schieghoven said:
No, this is not the correct starting point for gauge theory.
Starting point?
The potential in a gauge theory is, by definition, a connection 1-form which takes values in the Lie algebra of the gauge group.
Correct. It is a connection not vector. see post#18.
Under a coordinate transformation it transforms as
<br />
a_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} a_{\nu}<br />
i.e., as a 4-vector.
Very wrong;
1) Regarding the tensorial nature of potentials, gauge theories can only say that potentials CARRY a spacetime index. This does not mean that A_{\mu} itself is a vector. In general relativity, Reimannian connection (potential) carries three spacetime indices but it is not a rank-3 tensor. Like \Gamma^{\rho}{}_{\mu \nu}, the potential A_{\nu} is a connection. Therefore, only the difference between two such connections, (A_{2}^{\mu}-A_{1}^{\mu} \ , \Gamma_{2}-\Gamma_{1}), has a definite tensorial character.see post#18.
2) To understand the whole issue, please pay attention to the following;
Unlike
massive vector fields, non-trivial
massless vector field is not allowed by Lorentz group. The fundamental theorem on the massless representation of Lorentz group states:
"The only admissible vector field of mass zero is a gradient of a scalar field."
So, if V_{\mu} is a vector field
<br />
V_{\mu} \rightarrow \Lambda_{\mu}{}^{\nu} V_{\nu}<br />
the theorem says that V_{\mu} cannot be massless unless
<br />
V_{\mu} = \partial_{\mu} \phi<br />
where \phi is a scalar field. But this massless vector is not good for us; it implies an everywhere vanishing field tensor;
<br />
F_{\mu \nu} = ( \partial_{\mu}\partial_{\nu} - \partial_{\nu}\partial_{\mu}) \phi = 0<br />
Since we know that the electromagnetic field is massless, we therefore conclude that electromagnetism (or any other massless theory) cannot be described by a vector field. That is, according to the above mentioned theorem, the potential of any gauge theory cannot be a 4-vector field.
To those who still think that A_{\mu} is a 4-vector, I say this:
Do yourself a favour and study the representation theory of Lorentz group.
See J. Lopuszanski, "An Introduction to Symmetry and Supersymmtry", World Scientific,1991.
From Page 186;
"the vector potential A_{\mu} cannot be genuinely a vector with respect to the Lorentz transformations as a massless vector must necessarily be given by the gradient of a scalar of helicity zero. Remember that F_{\mu \nu} is composed of two terms of helicity 1 and -1. Thus under Lorentz transformation A_{\mu} does not transform as a vector, although F_{\mu \nu} does transform as a tensor."
The rule is sweet and simple:
"If A_{\mu} is massless, then it cannot be a Lorentz-vector"
3) Now that we know it is not a vector, we ask: How does the potential transform under Lorentz transformation? There are at least five different methods for deriving the transformation law! I will describe the shortest one.
In this method, the vital piece of information comes from the gauge principle which states that ANY two potentials are related by a gradient of a scalar
<br />
A_{2}^{\mu} = A_{1}^{\mu} + \partial^{\mu}\Omega<br />
From this we conclude that the difference between ANY two potentials (i.e., the object (V^{\mu}= A_{2}^{\mu} - A_{1}^{\mu}) is a trivial 4-vector (gradient of a scalar).
Thus, under Lorentz transformation, V_{\mu} transforms according to
<br />
\left( A_{2}^{\mu} - A_{1}^{\mu} \right) \rightarrow \Lambda^{\mu}{}_{\nu} \left( A_{2}^{\nu} - A_{1}^{\nu} \right) \ \ \ (1)<br />
Please note that this IS NOT a "gauge transformation followed by Lorentz transformation". I used the gauge principle only to INFER the tensorial (vector) nature of the object ( A_{2}^{\mu} - A_{1}^{\mu}).
Since Lorentz group does not admit massless vectors, we therefore conclude that Eq(1) is satisfied if and only if
<br />
A_{1}^{\mu} \rightarrow \Lambda^{\mu}{}_{\nu} A_{1}^{\nu} + \partial^{\mu}\lambda<br />
<br />
A_{2}^{\mu} \rightarrow \Lambda^{\mu}{}_{\nu} A_{2}^{\nu} + \partial^{\mu}\lambda<br />
Notice that the scalar function \lambda is completely arbitrary. This means that the above transformation law holds for ANY potential in ANY gauge (COVARIANT or NON-COVARIANT).
In QFT, the transformation of the OPERATOR A_{\mu} becomes
<br />
A_{\mu} \rightarrow U(\Lambda) A_{\mu} U^{-1}(\Lambda) = \Lambda_{\mu}^{\nu} A_{\nu} + \partial_{\mu}\lambda<br />
see Eq(14.25) in Bjorken & Drell, "Relativistic Quantum Fields".
and Eq(5.9.31) in Weinberg, "The Quantum Theory of Fields", VOL I.
For completely CLASSICAL description, see
K. Moriyasu, "An Elementary Primer for Gauge Theory", World Scientific.
On page 42, Moriyasu says
"Thus, the vector potential observed in the two frames are related by
<br />
A_{\mu}^{'} = L_{\mu}{}^{\nu}A_{\nu} - \partial_{\mu}\lambda \ \ \ III-22<br />
This shows that the vector potential does not transform like an ordinary vector under Lorentz transformation. ...
This interesting fact is well known in quantum field theory but it is rarely mentioned in ordinary electromagnetism."
I have been in the trade for some time and I have identified the sourses of confusion:
1) most students starts their post grad. course with little or no knowledge about the repretentation theory.
2) the fact (potential is not vector) is rarely addressed on the pedestrian level.
3) the misleading name "VECTOR" potential adds to the confusion.
4) in QED, A_{\mu} couples to a CONSERVED current. So, peopel speak of Lorentz ang gauge invariant interaction Lagrangian;
\mathcal{L} = A_{\mu}J^{\mu}
Here, most students jump to the conclusion;
"A_{\mu} must be a 4-vector"
Of course in such coupling, it is not a big sin to treat \mathcal{L} as FORMALLY Lorentz-invariant and A_{\mu} as a FORMAL vector.
QED can be defined in a non-covariant gauge, so it's convenient to define the **operator** a_\mu in this way, which perhaps is what Weinberg is doing. However, prior to quantisation, the potential in any gauge theory is certainly a vector.
Quantization means going from a C-number field on spacetime
<br />
A_{\mu}(x) = \int_{p} \ a(p)f_{\mu}(x;p) + a^{*}(p)f^{*}_{\mu}(x;p)<br />
to a Q-number field (operator-valued distribution) on spacetime
<br />
\hat{A}_{\mu}(x) = \int_{p} \ \hat{a}(p)f_{\mu}(x;p) + \hat{a}^{\dagger}(p)f^{*}_{\mu}(x;p)<br />
Notice that the spacetime index \mu is carried by the function f not by the operator a. So, my friend, quantization does not change the tensorial nature of the fields.
regards
sam
