will not put in the title what is not in the paper

Yes, and specifically, “I will not make sweeping claims in the title of a paper that are in no way substantiated by its contents.”

8. I will not pimp physics to the media

No – I would reserve the right to do this. Maybe I could agree to something like this: “I promise not to tell lies about my subject to science journalists.”

35. Newtonic Oath at Freedom of Science - November 13, 2007

[…] Oakley made good suggestions in Tommaso Dorigo’s blog. Nov.13, […]

36. Fred - November 13, 2007

T.

Geez, thanks for exposing me to the fact that my grandmothers are infinitely smarter than I am. I should have realized this long ago as they both enjoyed playing card games and drinking beer. A silver lining, though, as it is always a gift to be reminded of our grandparents in anyway possible. Somehow, I crawled towards the finish line until I met your words,

“This is the Goldstone theorem in a nutshell: The spontaneous breaking of a continuous symmetry of the lagrangian generates massless scalars.”

Now, the only thing I can think about is ‘ghost pain’, the phenomena experienced by amputees and others with related circumstances. Which brings me to this: do neurologists benefit from any work or results concluded by physicists?

p.s. The sombrero drawing should be added to your physic’s art collection

37. dorigo - November 13, 2007

Hi Coin,

sorry for my late answer. Here is what I can offer:

1) yes, goldstone bosons are the massless scalar fields that the lagrangian exhibits once a choice of the real vacuum has been made.
2) well, things are like I explained above. The lagrangian density contains a scalar field, but its excitations cannot be thought to describe particles because the vacuum is not the state of minimum energy and the particle spectrum of the theory is unclear. If a choice of the vacuum “hides” the symmetry, one realizes that the lagrangian describes massless scalars. Perhaps you are confusing the goldstone bosons, which appear after symmetry breaking, with gauge bosons that are there from the start in the lagrangian, and that acquire mass at the expense of the goldstones in the Higgs mechanism.
3) the goldstone bosons are “eaten” in the sense that they do not appear in the particle spectrum, and their corresponding degrees of freedom are used by gauge bosons to obtain mass. A massless particle has two degrees of freedom, a massive one has three.

I hope I was not too cryptic…
Cheers,
T.

38. dorigo - November 13, 2007

Hi Fred,

well, neurologists probably live happily without bothering with goldstone bosons. I do not really know about ghost pain in particle physics, although these are ghost particles for sure…

Cheers,
T.

39. Jonathan Vos Post - November 13, 2007

IMAGINARY IMAGINARY MASS, FORCE, ACCELERATION, AND MOMENTUM: PHYSICAL OR NONPHYSICAL?
Draft 6.0-Short [for ICCS 2004] of 27 April 2004 [8-page version + 3 pp. biblio …]

This is the short version of the paper, as in the Proceedings of the 4th International Conference on Complex Systems. There’s a sequel in the proceedings of the 5th (or was it 6th?). I chaired the two Physics sessions at the 7th ICCS 28 Oct-2 Nov 2007. The 8th will be may 2009.

Since this was presented and published, I’ve left Woodbury University, but Prof. Carmichael is still full-time Physics professor there; and our son Andrew Carmichael Post graduated age 18 (eighteen) with double B.S. in Mathematics and Computer Science, and is at USC Law School working towards at JD in Intellectual Property.

There is a very Science Fictional prediction of LHC results, not inappropriate for a paper reviewed by Prof. Gregory Benford, as well as Dr. George Hockney (JPL) and Prof. Michael Salamon (NASA HQ).

Since I’m not seeking tenure, I can write VERY speculative papers.

40. Qubit - November 13, 2007

I did not understand any of it! ??? But I still got to the end; I especially liked the pretty pictures. From them I realised that, your talking about the mechanics of a ballpoint pen, when the ball is pushed out you can balance the pen on its end, even if the symmetry of the pen is broken, here a picture of said pen on its end; http://bp2.blogger.com/_-mE1mHUrF8A/RzoWOADC0GI/AAAAAAAAADE/eqdaGc9nkik/s1600-h/Pen+on+tip.jpg
This is of course is your Mexican hat.

