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Aethelwulf
2012-May-11, 10:37 PM
So, what do physicists mean when they speak of a mass, or rather, a mass term? For a while, physicists tried to answer this question a number of ways.

There was such a thing as an electromagnetic mass [1] at one point in physics. The was the idea that the electromagnetic field interacted with a quantum object in such a way that it gave a contribution of mass to a system. This was called the self-energy of a system -- in fact, so mainstream the idea was at one point, it was in fact considered as an explanation for inertial mass itself.

It is quite curious to mention that there was even an electromagnetic inertia which was experienced by particles which where accelerated, this was explained through what is called the Larmor Equation, but given as an expression here:

\frac{2}{3}\frac{e^2}{c^3}a^2

This would in fact describe how much radiation a charged particle would give off as it was accelerated to higher and higher speeds. The faster you made your particle move at, the more power was required to bring it to that speed. This was in fact analogous to inertia which was experienced by particles with a charge: today, our new definition is that inertia is strictly a property of mass. As much as this idea has faided, some scientists still use the equations describing the EM-mass.

There maybe some important clues however that it may still have importance, especially since the current model, the Higgs Boson is now finally faiding into the past as scientists have narrowed the possible energy levels for the Higgs Boson to a very small corner. Some of these clues, may exist in the fact that charged massless bosons don't exist. The Yang Mills Equations once predicted massless charge bosons but the approach has been generally considered today as a false one. But why should this prediction and how it failed important?

Charge is considered by physicists as an ''intrinsic property'', something which is inherent with particles of mass. When we think about charge, we usually talk about electromagnetic charges e, which will sometimes be found as a coefficient to the electromagnetic potential in equations ieA. What is important here, is that the electric charge is always found with mass, so maybe electromagnetic mass contributions will arise from the shadows again? Who knows, but what does seem clear that the Higgs Boson is now failing our theory so we will need to revise the equations again. So what is a Higgs Boson? How does it enter our theories?

Our theory of Gauge Fields [2] where \phi is a Gauge Boson, has a structure in the equations which invariance. This has been called, Gauge Invariance and this invariance allows physicists to make certain transformations to fields. The laws of physics always remain the same under gauge invariance, indeed, that is what it's all about! It means there is no such thing as an absolute position in physics - the only thing which does count is relative positions. Another feature of the gauge fields is that they retain a symmetry of the theory.

A symmetry can be understood from the most simplest langrangian term

\partial\phi^{\dagger} \partial \phi

\partial \phi'(x) \rightarrow e^{i\theta}\phi(x)

In this transformation, the derivatives of \phi' are concerned only with \phi. This transformation will look like

\partial \phi'(x) = e^{i \theta} \partial \phi(x)

\partial \phi'^{\dagger}(x) = e^{-i \theta} \partial \phi^{\dagger}(x)

Therefore, when you multiply \partial \phi ' with \partial \phi^{\dagger} the e^{i\theta} and e^{-i\theta} will cancel out because that is how you would compute them with their conjugates. Thus

In other words, our field \theta is constant, and does not require the same derivatives as our boson. In return, we would just get

\partial \phi^{\dagger}\partial \phi

and viola! This was the most simplest demonstration of a symmetry conserving field. An even simpler demonstration would be:

\frac{d(x +C)}{dt}

is in fact simply the same as

\frac{dx}{dt}

this means that the equations where symmetry existed and what we have in these symmetries are extra constants always remain the same, just as our Gauge example above.

But what if \theta also was a function of position \theta (x)? Well let's see shall we.

