Sunday, December 27, 2015

Lagrangian formalism for fields higgs Goldstone The lagrangian density 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

http://www.scholarpedia.org/article/Lagrangian_formalism_for_fields


[PDF]4. Central Forces - damtp
www.damtp.cam.ac.uk/user/tong/.../four.pdf

University of Cambridge
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In this section we will study the three-dimensional motion of a particle in a central ... forces are of this form, solutions to this equation contain some of the most important results in ... If the particle travels with constant angular velocity ˙θ = ω then the ... Figure 12: The effective potential arising from the inverse square force law.

 

The Goldstone Theorem - Fine Structure

www.finestructure.com/2007/11/the-goldstone-theorem-quantum/

Nov 12, 2007 - Although this for-real-dummies post includes phrases like "...we ... http://dorigo.wordpress.com/2007/11/10/the-goldstone-theorem-for-real- ...


[PDF]symmetry breaking - Fermilab

home.fnal.gov/~felixyu/Ask_A_Scientist_Felix_Yu.pdf

Fermilab
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Nov 3, 2013 - broken weak symmetry. 42/48. Picture credit http://dorigo.wordpress.com/2007/11/10/the-goldstone-theorem-for-real-dummies/. Early times.

https://dorigo.wordpress.com/2007/11/10/the-goldstone-theorem-for-real-dummies/
The Goldstone Theorem for Real Dummies November 10, 2007 Posted by dorigo in mathematics, personal, physics, science.
trackback 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…
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