10
$\begingroup$

In Einstein's General Relativity we relate the effects of gravity with the curvature of the Levi-Civita connection on the spacetime manifold. Also, when we get the electromagnetic tensor $F = dA$ where $A$ is the $4$-potential, I've heard it's possible to show that $F$ is the curvature $2$-form on a certain principal bundle.

Thinking about that I asked a physicist I know about the other two fundamental forces and he said that they both can also be related to curvature $2$-forms of certain connections. In particular, he said that the structure group of the principal bundle on which we define the connection characterizes the force.

I'm just starting to study connections on fiber bundles, so that I yet don't understand the implications of all of this completely but what I'm trying to get is the intuition. What is the intuition on relating forces to connections? Also, why connections on principal bundles? I know that connections can be defined on arbitrary fiber bundles (Ehresmann Connection), so why forces require principal fiber bundles?

$\endgroup$
7
  • 1
    $\begingroup$ note that assuming ordinary general relativity, gravity is rather different and shouldn't be lumped together with the other forces (this gets somewhat alleviated in the teleparallel formulation); for a hand-wavy explanation why we care about principal connections: because we need to parallel transport particle properties (which relate to internal symmetries) $\endgroup$
    – Christoph
    Commented Apr 15, 2014 at 23:22
  • 2
    $\begingroup$ Thanks @Christoph for your point. Now, as I've said I'm just starting with connections. What is the difference in general relativity? I know that we stop conceiving gravity as a force and conceive it as a property of spacetime itself. But still, it is described by a connection right? Also, what is this idea of parallel transporting particle properties? Could you tell me where to read more about it? Thanks again. $\endgroup$
    – Gold
    Commented Apr 16, 2014 at 0:32
  • 2
    $\begingroup$ @user1620696: in classical Yang-Mills theory, you get equations of motion for particles from force equations (the generalization of the Lorentz-force law); in the Lagrangian formulation, this is achieved via minimal coupling, which requires a generalized charge (in the general case, a coadjoint orbit instead of just a number); this is not the case for gravity; another way in which gravity is different that the gauge symmetries in YM theory are vertical (leaving space-time alone), wheras in case of gravity, they are not $\endgroup$
    – Christoph
    Commented Apr 16, 2014 at 6:50
  • 1
    $\begingroup$ the idea of parallel transporting particle properties is just how I tend to think about it; if your fiber bundle isn't trivial, you need a notion of parallel transport to compare stuff 'over here' to stuff 'over there'; thus, connections (basically the infinitesimal version of parallel transport) appear naturally; in the Yang-Mills case, the 'stuff' you want to compare depends on some sort of internal symmetry, and we end up with principal connections $\endgroup$
    – Christoph
    Commented Apr 16, 2014 at 6:54
  • 1
    $\begingroup$ Forces do not "require principal bundles". Instead, it is the description of forces in terms of potentials that leads to gauge symmetries (adding total derivatives to the potential, roughly), and gauge symmetries are theories on principal bundles in the Lagrangian formulation. There's no intuition for this because no non-Abelian gauge theory classically appears in the description of a force, and the $\mathrm{U}(1)$ case looks far simpler than this general case. $\endgroup$
    – ACuriousMind
    Commented Aug 10, 2015 at 19:57

2 Answers 2

4
$\begingroup$

If you're looking for intuition about the relationship between forces and connections, I think the best you can do is to think long and hard about the following elementary example (adapted from Moriyasu's book 'Elementary Primer for Gauge Theory'):

Consider a vector at position $x$. Call its length $f(x)$. We want to know how this length changes as we go from $x$ to $x + dx$. (This could be done either by moving the vector itself or by shifting our coordinate system.) Call this new length $f(x + dx)$.

In elementary physics applications, when we measure a vector at two different points in this way, to see if its length has changed, we make a rather large and unwarranted assumption: we assume that the rulers we use to measure the two lengths are scaled the same.

