Gauge theory

Ben · 07-04-2009

It seems that Einstein once remarked of a colleague that "he could calculate, but he couldn't think". I find myself in an analogous situation here; I have a degree of understanding of the mathematics used in gauge theories, but I am struggling with their physical interpretation.

Here's what I've got so far; I would like you phys jocks to help me out.

We are talking spacetime, a manifold, for now. Let's say that a group of coordinate transformation is "rigid" if, for any element of the group at one point in the manifold, I must have the same transformation (group element) at all points.

Example: the Lorentz group of coordinate transformations. The allowed transformations are rotations, translations and boosts. But the Special Theory does not allow me to apply a boost here, a rotation there and a translation elsewhere. It seems that physics likes to work in a coordinate-independent manner, which is tantamount to asking that any transformation group is not rigid in the above sense.

Do I fully understand this last statement? No, so please help.

Anyway, it seems that the way to relax rigidity in the above sense is to adjoin to each point in spacetime a "copy" of the relevant transformation group. This is more like it, from my point of view.

The structure that results from the adjunction of a continuous group of transformations, ie. a Lie group to each point of a manifold is called a "principal bundle". Let's call this $P$ for now, and accept (it's not at all hard to show) that this is a manifold [it's essentially the product of 2 manifolds - the Lie group and spacetime]; let's call spacetime as the "base manifold" $M$ .

So, since a continuous group describes infinitesimal transformations, then for any $m \in M$ I may have an infinity of maps $m \mapsto p \in P$ . And further, for any open curve $C \in M$ , an uncountable number of possible curves $C' \in P$ .

This is over-long. How am I doing so far, tough guys?

Jay-qu · 07-05-2009

Quote:

Originally Posted by Ben

It seems that Einstein once remarked of a colleague that "he could calculate, but he couldn't think". I find myself in an analogous situation here; I have a degree of understanding of the mathematics used in gauge theories, but I am struggling with their physical interpretation.

This was a wise statement, as a teacher to often do I see kids rushing to find a formula that relates quantities rather than thinking up the desired relationship from first principles. Note that at this level of physics it can be somewhat impossible to do without much trial and error.

Quote:

We are talking spacetime, a manifold, for now. Let's say that a group of coordinate transformation is "rigid" if, for any element of the group at one point in the manifold, I must have the same transformation (group element) at all points.

This is a condition of a rigid transformation, but more importantly the space must remain undistorted under the transformation. Imagine transforming the position of your arm, from straight up to straight in front of you, it is only rigid if you dont bend or twist your arm.

Quote:

Example: the Lorentz group of coordinate transformations. The allowed transformations are rotations, translations and boosts. But the Special Theory does not allow me to apply a boost here, a rotation there and a translation elsewhere. It seems that physics likes to work in a coordinate-independent manner, which is tantamount to asking that any transformation group is not rigid in the above sense.

I dont follow the logic of your final sentence. Physics does like to be coordinate independent, this is because we believe there is no preferred position or inertial reference frame - the laws of physics should be the same in every frame.

For the Lorentz transformations, the rotations and translations are both rigid so long as it is a global transformation - applies the same everywhere in space. The boost however is not a rigid transformation. Boosting induces the effects of length contraction and time dilation, these are non-rigid effects.
Do I fully understand this last statement? No, so please help.

Quote:

The structure that results from the adjunction of a continuous group of transformations, ie. a Lie group to each point of a manifold is called a "principal bundle". Let's call this $P$ for now, and accept (it's not at all hard to show) that this is a manifold [it's essentially the product of 2 manifolds - the Lie group and spacetime]; let's call spacetime as the "base manifold" $M$ .

So, since a continuous group describes infinitesimal transformations, then for any $m \in M$ I may have an infinity of maps $m \mapsto p \in P$ . And further, for any open curve $C \in M$ , an uncountable number of possible curves $C' \in P$ .

I am not very familiar with the fibre bundle approach (a very mathematical approach) to gauge theories, that said I see what you are saying above but Im afraid I dont see the point of this argument. Did you have a question or comment about this argument?

J

Ben · 07-06-2009

Quote:

Originally Posted by Jay-qu

I am not very familiar with the fibre bundle approach (a very mathematical approach) to gauge theories, that said I see what you are saying above but Im afraid I dont see the point of this argument.

It's not an argument, I am just trying to get my head around this stuff, using the math tools at my disposal.

