Ok Quark. I'm sitting in an airport in Houston, Texas after a weekend tryst with my girlfriend in San Antonio. This is a $9.95 answer because that's the cost of internet access. And there's nothing better than eating barbeque, drinking Shiner Bock beer, and talking about physics.
But there are a bewildering number of other animals in the particle zoo, pi, tau, mu and so on. May I, as a first approximation, say something like this: the results of experiments with atom smashers show the existence of such particles. Moreover, since Earth is constantly bombarded by ultra-high energy "cosmic rays" (whatever these might be) resulting form the violent nuclear reactions in our Sun, one may assume these guys occur naturally when these cosmic rays interact with atoms in our upper atmosphere.
This is correct. As always, there are subtleties, but in general this is correct.
But, can you please explain what "gauge boson" is. What does the term gauge mean?
This may be a long forray
The long answer is that the term ``gauge boson'' comes from the fact that the physics is ``gauge invariant''. Essentially, the gauge boson is a consequence of the fact that the physics is invariant under a local symmetry.
The first thing to know is that all of the physics can be derived from the action, which is a functional of the fields in the problem. I think mathematicians call it a ``functor'', but I could be mistaken---either way, I don't want to confuse you by trying to speak your language and failing miserably!
The action is written in terms of fields (which are functions of space-time coordinates), and describes all of the interactions that a set of fields can have, as well as how the fields get from point A to point B. So the action is a function(al) of fields, and the fields are functions of space-time coordinates x. Typically we write the action as $$S[\phi(x)]$$. There is nothing particularly physical about the field---one can write a local expression to describe a particle, in terms of fields. The field itself should be thought of as an operator---it has no meaning by itself, and needs a good Hilbert space to act on. (I don't know if the definition of Hilbert space is the same as the definition you'd use---I would say that a Hilbert space is a vector space with a positive definite inner product.) Also, to keep things simple, we will work with scalar fields, which have spin 0.
Ok. So we have an action which tells us about our fields. It turns out that sometimes the fields have some symmetry associated with them. What does this mean? Well, for example, let me take a stupid example. Take the function $$f(x) = x^2$$. There is a symmetry of this function that we all probably learned after we learned to read---that is, $$x\rightarrow -x$$ leaves f alone. To use the parlance of a physicist (pretending to know something about math), we would say that there is a discrete symmetry $$\mathbb{Z}_2$$ of this function. Further, this symmetry is global in the sense that we have to apply this symmetry at every point at the same time, in order for the symmetry to make sense.
Now we can wave our hands and generalize this. Suppose the action is invariant under a symmetry which acts on fields. Let's take another stupid example. Suppose we have a boson, spin 0 particle with an easy action:
$$S[\phi] = \partial_{\mu}\phi\partial^{\mu}\phi^* -m^2\phi^*\phi + \lambda (\phi^*\phi)^2$$.
The $$\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}$$, and the indices, when they are repeated are automatically summed over. The first term is called the kinetic term, and the second two terms are interaction terms. (This is why I said earlier that it is always useful to think of things in terms of interactions. The action describing nuclear decay can be written in a similar way, with a terms connecting a neutron, a proton, an electon, and a neutrino.) Also, this action describes the interaction of TWO fields (again, a subtlety!), $$\phi$$ and $$\phi^*$$. Let's not worry right now that the two fields are related by complex conjugation, and treat them as independant fields.
Now let's ignore some interesting physics, and just look at the symmetries of this action. Clearly $$\phi \rightarrow -\phi$$ is a symmetry of this action. We can try and be a bit more sophisticated about this, by writing $$\phi \rightarrow e^{-i \alpha}\phi$$, and $$\phi^* \rightarrow e^{i \alpha} \phi^*$$. It is now clear that the phase is a GLOBAL phase, which means that the symmetry I talked about is a global symmetry. In other words, the symmetry does not depend on the space-time coordinate x. Further, the symmetry is good for ANY phase $$\alpha$$. What we have discovered is a GLOBAL U(1) smmetry, because $$e^{i\alpha}$$ is the generator of U(1) transformations. Here, I mean the group U(1), not the algebra (I think).
