# Mathematically rigorous way of expanding propagators?

1. Jan 19, 2012

### Hepth

Such propagators as found in HQET $\frac{i}{2 v \cdot k}$ come about from expanding the full propagator. I'm wondering what the method is to properly Taylor expand denominators that contain 4-dimensional dot products.

$$\frac{1}{2 v \cdot k + k^2}$$
If we treat the dot products as regular 1-D multiplications this expands to:

$$\frac{1}{2 k v}- \frac{1}{4 v^2} + \frac{k}{8 v^3} + ...$$

We can naively just say k^2 is small and drop it but I'm looking for something more general.

If we split it:
$$\frac{1}{g_{\mu \nu}} \frac{1}{2 v^{\mu} k^{\nu} + k^{\mu} k^{\nu}}$$
And expand first in $k^{\nu}$ we have
$$\frac{1}{g_{\mu \nu}} \frac{1}{k^{\nu}} \frac{1}{2 v^{\mu} + k^{\mu}}$$
Which is fine. Now, to expand the last term $\frac{1}{2 v^{\mu} + k^{\mu}}$ we can't introduce any new indices as this is already contracted. How can i rigorously expand this without introducing new indices, to any order.

The generic algebraic expansion is:
$$\frac{1}{2 v}-\frac{k}{4 v^2}+\frac{k^2}{8 v^3}+...$$
Now is there a reason I can directly ASSUME that vectors with odd powers will carry the index so this becomes:
$$\frac{1}{2 v^{\mu}}-\frac{k^{\mu}}{4 v^2}+\frac{k^2}{8 v^2 v^{\mu}}+...$$

This is the step that confuses me. We now have a $k^{\mu}$ in the numerator of the second term in the expansion, but since these indices are already contracted this is actually
$$\frac{1}{g_{\mu \nu} k^{\mu}} \frac{k^{\nu}}{4 v^2}$$

Which I'm having a hard time understanding. If I Just assume I can work with the nu as an arbitrary index then I can either multiply by $\frac{k^{\nu}}{k^{\nu}} or \frac{v^{\nu}}{v^{\nu}}$ to change the location of the index and continue on my way. I just felt like this wasn't a very competent way of doing things, as if its true I can let it be a free index to play with then operations like
$$\frac{1}{v \cdot k} = \frac{1}{v_{\mu} k^{\mu}} = \frac{1}{v_{\mu}} \frac{k_{\mu}}{k^2} = \frac{1}{v_{\mu}} \frac{k \cdot v}{k^2 v^{\mu}} = \frac{v \cdot k}{v^2 k^2} = \frac{v \cdot k}{(v \cdot k)^2}$$

which is VERY wrong. So how can I perform this expansion if I'm not free to use the index. OR is my first assumption about the odd indices being the ones carrying the index wrong.
Should it rather be that
$$\frac{k}{4 v^2} \rightarrow \frac{k^2}{ 4 v v k} \rightarrow \frac{k^2}{4 (v \cdot k) v^{\nu}}$$

What is the proper method?

2. Jan 20, 2012

### Finbar

What are your assumptions. Are the components of k small compared to v?

3. Jan 20, 2012

### Hepth

Yes. K is like a scaled momentum. V is a heavy quark velocity. K is actually k/M where M is large.

4. Jan 21, 2012

### Finbar

Yes this is what confused me v is a velocity so dimsionless. k is a ratio of a momenta with some mass. Why not treat it as a taylor expansion in 1/M then. I assume M is a scalar quantity? So I think your original expansion is the correct one.