# Krauss's interpretation of gauge theory

• I
• joneall

#### joneall

Gold Member
TL;DR Summary
Lawrence Krauss states that the connection term in EM gauge symmetry is the EM field and that it "... allows us to change our definition of electric charge from place to place." I don't see what it has to do with changing the definition (+ or -) of electric charge.
Lawrence Krauss, "The greatest story ever told ... so far", pp. 108-109. "Gauge symmetry in electromagnetism says that I can actually change my definition of what a positive charge is locally at each point of space without changing the fundamental laws associated with electric charge, as long as I also somehow introduce some quantity that helps keep track of this change of definition from point to point. This quantity turns out to be the electromagnetic field."

Introductions, like Robinson or Schwichtenberg, talk only about local U(1) without making it clear just what the U(1) transformation is changing. Sure, it's the field, but claiming that that is the charge escapes me.

What have I missed?

• ohwilleke
The idea is to make a global symmetry local. Let's start with the free Dirac equation, which follows from the action principle using the Lagrangian
$$\mathcal{L}=\bar{\psi} (\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi.$$
Obviously this Lagrangian does not change, when we introduce new fields by
$$\psi'(x)=\exp(-\mathrm{i} \alpha) \psi(x), \quad \bar{\psi}'(x)=\exp(+\mathrm{i} \alpha) \bar{\psi}(x).$$
According to Noether's theorem there's a conserved charge with the four-current density given by
$$j^{\mu}=\bar{\psi} \gamma^{\mu} \psi.$$
Now if you want to make the Lagrangian invariant under a "local" gauge transformation, where ##\alpha=\alpha(x)## is dependent on the spacetime argument, ##x##, you need to introduce a connection to define a covariant derivative, i.e., a vector field ##A_{\mu}## and the gauge-covariant derivative
$$\mathrm{D}_{\mu} \psi = (\partial_{\mu} + \mathrm{i} q A_{\mu}) \psi.$$
We want it to be covariant in the sense that for
$$\psi'(x)=\exp[-\mathrm{i} q \chi(x)] \psi(x)$$
you have
$$\mathrm{D}_{\mu}' \psi'(x)=\exp[-\mathrm{i} q \chi(x)] \mathrm{D}_{\mu} \psi(x).$$
From this you get
$$\mathrm{D}_{\mu}' \psi'(x) = (\partial_{\mu} + \mathrm{i} q A_{\mu}') \exp(-\mathrm{i} q \chi) \psi(x) = \exp(-\mathrm{i} q \chi) (\partial_{\mu} + \mathrm{i} q A_{\mu}' -\mathrm{i} q \partial_{\mu} \chi) \psi \stackrel{!}{=} \exp(-\mathrm{i} q \chi) (\partial_{\mu} + \mathrm{i} q A_{\mu}) \psi.$$
Thus you have to transform the "connection" (or more physically the gauge field) as
$$A_{\mu}' = A_{\mu} + \partial_{\mu} \chi.$$
Then your Lagrangian is invariant not only under global but also under local gauge transformations,
$$\psi'(x)=\exp[-\mathrm{i} q \chi(x)] \psi(x), \quad \bar{\psi}(x) = \exp[+\mathrm{i} q \chi(x)] \bar{\psi}(x), \quad A_{\mu}'=A_{\mu} +\partial_{\mu} \chi,$$
provided you use a gauge-covariant instead of a partial derivative,
$$\mathcal{L} = \bar{\psi} (\gamma^{\mu} \mathrm{i} \mathrm{D}_{\mu} -m) \psi.$$
Writing it out you get
$$\mathcal{L}=\bar{\psi} (\gamma_{\mu} \mathrm{i} \partial_{\mu} -m )\psi -q A_{\mu} \bar{\psi} \gamma^{\mu} \psi=\bar{\psi} (\gamma_{\mu} \mathrm{i} \partial_{\mu} -m )\psi -q A_{\mu} j^{\mu}.$$
To interpret ##A_{\mu}## as the four-potential of the electromagnetic field, all you have to do is to add a "kinetic term", and this should be a gauge-invariant addition to the Lagrangian.

The natural gauge-covariant object given the gauge-covariant derivative is the corresponding "curvature", which you get from the commutator of covariant derivatives,
$$(\mathrm{D}_{\mu} \mathrm{D}_{\nu} -\mathrm{D}_{\nu} \mathrm{D}_{\mu}) \psi = (\partial_{\mu} + \mathrm{i} q A_{\mu})(\partial_{\nu}+\mathrm{i} q A_{\nu}) \psi -(\mu \nu) = (\partial_{\mu} \partial_{\nu} \psi + \mathrm{i} q A_{\mu} \partial_{\nu} \psi + \mathrm{i} q A_{\nu} \partial_{\mu} \psi + \psi \mathrm{i} q \partial_{\mu} A_{\nu} - q^2 A_{\mu} A_{\nu} \psi) -(\mu \nu)=\mathrm{i} q (\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}) \psi=\mathrm{i} q F_{\mu \nu} \psi.$$
Indeed, you can easily show that
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
is invariant under gauge transformations, i.e.,
$$F_{\mu \nu}'=F_{\mu \nu}.$$
$$\mathcal{L}_{\text{gauge}}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}$$
to the Lagrangian.

