[alancalverd]

c^2t^2/(2R)-R/R
and
c^2t^2/(2R)-R/c/t

1.   Since ct = R, c²t²/(2R) = R/2, which is a length.   
R/R = 1, dimensionless, so you can't subtract it from c²t²/(2R).

2.   R/c/t = Rt/c = (ct x t)/c = t²,

so c²t²/(2R) - R/c/t is a length minus time squared, which is also meaningless.

We've been here many times before. Ignore the numbers, which are mostly arbitrary human choices. Check the dimensions - that's where the physics resides.

[Colin2B]

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.

Most of us are familiar with the field equations, but whether your correction is seen as valid will depend the logic of how you got there and you will have to show that in any paper. Alan beat me to making the same comment on your use of the basic motion equations, so whichever route you take in your explanation, you will still have to show us how you got there.

[timey (OP)]

@alancalverd
Colin gave me the maths c^2t^2/2R = distanceR+distance extra travelled due to acceleration. (Or so I understood anyway)

If this equation does = distance R+extra distance travelled due to acceleration, then - R leaves me with the distance travelled due to acceleration, and I divide this extra distance by distance R. Which in my mind a distance divided by a distance is a distance.  And it is a distance that I'm looking for. (Which I then use for another consideration altogether)

@Colin2B If you already understood from page 1 to page 4b of my video, that an acceleration to a constant speed can be represented as a straight line by adding an additional time dilation for background space. Where the changes of this time in this background space - that occurs concurrently with, and as a separate phenomenon to the timing of clocks in the gravity potential - are inversely proportional in the gravity potential to the changes of the timing of a clock in the gravity potential, then no need to further explain...
Other than to say again, that the c^2t^2/2R equation, and the process that I want to apply to it, are not involved in making that description, and the description that I do want to make with it, I cannot make until I've received your advice as to whether or not my continuation to that equation results in 0.5 metres.

So b/c I am now thoroughly confused my next question is:
Does c^2t^2/2R = distanceR+distance extra travelled due to acceleration?
If not, then could you please show me the mathematical means to arrive at the extra distance travelled due to the acceleration if R is the radius of the universe, and c2/R is the acceleration?

[alancalverd]

We know R = cT, by definition, if T is the age of the universe.

Now add an acceleration a= d²r/dt². The position of a point previously at R, after a time interval Δt, will be R + a(Δt)²/2. Basic equation of motion.

So what? Well, if a(Δt)²/2 > cΔt, the object will have disappeared beyond the Schwarzchild radius, so the mass of the observable universe will have decreased and (assuming gravity travels at no more than c) the universe will continue to expand as g will have decreased at the periphery.

So the requirement for an expanding universe is  a(Δt)²/2 > cΔt, simplifying to aΔt > 2c. All we need to do is to find a cause for a, however small its value, because if we wait long enough we will satisfy the inequality.

Now if an object is already at escape speed and you reduce g, it will accelerate. So if just one particle leaves the Schwarzchild radius, the observable universe will expand at an ever-increasing rate. In summary: the universe expands because it does.

[timey (OP)]

And could you please show me what is the mathematical means (when using R as the distance) to arrive at the extra distance travelled in 13.8 billion years, when travelling at speed c, when subject to acceleration c^2/R?

[alancalverd]

s = at²/2, as we said earlier, so 13.8 x  c⁴/2R²  billion light years.

Problem occurs if you start with R = 0, giving an infinite initial acceleration, but you can use the above approximation for short times where s<<R and  R>>0

[timey (OP)]

So:

c^2t^2/2R= extra distance travelled due to acceleration?
(Where t is the age of the universe.)

[Colin2B]

So:

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/R
Extra distance due to acceleration depends what you are comparing it to eg constant speed ct

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.

[timey (OP)]

So (assuming Smolins equation c^2/R is correct):

(c^2/R)t^2/2R= distance (ct)+extra distance travelled due to acceleration
(Where t is the age of the universe.)

And
(c^2/R)t^2/(2R)-R/R= 'a distance'

[alancalverd]

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.

yes indeed! Apologies.

I’m still not sure how Smolin derived c2/R as being the acceleration.

s = ½at² (equation of motion)  R = ct  (definition of Schwarzchild radius) so at any time t, R = ct = ½at², hence a = 2ct/t² =2c/t or 2c²/R.

Did Smolin give a reason for dividing by 2? Or explain how to prevent a→∞ if R = 0?

[alancalverd]

And
(c^2/R)t^2/(2R)-R/R= 'a distance'

No. R/R is dimensionless so you can't subtract it from 2R or anything with any dimension.

[timey (OP)]

Smolin said that c^2/R = an acceleration that matches the acceleration that the universe's expansion (as per the expanding theory) is now found to be accelerating at. And this acceleration is now thought associated with the cosmological constant that Einstein retracted from his theory of general relativity.