The other picture is this, http://bp1.blogger.com/_-mE1mHUrF8A/RzoWFwDC0FI/AAAAAAAAAC8/AYSYKwdlLRQ/s1600-h/pen+on+side.jpg
This is representation of the left hand picture of your two pictures that are together, the other being the right.

Qubit

There is no difference to the symmetries of each picture, even though one is laid on the table and the other is on its end, because the potential is the same for both states of the pen, unless on you are viewing it from the outside. Then one has a greater potential to be stood on its end, while the other has the potential to do the same, but the odds are the almost impossible.

So you could say that; there was not a real big bang, but rather one that was done to make you believe there was a big bang, when really it was done with birotechnics J

41. Qubit - November 13, 2007

“when the ball is pushed out”!

I meant to say; “when the ballpoint is inside the pen!”

42. Newtonic oath 2 at Freedom of Science - November 14, 2007

[…] Oakley writes: Renormalizability is a swamp. Yes — you can get the right answer by subtracting infinity […]

43. Coin - November 15, 2007

Perhaps you are confusing the goldstone bosons, which appear after symmetry breaking, with gauge bosons that are there from the start in the lagrangian, and that acquire mass at the expense of the goldstones in the Higgs mechanism.
3) the goldstone bosons are “eaten” in the sense that they do not appear in the particle spectrum, and their corresponding degrees of freedom are used by gauge bosons to obtain mass. A massless particle has two degrees of freedom, a massive one has three.

This actually clarifies a lot, thanks so much.

I guess the one thing that’s still confusing me is that we seem to be behaving really nonchalantly with regards to taking these fields apart and putting them back together. I’ve been kind of assuming that I’m supposed to be keeping two different notions of “field” separate, one the abstract, mathematical idea of “field” as a just a number associated with every point in space, and the other the particle idea of “field” which I’d just assumed described something more concrete, something that “exists”. I know that you can take a generic mathematical field and analyze it such that it behaves like a particle field, like these “phonon” things, but I assumed that this was just a formalism and that the phonon has no physical reality.

But, here you’re taking something that really is just a “mathematical” field, in the sense that it’s just a value which exists at every point in space– that is, the U(1) value taken by that degree of freedom at the bottom of the (I hate this term) “mexican hat”– and somehow we can act as if that value-at-every-point is an actual particle field. Even weirder, then we suddenly decide not to consider this U(1) value by itself, but consider it together with the vector values of a boson field, and say “hey, that’s the same set of degrees of freedom as a massive boson, let’s act like this is a massive boson”. Unless I’m totally missing something, it’s really surprising and interesting that quantum field theory is flexible enough to let you do that. If you can paste together degrees of freedom originating from different fields this way and say the result is a proper particle, then that seems to say phonons have as much physical reality as electrons…

Am I really understanding this right? Or is there some mathematics going on where I can’t see it that decides “why” the U(1) combines with the gauge boson? Thinking about it it seems like there must be, since it seems like there must be SOME reason why the U(1) symmetry decided to become the extra degree of freedom of the WZ gauge boson and not some other gauge boson. What determines which gauge symmetry “eats” the goldstone boson once the goldstone boson is hanging around?

Thanks again…

44. dorigo - November 16, 2007

Thank you Jonathan, unfortunately I have had no time to look at the paper yet, but I consider my blog an archive and I often dig out the stuff I had not seen before. I will get back to this.

Qubit, I am confused by your analogies no less than you have been by my post. How about declaring a draw and leaving it at that ? 😉

Hi Coin, your questions are meaningful, and I thank you for submitting them. I am not the best person to answer them, but what I can say is that if a lagrangian density describes some physical process, it does not matter what transformations you do on it: the physics will stay the same. We know that a quadratic term in a field corresponds to a mass term. We also know how to recognize terms that describe the coupling strength of three or four particles joining at a vertex. If we change coordinates unwittingly, we may hide the physics from plain view, but we won’t change it….

