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c^2t^2/(2R)-R/Randc^2t^2/(2R)-R/c/t
Are you familiar with the set of 10 equations in Einstein's general theory of relativity ........it would be a damn sight simpler for me to take you through these existing equations ......and describe my correction at source, than it would for me to explain to you the route that I am attempting.
So:c^2t^2/2R= extra distance travelled due to acceleration?(Where t is the age of the universe.)
s = at2/2, as we said earlier, so 13.8 x c4/2R2 billion light years.
Quote from: alancalverd on Today at 13:47:04 s = at2/2, as we said earlier, so 13.8 x c4/2R2 billion light years.@alancalverd Alan, did you mean c2/2R x 13.82 billion light years.
I’m still not sure how Smolin derived c2/R as being the acceleration.
And(c^2/R)t^2/(2R)-R/R= 'a distance'
Did Smolin give a reason for dividing by 2? Or explain how to prevent a→∞ if R = 0?
...and It's not minus R/R, it's minus R =distance/R
Quote from: timey on Today at 13:58:22So:c^2t^2/2R= extra distance travelled due to acceleration?(Where t is the age of the universe.)No, total distance travelled, from rest, at a given acceleration of c^2/RExtra distance due to acceleration depends what you are comparing it to eg constant speed ct
Quote from: timey on 08/02/2018 13:58:22So:c^2t^2/2R= extra distance travelled due to acceleration?(Where t is the age of the universe.)No, total distance travelled, from rest, at a given acceleration of c^2/RExtra distance due to acceleration depends what you are comparing it to eg constant speed ctQuote from: alancalverd on 08/02/2018 13:47:04s = at2/2, as we said earlier, so 13.8 x c4/2R2 billion light years. @alancalverd Alan, did you mean c2/2R x 13.82 billion light years.I’m still not sure how Smolin derived c2/R as being the acceleration.
(Back on lap top now and can see much better what is going on, the phone is restricting)Quote from: Colin2B on 08/02/2018 14:41:52Quote from: timey on Today at 13:58:22So:c^2t^2/2R= extra distance travelled due to acceleration?(Where t is the age of the universe.)No, total distance travelled, from rest, at a given acceleration of c^2/RExtra distance due to acceleration depends what you are comparing it to eg constant speed ctOh... craps! Well I don't want to calculate the distance of R travelled from rest.I want to say that:distance R = radius of observable universe, travelled at speed c + the acceleration.Then I want to subtract the distance travelled if travelling at c without the acceleration. ie: (ct) (where t is the age of the universe) from this c + the acceleration distance, leaving me with the distance travelled due to the acceleration.Then I divide this distance by the distance (ct) and this will give me a small distance.(edit corrections were made)