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- TL;DR Summary
- Does time dilation applied to the wave function lead to acceleration?
Background
While watching Does time cause gravity? from PBS Spacetime, i wondered if its possible to "derive" the geodesic equation
not from GR alone, but by assuming each particle is described by an extended wave function and the time evolution
of this wave is not constant but the rate varies depending on position, based on the time dilation.
Mathematical attempt
We start with the time dependent schrödinger equation for a free particle:
$$i\hbar\frac{\partial}{\partial t} \Psi(x, t) = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi(x, t)$$
Now we add a dilation factor in. As a simple approximation we assume a constant acceleration field
which would result in a linear dilation factor:
$$D(x) = \frac{1}{1+kx\epsilon }$$
Where ##k## is a vector pointing along the acceleration field and ##\epsilon ## is the linearized factor
which describes the strength of the field. The result ##D(x)## describes the relative clock rate
of something at the position ##x## as seen from the perspective of someone sitting at ##x=0##.
If you are "higher up" in the field you would have a higher clock rate ##(>1)## as seen by our observer,
if you are "lower" you would have a lower clock rate ##(<1)## as seen by the observer.
For a more accurate model one would have to derive the position dependent clock rate from e.g. the schwarzschild
metric, but as a linear approximation for small scales (size of a particle) it should be good enough. Since on
e.g. Earth the time dilation effect is very small, ##\epsilon## will be very small as well.
Now we modify our original equation to include the dilation rate:
$$i\hbar\frac{\partial}{\partial t} \Psi(x, t) = D(x)\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi(x, t)$$
Now we plug in the initial state of a single particle at rest and see what happens. Of course it will initially
spread out, but will it pick up momentum?
And if so, is this acceleration experienced consistent with the same acceleration we would get from "normal"
gravity?
While watching Does time cause gravity? from PBS Spacetime, i wondered if its possible to "derive" the geodesic equation
not from GR alone, but by assuming each particle is described by an extended wave function and the time evolution
of this wave is not constant but the rate varies depending on position, based on the time dilation.
Mathematical attempt
We start with the time dependent schrödinger equation for a free particle:
$$i\hbar\frac{\partial}{\partial t} \Psi(x, t) = \frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi(x, t)$$
Now we add a dilation factor in. As a simple approximation we assume a constant acceleration field
which would result in a linear dilation factor:
$$D(x) = \frac{1}{1+kx\epsilon }$$
Where ##k## is a vector pointing along the acceleration field and ##\epsilon ## is the linearized factor
which describes the strength of the field. The result ##D(x)## describes the relative clock rate
of something at the position ##x## as seen from the perspective of someone sitting at ##x=0##.
If you are "higher up" in the field you would have a higher clock rate ##(>1)## as seen by our observer,
if you are "lower" you would have a lower clock rate ##(<1)## as seen by the observer.
For a more accurate model one would have to derive the position dependent clock rate from e.g. the schwarzschild
metric, but as a linear approximation for small scales (size of a particle) it should be good enough. Since on
e.g. Earth the time dilation effect is very small, ##\epsilon## will be very small as well.
Now we modify our original equation to include the dilation rate:
$$i\hbar\frac{\partial}{\partial t} \Psi(x, t) = D(x)\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \Psi(x, t)$$
Now we plug in the initial state of a single particle at rest and see what happens. Of course it will initially
spread out, but will it pick up momentum?
And if so, is this acceleration experienced consistent with the same acceleration we would get from "normal"
gravity?