The equations of motion, following from the Euler-Lagrange equations, indeed give the inhomogeneous Maxwell equations in covariant form,
$$\partial_{\mu} F^{\mu \nu} = q \bar{\psi} \gamma^{\mu} \psi = q j^{\mu},$$
i.e., the electromagnetic current is
$$j_{\text{em}}=q j^{\mu}.$$

• PeroK, joneall and malawi_glenn
One can play this game with other global symmetries as well. Say if you have two generations of fermions and your Lagrangian is invariant under global SU(2) transformations. You can then "gauge" that symmetry by making it local. But in order to do so, you need to invoke gauge fields and covariant derivative.

As you saw in the post above, as long as you add a ##\partial_\mu \chi (x)## term to the gauge field, you will still describing the same physics. Now ##\chi(x)## depends on space-time so at different points in space-time, you can have different definitions of your gauge-field ##A_\mu##

Last edited:
• joneall and vanhees71
The idea is to make a global symmetry local. Let's start with the free Dirac equation, which follows from the action principle using the Lagrangian
$$\mathcal{L}=\bar{\psi} (\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi.$$
Obviously this Lagrangian does not change, when we introduce new fields by
$$\psi'(x)=\exp(-\mathrm{i} \alpha) \psi(x), \quad \bar{\psi}'(x)=\exp(+\mathrm{i} \alpha) \bar{\psi}(x).$$
According to Noether's theorem there's a conserved charge with the four-current density given by
$$j^{\mu}=\bar{\psi} \gamma^{\mu} \psi.$$
Now if you want to make the Lagrangian invariant under a "local" gauge transformation, where ##\alpha=\alpha(x)## is dependent on the spacetime argument, ##x##, you need to introduce a connection to define a covariant derivative, i.e., a vector field ##A_{\mu}## and the gauge-covariant derivative
$$\mathrm{D}_{\mu} \psi = (\partial_{\mu} + \mathrm{i} q A_{\mu}) \psi.$$
We want it to be covariant in the sense that for
$$\psi'(x)=\exp[-\mathrm{i} q \chi(x)] \psi(x)$$
you have
$$\mathrm{D}_{\mu}' \psi'(x)=\exp[-\mathrm{i} q \chi(x)] \mathrm{D}_{\mu} \psi(x).$$
From this you get
$$\mathrm{D}_{\mu}' \psi'(x) = (\partial_{\mu} + \mathrm{i} q A_{\mu}') \exp(-\mathrm{i} q \chi) \psi(x) = \exp(-\mathrm{i} q \chi) (\partial_{\mu} + \mathrm{i} q A_{\mu}' -\mathrm{i} q \partial_{\mu} \chi) \psi \stackrel{!}{=} \exp(-\mathrm{i} q \chi) (\partial_{\mu} + \mathrm{i} q A_{\mu}) \psi.$$
Thus you have to transform the "connection" (or more physically the gauge field) as
$$A_{\mu}' = A_{\mu} + \partial_{\mu} \chi.$$
Then your Lagrangian is invariant not only under global but also under local gauge transformations,
$$\psi'(x)=\exp[-\mathrm{i} q \chi(x)] \psi(x), \quad \bar{\psi}(x) = \exp[+\mathrm{i} q \chi(x)] \bar{\psi}(x), \quad A_{\mu}'=A_{\mu} +\partial_{\mu} \chi,$$
provided you use a gauge-covariant instead of a partial derivative,
$$\mathcal{L} = \bar{\psi} (\gamma^{\mu} \mathrm{i} \mathrm{D}_{\mu} -m) \psi.$$
Writing it out you get
$$\mathcal{L}=\bar{\psi} (\gamma_{\mu} \mathrm{i} \partial_{\mu} -m )\psi -q A_{\mu} \bar{\psi} \gamma^{\mu} \psi=\bar{\psi} (\gamma_{\mu} \mathrm{i} \partial_{\mu} -m )\psi -q A_{\mu} j^{\mu}.$$
To interpret ##A_{\mu}## as the four-potential of the electromagnetic field, all you have to do is to add a "kinetic term", and this should be a gauge-invariant addition to the Lagrangian.