\partial \phi' = (\partial \phi + i \phi \frac{\partial \theta}{\partial x})e^{i \theta}

and our conjugate would be

\partial \phi'^{\dagger} = (\partial \phi^{\dagger} - i \phi^{\dagger} \frac{\partial \theta}{\partial x})e^{-i \theta}

Multiplying the two, we need to factorize it

(\partial \phi + i\phi\frac{\partial \theta}{\partial x})(\partial \phi^{\dagger} - i\phi^{\dagger}\frac{\partial \theta}{\partial x})

which gives

=\partial \phi^{\dagger} \partial \phi + i(\phi \partial \phi^{\dagger} - \phi^{\dagger} \partial \phi)\frac{\partial \theta}{\partial x} + \phi^{\dagger} \phi (\frac{\partial \theta}{\partial x})^2

The reason, again why this equation turned out to be such a mess was because our \theta-field now depended on position, so the derivatives of our boson field also included it.

This motivated scientists to find a symmetry again in the equations, and to do so required the use of te Covariant Derivative which originally came form the work on fibre bundles.

To restore symmetry, we need to define the Covariant Derivative as

A_{\mu}' \rightarrow A_{\mu} - \partial_{\mu} \theta

Here, we can see our four-vector potential again A_{\mu} - you can basically build the electromagnetic fields form this. It's time component in A_0 but that is really not relevant right now. Our Covariant Derivative and the respective conjugate fields are usually denoted as

D_{\mu} \phi = \partial_{\mu} \phi + iA_{\mu} \phi

and

D_{\mu} \phi^{\dagger} = \partial_{\mu} \phi^{\dagger} - iA_{\mu} \phi^{\dagger}

So calculating it all together, we just define the whole thing again as:

D\phi' = (\partial \phi + i\phi \frac{\partial \theta}{\partial x})e^{i\theta} + i(A_{\mu} - \partial_{\mu} \theta) \phi e^{i \theta}

Well, with this, we can see straight away that some terms cancel out. The i\phi terms cancel, and \frac{\partial \theta}{\partial x} is in fact the same as \partial_{\mu} \theta. So what we are really left with is

D\phi' = D \phi e^{i\theta}

and so its conjugate is

D\phi^{\dagger} = D\phi^{\dagger} e^{-i\theta}

Again, the latter terms cancel out when you multiply these two together and so what you end up with is

D\phi^{\dagger}' D\phi' = D\phi^{\dagger}D\phi

and so by using the Covariant Derivative, we have been able to restore the lost U(1) symmetry where U(1) symmetries deal with rotations.

But when physicists talk about a mass, we don't want to retain symmetry in these fields. In fact, the very presence of a mass term will imply an explicit symmetry breaking. The process of course, is a little more complicated however. It involves another boson, called a Goldstone Boson, which can be thought of as a ground-state photon which lives in the minimum of a Mexican Hat potential. Something which exists in the minimum is the same as saying our system does not contain a mass term and so \phi=0. In such a potential, we may describe our field as

\phi = \rho e^{i\alpha}

Here, \rho is a deviation from the ground state. \rho is in fact our Higgs Boson and \alpha is our Goldstone Boson. If \alpha is a frozen (constant) field, then there are no changes in the equations. But if there is a deviation of the Goldstone Boson from the minimum of our potential, then we are saying that it costs energy to do so, and this energy is what we mean by particles like a photon obtaining a mass. In fact, the Goldstone Boson is gobbled up by the Higgs Boson which gives the system we speak about a mass. Our flucuation from the minimum has the identity f \ne 0 where f plays the role of mass.

Let's discuss this mass term in terms itself of the electromagnetic field tensor. Such a tensor looks like:

F_{\mu \nu}F^{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}

here, F_{\mu \nu}F^{\mu \nu} is in fact just F^2 and it makes up the langrangian. F_{\mu \nu} is an antisymmetric object with respect to swapping its indices. It is like a four dimensional curl.

It is Gauge Invariant, but checking that, remember our transformation

A' \rightarrow A - \partial \theta

plugging that into the tensor gives:

F_{\mu \nu}' = \partial_{\mu}A'_{\nu} - \partial_{\nu}A'_{\mu}

That is

\partial_{\nu} \partial_{\mu} A_{\nu} - \partial_{\mu} \partial_{\nu} \theta - \partial_{\mu} \partial_{\nu}A_{\mu} + \partial_{\nu} \partial_{\mu} \theta

Since the order of partial differentiation doesn't matter, the -\partial{\mu} \partial_\nu}\theta and \partial_{\nu}\partial_{\mu} \theta cancel and what you are left with is invariance.

