# Obtaining Euclidean action from Minkowski action

1. Jan 15, 2016

### spaghetti3451

The Euclidean classical action $S_{\text{cl}}[\phi]$ for a scalar field $\phi$ is given by

S_{\text{cl}}[\phi]=\int d^{4}x\ \bigg(\frac{1}{2}(\partial_{\mu}\phi)^{2}+U(\phi)\bigg).

This can be obtained from the action $S_{\text{Mn}}[\phi]$ in Minkowski space for the same scalar field $\phi$ as given by

S_{\text{Mn}}[\phi]=\int d^{4}x\ \bigg(\frac{1}{2}(\partial_{\mu}\phi)^{2}-U(\phi)\bigg).

To transform the action $S_{\text{Mn}}[\phi]$ in Minkowski space to the Euclidean classical action $S_{\text{cl}}[\phi]$, one needs to perform a Wick rotation. I need someone to show the derivation.

2. Jan 15, 2016

### samalkhaiat

Continue from real $x^{0}$ to $x^{0} = - i x_{4}$, and from $S_{Min}$ so obtained get $S_{Euc}$ using $S_{Euc} = - i S_{Min}$:
$$S_{M} = \int dx^0 \ d^3x \ \left( (1/2) (\frac{\partial \phi}{\partial x^{0}})^{2} - (1/2) (\nabla \phi)^{2} - U(\phi) \right)$$
$$S_{M} = -i \int dx_{4} \ d^3x \ \left( - (1/2) (\frac{\partial \phi}{\partial x_{4}})^{2} - (1/2) (\nabla \phi)^{2} - U(\phi) \right)$$
$$S_{M} = i \int dx_{4} \ d^3x \ \left( (1/2) (\frac{\partial \phi}{\partial x_{4}})^{2} + (1/2) (\nabla \phi)^{2} + U(\phi) \right) = i S_{E}$$
$$S_{E} = \int d^{4}x_{E} \ \left( (1/2) (\frac{\partial \phi}{\partial x_{\mu}})^{2} + U(\phi) \right)$$