# Potentials, connections and curvature

1. Sep 1, 2010

### Orbb

Hi everyone,

I have a question related about the relation between potentials, connections and curvature in gauge theories. In Newtonian physics, the common starting point is Newton's law, which determines the motion in terms of the derivative of the potential, i.e. sth. like

$$\ddot x_{\alpha} \propto \partial_{\alpha}\phi.$$

In electrodynamics, $$\phi$$ is contained in the vector potential $$A_{\mu}$$. Now in QED, the latter gives rise to a connection which enters e.g. the covariant derivative, $$D_{\mu}=\partial_{\mu}+iqA_{\mu}$$. The failure of these covariant derivatives to commute gives rise to the electromagnetic field tensor $$F_{\mu\nu}=\partial_{[\mu}A_{\nu]}$$, which is also referred to as the electromagnetic curvature.
Now in GR, what resembles the potential, is the metric, since in the Newtownian limit, the Newtownian potential is contained in it. However, it is not this potential which enters the corresponding covariant derivative. Instead, it's a derivative of the potential, which gives the Christoffel symbols, schematically, $$\partial g \propto \Gamma$$. Now this connection again enters covariant derivatives and their commutator yields the Riemann curvature tensor, roughly speaking.

So eventhough the overall structure is similar, the crucial differnce is while in QED, the potential enters the cov. derivative D, in GR it is the derivative of the potential entering D. Both times, D enters the equation of motion e.g. for matter fields in the same way and both times, the corresponding field strength/curvature is constructed in a similar fashion. Yet this difference - why?

I posed my question very sketchy, I hope I got across what I'm asking. Looking forward to your responses!

2. Sep 1, 2010

### Mentz114

Kinematics and Dynamics ?

3. Sep 1, 2010

### Orbb

Mentz, thanks for your answer, but could you elaborate on this a bit more? Are you referring to the dynamics of the electromagnetic and gravitational field, respectively, or of the dynamics of a particle propagating in a background given by the fields? I thought both EM and GR can be assigned a kinematical part and some dynamics. Apperently, I am confused.

4. Sep 1, 2010

### Mentz114

Sorry for my inadequate response. Surely the difference between QED and GR is so great that analogies are of little use. I can't answer your 'why ?'.

GR can be formulated in a similar way to QED by demanding local translational invariance instead of local phase invariance. In this case the gauge field arises naturally, and with some additional assumptions, the field equations of GR emerge. I'm in a rush but I'll post a reference later.

5. Sep 1, 2010

### Orbb

Are you referring to the use of torsion instead of the metric to provide the dynamical degrees of freedom?

Anyways, 'why?' may not be a good question. I was hoping of some physical interpretation of this mathematical difference.

6. Sep 1, 2010

### Mentz114

Yes.

Maybe the components of the metric are not potentials, even though in going over to the weak field limit g00 forms part of the emergent Newtonian potential. The other metric coefficients don't enter at all. So, I would look suspiciously at the mathematics of this transition, or the physical interpretation.

If the connection is treated as a velocity dependent 'potential', does that solve the problem ?

Last edited: Sep 1, 2010
7. Sep 1, 2010

### Orbb

Yes, that may be. But regarding the components I think it's just a manifestation of the fact that Newtonian gravity has fewer degrees of freedom. It is not really related to the hierarchical structure 'potential -> connection -> curvature' in terms of successive derivatives.

Do you mean the way the connection enters the geodesic equation for a particle, like $$\ddot x^{\alpha} \propto \Gamma^{\alpha}_{\mu\nu}\dot x^{\mu}\dot x^{\nu}$$? In that case, it looks to me rather like a velocity-dependent force, which again is the derivative of a potential, in that case the metric potential. But yes, maybe this issue is just about taking the identification $$g_{00}\to\phi$$ too seriously.

And maybe you could expand on what you meant by kinametics vs. dynamics? Anyways thanks for your responses!

8. Sep 1, 2010

### Mentz114

It's mostly that dynamics describes motion in terms of forces, and kinematics does not. Kinematics is the study of force-free motion. The force term in the Euler-Lagrange equations is $\partial \mathcal{L}/\partial x$, but this can be zero, giving a kinematic situation. If a velocity dependent field is present in the Lagrangian, then $\partial \mathcal{L}/\partial \dot{x}$ will have two terms, and one will offset the change in coordinate momentum, giving velocity dependent forces.

The same sort of thing happens in GR ( I'm sticking my neck out here). Kinematic systems have $\partial \mathcal{L}/\partial x=0$ and $\partial \mathcal{L}/\partial \dot{x}=P+F$ so the 'forces' are velocity dependent. I guess the distinction between kinematic and dynamic is not as clear cut as I thought.

True. I've worked out trajectories using this as if it was dynamics.

Anyhow, it's late and I may be rambling ...

Last edited: Sep 1, 2010
9. Sep 2, 2010

### bcrowell

Staff Emeritus
The OP referred to QED, but I don't think there are any purely quantum-mechanical issues involved. IMO this is really about the analogy between GR and classical E&M.

