Spacetime uncertainty makes quantum field theory finite
Fuente:
arXiv
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| Natura: | Preprint |
| Pubblicazione: |
2024
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| _version_ | 1866915082796007424 |
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| author | Cahill, Kevin |
| author_facet | Cahill, Kevin |
| contents | Since Einstein's equations $G_{ij} = 8π\, G \, T_{ij} \, / c^4 $ relate the metric $g_{ij}$ of spacetime to the energy-momentum tensor $T_{ij}$ which is a quantum field, the metric $g_{ij}$ must be a quantum field. And since the metric $g_{ij}(x)$ is the dot product $g_{ij}(x) = \partial_i p^α(x) \, \partial_j p_α(x)$ of the derivatives of the points $p(x)$ of spacetime, spacetime must be a quantum field. Its points have average values $\langle p(x) \rangle$ that obey general relativity and fluctuations $q(x) = p(x) - \langle p(x) \rangle$ that obey quantum mechanics. It is suggested that the fields of quantum field theory be regarded not as functions $ϕ(x)$ of their classical coordinates $x$ but as functions $ϕ(p(x))$ of their quantum coordinates $p(x)$. In empty flat spacetime where $p(x) = x + q(x)$ and $x = (t, \boldsymbol x)$, the Fourier exponentials $\exp(i k(x+q(x))$ averaged over normally distributed fluctuations $q(x)$ are gaussians $\exp(i kx -\ell^2 \boldsymbol k^2 - \ell^2 m^2/2)$. These gaussians make Feynman diagrams finite. The zero-point energy density of the vacuum also is finite -- but negative and too large to explain dark energy unless new bosons exist. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2406_09448 |
| institution | arXiv |
| publishDate | 2024 |
| record_format | arxiv |
| spellingShingle | Spacetime uncertainty makes quantum field theory finite Cahill, Kevin General Physics Since Einstein's equations $G_{ij} = 8π\, G \, T_{ij} \, / c^4 $ relate the metric $g_{ij}$ of spacetime to the energy-momentum tensor $T_{ij}$ which is a quantum field, the metric $g_{ij}$ must be a quantum field. And since the metric $g_{ij}(x)$ is the dot product $g_{ij}(x) = \partial_i p^α(x) \, \partial_j p_α(x)$ of the derivatives of the points $p(x)$ of spacetime, spacetime must be a quantum field. Its points have average values $\langle p(x) \rangle$ that obey general relativity and fluctuations $q(x) = p(x) - \langle p(x) \rangle$ that obey quantum mechanics. It is suggested that the fields of quantum field theory be regarded not as functions $ϕ(x)$ of their classical coordinates $x$ but as functions $ϕ(p(x))$ of their quantum coordinates $p(x)$. In empty flat spacetime where $p(x) = x + q(x)$ and $x = (t, \boldsymbol x)$, the Fourier exponentials $\exp(i k(x+q(x))$ averaged over normally distributed fluctuations $q(x)$ are gaussians $\exp(i kx -\ell^2 \boldsymbol k^2 - \ell^2 m^2/2)$. These gaussians make Feynman diagrams finite. The zero-point energy density of the vacuum also is finite -- but negative and too large to explain dark energy unless new bosons exist. |
| title | Spacetime uncertainty makes quantum field theory finite |
| topic | General Physics |
| url | https://arxiv.org/abs/2406.09448 |