(2/2) > I have no idea what's going on here. Why would you measure different points of the field with different coordinate systems and expect sensical results? I'm imagining a surveyor walking in a line starting from the origin: he takes a measurement at the origin, then at (1,0), then at (2,0), then at (3,0), etc. (Imagine that the underlying field is frozen in time so we aren't dealing with the Lagrangian yet.) Since we know the field equations we can predict what he'll measure at each of those points in the line.

> But if the coordinate system changes with every step, he's still moving in a straight line as seen from a bird flying overhead, but at his first step he's at (1,0), then at the next step (2,0) turns into (5,1), then at the next step (6,1) (aka (3,0)) turns into (12,-3), etc, because the coordinate system changes each step. It's still (1,0),(2,0),(3,0) if you measure in the original coordinate system. But the underlying field wouldn't change in that case. Sure, if you put (5,1) into the field equation you'll get a different result than if you put in (2,0), but if you're only changing the coordinate system then that has to be compensated for in the field equation itself and you're not going to get different results for the same physical point. I mean, you should get the same result if you do f((2,0), coordinate system a) as if you did f((5,1), coordinate system b)

So I guess a simpler way to see this is to note that changing coordinates should never affect the results of a physical theory. For example, if I measure things with the origin at x = 0, but you use x = 5 - all our measurements of position will differ by 5 units. But when we apply the equations of motion to some object we're both looking at - F = ma, our predictions of how the object will move will agree. My predictions will be the same as yours, but the positions will differ by 5 units. This is because the equation of motion, F = ma, does not care about translations in the coordinate system, the acceleration is the second derivative. If we had an equation like F = ma + x, then we would no longer be coordinate invariant, and the equation would be unphysical. It wouldn't make sense, you could look at the system in a different way (i.e. using different coordinates) and get completely different results.

> Why is it called that? I assume someone chose that name because it made sense to them for good reasons

This I'm not sure about - I haven't actually seen the reason mentioned anywhere, other than just an indication that it's for some historical reason.

> Wait, "our field" and "the new field"? What fields are those? We were talking about a field defined by the function ϕ(x,t) and thinking about its Lagrangian. We added a term A to the Lagrangian and that was it. What's the "new field"? Why does ϕ(x,t) have an interaction with it? Is A the new field?

Yup! A is the 'new field' - the terminology could be clearer here. So by requiring that our toy Lagrangian with phi be gauge invariant, we now have a need to introduce another field, A, and way this field appears in the Lagrangian (by multiplying by phi) is something that will end up acting like a force.

> What is "the way we're measuring it"? I think it's, basically, the coordinate system of the surveyor changes each step, so that's a different "way" of measuring it? I still don't see why changing the coordinate system makes new stuff pop out of the equation

Yes - by way of measuring it I mean that we are using a different coordinate system at each point, just to see what the effects of doing so are. And this is the magic bit - if we don't have this other field A, then the equation will produce different results depending on which coordinate system you use. So any theory without A will not consistently give the same results regardless of coordinate system.

> How do you expand it? Are you giving the definition of W(x, x+δx)? How'd you get that? What's O(δx^2)?

Power series expansion. So we are defining W(x, y) as a function that allows us to compare the field at x and y. We don't know what this function is, we just assume it exists. We then expand it out in powers of delta x. Since we are eventually going to take the limit, any higher order term - that involves delta x squared or higher - will be too small to matter, so we just care about the constant term and the linear term.

> Where'd the second set come from? Wait, I see, it's just saying that you can say that "multiplying a number by a 1x1 matrix" is the same thing as saying "you can compose a number with a rotation". It's literally the same thing, just said in a less clear manner. Does the new terminology get us anything useful?

Once we make the connection between a symmetry of our Lagrangian and some abstract group like SU(3), we can immediately bring in group theoretic results about that group. For example, since we know (from group theory) that SU(3) has eight generators, we can now use that result and infer that we need eight gauge bosons to make the theory gauge invariant.

> What? How do you know how many generators there are? Why does SU(2) have two generators but three gauge bosons?

Typo - will fix! Should be three generators.

> Yeah, I think the part I'm not getting is how changing coordinate systems affects the equation. I think I can see that if you insist on doing something ridiculous like this you'd need some math to correct for it and if the new correction functions are fields then it looks like new particles popping out, but I don't see how that doesn't result in an infinite number of new particles. Like, I can add a function f(x) = x^2 to the Lagrangian, then a g(x) = -x^2 to compensate for it, but those don't represent new particles do they? Why do those cancel out but A doesn't? I just don't see how changing coordinate systems results in different results

So anytime something gets added to the Lagrangian, you effectively have a new theory - it's a proposal. The point here is that you don't really have infinite flexibility in adding things to the Lagrangian - if you add complex scalar fields, then you must also add gauge bosons. So symmetry doesn't constrain everything - you still need to figure out what the Lagrangian should be, but it'll force you to add other fields as well to make things gauge invariant.

> Despite my questions, I think I have a better idea of what's going on. You have a function; it should spit out the same numbers when you rotate it; you need a function to correct for the rotation; in physics the new function looks like a particle. I can kinda sorta see how it works now. Thanks for the article!

Appreciate the thorough review! Lots of things that I should have been more thorough about - I will fix :) Thanks again.