Hi, its me again hoping to establish a quantum mechanical explanation for the red shift rather than recession. My last argument failed the units test as you all correctly pointed out. (thread - Apparent expansion of space alternative) So, let me run this past you again please.
This first bit is standard physics without controversy. I just want to get the Hubble recession rate into SI units.
Hubble rate of recession 73 km s^-1 Mpc^-1. This breaks down to:
73000m s^-1/ 30.857 x 10^21 m = 2.3675 x 10^-18s
The distance to the edge of the observable universe without invoking relativity would be the speed of light divided by this value.
C/2.3675 x 10^-18 = 1.266 x 10^26 m = the edge of the observable universe.
Now back to speculation:
I think we had established that the Heisenberg Uncertainty Principle when substituting the momentum for the momentum of a photon could be rewritten as Delta E. Delta. x > hc/(4Pi) where E is the energy and x is the distance, bearing in mind we are still really talking about momentum and distance.
Now, we are going to say that there is no uncertainty in the energy measurement and that all the uncertainty is in the distance measurement. Setting Delta E to unity we now have:
Delta x > hc/(4Pi. Delta E)
So Delta x > 1.58 x 10^-26 m
In fact, in any one interaction (say with a virtual electron positron pair) the uncertainty in the distance measurement can be anything between 0 and 1.58 x 10^-26 m but no greater. So we take the average uncertainty probability by dividing by 2 and get, the average uncertainty in the distance measurement extrapolated across the universe is 7.9 x 10^-27 m.
Now for a simple, but not scientifically exact, analogy. If I have a bag of coins and I lose 1/100th of them for every metre I travel. How far will I go before losing them all? The naïve answer is the inverse = 100m.
So now the argument runs: If I lose this tiny fraction of information 7.9 x 10-27 m for every for every metre of space, how far do I go before I lose the whole? The answer is the inverse 1.265 x 10^26 m. The edge of the observable universe.
OR a graph of 1- 7.9 x 10-27x cuts the x axis at 1.265 x 10^26 m.
So the simple linear red shift in our own locality of the universe is the ratio of distance travelled compared to the distance to the edge of the observable universe:
Eg What is the red shift at 10^25 m?
10^25 m/1.26 x 10^26 m = 0.079
However, we know the coin analogy is only correct for a short distance because it should really be an exponential loss and the same applies to the universe.
So, the function that covers all distances is:
Z = e(x/1.26x10^26) -1 where x is the distance travelled in metres and this gives the slightly higher red shift as observed at very large distances.
Are these scientifically sound steps or am I just manipulating numbers to get the answer I want? When you have put yourself in a hole it’s a bit difficult to see out of it. Your advice is appreciated.