However, you seem to be implying that the route to gauge theories through fibre bundles, specifically principal bundles, and their connections is not the only one.

Maybe therein lies my folly - I tend to try to imagine what it might mean for an mathematical abstraction to have some sort of physical reality. Maybe I should reverse my thinking, id est understand the physics and find the math to fit it?

That's next to impossible for me, given my background and temperament.

Quote:

Did you have a question or comment about this argument?

Then yes: please explain the "non-bundle" approach to gauge theories as you understand it. Maybe, just maybe, I can then fit that with the math that I sort of know.

Jay-qu · 07-06-2009

Ok, I will give it a shot. (sorry for the ugly TeX )

The gauge principle can be is used to usher in three of the four fundamental forces (so far). Back in the early 20th century Herman Weyl discovered that General Relativity was invariant under a change in length scale, he noted this as an invariance wrt change of gauge (size). Eventually the term became generalised to cover all types of invariance under changes.

The gauge theories of the forces happen to be the manifestation of internal symmetries of our particles. These internal symmetries can be represented by Lie groups. The simplest case is the invariance of the wavefunction with respect to the Lie group U(1). Physically this amounts to the fact that the phase of the wavefunction is irrelevant. We shall see this necessitates the introduction of the electromagnetic field. Beginning with the Schrodinger equation for a particle in free space
$i\partial _t \Psi = -\frac{\hbar^2}{2m}\nabla^2\Psi$
It is trivial to show that this equation will remain invariant under a global U(1) change of phase - $\Psi -> \Psi e^{i \theta}$ , where $\theta$ is a constant. This is because a U(1) transformation is just rotating the phase of the wavefunction and the phase does not effect the observed probability density $|\Psi|^2$ . But why stop at just global phase changes? What if we want the wavefunction to be invariant under an arbitrary transformation of the phase of the form
$\Psi -> \Psi e^{ie\theta(\vec{x},t)}$
When transforming the wavefunction by this arbitrary phase change the Schrodinger equation is changed. Let us now consider the Schrodinger equation in the presence of an electromagnetic field (in rationalised Gaussian units and $\hbar=c=1$
$-i\partial _t \Psi = [\frac{1}{2m}(\nabla + ie\vec{A})^2 - e\phi]\Psi,$

where the vector and scalar potentials ( $\vec{A}$ and $\phi$ ) are related to the electromagnetic fields via the following
$\vec{B}=\nabla \times \vec{A}$
$\vec{E}= -\nabla \phi-\partial_t\vec{A}$
Taking the Schrodinger equation and performing an arbitrary phase transformation yields a mess of terms that can be neatly factorised into (try this, it may take a page of algebra though

):
$-e^{ie\theta(\vec{x},t)}\partial _t \Psi = e^{ie\theta(\vec{x},t)}[\frac{1}{2m}(\nabla + ie(\vec{A}+\nabla \theta))^2-e(\phi-\partial _t \theta)]\Psi$
Cancelling the $e^{ie\theta(\vec{x},t)}$ terms we find that we can essentially summarise the changes as
$\vec{A} -> \vec{A}' = \vec{A}+\nabla \theta$
$\phi -> \phi ' = \phi-\partial_t \theta$
Taking these new potentials and substituting them into the equations that relate the potential to the E and B fields (above) will result in the electromagnetic fields being unchanged; and hence the physical situation is invariant under arbitrary phase (U(1)) transformations. The transformation of the form above is formally known as a gauge transformation. Here we have the freedom to choose a range of potentials that can be formed from any arbitrary choice of the function theta - and it will still describe the same physical situation. This is called gauge freedom.

In maintaining the gauge invariance of the Schrodinger equation we had to introduce the interaction terms for the EM field. This is known as the principle of minimal gauge invariance. You can then think of the interaction terms as been part of the derivative. In the relativistic formalism of Lagrangian dynamics the derivatives become gauge covariant derivatives
$D_{\mu} = \partial_{\mu} -ieA_{\mu}$
Here the gauge field $A_{\mu} = (\phi,\vec{A})$ .
In a similar way gauge theories of the strong and weak nuclear forces are able to be developed by using corresponding symmetries. These other internal symmetries are described by the lie groups SU(2) and SU(3) for the weak and strong force respectively.

