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And if so, is there any way to construct the metric explicitly?

Thanks,

Dave

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- Thread starter schieghoven
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In summary, the curvature tensor is a 2-form that corresponds to a metric in a given gauge. It is Abelian if the gauge group is Abelian, and non-Abelian if the gauge group is non-Abelian. It is possible to solve the equation R=dA + A^A for A, but this is not always straightforward.

- #1

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And if so, is there any way to construct the metric explicitly?

Thanks,

Dave

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I'm in a hurry so I'll give you a very dirty answer to decode.

The Riemann tensor is a gl(N,R) Lie-algebra valued 2-form, just like the field strength in Yang Mills theories (there, the algebra is su(N)). In the latter case, one may impose the recover the gauge connection from the field strength in a particular gauge, namely the Fock-Schwinger gauge (the calculation is done here: http://www.physics.thetangentbundle.net/wiki/Quantum_field_theory/gauge_theory/gauge_fixing [Broken])

Now, if you have some experience dealing with gravity as a gauge theory (see e.g. Nash & Sen, or Nakahara), then you have a connection from which you have enough data to reconstruct almost anything else.

The Riemann tensor is a gl(N,R) Lie-algebra valued 2-form, just like the field strength in Yang Mills theories (there, the algebra is su(N)). In the latter case, one may impose the recover the gauge connection from the field strength in a particular gauge, namely the Fock-Schwinger gauge (the calculation is done here: http://www.physics.thetangentbundle.net/wiki/Quantum_field_theory/gauge_theory/gauge_fixing [Broken])

Now, if you have some experience dealing with gravity as a gauge theory (see e.g. Nash & Sen, or Nakahara), then you have a connection from which you have enough data to reconstruct almost anything else.

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Thanks for the pointer to differential forms, it certainly seems to offer a new angle to the question. However, it seems to me that the link you provided about Fock-Schwinger gauge assumes the gauge group is Abelian, since it constructs a potential A which satisfies F=dA. For Yang-Mills or the gauge formulation of gravity, the field strength/curvature is related to the connection by R = dA + A^A.

In the non-abelian case, the problem is how to solve the equation R=dA + A^A for A, which I don't know how to do. I'm not even sure what condition to put on R to guarantee that a solution exists. (dR=0 is not generally true. Contrary to my earlier post, the Bianchi identities can't even be applied since we need A to do this.)

Cheers,

Dave

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The trick is that [tex]x^\mu A_\mu = 0[/tex], which is the actual gauge condition. Thus the equation presented is both Abelian and non-Abelian.

You had me worried there for a second =)

You had me worried there for a second =)

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Dave

[1] Cronstrom, Phys Lett B90 (1980) 267

[2] Kummer & Weiser, Z. Phys C31 (1986) 105

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Strange, the Fock-Schwinger gauge is ancient... it is, after all, named after Fock and Schwinger =)

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lbrits said:Strange, the Fock-Schwinger gauge is ancient... it is, after all, named after Fock and Schwinger =)

Perhaps you're right there, I'm not sure. Well, it's a reference, if not the original reference.

It occurred to me that I'm not even sure if the set of admissible curvatures forms a vector space. By this I mean that if R and R' are curvature tensors corresponding to metrics g and g' respectively, it's not clear to me that R+R' is also a curvature tensor, and if so, how its metric relates to g and g'. Perhaps its possible to prove or disprove this using the Fock-Schwinger gauge - I'll see whether I can get anywhere with it.

This is quite a contrast to electromagnetism, because in that context the sum of two EM fields also satisfies Maxwell's equations, and the set of EM field configurations is a vector space (a Hilbert space). I was interested to see whether a similar analysis can be done for gravitation; but this one may not go through quite as cleanly as I hoped!

Best,

Dave

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