0 Members and 1 Guest are viewing this topic.
Yes I do want to compare the extra distance travelled with the acceleration, to the original distance (wavelength Rscale) without acceleration, but I think in order to get the small amount of distance I'm looking for I need to divide the extra distance by the distance of wavelength Rscale. And in order to get the small amount of time I'm looking for I need to divide the extra distance by the speed of light. (Edit: Oh I don't think that is quite finished. Then perhaps divide this time by the age of the universe? Noting that all measurements are made via the observers clock.) In this manner the small acceleration is proportional to the small distance and the small time. Am I right?Ok, I get your notation right up until you say divided by 2 and divided by 2R. I'm assuming that R is the Rscale wavelength, but could benefit from some clarification as to this, and the use of 2.
What is, or why 2?
Quote from: timey on 01/02/2018 18:51:26What is, or why 2?I’ll do a diagram
...so the maths "at2/2" where a is c2/R, says acceleration multiplied by time squared, divided by 2.What is, or why 2?
So just to check - the distance that =c2t2/2R, that is the extra distance travelled due to the acceleration?
That tiny length of time should be just a tiny fraction of a second, if it isn't then I'm not doing the equation correctly.
R/c=t (age of universe)ct=R (distance)c2/R=a (acceleration = cosmological constant)
Not really sure about the construct of my method to say so, still running it through, and in all honesty could use some advice really.
to get away from the big numbers I used
R is the radius of the observable universe.