There is a gravitational potential $\phi$ in GR, but it only works in special cases. See http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html [Broken] , section 7.3.

Yes, in general the thing in GR that plays a role most closely analogous to a potential is simply the metric. The gauge transformations are changes of coordinates. You may want to look at sections 3.6.3, 5.6, and 5.9.4 at the link above, which has a table showing some analogies between GR and E&M. Although the table tries to make everything look strictly analogous, the analogy does fail in some ways. For example, the observable field E in E&M falls off like 1/r^2, whereas the observable thing in GR is the Riemann curvature tensor, which falls off like 1/r^3. This is a direct result of the fact that there's one more derivative built into R than there is in E.

I think if we could put together a straightforward explanation of why the observable in one case goes like 1/r^2 and in the other like 1/r^3, that would constitute an explanation of why the analogy betwen the two theories is inexact, as pointed out by the OP. I'll take a stab at that, but I don't know if I have the right insight or not. In E&M, the density of field lines is what's observable, and since we live in 3 spatial dimensions, the density of field lines falls off like 1/r^2. In GR, the density of field lines would be the Newtonian gravitational field, and the closest thing we have to that would be the Christoffel symbol. But the equivalence principle tells us that gravitational fields are not observable (which is why $\Gamma$ can't be a tensor). Only a gradient of the gravitational field can be observable, so we need one more derivative before we get to an observable quantity, and that goes like 1/r^3. There's probably some deeper insight to be had here, which is escaping me. Possibly it has to do with the fact that E&M takes placed on a fixed geometrical background, whereas in GR, the field is the geometry.

Last edited by a moderator: May 4, 2017
10. Sep 3, 2010

### Orbb

bcrowell, thanks for your insightful words about what is observable and what is gauge. It appears the clash is really that while GR and E&M can be expressed in a very similar language, the same mathematical objects in the respective theories are 'shifted' by one derivative with respect to one another regarding their physical interpretation. Looking at GR as a gauge theory of a SL(2,C) connection would make the structural similarity most manifest. Then it can be seen that while in E&M, the connection directly is also the potential and is a tensor, in GR the connection is already something that corresponds to a derivative of a potential (sth. like a field strength), and is not a tensor. So the actual gravitational field strength, namely the curvature, is one more derivative apart from what corresponds to the field strength in Newtownian gravity. So it also seems like an important difference between Newtownian and Einstein gravity. GR is 'one layer deeper'. I like your speculation that this is a manifestation of the fact that GR is the dynamics of the geometry, and not the dynamics of a field on top of it. In Newtownian gravity, the gravitational field still lives 'on top' of spacetime, just like the electromagnetic field.

11. Sep 3, 2010

In Kaluza-Klein theory electromagnetic potential is a part of the 5-dimensional metric tensor. This reproduces the standard GR-EM coupling.

12. Sep 8, 2010

### TrickyDicky

This thought-provoking comments suggest me some considerations about the difference between the EM field and the gravitational field, the fact that the latter isn't observable might lead to the question of whether there is any such thing as a "gravitational field" which can act as an independent entity, and this question could indeed be related to what you mention about the different geometrical background and the extra derivative needed to obtain the observable curvature.
But in case we couldn't consider the "gravitational field" as an active entity, this difference with the EM field could point to theoretical difficulties with the main expected observable outcome of the parallelism between the gravitational field and the EM field which is the detection of gravitational waves.
It seems as if the differences pointed out could be a theoretical barrier to the existence of these waves but then we would have to explain the Hulse-Taylor binary orbit decay some other way.

13. Sep 8, 2010

### bcrowell

Staff Emeritus
Historically there was a lot of debate about whether gravitational waves were real, or whether they were always just coordinate waves, or mathematical artifacts without physical reality. Some interesting history here: http://en.wikipedia.org/wiki/Sticky_bead_argument So the fact that you have doubts means that you're in good company :-)

Just because the gravitational field isn't observable, that doesn't necessarily mean gravitational waves aren't observable. Gravitational waves impose curvature on spacetime, and curvature is observable.

14. Sep 9, 2010

The intriguing question is why so many experts calculated and predicted them including the magnitudes of the observed effects (they are calculable, aren't they?), and yet the observations failed. What does it say about the experts?

15. Sep 9, 2010

### TrickyDicky

You are right, just note that in order to observe that curvature imposed by gravitational waves, we are assuming their existence in the first place and to do this you have to assume a gravitational field that acts independently of the spacetime geometry (as the EM field does against a background geometry) but this is contradicted with your assertion that in GR the field is the geometry, so in the end this is kind of a circular argument.

Certainly this is a doubtful issue, even Einstein changed his mind several times about the existence of these waves. I think there is reasons for that, and we cannot right now decide with complete confidence as there is good arguments for and against their existence.
Ultimately is an empirical question and fortunately we are reaching the technological point where we should be detecting something so it is probably a matter of a few years till we can confidently reach an informed conclusion.

If finally they end up being a mathematical artifact, that in no way would affect our confidence in GR as a theory, since following the above-mentioned reasoning actually they wouldn't follow from the theory.