Covariant gradient expansion

Due to the relevance of the argument, after a nice discussion with a contribution of Carl Brannen, I decided to pursue this matter further. Indeed, the only way to have a covariant formulation of a gradient expansion is adding a time variable and taking the true time variable Wick rotated. In this way, for d=1+1 wave equation you will use d=2+1 wave equation and so on. In d=3+1 you will use d=4+1 wave equation. Let me explain with some equations what I mean. I consider again d=1+1 case as

\partial^2_{tt}u-\partial^2_{xx}u=0

but, instead to apply a gradient expansion to it, I apply this to the equation

\partial^2_{tt}u-\Delta_2u=0

being \Delta_2 = \partial_{xx}+\partial_{yy}. As usual, I rescale time variable as t\rightarrow\sqrt{\lambda}t and I take a solution series

u=u_0+\frac{1}{\lambda}u_1+\frac{1}{\lambda^2}u_2+\ldots.

Now I will get the set of equations

\partial^2_{tt}u_0=0

\partial^2_{tt}u_1=\Delta_2u_0

\partial^2_{tt}u_2=\Delta_2u_1

and so on. Let us note that, in this case, we can introduce two new spatial variables as z=x+iy and \bar z=x-iy. These are conjugate variables as you know. So, already at the leading order I have solved my equation. Indeed, I note that

\Delta_2=\partial_z\partial_{\bar z}

and so the Laplacian has the solution f(z)+g(\bar z) being f and g arbitrary functions. In this case the gradient expansion gives immediately the exact result making its application trivial as should be. Indeed, I take t=0 in the perturbation series and put iy=t and I get

u=f(x+t)+g(x-t)

that is the exact solution. Nice, it works! This means that a quantum field theory using gradient expansion exists and it is a strong coupling expansion. This result is surely less trivial than the one obtained above.

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One Response to Covariant gradient expansion

  1. […] field theory and gradient expansion In a preceding post (see here) I showed as a covariant gradient expansion can be accomplished maintaining Lorentz invariance […]

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