SimonsCat
Registered Member
Spacetime uncertainty is constructed from the metric and it is given as
$$\delta L \delta t\ g_{tt} \geq 1 - \frac{2 G \hbar}{c^4} = \frac{L^2_P}{c}$$
I showed how this might be suggestive of a relationship between redshift and quantum mechanics via:
$$\delta L \delta t \geq \frac{(1 - \frac{2 G \hbar}{c^4})}{g_{tt}}$$
The metric is just
$$g_{tt} = 1 + \frac{\phi}{c^2}$$
expanding we get
$$1 + \phi_1 - \phi_2 - \frac{1}{2} \phi^2_1 - \phi_1\phi_2 + \frac{3}{2}\phi^2_2 +...$$
To first order we get
$$\delta L \delta t \geq \frac{(1 - \frac{2 G \hbar}{c^4})}{[\frac{\phi_1}{c^2} - \frac{\phi_2}{c^2}]} = \frac{(1 - \frac{2 G \hbar}{c^4})}{\frac{\delta \phi}{c^2}}$$
adopting $$\delta s \equiv c \delta t$$ we can theorize the fundamental length as
$$\delta L \geq \frac{(1 - \frac{2 G \hbar}{c^3})}{g_{tt}c\delta t} = \frac{(1 - \frac{2 G \hbar}{c^3})}{\sqrt{\delta s^2 - g_{xx}\delta x^2}}$$
this can be constructed by using
$$\delta s^2 = g_{tt} c^2\delta t^2 + g_{xx} dx^2$$
and so
$$g_{tt}c \delta t = \sqrt{\delta s^2 - g_{xx}\delta x^2}$$
Since this formalism is NOT coordinate free, there is clearly work to be done.
$$\delta L \delta t\ g_{tt} \geq 1 - \frac{2 G \hbar}{c^4} = \frac{L^2_P}{c}$$
I showed how this might be suggestive of a relationship between redshift and quantum mechanics via:
$$\delta L \delta t \geq \frac{(1 - \frac{2 G \hbar}{c^4})}{g_{tt}}$$
The metric is just
$$g_{tt} = 1 + \frac{\phi}{c^2}$$
expanding we get
$$1 + \phi_1 - \phi_2 - \frac{1}{2} \phi^2_1 - \phi_1\phi_2 + \frac{3}{2}\phi^2_2 +...$$
To first order we get
$$\delta L \delta t \geq \frac{(1 - \frac{2 G \hbar}{c^4})}{[\frac{\phi_1}{c^2} - \frac{\phi_2}{c^2}]} = \frac{(1 - \frac{2 G \hbar}{c^4})}{\frac{\delta \phi}{c^2}}$$
adopting $$\delta s \equiv c \delta t$$ we can theorize the fundamental length as
$$\delta L \geq \frac{(1 - \frac{2 G \hbar}{c^3})}{g_{tt}c\delta t} = \frac{(1 - \frac{2 G \hbar}{c^3})}{\sqrt{\delta s^2 - g_{xx}\delta x^2}}$$
this can be constructed by using
$$\delta s^2 = g_{tt} c^2\delta t^2 + g_{xx} dx^2$$
and so
$$g_{tt}c \delta t = \sqrt{\delta s^2 - g_{xx}\delta x^2}$$
Since this formalism is NOT coordinate free, there is clearly work to be done.