Basic Question about Gauge Transformations

[Baela]

TL;DR Summary

Normally a given action does not have an infinite number of gauge transformations but the following observation implies that an action with three or more fields has an infinite number of gauge transformations! What's going wrong?

Suppose we have an action ##S=S(a,b,c)## which is a functional of the fields ##a,\, b,\,## and ##c##. We denote the variation of ##S## wrt to a given field, say ##a##, i.e. ##\frac{\delta S}{\delta a}##, by ##E_a##.
Then ##S## is gauge invariant when
$$\delta S = \delta a E_a + \delta b E_b +\delta c E_c = 0 \tag{1}$$
This gives
$$\delta c = - (\delta a E_a + \delta b E_b)/E_c \tag{2}$$
From the above equation ##\delta c## can be obtained for arbitrary ##\delta a## and ##\delta b##. It is not necessary to have a relation b/w ##\,\delta a## and ##\delta b##. Doesn't this imply that there is an infinite number of gauge-transformations here? If yes, isn't that absurd?

 

[malawi_glenn]

Consider the simple case of ##U(1)## gauge transformation in scalar QED. The gauge transformations of the scalar field ##\phi (x)## reads ##\phi (x) \mapsto \phi (x) e^{\mathrm{i}\theta (x)} ##, now how many gauge transformations do you have here?... one? infintely many?

 

[renormalize]

And how does this work when you put the field ##c## "on shell"; i.e., ##E_{c}=0##? What does your eq.(2) mean then?

 

[Baela]

Consider the simple case of ##U(1)## gauge transformation in scalar QED. The gauge transformations of the scalar field ##\phi (x)## reads ##\phi (x) \mapsto \phi (x) e^{\mathrm{i}\theta (x)} ##, now how many gauge transformations do you have here?... one? infintely many?

That is one gauge transformation. The case that I am talking of in the original post needs at least three different fields in the action for there to be an infinite number of gauge transformations. If there were only two fields, ##a## and ##b##, then eq. (2) in the original post will read as
$$\delta b=-\delta a E_a/E_b\,.$$
Whatever arbitrary value ##\delta a## takes here, it will give only one value to ##\delta b## in order to satisfy the above equation. So it will count as only one gauge transformation.
But when there are three or more fields, we have
$$\delta c = - (\delta a E_a + \delta b E_b)/E_c\,. $$
When ##\delta a## takes an arbitrary value, ##\delta b## can take an infinite number of unrelated/unconstrained values, and we will get a solution for ##\delta c## for each of those cases. So the number of possible gauge transformations becomes infinite.

 

[Baela]

And how does this work when you put the field ##c## "on shell"; i.e., ##E_{c}=0##? What does your eq.(2) mean then?

Then eq. (2) becomes undefined. Such a gauge-transformation is undefined on-shell. But that does not mean that such a gauge transformation is not possible or not allowed.

 

[renormalize]

Then eq. (2) becomes undefined. Such a gauge-transformation is undefined on-shell. But that does not mean that such a gauge transformation is not possible or not allowed.

But one of the motivations for seeking a gauge-invariant action is to guarantee that the equations of motion for the various fields will themselves be gauge invariant. What is the value of a gauge transformation that is undefined on shell?

 

[Baela]

But one of the motivations for seeking a gauge-invariant action is to guarantee that the equations of motion for the various fields will themselves be gauge invariant. What is the value of a gauge transformation that is undefined on shell?

The equations of motion are invariant also under the gauge transformations which are undefined on-shell. ##E_a,\,E_b,## and ##E_c## can be shown to be invariant under such gauge transformations.

I would be happy to come to know of some reasoning which rules out the gauge-transformations that are undefined on-shell. But I need a good reason to justify the ruling out.

