Are Charge And Mass The Same Thing?

[Aethelwulf]

Is mass the same thing as charge?

 

 

Our theory of Gauge Fields where [math]\phi[/math] 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

 

[math]\partial\phi^{*} \partial \phi[/math]

 

Suppose we had a transformation

 

[math]\partial \phi' \rightarrow e^{i\theta}\phi(x)[/math]

 

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

 

[math]\partial \phi'(x) = e^{i \theta} \partial \phi(x)[/math]

 

[math]\partial \phi'^{*}(x) = e^{-i \theta} \partial \phi^{*}(x)[/math]

 

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

 

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

 

[math]\partial \phi^{*}\partial \phi[/math]

 

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

 

[math]\frac{d(x +C)}{dt}[/math]

 

is in fact simply the same as

 

[math]\frac{dx}{dt}[/math]

 

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 [math]\theta[/math] also was a function of position [math]\theta (x)[/math]? Well let's see shall we.

 

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

 

and our conjugate would be

 

[math]\partial \phi'^{*} = (\partial \phi^{*} - i \phi^{*} \frac{\partial \theta}{\partial x})e^{-i \theta}[/math]

 

Multiplying the two, we need to factorize it

 

[math](\partial \phi + i\phi\frac{\partial \theta}{\partial x})(\partial \phi^{*} - i\phi^{*}\frac{\partial \theta}{\partial x})[/math]

 

which gives

 

[math]=\partial \phi^{*} \partial \phi + i(\phi \partial \phi^{*} - \phi^{*} \partial \phi)\frac{\partial \theta}{\partial x} + \phi^{*} \phi (\frac{\partial \theta}{\partial x})^2[/math]

 

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

 

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

 

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

 

[math]A_{\mu}' \rightarrow A_{\mu} - \partial_{\mu} \theta[/math]

 

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

 

[math]D_{\mu} \phi = \partial_{\mu} \phi + iA_{\mu} \phi[/math]

 

and

 

[math]D_{\mu} \phi^{*} = \partial_{\mu} \phi^{*} - iA_{\mu} \phi^{*}[/math]

 

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

 

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

 

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

 

[math]D\phi' = D \phi e^{i\theta}[/math]

 

and so its conjugate is

 

[math]D\phi^{*} = D\phi^{*} e^{-i\theta}[/math]

 

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

 

[math]D \phi^{*'} D \phi ' = D \phi^{*} D \phi[/math]

 

and so by using the Covariant Derivative, we have been able to restore the lost [math]U(1)[/math] symmetry where [math]U(1)[/math] 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 [math]\phi=0[/math]. In such a potential, we may describe our field as

 

[math]\phi = \rho e^{i\alpha}[/math]

 

Here, [math]\rho[/math] is a deviation from the ground state. [math]\rho[/math] is in fact our Higgs Boson and [math]\alpha[/math] is our Goldstone Boson. If [math]\alpha[/math] 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 [math]f \ne 0[/math] where [math]f[/math] plays the role of mass.

 

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

 

[math]F_{\mu \nu}F^{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}[/math]

 

here, [math]F_{\mu \nu}F^{\mu \nu}[/math] is in fact just [math]F^2[/math] and it makes up the langrangian. [math]F_{\mu \nu}[/math] 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

 

[math]A' \rightarrow A - \partial \theta[/math]

 

plugging that into the tensor gives:

 

[math]F_{\mu \nu}' = \partial_{\mu}A'_{\nu} - \partial_{\nu}A'_{\mu}[/math]

 

That is

 

[math]\partial_{\nu} \partial_{\mu} A_{\nu} - \partial_{\mu} \partial_{\nu} \theta - \partial_{\mu} \partial_{\nu}A_{\mu} + \partial_{\nu} \partial_{\mu} \theta[/math]

 

Since the order of partial differentiation doesn't matter, the [math]-\partial{\mu} \partial_{\nu}\theta[/math] and [math]\partial_{\nu}\partial_{\mu} \theta[/math] 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

 

[math]D_{\mu}\phi D_{\mu} \phi^{*} - V(\phi^{*}\phi) + F_{\mu \nu} F^{\mu \nu}[/math]

 

One should keep in mind that the potential term here [math]V(\phi \phi^{*})[/math] 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 ([math]F^2[/math])-part, is this:

 

[math]\frac{M^2}{2}A_{\mu}A^{\mu}[/math]

 

This is just the same as [math]\frac{M^2 A^{2}_{\mu}}{2}[/math]. 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 [math]U(1)[/math], 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....Higgsless_model .

 

 

Studying Gauge theory, one thing seems quite clear - that being the presence of matter seems to be synonymous with the presence of charge. One can see this clearly when you study the equations which describe massless boson fields which satisfy a charge [math]e = 0[/math]. As we have seen, a mass term in an electromagnetic tensor equation would have the appearance of

 

[math]\frac{1}{2} M^2 A^{2}_{\mu}[/math]

 

If there is a symmetry we have

 

[math]\delta \mathcal{L} = 0[/math]

 

The langrangian can be symmetric but the vacuum may not be

 

[math]e_i|0> \ne 0[/math]

 

Where [math]e[/math] is the charge. In this case for symmetry breaking, Noethers theorem does not imply a conserved charge. In this terminology we say the ground state is in fact degenerate.

 

The only problem with idea of mass being invariant with charge is the theoretical models involving neutrino's. Neutrino's have a very very small mass but is said to contain no charge. This seems to be the only exception in nature - but how accurate are we is something to debated... We once thought the neutrino was massless but it turned out we where wrong. Neutrino's may have a non-zero magnetic moment however. Interestingly, a magnetic moment is caused by an intrinsic angular momentum [math]\hbar[/math].

 

The charge of a neutrino may just be very very small, which might be satisfied as

 

[math]\mu = \frac{e}{2M} \frac{\hbar}{2}[/math]

 

They are fermions afterall so if they did have a very small charge (corresponding) to it's vanishingly small mass, then it would have a magnetic moment equation similar to an electron.

 

However, if a scientist does not take my speculations on the neutrino seriously, I ask them to consider the ramifications of the symmetry breaking involved for systems. Symmetry breaking explicitely states that charge is not conserved so we are not talking about massless, chargless particles.

 

 

 

[ref]

 

 

http://arxiv.org/pdf/1102.0468v2.pdf

 

 

http://en.wikipedia.org/wiki/Charge_conservation

Edited November 26, 2012 by Aethelwulf

[Aethelwulf]

I read that paper wrong!

Edited November 26, 2012 by Aethelwulf

[Aethelwulf]

I thought it said that M and Q was invariant for a neutrino. It was saying it was non-invariant for Q and M, and that it may be a solution to the solar neutrino puzzle. So deleted it.

[Aethelwulf]

Here is an important paper towards the question of neutrino charge.

 

http://arxiv.org/pdf/hep-ph/0311176v1.pdf

 

(Though very short)