Tuesday, July 1, 2014

The Euclidean Hamiltonian and the Wick Rotation

Excerpts from The Revolution of Matter

Under the Wick Rotation the potential in the Lagrangian or the Hamiltonian for a system switches from positive to negative,

 $$ V_E( {R}^4 )\geq 0 \quad \quad \underset{\rightleftarrows }{\text{Wick}} \quad \quad  -   V_M({R}^{1,3})\leq 0$$

for instance the Minkowskian Hamiltonian switches from

$$ \mathcal{H}_M ( \mathbb{R}^{1,3} ) = T+V>0  $$


 $$ \mathcal{H}_M ( \mathbb{R}^{4} ) = T-V=0  $$

Obviously the total energies do not match.

$$\mathcal{H}_M(\mathbb{R}^{1,3})\geqslant  \mathcal{H}_E (\mathbb{R}^4) $$

Similarly in comparing the Euclidean Lagrangian $\mathcal{L}_E$ to the Minkowski Lagrangian $\mathcal{L}_M$ via the Wick rotation,

$$ \mathcal{L}_E( \mathbb{R}^4 )=\mathbb{T}+\mathbb{U}\geq 0$$


$\mathcal{L}_E$ is ordered positive semidefinite to $\mathcal{L}_M$

$$|   \mathcal {L}_E (\mathbb{R}^4) | \geqslant | \mathcal{L}_M(\mathbb{R}^{1,3})|$$

It can also be seen the Lagrangians do not match, therefore the Hamiltonian Actions derived from the Lagrangians do not match. To match the Hamiltonians of $\mathbb{R}^4$ to $\mathbb{R}^{1,3}$ requires an additional potential be subtracted from  $\mathbb{R}^{1,3}$, and the obvious field is gravity G,

$$\mathcal{H}_M(\mathbb{R}^{1,3})=T+V - G =0$$

This introduction of gravity follows on as Hartle-Hawking put it - "Indeed in a certain sense the total energy for a closed universe is always zero - the gravitational energy cancelling the matter energy." So along with the other potentials V for Electro-Weak and Strong which have charges who energies cancel out, the gravitational potential appears as an extra geometric term to maintain the problem Euclidean flatness in Minkowski spacetime. In effect the gravitational field is a gauge potential that is introduced to maintain zero global energy.

copyright © today