In general, though, we want our mathematical model to respect the fact that the vector is a physical object whose properties cannot depend on something so arbitrary as the scaling of a ruler. Any pattern or regularity that depends on the units on our ruler is not 'real'.

This means we now have a rule: we must let each point - each 'place' - have its own ruler for measuring lengths of vectors. But this presents us with a problem: if we measure the vector's length using different units from one point to another, how can we know whether it has really changed and if so, by how much?

In summary, when differentiating the length function $f(x)$, we're asking how the vector's length changes as we move from $x$ to $x + dx$. And there are two possible reasons it could change: a) the vector itself has actually stretched or shrunk (and so the change would show up even if we used the same ruler everywhere) or b) the vector hasn't really changed at all, it just LOOKS like it's changed because the ruler we use at $x + dx$ is scaled differently from the ruler we use at $x$. Of course, the change we see in the components could result from contributions from both effects.

The solution is to re-define what it means to take a derivative by building in a 'correction' term that tracks how the units on the ruler shrink or grow. If you don't include this 'correction' term, you'll end up detecting changes that didn't actually happen. You know they didn't happen because other observers, using equally valid rulers, didn't detect them. Why? Because they are not bedrock facts of the universe. They are accidental artifacts of your coordinate system.

In elementary calculus, where we assume the ruler is scaled the same everywhere, the derivative of the length of the vector is given by

$$f(x + dx) = f(x) + \partial_\mu f dx^\mu$$

To keep track of how the ruler's units change (which is the most basic, intuitive form of what is called a 'gauge transformation'), we introduce a 'scale factor' $S$. This tells us how much the a unit on the ruler has grown or shrunk. We define $S(x) = 1$, i.e. we use the scaling of the ruler at point x (our starting point) as our 'baseline'.

This means the scale factor at nearby point $x + dx$ is

$$ S(x + dx) = 1 + \partial_\mu S dx^\mu$$

Now it's pretty clear that the length of the vector at the point $x + dx$ is the product of $f(x + dx)$ and $S(x + dx)$. I.e. it's the change in the vector's actual size multiplied by the change in the units you're using to measure that size.

Doing some simple algebra, this product is

$$ Sf = f + \partial_\mu S fdx^\mu + \partial_\mu f dx^\mu$$

$$ = f + (\partial_\mu + \partial_\mu S)fdx^\mu$$

So notice that our differential operator is now no longer just $\partial_\mu$ but has a correction term $\partial_\mu S$. This last term is the connection.

I would keep this simple 'toy' example in mind throughout your studies of these topics. It will help keep you oriented. When you learn something new or are confused, ask what that object or relationship looks like in this simple picture. This will help clarify things.

Roughly, using a different ruler at each point generalizes to using different basis vectors, or frames, to describe our vector fields at each point in space-time. (However, the vectors are not space-time vectors. They live in various kinds of space 'attached' to space-time at each point.) These basis vectors are related to each other by generalized rotations. At each point, the basis is rotated independently of how it is rotated at every other point. This is analogous to the ruler being stretched by different amounts at each point.

We want to build into our mathematics a way of keeping track of this. That's exactly what the connection does. This is why you will often find the connection defined as a Lie-algebra-valued 1-form. The rotations of the basis vectors are represented as a group acting on the basis vectors. Lie algebra elements are basically 'infinitesimal' versions of group elements, so they model 'infinitesimal rotations'.

As to how all this relates to forces, it is a subtle point. The basic insight is that the existence of what we are used to calling 'forces' is equivalent to the fact that the paths of physical systems through various spaces do not depend on how we coordinatize those spaces. I.e. physical facts are independent of the basis we choose to use to convert those facts to numbers that we can manipulate mathematically.