There are some numerical factors left out of this, such as a quarter, but that really isn't all that important in this demonstration. And so, we may have a completely Gauge Invariant term

D_{\mu}\phi D_{\mu} \phi^{\dagger} - V(\phi^{\dagger}\phi) + F_{\mu \nu} F^{\mu \nu}

One should keep in mind that the potential term here V(\phi \phi^{\dagger}) is also manifestly invariant. What becomes interesting however, is the question of what cannot be added to this but still can remain Gauge Invariant. What we cannot add to the electromagntic part (F^2)-part, is this:

\frac{M^2}{2}A_{\mu}A^{\mu}

This is just the same as \frac{M^2 A^{2}_{\mu}}{2}. The reason why, is because a mass-term for a photon would typically look like this. This actually breaks local invariance. In nature there is only one massless spin-1 boson - the photon - then it seems that if we where to use a local gauge group that is less trivial than U(1), then the symmetry would be broken. It's interesting, that when we speak about symmetry breaking for a photon, we are in fact talking about an electromagnetic term which is squared in the Langrangian.

It seems, by no far shot, that physicists will not abandon the use of symmetry breaking to describe mass for particles, but it must be noted that our usual, and perhaps, most insightful way out of the problem, the Higgs Boson, may need some drastic changes. As I presented at the beginning of this thread, we have dealt with theories involving electromagnetic theories of mass. If this idea is not recollected, what will we turn to?

There are some alternatives. These actually go by the specific name of ''Higgless models''. We have actually a reasonable list of different theories which can be read about here: http://en.wikipedia.org/wiki/Higgsless_model

[1] - It seems that Einstein may well have been influenced by Thomson in his derivation of particles with mass unable to reach the speed of light. Thomson stated long before the development of modern relativity, ''... when v=c the energy becomes infinite, so that it would seem to be impossible to make a charged body move at a greater speed than that of light.'' Searle also stated this in the 1800's.

[2] - http://www.wiley-vch.de/books/sample/3527408355_c01.pdf

Aethelwulf
2012-May-11, 10:40 PM
Why has my tex code came up in this strange [img] thing?

Aethelwulf
2012-May-11, 10:49 PM
I just tried the equations on another site... they came up fine. I don't understand what is wrong here?

pzkpfw
2012-May-11, 11:00 PM
Tex tags are turned by the software into image tags, pointing (for a source for the image) to a site where the mark-up text is turned on the fly into an image.

e.g.

If you enter this: \mathcal{B} \subseteq \mathcal{A}

It gets turned into something like this: http://www.codecogs.com/eq.latex?\mathcal{B} \subseteq \mathcal{A}

Which makes this when you view: \mathcal{B} \subseteq \mathcal{A}

Not sure yet why some of your equations in post #1 don't work. Possibly a fault in them, or maybe that particular tex processor can't handle something in them.

Edit1: One that works one that doesn't

A: \partial \phi'(x) = e^{i \theta} \partial \phi(x)

B: \partial \phi'^{*}(x) = e^{-i \theta} \partial \phi*(x)

A: \partial \phi'(x) = e^{i \theta} \partial \phi(x)

B: \partial \phi'^{*}(x) = e^{-i \theta} \partial \phi*(x)

Edit2: I think that processor doesn't like use of "phi^{*}" or "phi*" - that seems a commonality in the tex tags that don't show correctly.

Aethelwulf
2012-May-11, 11:45 PM
hmmm... I wonder if it would prefer dagger notation... if that is indeed the problem.

Aethelwulf
2012-May-11, 11:45 PM
Yup, that seems to be the problem.. sigh... more work...

Aethelwulf
2012-May-11, 11:50 PM
Bravo pzkpfw, that seems to have fixed it. Thank you.