But we can do better. What about LOCAL symmetries? What if we want our symmetry to depend on local functions of the space-time coordinate x? We might be worried about this because, for example, we know that the Lorentz transformations are local---we might expect that there are other transformations that we can do that are also local.
So let's see what happens. Suppose we want to look for symmetries of the form $$\phi \rightarrow e^{i\alpha (x)}$$. We see that there are no real problems except in the derivatives in the kinetic term of the action. The transformation is now a local function of x, so the derivative acts on it in a non-trivial way. You might think that we are screwed, but as good theoretical physicists, we know that if at first we don't succeed, we should add shit until it works.
The way to add shit is the following: just work it out. We should see that the transformation $$\phi \rightarrow e^{i\alpha (x)}\phi$$ gives:
$$S = \left[\partial_{\mu}e^{-i\alpha(x)}\phi\right] \cdot
\left[\partial^{\mu}e^{i\alpha(x)}\phi^*\right]$$
$$\Rightarrow S = \partial_{\mu}\phi\partial^{\mu}\phi^* + \phi^{*}\phi \partial_{\mu}\alpha (x) \partial^{\mu}\alpha (x) -i \partial_{\mu}\phi\partial^{\mu}\alpha(x) + i\partial_{\mu}\phi^*\partial^{\mu}\alpha(x) + \cdots$$.
Now look what we have found! Not only do we have a scalar field, but the fact that we promoted our U(1) to a LOCAL symmetry gave us another interaction. In fact, it gave us several new interactions! The new field we have found is $$\partial_{\mu} \alpha(x)$$. We can just rename this to $$A_{\mu}$$. Because $$\alpha$$ is a local transformation (or, a GAUGE transformaiton), the fied $$A_{\mu}$$ is a GAUGE field. Specifically, we have discovered a theory with a charged scalar field (spin 0 boson, called a scalar because it transforms trivially (i.e. spin 0) under the Lorentz group) interacts with another boson, except this one with a (space-time) vector index. The field $$A_{\mu}$$ is called a vector boson.
You may ask, why are we justified in adding things to our action? Remember---we want to use math to describe physics. We only want to add the minimum amount of shit---this is called (by some) Occam's razor...the simplest solution is typically the correct one. Either way, you may be a bit more comfortable if I redefine what I mean by derivative. Remember---the real trouble came in when we looked at the trnasformation properties of the
derivative terms in the action. We can define a ``covariant derivative'' that transforms trivially under the symmetry:
$$\mathcal{D_{\mu}} \equiv \partial_{\mu} - i A_{\mu}$$
So I have probably gone a bit far in answering your question. BUT now (hopefuly) you know what gauge invariance is: it is a local symmetry of the action. Gauge invariance (which is my second favorite invariance) of the action IMPLIES (or, predicts, if you like) the existence of a gauge boson. A gauge boson always mediates a force. So, putting it all together, a ``force'' in nature is caused by the existence of a symmetry in the action, which IMPLIES the existence of a gauge boson.
What did you mean by "at high energies"? Are we talking about experiments here or what? What does it mean for a gauge boson to "become massive"? Surely a boson has mass or it doesn't?
Perhaps I got a bit ahead of myself. But let's consider our new, gauge invariant action, written in terms of the field $$A_{\mu}$$:
$$S = \partial_{\mu}\phi\partial^{\mu}\phi^* + \phi^{*}\phi A_{\mu}A^{\mu} -i \partial_{\mu}\phi A^{\mu} + i\partial_{\mu}\phi^*A^{\mu}-m^2\phi^*\phi + \lambda (\phi^*\phi)^2 $$.
A few things should pop out at you. First, there is no kinetic term for $$A$$. We can add that by hand (it would have fallen out more naturally if I had done this in another way). But there is also no MASS term for A. That is, A is a massless field, as opposed to $$\phi$$, which has a mass term.
This means that this boson, $$A$$ is
massless. This is good, because we WANT to describe massless bosons, like the photon. Now, as zephir pointed out, we COULD add a mass term for the photon, but there is ABSOLUTELY no reason to do so, and this is utterly something that we would do by hand. This is opposed to the kinetic term that we added (seemingly) by hand for the vector field $$A$$. But one can motivate that by doing all of this derivation in another manner.