 

[vanhees71]

A gauge "symmetry" describes a redundancy in the description of a physical situation. E.g., the electromagnetic field can be described by the electromgnetic potentials (or in relativistic notation, the four-potential, ##A^{\mu}##), but these are not uniquely determined by the equations of motion, i.e., any potential ##A_{\mu}'=A_{\mu} + \partial_{\mu} \chi##, where ##\chi## is an arbitrary scalar field, describes the same physical situation. The observables are necessarily gauge invariant. An example is the electromagnetic field, ##F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}##, and indeed the Maxwell equations for the electromanetic field has unique solutions for given sources and initial conditions.

 

[renormalize]

The equations of motion are invariant also under the gauge transformations which are undefined on-shell. ##E_a,\,E_b,## and ##E_c## can be shown to be invariant under such gauge transformations.

Can you post your demonstration that shows that, e.g., ##E_{c}## is invariant under your gauge transformations even when ##E_{c}=0##?

 

[Baela]

Can you post your demonstration that shows that, e.g., ##E_{c}## is invariant under your gauge transformations even when ##E_{c}=0##?

It'll take some time to show that. Will try to do it when I get enough time for it.

 

[Baela]

But one of the motivations for seeking a gauge-invariant action is to guarantee that the equations of motion for the various fields will themselves be gauge invariant. What is the value of a gauge transformation that is undefined on shell?

As per the answer given by Demystifier here: https://www.physicsforums.com/threa...er-gauge-transformations.1052060/post-6879952 , equations of motion of an action are not necessarily required to be invariant under the gauge transformations of the action.

So saying that the gauge transformations which do not leave the equations of motion invariant, are not real gauge transformations, is not correct.

 

[renormalize]

So saying that the gauge transformations which do not leave the equations of motion invariant, are not real gauge transformations, is not correct.

True, Demystifier makes a valid point: namely, for local gauge transformations that depend on more than one parameter, the equations-of-motion transform covariantly (i.e., they are multiplied by a collection of position-dependent parameters), rather than being invariant (multiplied by 1). But that makes no essential difference to the operational questions I previously put to you.

So, I revise post #6 to ask:
But one of the motivations for seeking a gauge-invariant action is to guarantee that the equations of motion for the various fields will themselves be gauge invariant or covariant. What is the value (i.e., usefulness) of a gauge transformation that is undefined on shell?

And alter post #9 to request:
Can you post your demonstration that shows that, e.g., ##E_{c}## is invariant or covariant under your gauge transformations even when ##E_{c}=0##?

 

[Baela]

Sorry about the long delay in my reply. I've had a very busy couple of months with deadlines to meet. After having met some of my deadlines, I have returned to reply to your questions as soon as I could manage.

So, I revise post #6 to ask:
But one of the motivations for seeking a gauge-invariant action is to guarantee that the equations of motion for the various fields will themselves be gauge invariant or covariant. What is the value (i.e., usefulness) of a gauge transformation that is undefined on shell?

Generally the main purpose of a gauge transformation is to keep the action invariant, not the equations of motion invariant. So to me at least, the value of a gauge transformation comes from its ability to keep the action invariant.

And alter post #9 to request:
Can you post your demonstration that shows that, e.g., ##E_{c}## is invariant or covariant under your gauge transformations even when ##E_{c}=0##?

I would like to modify something I said earlier. I said "The equations of motion are invariant also under the gauge transformations which are undefined on-shell." Reflecting on it a bit more, I see that a transformation which is undefined on-shell will simply not exist when the equations of motion hold. So it is incorrect to talk about the capability of such a transformation when the equations of motion hold. It is still a mystery to me about whether such gauge transformations are acceptable in the off-shell case.

True, Demystifier makes a valid point: namely, for local gauge transformations that depend on more than one parameter, the equations-of-motion transform covariantly (i.e., they are multiplied by a collection of position-dependent parameters), rather than being invariant (multiplied by 1). But that makes no essential difference to the operational questions I previously put to you.

Demystifier only said that generally equations of motion are covariant. It is not clear to me how you infer

for local gauge transformations that depend on more than one parameter, the equations-of-motion are multiplied by a collection of position-dependent parameters, rather than being invariant (multiplied by 1)

If you could clarify it by giving an example, that would be great. Thanks.