Cheers all,
T.

45. Guess Who - November 16, 2007

Coin, there is indeed no way to see from this post alone that the Goldstone boson combines with a gauge boson. To see that you must do a few things:

1) Add the pure electromagnetic Lagrangian to the Goldstone Lagrangian.

2) Couple the two, i.e. gauge the Goldstone Lagrangian, by changing the plain derivatives in it to covariant derivatives: derivative + gauge field term. Physically, this means that the complex scalar now carries electric charge and interacts with the electromagnetic field. (Without the Mexican hat potential, this model is just scalar electrodynamics, i.e. the complex scalars are like electrons and positrons without spin).

3) Do the same shift of scalar field coordinates as before, to some arbirary point in the potential valley, and notice what this does to the covariant derivative terms: terms containing the electromagnetic field squared times the constant scalar part squared pop out. Those are your new gauge field mass terms. (You also get other, messy interaction terms, but let’s not worry about those now.)

So you see, there is nothing arbitrary about how the complex scalar combines with the electromagnetic field. It does so because it carries electric charge. More generally, a scalar field carrying some gauge charge can give mass to the associated gauge field, if you give it a potential with multiple ground states.

And that’s the Higgs mechanism. Which probably means that I just ruined TD’s next post in this series… 😉

46. Coin - November 16, 2007

Dorigo/Guess Who, thanks!

47. dorigo - November 16, 2007

I associate in thanking GW; because he or she answered your question in a very clear way, while I failed to do that. Anyway, I am now into writing about the top quark, so I did not have the higgs mechanism in the agenda…

Cheers,
T.

48. Qubit - November 20, 2007

A “draw”, that’s ok dorigo, it was a joke. “Birotechnics”? 🙂 There was a smiley but it seems to have come out as a “_l”. I may not understand what you wrote, but I do understand the universe…

Later.

49. Qubit - November 26, 2007

In fact I just realised; declaring a Draw! That really is funny! :):):)

50. dorigo - November 27, 2007

Yes Qubit, draw as in chess, when one is stalemated, or better, when two players repeat their moves aimlessly, realizing they cannot make progress.

Cheers,
T.

51. vputz - December 14, 2007

Hm. I liked it. As a grad student taking QFT with 15 years since my undergraduate in physics, I’m finding that “big-picture-what-does-this-mean” is often missing in lectures, so frankly I found this quite handy.

52. dorigo - December 14, 2007

Thank you vputz. How come you spent so much time before entering grad school ? Anyway, good luck with your studies.

Cheers,
T.

53. Scott - March 30, 2008

Although, I’m definitely not your target audience, I very much appreciate the lecture that you have presented here. I would love to see more.

54. dorigo - March 30, 2008

Thank you Scott. Browse the site, or just hang around, and you will.
Cheers,
T.

55. Tia Miceli - April 28, 2008

Please explain the next step of eating the Nambu-Goldstone bosons! I’m a graduate student (2nd year in the US I just took the equivalent of your course last quarter in school) but I still don’t really get it. You explain things well! Please go on!

56. dorigo - April 28, 2008

Dear Tia,

if you browse around in this site, you will find more information on particle physics from an experimentalist point of view. The post above is some kind of exception, in the sense that it is devoted to a purely theoretical concept. It takes me much more effort to write about theory, but I do it time and again.

I am afraid I have no time to discuss the ins and outs of the Higgs mechanism (if that is what you were asking for) in the short period. I will, but it may take a while. My best advice is to hang around – there is a lot of Higgs physics in my blog.

In any case, thank you for the visit…

Cheers,
T.

57. Goldstone Theorem « Force - October 3, 2008