The simplest illustration of this that I've come across is in MTW's 'Gravitation.' We like to think of clocks as ticking at 'regular rates', i.e. with equal intervals of time between them, like a metronome. However, who is to say what a 'regular rate' is anyway? Suppose you use a standard clock, while I am using a 'crazy' clock, which ticks at highly irregular, unpredictable intervals. But who is to say that my clock is the crazy one? If I use the intervals of my clock to count out units, so that each tick of my clock is by definition one unit of time, then your clock is the 'crazy' one. Both our clocks are perfectly valid. They just don't agree. However, we must agree on the physical facts, mustn't we? But if I take a particle that you perceive as executing uniform motion, since my clock is 'crazy', it looks to me it's accelerating in all kinds of weird ways. Acceleration means force. So in order for me to explain what I am seeing as being the same thing you are seeing, I have to postulate a 'force.' This force is not a push or a pull, since evidently nothing is pushing or pulling the particle. (Otherwise you wouldn't perceive it as executing uniform motion.) This force arises from the fact that I am using a 'crazy clock.'

Another good example of this is polar coordinates in the plane. If you want to describe the trajectory of a billiard ball executing uniform motion across a plane using polar coordinates, you will get extra terms that you could characterize as 'forces'. This is because, by using polar coordinates, you thereby define motion along the coordinate lines (circles and radii) to be 'straight-line motion'. So a path that looks straight in cartesian coordinates actually looks accelerated in polar coordinates. I.e. a force is required to explain the deviation.

Sometimes you hear it said that Einstein showed that gravity is a 'fictional' force, a la the centrifugal force. In a sense that is true, if we take 'force' to mean 'a push or a pull that disturbs a particle from straight-line motion through a flat space-time'. Einstein showed that gravity is not such a push or a pull but rather is undisturbed motion through a non-flat space-time.

This curvature shows up in the fact that you cannot find a coordinatization in which you can use the same set of basis vectors everywhere in space-time. Curvature means you must rotate your basis vectors at some point. But now we come back to the notion of connection illustrated above: this means the connection 'correction' term can't be zero everywhere. It can always be zero locally, but as you move out from a very small region around your point of interest, in a curved space, you'll always find that the connection coefficients (the $\Gamma$ terms in GR) take on non-zero values. What this means is that their derivatives do not vanish. Roughly speaking, in a curved space, the connection can't be zero everywhere. Eventually you have to keep track of how the basis vectors are changing.

GR's status as a gauge theory is contested, as I understand it. However, it is true that the other forces of nature have been given a similar treatment as gravity in that they are understood to be similarly 'fictional' in the sense that they 'arise' from the need for different coordinate descriptions of the same physical facts to be mutually compatible.

You'll often hear it said that 'local gauge symmetries (i.e. local rulers) give rise to forces/interactions.' This is incorrect, or at least incomplete. Local gauge symmetries and forces imply each other. They are equivalent, flip sides of a coin. Starting with the demand for local gauge symmetry to hold, we can derive forces. But starting with forces, we can derive local gauge symmetry. It's not so much that one causes the other as that they are manifestations of the same thing. As you rotate around a circle, the sine function getting smaller doesn't 'cause' the cosine function to get bigger. Given the constant radius defining the circle, each change implies the other.

Honestly, this is a lot of words to say things that are almost mind-numbingly simple. But at its heart, the notion of connections and forces is not much more complicated than this. It's just that the objects get more numerous and abstract and the notations get messier and less consistent.

As to why all this is best described using principal bundles, I have no satisfying answer. Why is ordinary space best modeled using Euclidean 3-space? Why does classical mechanics use phase space? Why does SR use Minkowski space? Why does quantum mechanics use Hilbert spaces? The best answer is perhaps that this is what mathematicians have found that best does the job.

$\endgroup$
0
$\begingroup$

Following Cunningham's Law, here is a probably somewhat wrong answer:

When spacetime/our gauge bundle is flat, we do not really need connections. Even if we write down our theory in general terms which includes connections, there is a choice of local coordinate systems where the connection vanishes everywhere. The dynamics in these systems is trivial.

We only get nontrivial dynamics when spacetime/our gauge bundle is non-flat. In this case, we can't get rid of the connections by changing our local coordinate systems. Hence nontrivial dynamics always imply that we need to use connections.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.