Ben · 07-07-2009

Quote:

Originally Posted by Jay-qu

Beginning with the Schrodinger equation for a particle in free space
$i\partial _t \Psi = -\frac{\hbar^2}{2m}\nabla^2\Psi$

Missing an hbar on the LHS? Not important. But I am slightly troubled by the absence of a potential term in your Hamiltonian. Is this to do with the qualification "in free space"?

Quote:

But why stop at just global phase changes? What if we want the wavefunction to be invariant under an arbitrary transformation $\Psi -> \Psi e^{ie\theta(\vec{x},t)}$

Yeah, I have something similar here, but without $e$ in the exponent. Is this strictly required? And why have you vectorized the spatial component x? (if that's what it is)

Quote:

When transforming the wavefunction by this arbitrary phase change the Schrodinger equation is changed.

OK, so I got, as $\Psi \mapsto \Psi e^{i \theta(x,t)} = \Psi'$ then, after a bit of manipulation, using God's units, and ditching the $e$ in the exponent

$i\partial_t \Psi -\partial_t\theta(x,t) = \frac{1}{2m}[\nabla+ i\nabla\theta(x,t)]^2 \Psi$ where I have also "cancelled" the exponential factor. This is clearly nothing to do with Erwin.

Quote:

we can essentially summarise the changes as
$\vec{A} -> \vec{A}' = \vec{A}+\nabla \theta$
$\phi -> \phi ' = \phi-\partial_t \theta$

Um. I got a mass term in there - $\phi' = \phi -2m \partial_t \theta$ Do I need to worry about that?

Also, I think I may have an argument that says that your `vector potential' $A$ is actually a connection 1-form on the $U(1)$ bundle. Leave it with me a while.

Jay-qu · 07-07-2009

Quote:

Originally Posted by Ben

Missing an hbar on the LHS? Not important. But I am slightly troubled by the absence of a potential term in your Hamiltonian. Is this to do with the qualification "in free space"?
Yeah, I have something similar here, but without $e$ in the exponent. Is this strictly required? And why have you vectorized the spatial component x? (if that's what it is)
OK, so I got, as $\Psi \mapsto \Psi e^{i \theta(x,t)} = \Psi'$ then, after a bit of manipulation, using God's units, and ditching the $e$ in the exponent

Yes, take free space to mean in the absence of a potential.

I must apologise for the few mistakes I have made, firstly I have not really been consistent with my factors or hbar and e.

Having the x as a vector simply saves me writing f(x,y,z) everywhere.

Quote:

Um. I got a mass term in there - $\phi' = \phi -2m \partial_t \theta$ Do I need to worry about that?

Also, I think I may have an argument that says that your `vector potential' $A$ is actually a connection 1-form on the $U(1)$ bundle. Leave it with me a while.

Hmm Im not sure about the mass term.. I think I have made a mistake above and the mass term should not appear in the gauge transformation - as it will not be an invariant with it there..

Ben · 07-08-2009

Quote:

Originally Posted by Jay-qu

I must apologise for the few mistakes I have made, firstly I have not really been consistent with my factors or hbar and e.

No worries, my pedantry is infamous; I actually read you well.

Anyway, I thank you for your help.

I have been working on an argument that the vector potential $A$ is a 1-form connection for a $U(1)$ theory, and indeed there is a generalization of $A$ that extends to $SU(2)$ and $SU(3)$ theories.

It's a geometric argument, which I guess is not appropriate here; your approach is algebraic, which I find tough going.

Thanks again

Jay-qu · 07-08-2009

You may post it here, I would like to see it. I am less familiar with the geometrical approach, so I think it would benefit us both to see gauge theories in both representations.

Ben · 07-10-2009

Quote:

Originally Posted by Jay-qu

You may post it here, I would like to see it. I am less familiar with the geometrical approach, so I think it would benefit us both to see gauge theories in both representations.

Say what, I am struggling to deliver. I think this subject may be too hard for me; though I am reasonably comfortable with the geometry, I am finding it hard to make it into any sort of physics (as it is currently understood).

Sorry to raise expectations, though I will keep trying.

PS. Jay-qu - watching the Ashes? Your guys are awesome, I regret to say

Jay-qu · 07-10-2009

You shouldnt limit yourself by saying its to hard. You only need more time and different perspectives, its not easy but not impossible - you will get it eventually.

Sorry I dont really follow the cricket. Though I do find the whole ashes history quite funny (especially Douglass Adams' take on it

)

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