Photon propogator = metric tensor?

1. Jun 12, 2010

Pengwuino

This is a bit of a novice question I'm sure, but I was reading through Griffiths introductory particle physics text and in his section on QED, he states that the propagator of the photon is $$\frac{ig_{\mu \nu}}{q^2}$$ but I can't see where he gets this from. Where does the metric come from? Can someone explain this? It seems completely out of the blue. Thanks!

2. Jun 13, 2010

haael

The metric tensor appears quite often. Treat it like an identity matrix of a kind. The identity matrix itself is not Lorentz-invariant, so the metric tensor jumps in the place.

3. Jun 13, 2010

Pengwuino

Why is specifically the metric tensor used as the propagator for the photon though? As opposed to some other tensor.

4. Jun 13, 2010

haael

"Some other tensor" would mean, that the propagator depends on something. Alas, it doesn't. It depends only on the shape of spacetime.

Imagine that the metric tensor is an identity matrix.

The metric tensor is the default tensor, that goes into a formula when no other tensor does. It's the simplest Lorentz invariant 2nd order tensor. Like an identity matrix in plain vector algebra.

5. Jun 14, 2010

haushofer

Here's a intuïtive explanation:

The photon propagator is a second rank symmetric tensor. The only two tensors available which obey this demand are the two tensors

$$g_{\mu\nu}, \ \ \ \ k_{\mu}k_{\nu}$$

where k is a momentum vector. The so-called Ward identities show that the last term doesn't come in for physical processes and is thus a gauge-term. What remains effectively is the metric tensor. You could look at chapters 2.7 and 3.4 of Zee's QFT book :)

6. Jun 17, 2010

samalkhaiat

1) The propagator can be obtained by inverting the differential operator contained in the action integral. That is. If you can put the action in the form;

$$S = \int \ dx \ \phi_{i}(x)D_{ij}(\partial) \phi_{j}(x)\ \ \ (1)$$

and, if the differential operator $D_{ij}$ has an inverse, then the propagator for the field $\phi$ (in the momentum space: $\partial \rightarrow ik$) is given by;

$$P_{ij}(k) = D^{-1}_{ij}(k)\ \ \ (2)$$

2) It is clear from eq(1) that a vector field implies a tensor propagator, hence the appearance of the metric tensor in the “photon” propagator. See below.
3)Without a gauge-fixing term, the photon has no propagator, because the differential operator for Maxwell equation has no inverse;

$$S = -1/4 \int \ dx \ (F_{ab})^{2} = 1/2 \int \ dx \ A^{a}\partial^{b}(\partial_{b}A_{a}-\partial_{a}A_{b})$$

or,

$$S = 1/2 \int \ dx\ A^{a}(g_{ab}\partial^{2} - \partial_{a}\partial_{b})A^{b}$$

Notice that

$$D_{ab}(k) = -g_{ab}k^{2} + k_{a}k_{b}$$

has no inverse. However, in the gauge $\partial_{a}A^{a} = ik_{a}A^{a}= 0$, we find

$$D_{ab}(k) = -g_{ab}k^{2}$$

which has an inverse ; the photon propagator,

$$P^{ab} = - g^{ab}/k^{2}$$

regards

sam

Last edited: Jun 17, 2010