The natural gauge-covariant object given the gauge-covariant derivative is the corresponding "curvature", which you get from the commutator of covariant derivatives,
$$(\mathrm{D}_{\mu} \mathrm{D}_{\nu} -\mathrm{D}_{\nu} \mathrm{D}_{\mu}) \psi = (\partial_{\mu} + \mathrm{i} q A_{\mu})(\partial_{\nu}+\mathrm{i} q A_{\nu}) \psi -(\mu \nu) = (\partial_{\mu} \partial_{\nu} \psi + \mathrm{i} q A_{\mu} \partial_{\nu} \psi + \mathrm{i} q A_{\nu} \partial_{\mu} \psi + \psi \mathrm{i} q \partial_{\mu} A_{\nu} - q^2 A_{\mu} A_{\nu} \psi) -(\mu \nu)=\mathrm{i} q (\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}) \psi=\mathrm{i} q F_{\mu \nu} \psi.$$
Indeed, you can easily show that
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$$
is invariant under gauge transformations, i.e.,
$$F_{\mu \nu}'=F_{\mu \nu}.$$
$$\mathcal{L}_{\text{gauge}}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}$$
to the Lagrangian.

The equations of motion, following from the Euler-Lagrange equations, indeed give the inhomogeneous Maxwell equations in covariant form,
$$\partial_{\mu} F^{\mu \nu} = q \bar{\psi} \gamma^{\mu} \psi = q j^{\mu},$$
i.e., the electromagnetic current is
$$j_{\text{em}}=q j^{\mu}.$$
i like this succint explanation of this gauging process and of the Noether conservation. But I don't see how it says that what the transformation represents is the electric charge.

That is a popular science book by Krauss. Keep that in mind.
I think what he is trying to convey is that electric charge conservation is due to global U(1) symmetry, you can multiply your fields with ##e^{\mathrm{i}\theta}## and the corresponding charge-conjugate fields by ##e^{-\mathrm{i}\theta}##. Because of this symmetry, you can change the convetion of what is a positive charge, and it won't affect any physical law or measurment. But you have to do that change everywhere in the entire space-time. Now the idea with changing this locally, with a space-time dependent phase, you need some way to communicate this change to all other regions of space time that "hey here we are using the convetion of proton being negatively charged". This information is connected via the gauge-field, which is massless since it needs infinite range (well the masslessness of the gauge field comes naturally from the required gauge-transformation law).
I would not bother too much trying to put things that are written in popular science books too literally.

• apostolosdt, joneall and Demystifier
i like this succint explanation of this gauging process and of the Noether conservation. But I don't see how it says that what the transformation represents is the electric charge.
I don't know, what you mean by "the transformation represents the electric charge". Electric charge is the conserved quantity due to the invariance under global changes of the phase of the fields representing charged particles. If you make this symmetry local you get a gauge theory. The conservation of charge is of course still valid and a necessary condition for the consistency of the gauge theory.

• joneall and malawi_glenn
I don't know, what you mean by "the transformation represents the electric charge". Electric charge is the conserved quantity due to the invariance under global changes of the phase of the fields representing charged particles. If you make this symmetry local you get a gauge theory. The conservation of charge is of course still valid and a necessary condition for the consistency of the gauge theory.
I think he is trying to make sense of professor Krauss description of gauge-theory in that pop sci book
"Gauge symmetry in electromagnetism says that I can actually change my definition of what a positive charge is locally at each point of space without changing the fundamental laws associated with electric charge, as long as I also somehow introduce some quantity that helps keep track of this change of definition from point to point. This quantity turns out to be the electromagnetic field"

• PeroK and joneall
It's my experience that to understand pop-sci writing you need to read the original papers the writing is based on ;-)). It is very difficult to write good popular-science texts, and I don't blame anybody trying to do that, but sometimes I really wonder, how experts in their field come up with such distortions of the subject.

In the Abelian case the electric charge is a gauge invariant quantity. So it's neither changed by global nor local gauge transformations. The charge as an operator generates (global) gauge transformations.

In the non-Abelian case the charge operators form a representation of the gauge group's Lie algebra and transform according to this representation.

• joneall
Ok, guys. I get it now. Thanks for all your help.

• malawi_glenn
Ok, guys. I get it now. Thanks for all your help.
You should check out gauge theory and finance book btw :) by Schwichtenberg

• Demystifier
Thanks, I have. Guess I need to re-read it. :)

• malawi_glenn
Hm, is this along the lines of this?

https://arxiv.org/abs/1901.10420

For me, that's an example for how you can confuse an admittedly somewhat subtle but understandable mathematical subject by didactics ;-)).

• Demystifier and dextercioby
Hm, is this along the lines of this?

https://arxiv.org/abs/1901.10420

For me, that's an example for how you can confuse an admittedly somewhat subtle but understandable mathematical subject by didactics ;-)).
The book "finance from physics" is somewhat based on that paper yes. I'd say its in between pop sci and "the real deal."
Though gauge theories are used in financial models, I have "Quantum Field Theory for Economics and Finance" by Ehsan Baaquie on my "to read list"

• Demystifier and vanhees71