...and It's not minus R/R, it's minus R =distance/R

[Colin2B]

Did Smolin give a reason for dividing by 2? Or explain how to prevent a→∞ if R = 0?

I don't know. @timey found it in a book so I don't understand why it's double the calc

[timey (OP)]

Smolin didn't mention anything about /2.

The only place /2 has entered into the discussion is via the /2R, that is in the equation that 'you' provided me as a means to calculate the extra distance travelled.

[alancalverd]

...and It's not minus R/R, it's minus R =distance/R

That is completely meaningless!

Worth reminding you of the order in which expressions are calculated: BODMAS.

First, evaluate anything in Brackets

then "Of" e.g. "40% of ....." (you won't find this in most physics texts)

then Divisions.Multiplications, Additions and Subtractions, in that order

so ((4+5)/(1 + 4 -2)) x 6 + 2 = ((9)/(3)) x 6 +2 = 3 x 6 +2 = 18 + 2= 20. Note that some additions and subtractions were inside the brackets, so we tackled them first. Once we have evaluated the brackets and got rid of the division, the multiplication and addition were unequivocal.

So when proposing a calculation, first sort out your physics, then make sure your calcs are ordered by BODMAS, then check that the dimensions give you what you need  - speed,length, time, mass, whatever, without adding apples and pears. Finally, put in the numbers. Remember that the numbers don't matter much. Famous exchange from history:

Fred Hoyle  "And there we have the luminosity of the galaxy...... apart from a factor of 10²⁶"

Student "Multiply or divide?"

Fred "Er...um....I don't think it affects the physics either way"

[timey (OP)]

(Back on lap top now and can see much better what is going on, the phone is restricting)

Quote from: timey on Today at 13:58:22
So:

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/R
Extra distance due to acceleration depends what you are comparing it to eg constant speed ct

Oh... 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)

[jeffreyH]

So:

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/R
Extra distance due to acceleration depends what you are comparing it to eg constant speed ct

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.

Well you have length^2/(time^2 x length) = c^2/R. That looks an awful lot like acceleration. Whether or not the value is valid is another matter entirely.

[timey (OP)]

(Back on lap top now and can see much better what is going on, the phone is restricting)

Quote from: timey on Today at 13:58:22
So:

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/R
Extra distance due to acceleration depends what you are comparing it to eg constant speed ct

Oh... 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)

So is that going to be:
(R/t+c^2/R)xt^2/(2R)-R= distance travelled due to c^2/R  ?
where R is the radius of the observable universe, and t is the age of the universe.

Or is it:
R/t+(c^2/Rxt^2)/(2R)-R=distance travelled due to c^2/R  ?

Or on the basis that
s = ½at^2

is it:

½(c^2/R)t^2=distance travelled due to c^2/R  ?

And if not, then please, how do I notate it?

[timey (OP)]

@alancalverd
Me: Do you know the way to the shop?
You: Yes
Me: Could you give me directions please?
You: Certainly. Buy a map and apply map reading skills.
Me: To buy a map, clearly I'd need to get to the shop.  Can't you just tell me how to get there?

On the basis that I think it important that people receive feedback on personal interactions I will now tell you that I have not slept a wink, feel quite ill with frustration, and have been crying a lot.  I do not understand your purpose in being glib, or continuously bring up expanding universe theory, when I have told you my consideration is contracting.
In giving you this feedback, I ask you if there is potential for a change in narrative, where it is my hope that your intention is not to be a deconstructive force, or cause upset.

@Colin2B
Having given the matter about 9 hours thought now, if that equation c^2t^2/2R = distance, and ct=R, then distance=c^2t^2/2R has to be the extra distance travelled due to the acceleration, in addition to distance R, doesn't it?

[alancalverd]

My only concern is to make sure that your hypothesis is summarised in equations that make sense, so that others can see your point. The first thing any physicist will (or should) do when faced with a novel equation is to ask whether it is dimensionally balanced. If not, there may be a typo or a misunderstanding, and he won't be able to discern the author's intentions. So let's get the typos out of the way and ensure that your ideas are expressed unequivocally.

We have given you the basic equations of motion in standard form. We agree that R =ct by definition. I am unclear about the proposed value of "the acceleration" - where does this come from? And what exists at this "extra distance"? Happy to have a reference to an earlier post if you have specified these.

Late edit: I have now picked up the reference that Smolin says the observed acceleration is c²/R. Mea culpa.

So at time t + Δt we have R + ΔR = R + c²(Δt)²/2R +vΔt where v is the "initial" velocity at t

Hence ΔR = c²(Δt)²/2R + Δt ∫(c²t/R) t (I think!)

Will give it more thought later - off to work!